├── .gitignore ├── README.md ├── english.code.tex ├── english.tex ├── general.tex ├── math.code.tex ├── math.tex ├── preamble.tex ├── review.code.tex ├── rv-utils.sty ├── sdutil.sty ├── showdata.sty └── update.py /.gitignore: -------------------------------------------------------------------------------- 1 | *.pdf 2 | *.aux 3 | *.fdb_latexmk 4 | *.fls 5 | *.log 6 | *.out 7 | *.synctex.gz 8 | *.wq 9 | *.xdv 10 | *.toc 11 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | 自制错题模板，主要提供可方便跳转的交叉引用。 2 | -------------------------------------------------------------------------------- /english.code.tex: -------------------------------------------------------------------------------- 1 | \input preamble.tex 2 | \input review.code.tex 3 | \geometry{ 4 | paperwidth=250.6mm, 5 | paperheight=174.1mm, 6 | left=2cm, 7 | right=2cm, 8 | top=2.5cm, 9 | bottom=2.5cm, 10 | } 11 | \ctexset{ 12 | subsection/numbering=false, 13 | subsection/format=\Large\bfseries\sffamily\color{teal}, 14 | } 15 | 16 | \fancyhf{} 17 | \lhead{\itshape\rightmark} 18 | \rhead{\thepage} 19 | \pagestyle{fancy} 20 | \renewcommand{\headrulewidth}{0pt} 21 | \setlength\headheight{15pt} 22 | 23 | \etocmulticolstyle[2]{ 24 | \noindent\bfseries\Large 25 | \leaders\hrule height1pt\hfill 26 | \MakeUppercase{Table of Contents} 27 | } 28 | 29 | \tcbset{ 30 | mark/.style={ 31 | } 32 | } 33 | \newtcolorbox{wordcard}[2][]{ 34 | fonttitle=\raggedright\sffamily\zihao{0}, 35 | fontupper=\itshape\zihao{0}, 36 | height=\textheight, 37 | width=\textwidth, 38 | boxrule=2pt, 39 | arc=3mm, 40 | boxsep=1em, 41 | enhanced, 42 | overlay={ 43 | \node[ 44 | anchor=south east, 45 | shift={(-1em, 1em)}, 46 | font=\itshape\zihao{4}, 47 | ] at (interior.south east) {#2}; 48 | }, 49 | #1 50 | } 51 | 52 | % \title{\today 测试} 53 | \author{} 54 | \date{} 55 | 56 | \ExplSyntaxOn 57 | \clist_new:N \g__en_info_clist 58 | \clist_new:N \g__en_review_point_clist 59 | \clist_new:N \g__saved_star_word_clist 60 | \clist_new:N \g__saved_star_complete_word_clist 61 | \clist_new:N \g__saved_star_word_meaning_clist 62 | \clist_new:N \g__saved_star_word_info_clist 63 | \int_new:N \l__en_word_count_int 64 | \int_new:N \g__en_review_int % for \show_word depracated 65 | \int_new:N \g__en_wordlist_count_int 66 | \int_new:N \g__en_star_word_count_int 67 | 68 | % \clist_gset:Nn \g__en_review_point_clist {2,4,7,15,30} 69 | \clist_gset:Nn \g__en_review_point_clist {3,10,30} 70 | \int_gzero:N \g__en_wordlist_count_int 71 | \int_gzero:N \g__en_star_word_count_int 72 | 73 | \AtBeginDocument 74 | { 75 | \bool_if:NTF \g__review_today_bool 76 | { 77 | \title{\today 复习} 78 | } 79 | { 80 | \title{\g__review_date_tl 复习} 81 | } 82 | \maketitle 83 | \pdfbookmark[1]{目录}{toc} 84 | \tableofcontents 85 | \clearpage 86 | } 87 | \AtEndDocument 88 | { 89 | \en_show:N \l__en_date_tl 90 | } 91 | 92 | \prg_new_protected_conditional:Nnn \en_review_if:n { T, F, TF } 93 | { 94 | \review_if:NnTF \g__en_review_point_clist { #1 } 95 | { \prg_return_true: } 96 | { \prg_return_false: } 97 | } 98 | 99 | % #1 日期 #2 题目信息比如11年真题 100 | \NewDocumentEnvironment {wordlist} { m m +b } 101 | { 102 | \group_begin: 103 | \en_review_if:nT { #1 } 104 | { 105 | \init_word_list:nn { #1 } { #2 } 106 | } 107 | \tl_clear_new:N \l__en_date_tl 108 | \tl_clear_new:N \l__en_info_tl 109 | \tl_set:Nn \l__en_date_tl { #1 } 110 | \tl_set:Nn \l__en_info_tl { #2 } 111 | #3 112 | \group_end: 113 | } {} 114 | 115 | \cs_new_protected:Nn \init_word_list:nn 116 | { 117 | \int_gincr:N \g__en_wordlist_count_int 118 | \clist_gput_right:Nn \g__en_info_clist { #2 } 119 | % list word 120 | \clist_gclear_new:c 121 | { g__saved_word_info_\int_use:N \g__en_wordlist_count_int _clist } 122 | \clist_gclear_new:c 123 | { g__saved_word_\int_use:N \g__en_wordlist_count_int _clist } 124 | \clist_gclear_new:c 125 | { g__saved_complete_word_\int_use:N \g__en_wordlist_count_int _clist } 126 | \clist_gclear_new:c 127 | { g__saved_word_meaning_\int_use:N \g__en_wordlist_count_int _clist } 128 | } 129 | 130 | % #1 date you do not remember again #2 题目编号 131 | % #3 word, #4 音标, #5 词性, #6 解释 132 | \NewDocumentCommand { \word } { O{} D(){} m D(){} O{} +m } 133 | { 134 | \tl_if_blank:nF { #1 } 135 | { 136 | \en_review_if:nT { #1 } 137 | { 138 | \save_star_word_piece:nnnnn 139 | { #2 } { #3 } { #4 } { #5 } { #6 } 140 | } 141 | } 142 | \en_review_if:nT { \l__en_date_tl } 143 | { 144 | \save_word_piece:nnnnn 145 | { #2 } { #3 } { #4 } { #5 } { #6 } 146 | } 147 | } 148 | 149 | \cs_new_protected:Nn \save_star_word_piece:nnnnn 150 | { 151 | \int_gincr:N \g__en_star_word_count_int 152 | \save_word_info:nn { g__saved_star_word_info_clist } { #1 } 153 | \save_word:nn { g__saved_star_word_clist} { #2 } 154 | \save_comlete_word:nnnn { g__saved_star_complete_word_clist } 155 | { #2 } { #3 } { #4 } 156 | \save_word_meaning:nn { g__saved_star_word_meaning_clist } { #5 } 157 | } 158 | 159 | \cs_new_protected:Nn \save_word_piece:nnnnn 160 | { 161 | \save_word_info:nn 162 | { g__saved_word_info_\int_use:N \g__en_wordlist_count_int _clist } 163 | { #1 } 164 | \save_word:nn 165 | { g__saved_word_\int_use:N \g__en_wordlist_count_int _clist } 166 | { #2 } 167 | \save_comlete_word:nnnn 168 | { g__saved_complete_word_\int_use:N \g__en_wordlist_count_int _clist } 169 | { #2 } { #3 } { #4 } 170 | \save_word_meaning:nn 171 | { g__saved_word_meaning_\int_use:N \g__en_wordlist_count_int _clist } 172 | { #5 } 173 | } 174 | 175 | \cs_new_protected:Nn \save_word_info:nn 176 | { 177 | \clist_gput_right:cx { #1 } 178 | { 179 | \l__en_info_tl\tl_if_blank:nF { #2 } { -#2 } 180 | } 181 | } 182 | 183 | \cs_new_protected:Nn \save_word:nn 184 | { 185 | \clist_gput_right:cn { #1 } { #2 } 186 | } 187 | 188 | \cs_new_protected:Nn \save_comlete_word:nnnn 189 | { 190 | \clist_gput_right:cn { #1 } 191 | { 192 | {\sffamily\bfseries#2} 193 | \tl_if_blank:nF { #3 } { \textipa{[#3]} } 194 | \quad{\normalfont\itshape{\tl_if_blank:nF { #4 } { #4.~ }}} 195 | } 196 | } 197 | 198 | \cs_new_protected:Nn \save_word_meaning:nn 199 | { 200 | \clist_gput_right:cn { #1 } { \itshape#2 } 201 | } 202 | 203 | \cs_new_protected:Nn \en_show:N 204 | { 205 | % \show_word: 206 | \en_if_show:NN \g__saved_star_word_clist \show_review_star_word: 207 | \en_if_show:NN \g__en_info_clist \show_review_word: 208 | \en_if_show:NN \g__saved_star_word_clist \show_star_word_card: 209 | \en_if_show:NN \g__en_info_clist \show_word_card: 210 | } 211 | 212 | \cs_new_protected:Nn \en_if_show:NN 213 | { 214 | \clist_if_empty:NF #1 { #2 } 215 | } 216 | 217 | \cs_new_protected:Nn \show_review_star_word: 218 | { 219 | \subsection*{易错单词测试} 220 | \addcontentsline{toc}{section}{易错单词测试} 221 | \begin{baselist}[itemsep=1em] 222 | \int_step_inline:nn { \g__en_star_word_count_int } 223 | { 224 | \item \customlabel{review_star_word:##1} 225 | { 226 | \sffamily\bfseries\huge 227 | \hyperref[star_card:##1]{ 228 | \clist_item:Nn \g__saved_star_word_clist { ## 1 } 229 | } 230 | } 231 | } 232 | \end{baselist} 233 | \clearpage 234 | } 235 | 236 | \cs_new_protected:Nn \show_review_word: 237 | { 238 | \phantomsection 239 | \addcontentsline{toc}{section}{文章单词测试} 240 | \int_step_inline:nn {\g__en_wordlist_count_int} 241 | { 242 | \customlabel{review:##1} 243 | \subsection*{ 244 | \clist_item:Nn \g__en_info_clist { ##1 } 245 | } 246 | \markright{ 247 | \clist_item:Nn \g__en_info_clist { ##1 } 248 | } 249 | \addcontentsline{toc}{subsection}{ 250 | \clist_item:Nn \g__en_info_clist { ##1 } 251 | } 252 | \begin{baselist}[itemsep=1em] 253 | \int_zero:N \l__en_word_count_int 254 | \clist_map_inline:cn { g__saved_word_##1_clist } 255 | { 256 | \int_incr:N \l__en_word_count_int 257 | \item \customlabel{review_word:##1_\int_use:N \l__en_word_count_int} 258 | { 259 | \sffamily\bfseries\huge 260 | % ####1 261 | \hyperref[card:##1_\int_use:N \l__en_word_count_int]{####1} 262 | } 263 | } 264 | \end{baselist} 265 | \clearpage 266 | } 267 | } 268 | 269 | \cs_new_protected:Nn \show_next_review: 270 | { 271 | 272 | } 273 | 274 | \cs_new_protected:Nn \show_star_word_card: 275 | { 276 | \phantomsection 277 | \addcontentsline{toc}{section}{易错单词卡片} 278 | \int_step_inline:nn 279 | { \g__en_star_word_count_int } 280 | { 281 | \customlabel{star_card:##1} 282 | \begin{wordcard}[ 283 | title={ 284 | \hyperref[review_star_word:##1]{ 285 | \clist_item:cn 286 | { g__saved_star_complete_word_clist } 287 | { ##1 } 288 | } 289 | }, 290 | colframe=red!30, 291 | coltitle=black!80, 292 | ]{ 293 | \clist_item:cn 294 | { g__saved_star_word_info_clist } 295 | { ##1 } 296 | } 297 | \clist_item:cn 298 | { g__saved_star_word_meaning_clist } 299 | { ##1 } 300 | \end{wordcard} 301 | \clearpage 302 | } 303 | } 304 | 305 | \cs_new_protected:Nn \show_word_card: 306 | { 307 | \phantomsection 308 | \addcontentsline{toc}{section}{文章单词卡片} 309 | \int_step_inline:nn {\g__en_wordlist_count_int} 310 | { 311 | \phantomsection 312 | \markright{ 313 | \clist_item:Nn \g__en_info_clist { ##1 } 314 | } 315 | \addcontentsline{toc}{subsection}{ 316 | \clist_item:Nn \g__en_info_clist { ##1 } 317 | } 318 | \int_step_inline:nn 319 | { \clist_count:c { g__saved_complete_word_##1_clist } } 320 | { 321 | \customlabel{card:##1_####1} 322 | \begin{wordcard}[ 323 | colframe=teal, 324 | coltitle=white, 325 | title={ 326 | \hyperref[review_word:##1_####1]{ 327 | \clist_item:cn 328 | { g__saved_complete_word_##1_clist } 329 | { ####1 } 330 | } 331 | }, 332 | ]{ 333 | \clist_item:cn 334 | { g__saved_word_info_##1_clist } 335 | { ####1 } 336 | } 337 | \clist_item:cn 338 | { g__saved_word_meaning_##1_clist } 339 | { ####1 } 340 | \end{wordcard} 341 | \clearpage 342 | } 343 | } 344 | } 345 | 346 | \NewDocumentEnvironment {baselist} { O{} +b } 347 | { 348 | \begin{multicols}{2} 349 | \begin{enumerate}[ 350 | font=\sffamily\huge, 351 | #1 352 | ] 353 | #2 354 | % \item[] \vspace*{\fill} 355 | \end{enumerate} 356 | \end{multicols} 357 | } {} 358 | 359 | \NewDocumentCommand {\enhint} { +m } 360 | { 361 | {\\\normalfont（#1）} 362 | } 363 | 364 | \NewDocumentCommand {\enem} { +m } 365 | { 366 | {\sffamily#1} 367 | } 368 | 369 | % 暂时不用 370 | % \cs_new_protected:Nn \show_word: 371 | % { 372 | % \phantomsection 373 | % \addcontentsline{toc}{section}{单词列表} 374 | % \int_gzero:N \g__en_review_int 375 | % \int_step_inline:nn {\g__en_wordlist_count_int} 376 | % { 377 | % \subsection*{##1} 378 | % \markright{##1单词列表} 379 | % \en_review_if:nTF { ##1 } 380 | % { 381 | % \int_gincr:N \g__en_review_int 382 | % \bookmarksetup{color=red} 383 | % \addcontentsline{toc}{subsection}{ 384 | % \texorpdfstring{\color{red}##1单词列表}{##1单词列表} 385 | % } 386 | % \label{wordlist:##1} 387 | % \reviewtoday{\hyperref[review:##1]{前往测试}} 388 | % } 389 | % { 390 | % \bookmarksetup{color=black} 391 | % \addcontentsline{toc}{subsection}{##1单词列表} 392 | % } 393 | % \begin{baselist}[font=\sffamily\Large] 394 | % \int_step_inline:nn 395 | % { \clist_count:c { g__saved_complete_word_##1_clist } } 396 | % { 397 | % \item 398 | % \begingroup 399 | % \Large 400 | % \clist_item:cn 401 | % { g__saved_complete_word_##1_clist } 402 | % { ####1 } 403 | % \par 404 | % \clist_item:cn 405 | % { g__saved_word_meaning_##1_clist } 406 | % { ####1 } 407 | % \endgroup 408 | % } 409 | % \end{baselist} 410 | % \en_review_if:nT { ##1 } { \show_next_review: } 411 | % } 412 | % \clearpage 413 | % } 414 | 415 | \ExplSyntaxOff 416 | \endinput 417 | -------------------------------------------------------------------------------- /english.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt]{ctexart} 2 | \input english.code.tex 3 | \setdate{2020-11-23} 4 | 5 | \begin{document} 6 | \begin{wordlist}{2020-08-27}{12-完形} 7 | \word[2020-08-29]{judgments}[n]{评判} 8 | \word[2020-08-31]{justice}[n]{法官} 9 | \word{guardian}[n]{保护者；守卫者} 10 | \word{the rule of law}{法治} 11 | \word{impartial}[a]{公正的，中立的} 12 | \word(1){cannot emphasize \dots too much/\penalty50 enough}{再……强调都不为过} 13 | \word(2){lest}{唯恐；以免\enhint{表\enem{目的}，其后从句一般要用\enem{虚拟语气}}} 14 | \word[2020-09-11](3){restore}[v]{恢复} 15 | \word{eliminate}[v]{消除，消灭} 16 | \word{code of conduct}{行为准则} 17 | \word[2020-08-29]{judiciary}(dZU"dIS@ri)[n]{司法部，司法系统} 18 | \word(4){challenge}[v]{挑战，质疑} 19 | \word[2020-11-02](4){compromise}[v]{违背；达不到；使遭受危险（或怀疑）} 20 | \word(4){accept sth. as sth.}{相信/认为某事为真} 21 | \word[2020-09-03](5){be bound by}{受（准则、协议、法规等）约束} 22 | \word(5){advance career/knowledge}{取得进步} 23 | \word[2020-09-21](5){advance}[v]{提出（建议或提案等）} 24 | \word(6){resistant (to)}[a]{对……抵抗的，对抗的} 25 | \word(6){immune (to)}[a]{免疫的；不受……影响的} 26 | \word[2020-08-29](6){prone (to)}[a]{很可能做（遭受）……的\enhint{to后面若为名词，多为 27 | \enem{不好之事}}} 28 | \word(7){resort (to)}[v]{诉诸；求助于} 29 | \word(7){apply (to)}[v]{使用；应用；\enem{适用}；有效} 30 | \word(8){evade}[v]{回避谈论；逃避} 31 | \word(9){line}[n]{界限，边界\enhint{常用于指“两国/地之间的分界线”}} 32 | \word{framer}[n]{制定者} 33 | \word[2020-08-29]{envision}[v]{展望；想象} 34 | \word(14){precisely}[adv]{正是、恰恰是} 35 | \word(10){regard/see/envision as}{固定搭配，引出观点} 36 | \word(12){upset}[v]{使苦恼；打败} 37 | \word(13){cultivate}[v]{结交；获得} 38 | \word[2020-09-26]{liberty}[n]{自由} 39 | \word{property}[n]{财产} 40 | \word[2020-08-29]{split along sth.}{在……上有分歧} 41 | \word[2020-08-31]{ideological}[a]{意识形态的} 42 | \word(15){division}[n]{部门；分歧} 43 | \word(15){convention}[n]{习俗；惯例} 44 | \word(16){exclude}[v]{将……排除在外；故意不考虑} 45 | \word(16){shape}[v]{对某物的发展有重要影响} 46 | \word(17){dismiss}[v]{对……不屑一顾；不予考虑} 47 | \word[2020-08-29](17){distort}[v]{扭曲，使变形；曲解} 48 | \word{be ranked as unjust}{公正度方面参差不齐} 49 | \word[2020-08-29](18){exploit}[v]{利用，开发} 50 | \word[2020-11-20](18){address}[v]{对付；发表演讲，向……讲话} 51 | \word(19){agreeable}[a]{令人愉快的；同意的；可以接受的} 52 | \word[2020-08-31](19){amiable}[a]{和蔼可亲的；亲切友好的} 53 | \word[2020-08-29](19){accountable}[a]{负有责任的；应作出解释的} 54 | \word(20){in a word}{简言之，总之\enhint{用于对前文提及的多个观点进行总结}} 55 | \end{wordlist} 56 | \begin{wordlist}{2020-08-28}{12-阅读1} 57 | \word[2020-08-30]{contend}[v]{主张，坚决认为} 58 | \word[2020-08-30]{group dynamics}{群体互动} 59 | \word[2020-10-02](21){obstacle}[n]{阻碍} 60 | \word{recipient}[n]{接受者} 61 | \word[2020-08-30]{a host of}{大量} 62 | \word{promising}[a]{有前景的} 63 | \word[2020-10-17]{perceptive}[a]{敏锐的，有洞察力的} 64 | \word{critique}[n]{批评，评论} 65 | \word[2020-08-30]{lameness}[n]{缺陷，不足之处} 66 | \word[2020-09-10]{spot-on}[a]{准确的} 67 | \word[2020-09-18]{mobilize}[v]{调动，动员} 68 | \word[2020-09-25]{demonstrate}[v]{证实} 69 | \word[2020-08-30]{flawed}[a]{错误的，有缺陷的} 70 | \word{billboard}[n]{广告牌} 71 | \word[2020-08-30]{glaring}[a]{明显的（负面的事物）} 72 | \word[2020-08-30]{a glaring flaw}{明显的缺陷} 73 | \word(23){adequately}[ad]{充分地} 74 | \word[2020-09-05]{probe}[v]{探究} 75 | \word(23){illustrate}[v]{说明} 76 | \word[2020-09-01]{exert}[v]{施加，施以} 77 | \word{exert influence on sth.}{对某事产生影响} 78 | \word[2020-08-30]{subtle}[a]{不易察觉的，微妙的} 79 | \word[2020-08-30](24){imitation}[n]{模仿} 80 | \word[2020-09-29]{bureaucrat}(bjU@r@kr\ae t)[n]{官员、官僚} 81 | \word[2020-09-07]{steer}[v]{引导} 82 | \word[2020-11-05]{virtuous}("vE:tjU@s)[a]{有道德的} 83 | \word[2020-08-30]{tactic}[n]{方法，策略} 84 | \word[2020-08-30]{enginieer}[v]{策划，设计} 85 | \word[2020-09-27](25){profound}[a]{深刻的} 86 | \word(25){questionable}[a]{可疑的} 87 | \end{wordlist} 88 | \begin{wordlist}{2020-08-29}{12-阅读2} 89 | \word{provoke}[v]{激起} 90 | \word[2020-08-31]{outrage}[n]{义愤，愤怒} 91 | \word{longstanding}[a]{长时间的} 92 | \word[2020-08-31]{renege on}{背信弃义，食言} 93 | \word{a longstanding commitment to}{对……的一项长期承诺} 94 | \word[2020-09-02]{abide by}{遵守} 95 | \word(26){condemn}[v]{谴责} 96 | \word[2020-09-13](26){reaffirm}[v]{重申} 97 | \word(26){dishonor}[v]{违背} 98 | \word[2020-09-06](26){secure}[v]{获取} 99 | \word[2020-08-31]{desperate}[a]{不顾一切的，孤注一掷的} 100 | \word{a desperate effort}{拼死努力} 101 | \word[2020-09-02]{state regulator}{州监管机构} 102 | \word[2020-08-31]{extension}[n]{延期} 103 | \word{extension of the license}{许可证延期} 104 | \word[2020-08-31]{subject}[a]{取决于；视……而定} 105 | \word[2020-10-31]{be subject to}{须遵守某事物，服从，受……支配，取决于，经历，遭受} 106 | \word[2020-09-04]{enrage}[v]{激怒} 107 | \word[2020-09-08]{obscure (a.)}[a]{不分明的，费解的} 108 | \word[2020-08-31]{precedent-setting}[a]{开创先例的} 109 | \word{patchwork}[n]{拼缝物，补丁} 110 | \word{go to war with}{开战} 111 | \word[2020-09-28]{pledge}[v]{保证给予、正式承诺} 112 | \word{pledge to do}{保证，发誓} 113 | \word(30){defy}[v]{蔑视、藐视、违抗} 114 | \word(30){withdraw}[v]{撤回} 115 | \end{wordlist} 116 | \begin{wordlist}{2020-08-30}{12-阅读3} 117 | \word[2020-09-01]{context}[n]{背景，环境} 118 | \word{misinterpretation}[n]{误解} 119 | \word{self-deception}[n]{自欺} 120 | \word(31){deceptiveness}[n]{欺骗} 121 | \word[2020-11-16]{abound}[v]{充满} 122 | \word{protoscience}[n]{源科学} 123 | \word{stake}[v]{用桩支撑，标桩} 124 | \word{stake (out) a claim}{公开声明对……拥有所有权} 125 | \word{mining}[n]{采矿} 126 | \word[2020-09-03]{scrutiny}[n]{仔细而彻底的检查} 127 | \word(32){inspection}[n]{检查，彻查} 128 | \word[2020-10-14]{interaction}[n]{相互作用，相互影响} 129 | \word{confrontation}[n]{相互对抗} 130 | \word[2020-09-01]{prevailing}[a]{流行的} 131 | \word{confirmation}[n]{确认} 132 | \word{paradox}[n]{悖论，自相矛盾的话} 133 | \word{duplication}[n]{重复，复制} 134 | \word{modification}[n]{修改，修饰} 135 | \word[2020-09-29]{refutation}[n]{辩驳，反驳} 136 | \word{novelty}[n]{新奇的事物} 137 | \end{wordlist} 138 | \begin{wordlist}{2020-08-31}{12-阅读4} 139 | \word{unionist}[n]{工会会员} 140 | \word{civil servant}[n]{公务员} 141 | \word(36){membership}[n]{会员数量} 142 | \word{thriving}[n]{壮大，昌盛} 143 | \word{left-of-centre politics}{中左政治} 144 | \word{Labor Party}{工党} 145 | \word{as its name implies}{顾名思义} 146 | \word{tie}[n]{关联} 147 | \word{shut things down}{举行罢工} 148 | \word{unionism}[n]{贸易；工会主义} 149 | \word[2020-09-02](37){prudent}[a]{谨慎的} 150 | \word{fearsome}[a]{可怕的，害怕的} 151 | \word{backload}[v]{回程装载，变相提高} 152 | \word[2020-09-18]{vigorously}[ad]{强烈地} 153 | \word{oppose}[v]{反对，反抗} 154 | \word[2020-09-20]{notoriously}[ad]{出名地} 155 | \word[2020-09-02]{charter schools}{特许学校} 156 | \word[2020-11-01]{drawn-out}[a]{旷日持久的} 157 | \word{hardline}[a]{强硬的，不妥协的} 158 | \word[2020-09-04]{clamp down}{镇压} 159 | \word[2020-09-02]{rally against}{召集人群以反对……} 160 | \word(38){dominant}[a]{占据主导地位的} 161 | \word{norms of culture}{文化标准} 162 | \word{civil services}{行政部门} 163 | \word{stay put}{保持不变} 164 | \word(40){appreciation}[n]{欣赏} 165 | \end{wordlist} 166 | \begin{wordlist}{2020-09-01}{12-七选五} 167 | \word[2020-10-01]{tales}[n]{轶事，故事，秘闻} 168 | \word[2020-09-05]{fleeting}[a]{短暂的，飞逝的} 169 | \word{marvel}[n]{奇迹} 170 | \word{inheritor}[n]{（生活或思想的）继承人} 171 | \word{warrior}[n]{勇士} 172 | \word{entrepreneur}(""Ontr@pr@"nE:{(r)})[n]{企业家} 173 | \word[2020-09-05]{commercial agenda}{商业模式} 174 | \word{beaver}[n]{海狸} 175 | \word[2020-09-03]{superfluous}(su:"pE:flUEs)[a]{多余的} 176 | \word{move through}{穿过，通过} 177 | \word{move through the world}{比喻在世上生存} 178 | \word{dam}[n]{大坝} 179 | \word[2020-10-07]{advent}[n]{出现，到来} 180 | \word[2020-09-05]{tap}[n]{水龙头} 181 | \word[2020-09-07]{adhere}[v]{粘着，依附，追随} 182 | \word[2020-09-05]{passive}[a]{被动的} 183 | \word{strip}[v]{出去，剥夺} 184 | \word[2020-09-03]{constituent}(k@n"stItSU@nt)[n]{成分，构成} 185 | \word{millennium}(mI"lEnI@m)[n]{千禧年} 186 | \word{embed}[v]{使嵌入} 187 | \word[2020-09-27]{mount}[v]{发起，施加} 188 | \end{wordlist} 189 | \begin{wordlist}{2020-09-10}{05-完形} 190 | \word(3){not very great in ability}{能力有局限} 191 | \word(4){ignore}[v]{传递否定之意\enhint{强调\enem{主观上的故意不理睬或忽视}}} 192 | \word(5){though}[ad]{可是} 193 | \word(7){distinguish}[v]{辨别；看清，认出，听出\enhint{强调（\enem{从很多内 194 | 容中精准清晰地}）辨识区分出}} 195 | \word(8){dissolve}[v]{溶解} 196 | \word[2020-09-17](8){disperse}[v]{散开，分散} 197 | \word[2020-10-16](8){diffuse}[v]{使（气体等）扩散；传播} 198 | \word{generate}[v]{产生；使生成} 199 | \word(9){when}{相反，却\enhint{可以表转折，但不同于对比，对比重在客观呈现两 200 | 种对立的情况，无意凸显其中任意一种}} 201 | \word(9){whereas}{而\enhint{用以对比或比较两个事实}} 202 | \word(11){impulse}[n]{冲动，突然的欲望；动力，冲力} 203 | \word(14){incompetent}[a]{无能力的，不胜任的} 204 | \word(15){summon}[v]{召集；召唤} 205 | \word(16){nevertheless}{然而，不过} 206 | \word[2020-09-14](18){repel}[v]{推开，赶走，驱除；使厌恶} 207 | \word(19){identifiable}[a]{可识别的} 208 | \end{wordlist} 209 | \begin{wordlist}{2020-09-10}{05-阅读1} 210 | \word{outrage}(aUtreIdZ)[v]{激起义愤；使震怒} 211 | \word[2020-09-27]{grievance}[n]{不满} 212 | \word(21){justify an assumption}{证明一种假定} 213 | \word(21){make a comparison}{比较发现相同或相异之处} 214 | \word(21){pose a contrast}{对比明显差异} 215 | \word[2020-09-12](22){resent}[v]{厌恶} 216 | \word{counterpart}[n]{与对方地位想当的人/与另一方作用相同的物} 217 | \word{capuchin}("k\ae pju""tSIn)[n]{（南美的）卷尾猴} 218 | \word{readily}[ad]{乐意地，心甘情愿地} 219 | \word(23){inclined to}{更倾向于} 220 | \word[2020-09-12](23){temperament}[n]{性情} 221 | \word{chamber}[n]{房间} 222 | \word[2020-09-19]{reluctant}[a]{不情愿的} 223 | \word[2020-09-14]{toss}[v]{投掷} 224 | \word[2020-09-14]{resentment}(rI"zentm@nt)[n]{愤怒} 225 | \word(24){co-operative}[a]{合作的} 226 | \word{preserve}[n]{[单数]（某群体）独有或专用的活动} 227 | \word[2020-11-01]{abundantly}[ad]{极其，非常} 228 | \word{righteous indignation}{义愤} 229 | \end{wordlist} 230 | \begin{wordlist}{2020-09-10}{05-阅读2} 231 | \word[2020-09-14]{lobby}[n]{游说者} 232 | \word[2020-09-14]{buy}[v]{相信（尤指不大可能的事），承认接受} 233 | \word(26){argument}[n]{争吵；论据；内容提要；论证} 234 | \word[2020-10-05]{parallel}[n]{情形} 235 | \word[2020-09-17]{preface}[n]{前言} 236 | \word[2020-10-10]{critical}[a]{至关重要的} 237 | \word[2020-09-14]{panel}[n]{专家团} 238 | \word[2020-11-02]{enlist}(In"lIst)[v]{谋求某人帮助} 239 | \word[2020-10-10]{quarter}[n]{方面} 240 | \word{fume}[n]{[通常作复数]（气味强烈的）烟，气} 241 | \word{paralysis}[n]{瘫痪} 242 | \word{paralysis by analysis}{分析导致瘫痪} 243 | \word{bill}[n]{议案} 244 | \word{steward}("stju:@d)[n]{保护者} 245 | \word[2020-10-23]{initiative}[n]{先行措施；倡议} 246 | \word[2020-10-10]{fashion}[v]{形成，塑造} 247 | \word{sound}[a]{完好的，健康的，无损伤的} 248 | \word{take the legislative initiative}{行使立法动议权} 249 | \word{fashion conservation measures}{制定保护措施} 250 | \word{power plants}{发电厂} 251 | \word{be environmentally sound}{对环境无害，有益环境} 252 | \word[2020-09-27](30){negligence}[n]{忽视} 253 | \end{wordlist} 254 | \begin{wordlist}{2020-09-14}{05-阅读3} 255 | \word[2020-09-29]{suspend}[v]{暂停} 256 | \word[2020-10-01]{formulate}[v]{创立（新观点、理论等）} 257 | \word{disguise}(dIs"gaIz)[v]{伪装，假扮} 258 | \word{disguised}[a]{伪装的，假扮的} 259 | \word[2020-09-16]{limbic}[a]{边缘的，脑边缘系统的} 260 | \word[2020-11-06]{cortex}[n]{（脑或肾的）皮层，皮质} 261 | \word{prefrontal}[a]{（脑）额叶前部的，前额的} 262 | \word[2020-11-02](31){course}[n]{过程，进程} 263 | \word(31){be susceptible to}{容易受到……影响} 264 | \word{vivid}[a]{清晰的} 265 | \word{work through}{克服} 266 | \word(33){aggravate}[v]{加重} 267 | \word{visualize}[v]{想象} 268 | \end{wordlist} 269 | \begin{wordlist}{2020-09-17}{05-阅读4} 270 | \word[2020-09-19]{aspire to}[v]{渴望、立志做……} 271 | \word{see A as responsible for B}{认为A应对B负责，认为A是B的主要原因} 272 | \word{counter-culture}[n]{反文化运动} 273 | \word[2020-09-19]{command}[v]{驾驭，控制} 274 | \word[2020-09-28]{controversialist}(""kOntr@"vE:S@lIst)[n]{善辩者，好辩者} 275 | \word{public figure}{公众人物} 276 | \word[2020-09-30]{permissive}[a]{放任的，纵容的，姑息的} 277 | \word[2020-11-17]{cult}[n]{狂热推崇} 278 | \word{the modestly educated}{受教育不太多的人} 279 | \word{sought}[v]{seek过去分词，寻求} 280 | \word{triumph over}{胜过} 281 | \word[2020-10-14]{authentic}[a]{可靠的，真实的} 282 | \word{spell}[v]{导致，招致（尤指坏事）} 283 | \word{elevated}[a]{提高的} 284 | \word[2020-09-19]{performative}[a]{表述行为的} 285 | \word[2020-09-30]{spontaneity}(""spont@"ni:ItI)[n]{由内心所驱使的行为} 286 | \word[2020-09-19](37){denote}[v]{表示} 287 | \word{put pen to paper}{下笔，写字} 288 | \word{subtitle}[n]{副标题} 289 | \word[2020-09-30]{propose}[v]{提议} 290 | \word{radical}[v]{彻底的} 291 | \word{grieve}[v]{哀叹} 292 | \word{paper plates}{纸盘} 293 | \word(39){contempt}[v]{鄙视} 294 | \word(40){humble}[a]{卑微的} 295 | \word(40){noble}[a]{高贵的} 296 | \end{wordlist} 297 | \begin{wordlist}{2020-09-19}{06-阅读1} 298 | \word{democratic}[a]{大众的} 299 | \word{democratize}(dI"mokr@taIz)[v]{（使）大众化} 300 | \word{uniformity}[n]{同样；一致性} 301 | \word[2020-09-21]{discourse}("dIskO:s)[n]{（严肃的）谈话，交谈} 302 | \word{casualness}("k\ae ZjU@lnIs)[n]{随意，随便} 303 | \word[2020-10-06]{deference}[n]{顺从，尊重} 304 | \word[2020-09-21]{cater to}{迎合} 305 | \word[2020-09-21](21){monopolize}[v]{垄断，独占} 306 | \word(22){owe its emergence to}{出现的原因是……} 307 | \word{poisonous}[a]{有毒的} 308 | \word{elevate}[v]{提升} 309 | \word{indice}[n]{指标} 310 | \word[2020-09-21]{unprecedented}(2n"presIdentId)[a]{空前的} 311 | \word{at unprecedented levels}{达到前所未有的水平} 312 | \word{Census}[n]{人口普查} 313 | \word{bilingual}(""baI"lINgw@l)[a]{会两种语言的} 314 | \word{Hispanic}[n]{西班牙裔} 315 | \word[2020-10-02]{intermarriage}{异族通婚} 316 | \word{assimilative power}(@"sImIl@tIv){同化力量} 317 | \word{deteriorating}(dI"tI@rI@""reItIN)[a]{退化的} 318 | \word[2020-10-25]{divisive}[a]{引起分裂的，造成不和的} 319 | \word[2020-09-25]{seethe}(si:D)[v]{翻滚，涌动，翻腾} 320 | \word{seething anger}{涌动的怒火} 321 | \word{social indices}{社会指标} 322 | \word(25){absorption}[a]{同化} 323 | \end{wordlist} 324 | \begin{wordlist}{2020-09-19}{06-阅读2} 325 | \word[2020-09-25]{hostile}[a]{敌对的} 326 | \word{superb}[a]{极好的} 327 | \word{worthy}[a]{有价值的，值得尊敬的} 328 | \word{sandal}[n]{凉鞋} 329 | \word[2020-09-23]{deliciously}[ad]{非常令人愉悦地，有趣地} 330 | \word{cocktail}[n]{鸡尾酒} 331 | \word[2020-09-25]{lounge}[n]{（酒吧中的）雅间，沙龙酒吧} 332 | \word{rocket}[v]{飞涨} 333 | \word{dedicated}[a]{专注的} 334 | \word[2020-10-23]{lean}[a]{瘦的} 335 | \word{clientele}(""kli:@n"tel)[n]{（统称）顾客，主顾，客户} 336 | \word[2020-09-29]{pointed}[a]{尖锐的，尖刻的，锐利的} 337 | \word[2020-09-23]{flagstone}[n]{石板（方形，用于铺地面等）} 338 | \word[2020-11-05]{detached}[a]{中立的} 339 | \end{wordlist} 340 | \begin{wordlist}{2020-09-20}{06-阅读3} 341 | \word{game}[n]{猎物，游戏，比赛} 342 | \word{easy game}{容易被捕获的猎物} 343 | \word(31){vulnerable}[a]{易受伤的，脆弱的，敏感的} 344 | \word(31){outlive}[v]{比……活的长} 345 | \word[2020-09-24]{exploitation}[n]{开发，开采} 346 | \word(32){stock}[n]{储备物，备用物；供应物} 347 | \word(32){catch size}{捕捞量} 348 | \word{conservative}[a]{保守的} 349 | \word{proportion}[n]{比例} 350 | \word{vessel}[n]{船，舰} 351 | \word[2020-10-01]{prey}[n]{被（其他动物或人）捕食的动物，猎物} 352 | \word{saturate}[v]{使充满，使充斥} 353 | \word{a high proportion of}{高比例的} 354 | \word{be saturated with}{充满，处于饱和状态} 355 | \word[2020-09-22]{fish stock}{鱼类资源量} 356 | \word{baseline}[n]{基础，起点，基线} 357 | \word[2020-09-22]{a bad way to do business}{有害的经营方式} 358 | \end{wordlist} 359 | \begin{wordlist}{2020-09-20}{06-阅读4} 360 | \word[2020-09-30]{phony}[a]{假（冒）的，伪造的，欺骗的} 361 | \word[2020-10-01]{daffodil}("d\ae f@dIl)[n]{水仙} 362 | \word{ideology}(""aIdI"ol@dZI)[n]{思想体系，思想意识} 363 | \word[2020-10-05]{literacy}[n]{识字，读写能力，有文化，素养} 364 | \word{worm}[n]{蠕虫} 365 | \word[2020-10-12]{bummer}[n]{令人失望或不愉快的局面或情形} 366 | \word[2020-09-26]{beam}[v]{笑容满面，眉开眼笑} 367 | \word[2020-09-26]{arthritis}(A:"TraItIs)[n]{关节炎} 368 | \word{have an agenda}{有一个目的} 369 | \word[2020-10-10]{illusion}(I"lu:Zn)[n]{假象} 370 | \word[2020-09-22]{clove}[n]{丁香} 371 | \end{wordlist} 372 | \begin{wordlist}{2020-09-21}{06-完形} 373 | \word[2020-09-23](2){cope}[v]{应付，处理} 374 | \word[2020-09-25](2){retain}[v]{保持；记住} 375 | \word(6){range from A to B}{（在一定范围内）变化，变动} 376 | \word(6){sb. differs (with sb.) about/on/as to/over sth.}{某人在某问题上意见 377 | 不同} 378 | \word(7){now that}{既然，由于} 379 | \word(7){except that}{除了……之外都；要不是由于，只可惜} 380 | \word{bulk}[n]{大部分} 381 | \word[2020-09-23]{address}[v]{设法解决，处理} 382 | \word{package deal}{一揽子交易，整批交易} 383 | \word(12){lodging}[n]{寄宿处；租住的房间} 384 | \word[2020-09-27](13){stroll}[v]{溜达，闲逛\enhint{强调一种悠闲、放松的状态}} 385 | \word(14){whereas}{然而，但是} 386 | \word(14){while}{而，然而；虽然\enhint{位于句首时，用法常等同于although，引 387 | 导让步逻辑}} 388 | \word[2020-09-25](16){turn around}{使转身；\enem{使好转}，扭转} 389 | \word(16){turn over}{使翻转；翻身} 390 | \word(16){turn on}{打开，启动} 391 | \word(16){turn up}{调高} 392 | \word[2020-09-27](17){complementary}[a]{互补的，补充的} 393 | \word{as sb. puts it}{正如某人所说} 394 | \word[2020-09-28](19){interpret}[v]{解释，说明} 395 | \word(20){supervision}[n]{监督，管理} 396 | \word[2020-09-23](20){manipulation}[n]{操纵} 397 | \word(20){regulation}[n]{管理} 398 | \word[2020-09-23](20){coordination}[n]{协调，配合} 399 | \end{wordlist} 400 | \begin{wordlist}{2020-09-27}{07-阅读1} 401 | \word[2020-09-29]{tournament}[n]{锦标赛} 402 | \word{the ranks}{（组织或团体的）成员} 403 | \word[2020-10-27]{stamina}[n]{耐力} 404 | \word(22){hysteria}[n]{歇斯底里\enhint{强调一种近乎病态的情绪失控、不可理喻 405 | 的极端行为，多含贬义色彩}} 406 | \word{swamp}[v]{（骤然）淹没，浸没} 407 | \word{encode}[v]{编码} 408 | \word[2020-10-14]{entail}[v]{使必须（做某事）；需要（做某事）} 409 | \word(23){process}[v]{处理} 410 | \word{pursuit}[n]{事业，消遣，爱好} 411 | \word[2020-10-14]{biographical}[a]{传记的，关于某人生平的} 412 | \word{startling}[a]{令人吃惊的} 413 | \word[2020-09-29]{trait}[n]{特征，个人特征} 414 | \word(24){overlook}[v]{忽略} 415 | \word[2020-10-27](25){proverb}[n]{谚语} 416 | \word(25){Faith will move mountains.}{精诚所至，金石为开} 417 | \word[2020-10-27](25){{Like father, like son.}}{有其父必有其子} 418 | \end{wordlist} 419 | \begin{wordlist}{2020-09-27}{07-阅读2} 420 | \word{supplement}[n]{（报纸、杂志等的）副刊} 421 | \word[2020-10-27]{column}[n]{（报纸、杂志的）专栏（文章）} 422 | \word[2020-09-29]{analogy}[n]{类似，类比，类推} 423 | \word[2020-10-12]{deduce}[v]{推理，推断，演绎} 424 | \word[2020-10-04]{field}[v]{处理，应付（问题或意见）} 425 | \word[2020-10-04]{elude}[v]{难倒} 426 | \word{variation}[n]{变体} 427 | \word[2020-10-30]{chronological}[a]{按时间顺序的} 428 | \word(28){computational}[a]{计算的} 429 | \word{verbal}[a]{词汇的，文字的} 430 | \word{predict}[v]{预示} 431 | \word[2020-10-30]{toil}[v]{熬过，艰难度过} 432 | \end{wordlist} 433 | \begin{wordlist}{2020-09-29}{07-完形} 434 | \word{regime}[n]{统治制度，政权，政体} 435 | \word[2020-10-30]{viable}[a]{能独立发展的，能独立生活的} 436 | \word{Iberian}[a]{伊比利亚的} 437 | \word[2020-10-06]{representative government}{代议制政府} 438 | \word(1){native}[n]{本地人，当地人\enhint{在某地出生并长大的原住民}} 439 | \word[2020-10-30](2){integrate}[v]{合并，结合} 440 | \word(2){typical}[a]{典型的\enhint{侧重说明某类人群或事物中最具代表性的特征}} 441 | \word{rally}[v]{召集；集合} 442 | \word(11){allow}[v]{允许，准许，承认\enhint{可作正式用语，表示（法律上）接 443 | 受，任课，承认}} 444 | \word(11){grant}[v]{同意，准予（某人提出的要求）\enhint{多表同意\enem{他人的 445 | 请求}；也可表承认，强调某情况属实，但不影响自己的想法}} 446 | \word{egalitarian}(I""g\ae lI"te@rI@n)[a]{平等主义的} 447 | \word[2020-10-16]{temper}[v]{使缓和，使温和} 448 | \word[2020-10-01]{sentiment}[n]{观点，意见} 449 | \end{wordlist} 450 | \begin{wordlist}{2020-10-01}{08-阅读1} 451 | \word[2020-10-31]{psychiatrist}(saI"kaI@trIst)[n]{精神科医生，精神科专家} 452 | \word{trigger}[n]{诱因} 453 | \word{ovary}[n]{卵巢} 454 | \word{female reproductive organs}{女性生殖器官} 455 | \word{dose}[n]{剂量} 456 | \word{chronic}[a]{（尤指疾病）长期的，慢性的} 457 | \word{interpersonal}[a]{人与人之间的，人际的} 458 | \word{domestic}[a]{家庭的；家务的} 459 | \word{one-shot deal}{一次性事件} 460 | \word[2020-10-05]{wear-and-tear}{损耗，磨损} 461 | \word{combat}[n]{战斗} 462 | \word{physical violence}{身体上的暴力行为} 463 | \word[2020-10-07]{devastating}[a]{毁灭性的} 464 | \word{paycheck}[n]{付薪水的支票} 465 | \word{live from paycheck to paycheck}{盼着下一张工资单过日子} 466 | \word(24){household expenses}{家庭开支} 467 | \word{obligation}[n]{责任} 468 | \word{diffuse stress}{缓解压力} 469 | \end{wordlist} 470 | \begin{wordlist}{2020-10-01}{08-阅读2} 471 | \word[2020-10-03]{endeavour}[n]{尽力，竭力} 472 | \word{issue a report}{发布报告} 473 | \word[2020-10-03]{scientific endeavour}{科学探索} 474 | \word[2020-11-06](27){considerably}[ad]{非常} 475 | \word{site-licensing agreement}{网站许可协议} 476 | \word{repository}{存储库} 477 | \word[2020-10-07]{hybrid}[n]{（不同事物的）混合物} 478 | \word(28){facilitate}[v]{便利} 479 | \end{wordlist} 480 | \begin{wordlist}{2020-10-03}{07-阅读3} 481 | \word[2020-10-05]{fair play}{公平竞争} 482 | \word{count on}{依赖} 483 | \word{pink slip}{解聘通知书} 484 | \word{stripe}(straIp)[n]{类型，类别} 485 | \word{parachute}("p\ae r@Su:t)[n]{降落伞，引申为保护伞} 486 | \word{debate}[v]{讨论} 487 | \word{in times of financial setback}{经济困难时期} 488 | \word{lay off/get laid off}{解雇/被解雇} 489 | \word[2020-10-07]{make up sb. with sth.}{弥补} 490 | \word[2020-10-19]{weather}[v]{经受住，安全渡过（困难）} 491 | \word{bear}[v]{承担} 492 | \word[2020-10-07]{deductible}[n]{保险扣除条款，指代“免赔额”} 493 | \word[2020-10-27]{demographics}[n]{人口统计数据} 494 | \word{odds}[n]{（事情发生的）可能性} 495 | \word[2020-10-09]{attendant}[a]{伴随的，随之而来的} 496 | \word{eightfold}[ad./a]{八倍地（的），八重地（的）} 497 | \word{fashionable}[a]{流行的} 498 | \word{stock market fluctuation}{股市波动} 499 | \word{wholesale}[a]{大批的，大规模的} 500 | \word[2020-10-05]{fallout}[n]{后果，余波} 501 | \word[2020-10-07](34){outweigh}[v]{超过} 502 | \word(35){cliff}[n]{悬崖} 503 | \end{wordlist} 504 | \begin{wordlist}{2020-10-09}{07-阅读4} 505 | \word[2020-11-08]{accounting}[n]{财务} 506 | \word[2020-10-24]{compliance}[n]{服从，遵守} 507 | \word{governance}[n]{治理} 508 | \word{earn}[v]{招惹} 509 | \word{suite}[n]{（要人的）（一批）随员，随从} 510 | \word[2020-10-13]{put right}{纠正} 511 | \word{nasty headlines}{负面头条} 512 | \word{heads rolling}{重大的人事变故} 513 | \word{left}[v]{leave过去分词，遗留} 514 | \word{odd}[a]{古怪的} 515 | \word[2020-10-26](36){statement}[n]{声明，陈述} 516 | \word[2020-11-20]{intricate}[a]{复杂精细的} 517 | \word[2020-10-19]{peer into}{检查} 518 | \word{asset}[n]{资产} 519 | \word[2020-10-11]{redundancy}[n]{冗余} 520 | \word[2020-10-11](38){due}[a]{应得的，应有的} 521 | \word[2020-10-11](39){perceive}[v]{感知，察觉，意会} 522 | \word{justify}[v]{证明……有理、正当} 523 | \word[2020-10-16]{overshadow}[v]{使扫兴，使打丢脸面} 524 | \word[2020-10-11]{go astray}{丢失；被盗} 525 | \word{do the rounds}{为寻求政治支持而拜访，引申为一直处于审议的过程中} 526 | \word[2020-10-11]{put sb. on notice}{警示某人，警告某人} 527 | \word(40){be essential to}{对……极其重要} 528 | \end{wordlist} 529 | \begin{wordlist}{2020-10-10}{08-完形} 530 | \word{controversy}[n]{争论，辩论，论战} 531 | \word{bacterial}{细菌} 532 | \word(1){be prepared to do}{愿意做（尤指难事）；准备好做} 533 | \word(1){be obliged to do}{（因法律、义务、形势等）被迫做} 534 | \word(2){rare bird}{罕见、稀有的人或事物} 535 | \word[2020-10-12](4){subsequently}[ad]{随后} 536 | \word[2020-11-09]{tremble}[v]{极担心，焦虑，恐惧} 537 | \word{bring about}{引起，导致} 538 | \word[2020-10-12](8){in progress}{（事情）进行中} 539 | \word(8){in need}{贫困/困难中的} 540 | \word(8){in question}{正被讨论的} 541 | \word(10){mean value}{平均值} 542 | \word{elite}[n]{（统称）掌权人物，社会精英} 543 | \word[2020-10-21]{renowned}[a]{著名的} 544 | \word[2020-11-13]{nasty}[a]{危险的；严重的} 545 | \word{genetic isolation}{遗传隔离，指由于个体基因组成差异，不能产生后代，导 546 | 致生殖隔离} 547 | \word[2020-10-25](11){indefinitely}[ad]{不明确地；无期限地} 548 | \word(12){career}[n]{事业；成就} 549 | \word(13){affirm}[v]{肯定，证实} 550 | \word[2020-11-09](13){approve}[v]{赞成，同意} 551 | \word(16){value}[v]{尊重，重视} 552 | \word(18){woven}[v]{编，织；编造} 553 | \word(19){converted sb. to sth.}{使转变（信仰等）} 554 | \word[2020-10-25](19){subject sb. to sth.}{使遭受，经历} 555 | \word(20){paradoxical}[a]{似是而非的，自相矛盾的} 556 | \word[2020-11-21](20){incompatible}[a]{不相容的，不兼容的，不能共存的} 557 | \end{wordlist} 558 | \begin{wordlist}{2020-10-10}{08-阅读3} 559 | \word[2020-10-28]{obscure (v.)}[v]{遮掩；遮蔽} 560 | \word[2020-10-19]{protein}[n]{蛋白质} 561 | \word[2020-10-25]{secular}[a]{长期的} 562 | \word{secular trend}{长期趋势} 563 | \word{substantial}[a]{大量的，坚固的，实质的} 564 | \word{bipedal}[a]{两足动物的} 565 | \word(33){non-American}[n]{非美国人} 566 | \word[2020-10-12]{by and large}{总的来说，大体上} 567 | \word(35){cyclic}[a]{循环的} 568 | \end{wordlist} 569 | \begin{wordlist}{2020-10-10}{08-阅读4} 570 | \word[2020-10-12]{chop}[v]{砍} 571 | \word[2020-10-25]{make compromise}{做出妥协} 572 | \word{from the bottom up}{全方位地，彻底地} 573 | \word[2020-10-14]{infancy}[n]{婴儿期，初期} 574 | \word(37){confront}[v]{面对；遭遇；比较} 575 | \word[2020-10-25](37){delicate situation}{微妙的处境} 576 | \word{bedrock}[n]{基石，牢固基础} 577 | \word{part}[v]{分裂，分离} 578 | \word{inflate}[v]{使膨胀，使暴涨} 579 | \word[2020-10-25]{formula}[n]{方案} 580 | \word[2020-10-12]{in office}{就职} 581 | \word{extend}[v]{扩大} 582 | \word[2020-10-12]{Revolutionary War}{独立战争} 583 | \end{wordlist} 584 | \begin{wordlist}{2020-10-13}{09-完形} 585 | \word{fruit-fly}[n]{果蝇} 586 | \word{dim bulb}{暗淡的灯泡；愚笨的人} 587 | \word[2020-11-19](3){stabler}[a]{更稳定的，更稳固的} 588 | \word[2020-10-28]{upkeep}[n]{保养维修（费）；喂养的成本} 589 | \word{instinct}[n]{天分} 590 | \word[2020-10-28]{wistful}[a]{渴望的；沉思的；依依不舍的；惆怅的} 591 | \word{leave sb. in the dust}{使望尘莫及} 592 | \word(9){adaptive value}{适应性价值，适应性方面的益处} 593 | \word{operant conditioning}{操作条件反射} 594 | \word[2020-10-28]{inconclusive}[a]{无明确结果的} 595 | \word(19){equivalent}[a]{（价值、数量、意义、重要性等）相等的} 596 | \word(20){in time}{过段时间，迟早；及时} 597 | \word[2020-11-12](20){better still}{更好的是} 598 | \end{wordlist} 599 | \begin{wordlist}{2020-10-18}{09-阅读2} 600 | \word{boost}[v]{增强，使增长} 601 | \word[2020-11-17]{shell out}{付一大笔款；还账} 602 | \word{over-the-counter}{柜台的，场外的} 603 | \word[2020-11-19]{rage}[n]{时尚，流行} 604 | \word[2020-10-24]{swab}[v]{蘸取} 605 | \word[2020-10-20]{saliva}(s@"laIv@)[n]{口水} 606 | \word[2020-10-20]{hawk}[v]{沿街叫卖，兜售} 607 | \word[2020-10-20]{chromosome}("kr@Um@s@Um)[n]{染色体} 608 | \word[2020-11-02]{mitochondrial}("mIt@"kondrI@l)[a]{线粒体的} 609 | \word(28){bloodline}[n]{血统谱系} 610 | \word{lineage}[n]{血统，家系，世系} 611 | \word{lump}[v]{混在一起} 612 | \word{lump together}{把……归并一起或合起来考虑} 613 | \word{patent}[v]{取得……的专利权；授予专利} 614 | \word[2020-10-20](29){overlap}[v]{与……重叠} 615 | \end{wordlist} 616 | \begin{wordlist}{2020-10-26}{09-阅读3} 617 | \word{formal education}{正规教育} 618 | \word{consistently}[ad]{始终如一地} 619 | \word(31){groundless}[ad]{无根据的} 620 | \word[2020-10-28](31){downgrade}[v]{低估} 621 | \word[2020-11-01]{deride}[v]{嘲笑} 622 | \word{workforce}[n]{劳动力} 623 | \word{disciplined}[a]{纪律严明的} 624 | \word[2020-10-30]{suspect}[v]{推测，猜想} 625 | \word{trap}[n]{困境} 626 | \word{substantially}[ad]{大幅度；本质上地} 627 | \end{wordlist} 628 | \begin{wordlist}{2020-10-29}{09-阅读4} 629 | \word[2020-11-02]{minister}[n]{牧师；部长} 630 | \word[2020-10-31]{intellectual}[n]{知识分子，脑力劳动者} 631 | \word[2020-11-04]{preoccupation}[n]{使人全神贯注的事物} 632 | \word{Puritan}[n]{清教徒（16和17世纪基督教教派成员，主张简化宗教仪式）} 633 | \word[2020-11-02]{innovation}[n]{革新} 634 | \word[2020-11-02]{distinctive}[a]{独特的} 635 | \word{scene}[n]{（尤指不愉快事件发生的）地点，现场} 636 | \word[2020-11-06]{virtuosity}[n]{技艺} 637 | \word{impressive}[a]{令人印象深刻的} 638 | \word{crown}[n]{(the Crown)王国；王冠} 639 | \word{tailor}[n]{裁缝} 640 | \word{sign}[n]{迹象、征兆} 641 | \word[2020-10-31]{sermon}[n]{说教} 642 | \word[2020-11-16]{dependent}[n]{仆人，侍从} 643 | \word[2020-10-31]{let alone}{更不用说} 644 | \word[2020-11-18]{commitment}[n]{（对工作或某活动）献身，投入；承诺} 645 | \word{clergyman}[n]{牧师} 646 | \word{religious commitment}{宗教信仰} 647 | \word[2020-10-31]{main end}{主要目的} 648 | \word(40){illusory}[a]{虚幻的} 649 | \end{wordlist} 650 | 651 | \begin{wordlist}{2020-11-01}{10-完形} 652 | \word{supervise}[v]{管理，指导} 653 | \word{plant}[n]{工厂} 654 | \word[2020-11-03]{shop factory floor}{车间} 655 | \word(1){extract}[v]{提取} 656 | \word[2020-11-03](4){mischievous}[a]{调皮的，恶作剧的；恶意的} 657 | \word[2020-11-03](4){perplexing}[a]{令人费解的} 658 | \word(5){account}[n]{描述，叙述，说明} 659 | \word(7){in case that}{以防万一；如果，假使} 660 | \word[2020-11-03]{in store}{即将到来，必将发生} 661 | \word(11){be conveyed to}{被运送，传递到} 662 | \word{peculiar}[a]{特有的，特殊的} 663 | \word{interpretation}[n]{理解，解释} 664 | \word{plateau}[n]{稳定器，停滞期} 665 | \word{slacken (off)}[v]{（使）放慢、减缓、萧条} 666 | \word{alleged}[a]{声称的，所谓的} 667 | \word[2020-11-21]{pin sth. down}{确切说明（或理解）} 668 | \word{diligent}[a]{勤奋的} 669 | \word[2020-11-03](14){desputable}[a]{有争议的，值得商榷的} 670 | \word(16){duly}[ad]{如期地；适时地；按时地} 671 | \word[2020-11-13](17){cease}[v]{停止} 672 | \end{wordlist} 673 | 674 | \begin{wordlist}{2020-11-01}{10-阅读1} 675 | \word{inexorable}[a]{不可阻挡的；无法改变的} 676 | \word[2020-11-03]{coverage}[n]{新闻报道} 677 | \word{quarter-century}[n]{25年，1/4个世纪} 678 | \word{marvel}[v]{感到惊奇，大为赞赏} 679 | \word{deem}[v]{（尤用于被动语态）认为；视为} 680 | \word{to the point of}{达到……的程度} 681 | \word{criticism}[n]{批评，批判} 682 | \word{calling}[n]{天职，使命感} 683 | \word{be taken for granted}{被视作理所当然} 684 | \word{in detail and at length}{十分详尽} 685 | \word[2020-11-05]{be tempted to}{不禁} 686 | \word[2020-11-03]{daily press}{日报} 687 | \word[2020-11-05]{contempt}[n]{轻视，藐视} 688 | \word[2020-11-09](22){elaborate}[a]{缜密的，雕饰的，精心的} 689 | \word(22){radical}[a]{激进的} 690 | \word[2020-11-03](23){contemptible}[a]{可鄙的，可轻视的} 691 | \word(24){tempted}[a]{吸引人的} 692 | \word{virtually}[ad]{几乎；差不多；事实上；实际上} 693 | \word[2020-11-03]{stylist}[n]{语言风格优美的人；文体家} 694 | \word{autobiography}[n]{自传；自传体小说} 695 | \word[2020-11-03]{save}[prep./conj]{除了} 696 | \word{revival}[n]{复兴} 697 | \word{prose}[n]{散文} 698 | \word[2020-11-05]{unholster}(2p"h@Ulst@)[v]{给（椅子等）做软垫等；（用织物）装饰（家具）； 699 | 引申为对文章加以修饰} 700 | \word{amateur}[a]{业余的} 701 | \word[2020-11-07]{in headlong retreat}{急速衰落} 702 | \word(24){appeal (to)}[v]{吸引} 703 | \word(24){dispute}[n]{争议} 704 | \word(25){mournful}[a]{悲哀的} 705 | \word[2020-11-17](25){prominent}[a]{杰出的} 706 | \end{wordlist} 707 | 708 | \begin{wordlist}{2020-11-01}{10-阅读2} 709 | \word[2020-11-03]{allocation}[n]{配置，分配} 710 | \word{authorize}[v]{批准；认可} 711 | \word[2020-11-07]{abuzz}[a]{嘈杂的，叽叽喳喳的} 712 | \word{scale sth. back}{减少（数量）；缩小（规模或范围）} 713 | \word{intellectual property}{知识财产} 714 | \word(26){authorization}{授权} 715 | \word[2020-11-05](27){dimiss}[v]{解雇，辞退，黜免，驳回} 716 | \word{curb}[n]{起约束或控制作用的事物} 717 | \word{claim}[n]{专利申请} 718 | \word{about-face}[n]{政策或意见的180度转变} 719 | \word{pool}[v]{集中（金钱、资源等）} 720 | \word[2020-11-03]{filing}[n]{卷宗} 721 | \word{established}[a]{地位稳固的} 722 | \word{stake out exclusive rights to sth.}{宣布对……的独占权} 723 | \word{established companies}{老牌公司} 724 | \word[2020-11-05]{beat sb. to the punch}{先下手为强} 725 | \word{mutual-fund}{共同基金} 726 | \word(28){goodwill}[n]{善意} 727 | \word(28){hostility}[n]{敌意} 728 | \word(28){dignity}[n]{尊严} 729 | \word{hedge (v.)}[v]{躲闪；（商业）以套期保值避免或减轻损失} 730 | \word{hedge (n.)}[n]{树篱，障碍，阻碍} 731 | \word{issue}[v]{发布} 732 | \word{in the wake of}{随某事物之后到来} 733 | \word(30){prevailing}[a]{普遍的} 734 | \end{wordlist} 735 | 736 | \begin{wordlist}{2020-11-02}{10-阅读3} 737 | \word[2020-11-20]{epidemic}[n]{（社会观念等的）泛滥，蔓延；流行病} 738 | \word[2020-11-19]{informed}[a]{见识广博的} 739 | \word[2020-11-21]{compelling}[a]{令人信服的} 740 | \word{social epidemics}{社会流行潮} 741 | \word(31){cite}[v]{引用} 742 | \word{plausible}[a]{（借口或解释）有道理的} 743 | \word{embrace}[v]{乐意采纳（建议等）} 744 | \word[2020-11-19]{cursory}[a]{粗略的；仓促的} 745 | \word{anecdotal}[a]{轶事的；趣闻的；传闻的} 746 | \word{a cursory search}{粗略调查} 747 | \word{anecdotal evidence}{轶事证据} 748 | \word{fit nicely with}{与……高度契合} 749 | \word(32){prevalent}[a]{流行的} 750 | \word{required}[a]{必要的，必需的} 751 | \word{outsize}[a]{特大的} 752 | \word{presence}[n]{存在感；影响力} 753 | \word{cascade}[n]{瀑布；大量倾泻的东西} 754 | \word{propagete}[v]{传播；宣传} 755 | \word{resistant}[a]{抵抗的，有抵抗力的} 756 | \word{simulation}[n]{模拟；仿真} 757 | \word{the dynamics of social influence}{社交影响的动态变化} 758 | \end{wordlist} 759 | 760 | \begin{wordlist}{2020-11-02}{10-阅读4} 761 | \word{moan}[v./n]{抱怨} 762 | \word{fetch}[v]{售得，卖得（某价）} 763 | \word{blame sb. for sth.}{因为某事而责备某人} 764 | \word{toxic}[a]{有毒的} 765 | \word{capital market}{资本市场} 766 | \word{carry \ldots at prices}{以……价格出售} 767 | \word[2020-11-04]{bank share}{银行股票} 768 | \word[2020-11-12]{a bruising encounter with}{与……发生激烈冲突} 769 | \word{rush through}{匆匆通过} 770 | \word{cry out against}{强烈反对} 771 | \word[2020-11-04]{vacuum}[n]{真空} 772 | \word{overvalue}[v]{高估} 773 | \word[2020-11-04]{overstate}[v]{夸大} 774 | \word[2020-11-09]{extent}[n]{程度，范围，规模} 775 | \word{bargain}[n]{减价品；便宜货} 776 | \word{combative}[a]{好斗的，好战的} 777 | \word{cleaning up}{清理，整顿} 778 | \word[2020-11-06]{stock option}{股票期权} 779 | \word{hostility}[n]{（对思想、计划或情形的）愤怒反对、愤怒反抗} 780 | \end{wordlist} 781 | 782 | \begin{wordlist}{2020-11-03}{11-完形} 783 | \word{blood vessel}{血管} 784 | \word[2020-11-05](5){measurable}[a]{显著的} 785 | \word[2020-11-13](5){renewable}[a]{可再生的} 786 | \word[2020-11-05]{tone}[n]{（肌肉）结实、强壮} 787 | \word{conceivably}[ad]{可想象的，可信的} 788 | \word(10){physical}[a]{身体的} 789 | \word(10){mental}[a]{心理的，精神的} 790 | \word[2020-11-05](10){subconscious}[a]{潜意识的} 791 | \word{contract}[v]{（使某物）紧缩或收缩} 792 | \word(14){precede}[v]{在……之前发生} 793 | \word(16){fetch}{获取，拿来，取来} 794 | \word(17){indifferent}[a]{漠不关心的} 795 | \word(20){conversely}[ad]{相反地} 796 | \end{wordlist} 797 | 798 | \begin{wordlist}{2020-11-05}{11-阅读3} 799 | \word{rough}[a]{粗糙的，大致的} 800 | \word[2020-11-07]{approach}[v]{着手处理（事务、难题等）} 801 | \word[2020-11-07]{promote sth. to sb.}{将某物推荐给某人} 802 | \word[2020-11-07]{initiator}[n]{创始人，发起人} 803 | \word{traffic}[n]{流量} 804 | \word{complementary}[a]{补充的，附加的} 805 | \word{hostage}[n]{人质，引申为劫持物} 806 | \word[2020-11-09]{stakeholder}[n]{利益相关者} 807 | \word{allegation}[n]{说法，指控} 808 | \word{learning curve}{学习曲线} 809 | \word{steep}[a]{（上升或下降）突然的；急剧的} 810 | \word(35){alternative}[n]{替代} 811 | \word[2020-11-15](35){dominance}[n]{支配地位} 812 | \end{wordlist} 813 | 814 | \begin{wordlist}{2020-11-05}{11-阅读4} 815 | \word[2020-11-09]{provocative}[a]{挑衅的} 816 | \word{soul-crushingly}[ad]{摧残灵魂地} 817 | \word[2020-11-18]{dampen}[v]{使扫兴，使沮丧} 818 | \word{dampen one's moods}{使人心情沮丧} 819 | \word{mood}[n]{情绪} 820 | \word{intense}[a]{激烈的，强烈的} 821 | \word[2020-11-15]{gratification}[n]{满足} 822 | \word[2020-11-12](36){retrospect}[n]{回顾} 823 | \word(37){value}[v]{重视} 824 | \word[2020-11-09]{procreation}[n]{生育} 825 | \word{provoke}[v]{煽动，激励} 826 | \word{gape}[v]{张开；裂开} 827 | \word{round-the-clock}{日夜不停，夜以继日的} 828 | \word{intensify}[v]{强化，加剧} 829 | \end{wordlist} 830 | 831 | \begin{wordlist}{2020-11-17}{17-阅读1} 832 | \word{authority}[n]{[复数形式]当局，官方} 833 | \word{drag}[n]{累赘，拖累，绊脚石} 834 | \word[2020-11-20]{infuriating}[a]{使人极为生气（愤怒）的} 835 | \word{screener}[n]{（机场）安检员} 836 | \word{expedited}[a]{加快的} 837 | \word{eligible}[a]{有资格的，合格的，具备条件的} 838 | \word[2020-11-20]{sticker shock}{（令人瞠目结舌的高价所引起的）标价冲击波} 839 | \word{finance}[v]{提供资金} 840 | \word[2020-11-20]{get somewhere/anywhere/nowhere}{有所（无所）进展} 841 | \word(24){reluctance}[n]{勉强，不愿} 842 | \end{wordlist} 843 | \begin{wordlist}{2020-11-18}{17-阅读2} 844 | \word[2020-11-21]{feat}[n]{技艺，武艺，功绩} 845 | \word{be at issue}{（讨论、问题的焦点；在争论中，有分歧的} 846 | \word{the bulk of sth.}{主体，大部分} 847 | \word{vocal}[a]{大声表达的，直言不讳的，激烈表达意见的} 848 | \word{astronomer}[n]{天文学家} 849 | \word[2020-11-21]{relic}[n]{遗迹，遗风，遗俗} 850 | \word{Polynesian}[n]{波利尼西亚人} 851 | \word{starry}[a]{布满星星的} 852 | \word{primal}[a]{原始的，最初的，根本的} 853 | \word{archaeological}[a]{考古（学）的} 854 | \word{make compromises}{做出让步} 855 | \end{wordlist} 856 | \begin{wordlist}{2020-11-19}{10-2-阅读1} 857 | \word{bull}[n]{（股票市场上）买空者} 858 | \word{bull run}{牛市态势} 859 | \word{file for}{申请} 860 | \word{bewilderingly}[ad]{令人费解地} 861 | \word{lose momentum}{失去发展势头} 862 | \word{rise besilderingly}{急剧增长} 863 | \word{enormous egos}{自负，以自我为中心} 864 | \word{overheated}[a]{过热的} 865 | \word{become unfashionable}{变得不合时宜} 866 | \word{Impressionist}[n]{印象派} 867 | \word{slump}[n]{（价格、数值等）暴跌、不景气、萧条} 868 | \end{wordlist} 869 | 870 | \begin{wordlist}{2020-11-20}{10-2-阅读2} 871 | \word{suburban}[a]{郊区的} 872 | \word{address a gathering}{在聚会上演讲} 873 | \word{episode}[n]{插曲} 874 | \word{crystallize}[v]{使明确化、使具体化} 875 | \word{virtual}[a]{实际上的、几乎的} 876 | \word{amount to}{相当于；总计} 877 | \word{tangible}[a]{确凿的，实在的} 878 | \word{in short}{简言之} 879 | \end{wordlist} 880 | \begin{wordlist}{2020-11-21}{10-2-阅读3} 881 | \word{tropical}[a]{热带的} 882 | \word{subtle}[a]{微妙的} 883 | \word{moisturizer}[n]{润肤膏} 884 | \word{disinfecting}[a]{消毒的} 885 | \word{shrewd}[a]{精明的} 886 | \word{cavity-preventing}{口腔预防} 887 | \word{scrub}[n]{擦洗} 888 | \word{spring}[n]{泉、山泉} 889 | \word{bottle the springs}{将泉水装瓶} 890 | \word{slip A in between B and C}{将A悄悄放在B和C之间} 891 | \word{viable}[a]{可行的} 892 | \word{ruthless}[a]{无情的、残忍的} 893 | \end{wordlist} 894 | \end{document} 895 | -------------------------------------------------------------------------------- /general.tex: -------------------------------------------------------------------------------- 1 | \documentclass{ctexart} 2 | \edef\mydate{2020-11-23} 3 | 4 | \usepackage{showdata} 5 | \usepackage[balancingshow]{multicol} 6 | \usepackage{xcolor} 7 | \usepackage[margin=2.5cm]{geometry} 8 | \usepackage{enumitem} 9 | \usepackage{tikz} 10 | \usetikzlibrary{shapes.geometric} 11 | \usepackage{hyperref} 12 | \hypersetup{ 13 | bookmarksopen, 14 | bookmarksnumbered, 15 | bookmarksopenlevel=2, 16 | } 17 | 18 | \setCJKmainfont{Source Han Serif SC} 19 | \setCJKsansfont{Source Han Sans SC} 20 | 21 | \definecolor{label}{HTML}{ac8295} 22 | % \definecolor{error}{HTML}{4d243d} 23 | \definecolor{error}{HTML}{f53473} 24 | \newcommand\mylabel[1]{ 25 | \begin{tikzpicture}[baseline=(s.south)] 26 | \node[scale=0.6, star, star point height=0.5ex, star point ratio=2, 27 | fill=#1] (s) {}; 28 | \end{tikzpicture} 29 | } 30 | \SetEnumitemKey{multicol}{ 31 | topsep=0pt, partopsep=0pt, 32 | before=\begin{multicols}{5}, 33 | after=\end{multicols}, 34 | } 35 | 36 | \sdsetdomain{politics} 37 | \sdsetupsub{politics}{ 38 | prop={part, index, error, color}, 39 | global-group={single, multi} 40 | } 41 | 42 | % \NewsdReviewPoint{default}{2, 4, 7, 15, 30, 60} 43 | \NewsdReviewPoint{default}{3, 10, 30} 44 | % #1 * 单选，否则多选 #2 date #3 part number 45 | \NewsdBatchEnvironment { politics } { s m m +b } 46 | { 47 | % #1 date #2 error times #3 index 48 | \DeclareDocumentCommand { \addto } { O{#2} D(){1} m } 49 | { 50 | \sdIfReviewT{default}{##1}{\mydate} 51 | { 52 | \def\mysave####1{ 53 | \sdsave*{####1}{part}{#3} 54 | \sdsave*{####1}{index}{##3} 55 | \ifnum##2>1\sdsave*{####1}{error}{##2} 56 | \sdsave*{####1}{color}{\protect\mylabel{error}} 57 | \else\sdsave*{####1}{error}{} 58 | \sdsave*{####1}{color}{\protect\mylabel{label}} 59 | \fi 60 | } 61 | \IfBooleanTF { #1 } 62 | { \mysave{single} } 63 | { \mysave{multi} } 64 | } 65 | } 66 | #4 67 | } 68 | 69 | \newcommand\setpolitics[2]{ 70 | \sdsetshowstyle*{politics}{#1}{index} 71 | { 72 | \section{\mydate 政治#2选复习} 73 | \sdIfEmptyPropF{\begin{itemize}[multicol]} 74 | } 75 | { 76 | \sdIfEmptyPropF{\end{itemize}} 77 | } 78 | { 79 | \item[\sduse{color}] \sduse{part}-\sduse{index}% 80 | \textsubscript{% 81 | {\color{error}\itshape\bfseries\sduse{error}} 82 | } 83 | } 84 | } 85 | \setpolitics{single}{单} 86 | \setpolitics{multi}{多} 87 | 88 | \AtEndDocument{ 89 | \sdshow*{politics}{single} 90 | \sdshow*{politics}{multi} 91 | } 92 | 93 | \begin{document} 94 | \begin{sdpolitics}*{2020-10-27}{1} 95 | \addto[2020-10-29](2){2} 96 | \addto[2020-10-31](2){3} 97 | \addto{4} 98 | \addto{5} 99 | \addto{7} 100 | \addto{12} 101 | \addto{19} 102 | \addto{20} 103 | \addto[2020-10-31](3){25} 104 | \end{sdpolitics} 105 | 106 | \begin{sdpolitics}{2020-10-28}{1} 107 | \addto[2020-10-30](2){1} 108 | \addto[2020-10-30](2){3} 109 | \addto[2020-11-01](3){8} 110 | \addto{9} 111 | \addto[2020-11-01](3){13} 112 | \addto{14} 113 | \addto{21} 114 | \addto{22} 115 | \addto[2020-11-01](3){24} 116 | \addto[2020-10-30](2){29} 117 | \end{sdpolitics} 118 | 119 | \begin{sdpolitics}*{2020-10-29}{1} 120 | \addto[2020-10-31](2){34} 121 | \addto[2020-10-31](2){36} 122 | \addto{45} 123 | \addto[2020-11-09](3){52} 124 | \addto[2020-10-31](2){59} 125 | \addto{64} 126 | \addto{67} 127 | \end{sdpolitics} 128 | 129 | \begin{sdpolitics}{2020-10-30}{1} 130 | \addto{33} 131 | \addto[2020-11-06](4){35} 132 | \addto[2020-11-01](3){36} 133 | \addto[2020-11-01](3){38} 134 | \addto[2020-11-01](3){39} 135 | \addto[2020-11-06](4){41} 136 | \addto{42} 137 | \addto{43} 138 | \addto[2020-11-03](4){45} 139 | \addto[2020-11-01](3){46} 140 | \addto[2020-11-01](3){47} 141 | \end{sdpolitics} 142 | 143 | \begin{sdpolitics}{2020-10-31}{1} 144 | \addto[2020-11-04](3){48} 145 | \addto[2020-11-04](3){49} 146 | \addto{51} 147 | \addto{53} 148 | \addto[2020-11-02](2){54} 149 | \addto[2020-11-04](3){55} 150 | \addto[2020-11-10](3){57} 151 | \addto{58} 152 | \addto{59} 153 | \addto[2020-11-02](2){60} 154 | \addto[2020-11-02](2){61} 155 | \end{sdpolitics} 156 | 157 | \begin{sdpolitics}*{2020-11-01}{1} 158 | \addto{68} 159 | \addto[2020-11-06](3){70} 160 | \addto{72} 161 | \addto{74} 162 | \addto{79} 163 | \addto{81} 164 | \addto[2020-11-03](2){84} 165 | \addto{87} 166 | \end{sdpolitics} 167 | 168 | \begin{sdpolitics}{2020-11-02}{1} 169 | \addto[2020-11-18](4){64} 170 | \addto{67} 171 | \addto{69} 172 | \addto[2020-11-08](3){71} 173 | \addto[2020-11-04](2){72} 174 | \addto[2020-11-06](3){73} 175 | \addto{75} 176 | \addto[2020-11-08](4){78} 177 | \addto[2020-11-04](2){80} 178 | \addto[2020-11-06](3){83} 179 | \addto[2020-11-06](3){84} 180 | \end{sdpolitics} 181 | 182 | \begin{sdpolitics}{2020-11-03}{1} 183 | \addto[2020-11-06](3){88} 184 | \addto[2020-11-06](3){89} 185 | \addto[2020-11-07](2){90} 186 | \addto{91} 187 | \addto[2020-11-06](3){92} 188 | \addto{93} 189 | \addto{94} 190 | \addto[2020-11-08](5){95} 191 | \addto[2020-11-13](2){96} 192 | \addto[2020-11-06](3){97} 193 | \addto{99} 194 | \end{sdpolitics} 195 | 196 | \begin{sdpolitics}*{2020-11-04}{1} 197 | \addto[2020-11-06](3){97} 198 | \addto[2020-11-08](2){100} 199 | \addto[2020-11-06](3){109} 200 | \addto[2020-11-06](3){114} 201 | \end{sdpolitics} 202 | 203 | \begin{sdpolitics}{2020-11-05}{1} 204 | \addto[2020-11-07](4){100} 205 | \addto{101} 206 | \addto[2020-11-13](6){104} 207 | \addto[2020-11-07](4){106} 208 | \addto{107} 209 | \addto[2020-11-10](2){108} 210 | \addto[2020-11-10](5){109} 211 | \addto[2020-11-10](2){113} 212 | \addto{114} 213 | \end{sdpolitics} 214 | 215 | \begin{sdpolitics}{2020-11-06}{1} 216 | \addto[2020-11-08](2){115} 217 | \addto[2020-11-09](2){116} 218 | \addto[2020-11-09](2){118} 219 | \addto{119} 220 | \addto{122} 221 | \addto[2020-11-09](2){123} 222 | \addto{124} 223 | \addto{126} 224 | \addto[2020-11-09](2){127} 225 | \end{sdpolitics} 226 | 227 | \begin{sdpolitics}*{2020-11-07}{1} 228 | \addto[2020-11-17](2){120} 229 | \addto[2020-11-17](2){121} 230 | \addto{127} 231 | \addto{128} 232 | \addto{131} 233 | \addto{136} 234 | \addto{137} 235 | \addto[2020-11-20](3){139} 236 | \addto[2020-11-09](2){140} 237 | \addto{146} 238 | \addto{150} 239 | \addto{151} 240 | \addto[2020-11-17](2){152} 241 | \end{sdpolitics} 242 | 243 | \begin{sdpolitics}*{2020-11-08}{1} 244 | \addto{153} 245 | \addto{154} 246 | \addto{155} 247 | \addto[2020-11-14](3){156} 248 | \addto[2020-11-11](2){159} 249 | \addto{160} 250 | \addto[2020-11-14](3){165} 251 | \addto[2020-11-18](2){166} 252 | \addto{167} 253 | \addto[2020-11-14](3){171} 254 | \addto{172} 255 | \addto[2020-11-14](3){174} 256 | \addto{178} 257 | \end{sdpolitics} 258 | 259 | \begin{sdpolitics}*{2020-11-08}{1} 260 | \addto{184} 261 | \addto[2020-11-11](2){190} 262 | \end{sdpolitics} 263 | 264 | \begin{sdpolitics}*{2020-11-09}{1} 265 | \addto{191} 266 | \addto{193} 267 | \addto{195} 268 | \addto[2020-11-12](2){196} 269 | \addto{197} 270 | \addto{198} 271 | \addto[2020-11-12](2){199} 272 | \addto[2020-11-12](2){201} 273 | \end{sdpolitics} 274 | 275 | \begin{sdpolitics}{2020-11-09}{1} 276 | \addto{199} 277 | \addto[2020-11-20](2){204} 278 | \addto[2020-11-12](2){205} 279 | \addto{206} 280 | \addto[2020-11-12](2){207} 281 | \addto{209} 282 | \addto[2020-11-20](2){211} 283 | \addto{216} 284 | \addto[2020-11-12](2){221} 285 | \addto{223} 286 | \addto{224} 287 | \end{sdpolitics} 288 | 289 | \end{document} 290 | -------------------------------------------------------------------------------- /math.code.tex: -------------------------------------------------------------------------------- 1 | \usepackage{nicematrix} 2 | \usepackage{empheq} 3 | \usepackage{siunitx} 4 | \usepackage[bodytextleadingratio=2]{zhlineskip} 5 | \input preamble.tex 6 | \input review.code.tex 7 | \usetikzlibrary{arrows.meta} 8 | \usepackage{paracol} 9 | \usepackage{pgfplots} 10 | \usepackage{subcaption} 11 | \usepgfplotslibrary{fillbetween} 12 | \usetikzlibrary{patterns} 13 | \pgfplotsset{compat=newest} 14 | \everymath{\displaystyle} 15 | \allowdisplaybreaks 16 | \setlength\parindent{0pt} 17 | 18 | \geometry{ 19 | left=2cm, 20 | right=2cm, 21 | top=2.5cm, 22 | bottom=2.5cm, 23 | } 24 | 25 | \ctexset{ 26 | section/numbering=false, 27 | section/format=\huge\bfseries\sffamily\color{teal}, 28 | } 29 | 30 | \pgfplotsset{ 31 | math/.style={ 32 | set layers=standard, 33 | axis on top, 34 | axis lines=middle, 35 | unit vector ratio*=1 1 1, 36 | width=.8\linewidth, 37 | xlabel=$x$, 38 | ylabel=$y$, 39 | xlabel style={at={(current axis.right of origin)}, anchor=north}, 40 | ylabel style={at={(current axis.above origin)}, anchor=east}, 41 | enlargelimits={abs={0.5}}, 42 | tick align=inside, 43 | /tikz/every pin/.style={ 44 | inner sep=1pt, font=\small 45 | }, 46 | /tikz/every pin edge/.style={thick}, 47 | % xticklabel style={text height=2ex}, 48 | % ticks=none, 49 | } 50 | } 51 | 52 | \newlist{queslist}{enumerate}{2} 53 | \setlist[queslist]{ 54 | label=(\arabic*), 55 | noitemsep, 56 | topsep=2pt, 57 | } 58 | \newlist{solulist}{enumerate}{2} 59 | \setlist[solulist]{ 60 | label=(\arabic*), 61 | topsep=2pt, 62 | } 63 | 64 | \newcommand{\mathproof}{\par{\kaiti 证明：}} 65 | 66 | \newcommand{\at}[2][]{#1|_{#2}} 67 | 68 | \newcommand{\mathblank}{\CJKunderline[hidden]{四个汉字}} 69 | 70 | \newcommand{\mathmethod}{% 71 | \stepcounter{method}% 72 | {\bfseries\sffamily \par\hspace{2em}方法\zhnum{method}}\quad% 73 | } 74 | 75 | \tcbset{ 76 | math base/.style args={#1:#2}{ 77 | colframe=#1, colback=#2, 78 | size=small, on line, 79 | fontupper=\sffamily, 80 | } 81 | } 82 | 83 | \newtcbox{\mathtype}{math base={teal:teal!30!cyan}} 84 | \newtcbox{\mathdate}{math base={teal:cyan!70!pink}} 85 | \newtcbox{\mathlink}{math base={teal:teal!60!yellow}} 86 | \newtcbox{\mathnumber}{math base={teal:red!40}} 87 | 88 | \newtcolorbox{mathideabox}[1][]{ 89 | colframe=teal, colback=teal!10, breakable, title=#1, 90 | fonttitle=\bfseries\sffamily, 91 | } 92 | 93 | \newtcolorbox[use counter=step]{step}[1]{ 94 | title={Step \arabic{step}: #1}, 95 | breakable, 96 | } 97 | 98 | \newcounter{step} 99 | \newcounter{method} 100 | 101 | \ExplSyntaxOn 102 | 103 | \AtBeginDocument{ 104 | \title{\g__review_date_tl 复习} 105 | \author{} 106 | \date{} 107 | \maketitle 108 | } 109 | 110 | \AtEndDocument{ 111 | \math_show: 112 | } 113 | 114 | \cs_new_protected:Nn \math_def_seq:n 115 | { 116 | \seq_new:c { g__math_saved_all_#1_seq } 117 | \seq_new:c { g__math_saved_review_#1_seq } 118 | \seq_new:c { g__math_saved_more_#1_seq } 119 | } 120 | 121 | \tl_new:N \l__math_date_tl 122 | \tl_new:N \l__math_type_tl 123 | \tl_new:N \l__math_info_tl 124 | \tl_new:N \l__math_content_ques_tl 125 | \tl_new:N \l__math_number_style_tl 126 | \int_new:N \l__math_mistake_number_int 127 | \int_new:N \g__math_count_int 128 | \dim_new:N \l__math_ques_after_title_dim 129 | \dim_new:N \l__math_solu_item_sep_dim 130 | \dim_new:N \l__math_after_solu_dim 131 | \dim_new:N \l__math_lable_width_dim 132 | \dim_new:N \l__math_label_sep_to_number_dim 133 | \dim_new:N \l__math_label_sep_dim 134 | \bool_new:N \l__math_show_all_bool 135 | \bool_new:N \l__math_is_answer_bool 136 | \clist_new:N \l__math_review_point_clist 137 | 138 | \math_def_seq:n { date } 139 | \math_def_seq:n { type } 140 | \math_def_seq:n { info } 141 | \math_def_seq:n { content_ques } 142 | \math_def_seq:n { content_solu } 143 | \math_def_seq:n { mistake_number } 144 | 145 | \clist_set:Nn \l__math_review_point_clist { 2,4,7,15,30 } 146 | \bool_set_false:N \l__math_show_all_bool 147 | 148 | \keys_define:nn { mathques } 149 | { 150 | show~all . bool_set:N = \l__math_show_all_bool, 151 | show~all . default:n = { true }, 152 | number~style . tl_set:N = \l__math_number_style_tl, 153 | review~point . clist_set:N = \l__math_review_point_clist, 154 | after~title . dim_set:N = \l__math_ques_after_title_dim, 155 | before~solu . dim_set:N = \l__math_solu_item_sep_dim, 156 | after~solu . dim_set:N = \l__math_after_solu_dim, 157 | label~sep~to~number . dim_set:N = \l__math_label_sep_to_number_dim, 158 | label~sep . dim_set:N = \l__math_label_sep_dim, 159 | label~width . dim_set:N = \l__math_lable_width_dim, 160 | } 161 | 162 | \NewDocumentCommand { \mathsetup } { +m } 163 | { 164 | \keys_set:nn { mathques } { #1 } 165 | } 166 | 167 | \prg_new_protected_conditional:Nnn \math_review_if:n { T, F, TF } 168 | { 169 | \review_if:NnTF \l__math_review_point_clist { #1 } 170 | { \prg_return_true: } 171 | { \prg_return_false: } 172 | } 173 | 174 | % #1 date #2 seq name #3 data to save 175 | \cs_new_protected:Nn \math_save:nnn 176 | { 177 | \bool_if:NT \l__math_show_all_bool 178 | { 179 | \seq_gput_right:cn { g__math_saved_all_#2_seq } { #3 } 180 | } 181 | \math_review_if:nT { #1 } 182 | { 183 | \seq_gput_right:cn { g__math_saved_review_#2_seq } { #3 } 184 | } 185 | } 186 | 187 | % #1 做错次数 #2 做错日期 #3 题目类型 188 | \NewDocumentEnvironment { mathques } { O{} D(){1} m m +O{} +b } 189 | { 190 | \tl_clear_new:N \l__math_date_tl 191 | 192 | \tl_set:Nn \l__math_date_tl { #3 } 193 | 194 | #6 195 | 196 | \math_save:nnn { #3 } { date } { #3 } 197 | \math_save:nnn { #3 } { type } { #4 } 198 | \math_save:nnn { #3 } { info } { #5 } 199 | \math_save:nnn { #3 } { mistake_number } { #2 } 200 | } { } 201 | 202 | \NewDocumentEnvironment { ques } { +b } 203 | { 204 | \math_save:nnn { \l__math_date_tl } { content_ques } { #1 } 205 | } { } 206 | 207 | \NewDocumentEnvironment { solu } { +b } 208 | { 209 | \math_save:nnn { \l__math_date_tl } { content_solu } 210 | { 211 | \setcounter{step}{0} 212 | \setcounter{method}{0} 213 | #1 214 | } 215 | } { } 216 | 217 | \cs_new_protected:Nn \math_show: 218 | { 219 | \math_show_content_template:nn { review } { \g__review_date_tl } 220 | \bool_if:NT \l__math_show_all_bool 221 | { 222 | \math_show_content_template:nn { all } { 所有 } 223 | } 224 | } 225 | 226 | % #1 all/review #2 date/所有 227 | \cs_new_protected:Nn \math_show_content_template:nn 228 | { 229 | \section{#2复习题目} 230 | \bool_set_false:N \l__math_is_answer_bool 231 | \math_show_content:nn { #1 } { ques } 232 | \clearpage 233 | \section{#2题目答案} 234 | \bool_set_true:N \l__math_is_answer_bool 235 | \math_show_content:nn { #1 } { solu } 236 | \clearpage 237 | } 238 | 239 | % #1 all/review #2 ques/solu 240 | \cs_new_protected:Nn \math_show_content:nn 241 | { 242 | \int_gzero:N \g__math_count_int 243 | \seq_map_inline:cn { g__math_saved_#1_content_#2_seq } 244 | { 245 | \int_gincr:N \g__math_count_int 246 | \math_set:nn { #1 } { \g__math_count_int } 247 | \cs_if_exist_use:c { math_wrap_content_#2:nn } { #1 } { ##1 } 248 | \bool_if:NT \l__math_is_answer_bool 249 | { 250 | \vspace{\l__math_after_solu_dim} 251 | } 252 | } 253 | } 254 | 255 | % #1 all/review/more #2 int 256 | \cs_new_protected:Nn \math_set:nn 257 | { 258 | \tl_set:Nn \l__math_date_tl 259 | { \math_seq_item:nnn { #1 } { date } { #2 } } 260 | \tl_set:Nn \l__math_type_tl 261 | { \math_seq_item:nnn { #1 } { type } { #2 } } 262 | \tl_set:Nn \l__math_info_tl 263 | { \math_seq_item:nnn { #1 } { info } { #2 } } 264 | \tl_set:Nn \l__math_content_ques_tl 265 | { \math_seq_item:nnn { #1 } { content_ques } { #2 } } 266 | \int_set:Nn \l__math_mistake_number_int 267 | { \math_seq_item:nnn { #1 } { mistake_number } { #2 } } 268 | } 269 | 270 | % #1 all/review/more #2 date/type... #3 int 271 | \cs_new:Nn \math_seq_item:nnn 272 | { 273 | \seq_item:cn { g__math_saved_#1_#2_seq } { #3 } 274 | } 275 | 276 | % set label and bookmark and number 277 | % #1 all/review #2 ques/solu 278 | \cs_new_protected:Nn \math_common:nn 279 | { 280 | \ignorespaces 281 | \customlabel{#1_#2:\int_use:N \g__math_count_int} 282 | \pdfbookmark[2]{ 283 | \int_use:N \g__math_count_int. 284 | \space\l__math_date_tl~\l__math_type_tl 285 | }{ 286 | #1_#2_number:\int_use:N \g__math_count_int 287 | } 288 | \makebox[\l__math_lable_width_dim][l]{ 289 | \l__math_number_style_tl 290 | \int_use:N \g__math_count_int. 291 | } 292 | \hspace{\l__math_label_sep_to_number_dim} 293 | } 294 | 295 | % #1 all/review... #2 content 296 | \cs_new_protected:Nn \math_wrap_content_ques:nn 297 | { 298 | \begin{tcolorbox}[ 299 | % phantom={ 300 | % } 301 | ] 302 | \math_common:nn { #1 } { ques } 303 | \mathdate{ \l__math_date_tl } 304 | \hspace{\l__math_label_sep_dim} 305 | \mathtype{ \l__math_type_tl } 306 | \hspace{\l__math_label_sep_dim} 307 | \mathlink{ 308 | \hyperref[#1_solu:\int_use:N \g__math_count_int]{ 309 | 前往答案 310 | } 311 | } 312 | \int_compare:nT { \l__math_mistake_number_int > 1 } 313 | { 314 | \hfill 315 | \mathnumber{ \int_use:N \l__math_mistake_number_int } 316 | } 317 | \par\vspace{\l__math_ques_after_title_dim} 318 | #2 319 | \end{tcolorbox} 320 | } 321 | 322 | \cs_new_protected:Nn \math_wrap_content_solu:nn 323 | { 324 | \math_common:nn { #1 } { solu } 325 | \mathtype{ \l__math_type_tl } 326 | \hspace{\l__math_label_sep_dim} 327 | \mathlink{ 328 | \hyperref[#1_ques:\int_use:N \g__math_count_int]{ 329 | 返回题目 330 | } 331 | } 332 | \par\nopagebreak\vspace{\l__math_ques_after_title_dim} 333 | {\sffamily\bfseries 题目：}\l__math_content_ques_tl 334 | \tl_if_blank:eF { \l__math_info_tl } 335 | { 336 | \par\vspace{\l__math_solu_item_sep_dim} 337 | {\sffamily\bfseries 错因：} 338 | {\kaiti\color{red} \l__math_info_tl} 339 | } 340 | \par\vspace{\l__math_solu_item_sep_dim} 341 | {\sffamily\bfseries 答案：} 342 | #2 \par 343 | } 344 | 345 | \NewDocumentEnvironment { multichoice } { O{2} +b } 346 | { 347 | \begin{tasks}[label=(\Alph*), label-width=2em](#1) 348 | #2 349 | \end{tasks} 350 | } {} 351 | 352 | \NewDocumentCommand { \answer } { +m } 353 | { 354 | \bool_if:NTF \l__math_is_answer_bool 355 | { 356 | \tcbox[ 357 | colframe=teal, colback=teal!10, on~line, 358 | arc=0pt, outer~arc=0pt, 359 | boxsep=0pt, top=4pt, bottom=4pt, left=2pt, right=2pt, 360 | boxrule=0pt, toprule=1pt, bottomrule=1pt, 361 | every~box/.style={ 362 | highlight~math~style={colframe=red, colback=red!10} 363 | } 364 | ]{#1} 365 | } 366 | { #1 } 367 | } 368 | 369 | \NewDocumentCommand { \mathhint } { O{} +m } 370 | { 371 | \tcbset{ 372 | highlight~math~style={ 373 | colframe=teal, colback=teal!10, 374 | boxrule=0.6pt, size=small, 375 | } 376 | } 377 | \str_case:nnF { #1 } 378 | { 379 | { r } { \rlap{\tcbhighmath{#2}} } 380 | { l } { \llap{\tcbhighmath{#2}} } 381 | } 382 | { 383 | \tcbhighmath{#2} 384 | } 385 | } 386 | 387 | \newtcbox{\mathideai}{ 388 | colframe=teal, colback=teal!10, on~line, size=small 389 | } 390 | 391 | \NewDocumentCommand { \mathidea } { O{} +m } 392 | { 393 | \str_case:nnF { #1 } 394 | { 395 | { r } { \rlap{\mathideai{#2}} } 396 | { l } { \llap{\mathideai{#2}} } 397 | } 398 | { 399 | \mathideai{#2} 400 | } 401 | } 402 | 403 | \NewDocumentCommand { \matherror } { s m } 404 | { 405 | \tcbset{matherror/.style={ 406 | colframe=red, colback=red!10, boxrule=0.6pt, size=small}} 407 | \IfBooleanTF { #1 } 408 | { \tcbox[matherror, on~line, left=1pt, right=1pt]{#2} } 409 | { 410 | \tcbset{highlight~math~style=matherror} 411 | \tcbhighmath{#2} 412 | % \tcbhighmath[boxrule=0.6pt, size=small]{#2} 413 | } 414 | } 415 | 416 | \NewDocumentCommand { \mathmark } { O{red} D(){tmp} O{} m } 417 | { 418 | \tcbset{mathmark/.style={ 419 | colframe=#1, colback=white, boxrule=0.6pt, size=small, enhanced 420 | }} 421 | \tcbset{highlight~math~style=mathmark} 422 | \tcbhighmath[remember~as=#2, overlay={#3}]{#4} 423 | % \tcboxmath[mathmark, remember~as=#2]{#3} 424 | % \tcboxmath[remember~as=#2]{#3} 425 | } 426 | 427 | \ExplSyntaxOff 428 | 429 | \mathsetup{ 430 | show all = false, 431 | review point = {2, 4, 7, 14, 30}, 432 | number style = { \zihao{3}\itshape\bfseries }, 433 | after title = 1em, 434 | before solu = 1em, 435 | after solu = 1cm, 436 | label sep to number = 1em, 437 | label sep = 1em, 438 | label width = 1cm, 439 | } 440 | \endinput 441 | -------------------------------------------------------------------------------- /math.tex: -------------------------------------------------------------------------------- 1 | \documentclass{ctexart} 2 | \input math.code.tex 3 | % 可以设置复习日期，默认\today 4 | \setdate{2020-11-23} 5 | % 显示所有题目 6 | \mathsetup{ 7 | % show all, 8 | % 设置复习时间点，默认2, 4, 7, 14, 30 9 | review point = { 3, 10, 30 } 10 | } 11 | 12 | 13 | \begin{document} 14 | 15 | \begin{mathques}{2020-09-05}{极限} 16 | \begin{ques} 17 | 极限$\lim_{n \to \infty} \left( \frac{n+1}{n-2} \right) ^n=$\mathblank. 18 | \end{ques} 19 | \begin{solu} 20 | \begin{align*} 21 | \lim_{n \to \infty} \left( \frac{n+1}{n-2} \right) ^n 22 | &= \lim_{n \to \infty} \frac{\left( 1 + \frac{1}{n} \right) ^n}{\left( 1 - 23 | \frac{2}{n} \right) ^n}\\ 24 | & \mathhint{\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) ^n = e} 25 | \\ 26 | & = \frac{\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) ^n}{\lim_{n 27 | \to \infty} \left( 1 - \frac{2}{n} \right) ^n}\\ 28 | & = \frac{e}{e^{-2}} = e^3 29 | \end{align*} 30 | \end{solu} 31 | \end{mathques} 32 | 33 | \begin{mathques}{2020-09-05}{极限} 34 | \begin{ques} 35 | 求极限$\lim_{n \to \infty} (\sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}})$. 36 | \end{ques} 37 | \begin{solu} 38 | \begin{align*} 39 | \lim_{n \to \infty} (\sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}}) 40 | &= \lim_{n \to \infty} \frac{(n + \sqrt{n}) - (n - \sqrt{n})}{(n + 41 | \sqrt{n}) + (n - \sqrt{n})}\\ 42 | &= \lim_{n \to \infty} \frac{2}{\sqrt{1 + \frac{1}{\sqrt{n}}} + \sqrt{1 - 43 | \frac{1}{\sqrt{n}}}}\\ 44 | &= \frac{2}{\matherror{\sqrt{1} + \sqrt{1}}} = 1 45 | \end{align*} 46 | \end{solu} 47 | \end{mathques} 48 | 49 | \begin{mathques}{2020-09-05}{极限} 50 | \begin{ques} 51 | 设$x_{n + 1} = \sqrt{2 + x_n} (n = 1, 2, \dotsc), x_1 = \sqrt{2}$, 证明 52 | $\lim_{n \to \infty} x_n$ 存在, 并求$\lim_{n \to \infty} x_n$. 53 | \end{ques} 54 | \begin{solu} 55 | \mathidea{单调有界数列必有极限} 56 | \begin{step}{证明单调性(3种方法)} 57 | \begin{enumerate} 58 | \item 令$f(x) = \sqrt{2 + x}$ 证明$f(x)$的单调性 59 | \item 作差$x_{n} - x_{n - 1} = \sqrt{2 + x_{n - 1}} - x_{n - 1} = \frac{2 60 | + x_{n - 1} - x_{n - 1}^2}{\sqrt{2 + x_{n - 1}} + x_{n - 1}}$\par 61 | $2 + x_{n - 1} - x_{n - 1}^2 > 2 + x_{n - 1} - 2x_{n - 1} = 2 - x_{n 62 | - 1} > 0$\par 63 | \mathidea{方法2需要先进行Step 2证明上界为2才行} 64 | \item 证明$x_{n + 1} - x_n$与$x_{n} - x_{n - 1}$同号\par 65 | $x_{n + 1} - x_n = \sqrt{2 + x_n} - \sqrt{2 + x_{n - 1}} = \frac{x_n 66 | - x_{n - 1}}{\sqrt{2 + x_n} + \sqrt{2 + x_{n - 1}}}$\par 67 | 又$x_2 - x_1 = \sqrt{2 + \sqrt{2}} - \sqrt{2} > 0$，故$x_n$单调 68 | \end{enumerate} 69 | \end{step} 70 | \begin{step}{证明有界性} 71 | \begin{mathideabox} 72 | \itshape 当界值难以看出来时，可以先假定有界，得出有极限然后求出根据递 73 | 推公式求出极限得出界值再反过来想怎么证明（见Step 3） 74 | \end{mathideabox} 75 | $x_1 = \sqrt{2} < 2$, 设$x_k < 2$, 则$x_{k + 1} = \sqrt{2 + x_k} < 2$, 76 | 故$x_n < 2$有上界 77 | \end{step} 78 | \begin{step}{求极限} 79 | $x_n > 0$, 设$\lim_{n \to \infty} x_n = A > 0$, 则$A = \sqrt{2 + A} \Rightarrow A = 2$ 80 | \end{step} 81 | \end{solu} 82 | \end{mathques} 83 | 84 | \begin{mathques}(3){2020-09-22}{极限} 85 | \begin{ques} 86 | 求极限$\lim_{x \to 0^+} \frac{x^x - (\sin x)^x}{x^2 \ln (1 + x)}$. 87 | \end{ques} 88 | \begin{solu} 89 | \begin{align*} 90 | \text{原式} & = -\matherror{\lim_{x \to 0^+} x^x} \lim_{x \to 0^+} 91 | \frac{\left( \frac{\sin x}{x} \right) ^x - 1}{x^3}\\ 92 | &\mathhint{e^x - 1 \sim x (x \to 0)}\\ 93 | &= -\lim_{x \to 0^+} \frac{\ln \frac{\sin x}{x}}{x^2}\\ 94 | &= -\lim_{x \to 0^+} \frac{\ln \left( 1 + \frac{\sin x - x}{x} \right) }{x^2} 95 | \\ 96 | &\mathhint{\ln (1+x) \sim x (x \to 0)}\\ 97 | &= -\lim_{x \to 0^+} \frac{\sin x - x}{x^3}\\ 98 | &= -\lim_{x \to 0^+} \frac{x - \frac{1}{3!} x^3 + o(x^3) - x}{x^3} = \frac{1} 99 | {6} 100 | \end{align*} 101 | \end{solu} 102 | \end{mathques} 103 | 104 | \begin{mathques}(2){2020-09-08}{极限} 105 | \begin{ques} 106 | 求极限$\lim_{x \to 0} \left( \frac{a_1^x + a_2^x + \dotsb + a_n^x}{n} \right) 107 | ^{\frac{n}{x}}$，其中$a_i > 0, i = 1, 2, \dotsc, n$. 108 | \end{ques} 109 | \begin{solu} 110 | 因为$\lim _{x \to 0}a_i^x = 1$，所以原极限是“$1^{\infty}$”型未定式 111 | \mathmethod 使用\emph{洛必达法则}求极限 112 | \begin{align*} 113 | \lim_{x \to 0} \left( \frac{a_1^x + a_2^x + \dotsb + a_n^x}{n} \right) 114 | ^{\frac{n}{x}} &= \exp \left\{ \lim_{x \to 0}\frac{n}{x} \ln \left( 115 | \frac{a_1^x + a_2^x + \dotsb + a_n^x}{n} \right) \right\}\\ 116 | &\mathhint{\text{洛必达}\frac{0}{0}}\\ 117 | &= \exp \left\{ \lim_{x \to 0} n \cdot \frac{a_1^x \ln a_1 + a_2^x \ln a_2 118 | + \dotsb + a_n^x \ln a_n}{a_1^x + a_2^x + \dotsb + a_n^x} \right\} \\ 119 | &=a_1a_2\dotsi a_n 120 | \end{align*} 121 | \mathmethod 凑成第二个重要极限（“$1^{\infty}$”型未定式极限都可以凑成第二个重 122 | 要极限 123 | \begin{align*} 124 | \lim_{x \to 0} \left( \frac{a_1^x + a_2^x + \dotsb + a_n^x}{n} \right) 125 | ^{\frac{n}{x}} &= 126 | \lim_{x \to 0} \left( 1 + \frac{a_1^x + a_2^x + \dotsb + a_n^x - n}{n} 127 | \right) ^{\frac{n}{a_1^x + a_2^x + \dotsb + a_n^x - n} 128 | \cdot \frac{a_1^x + a_2^x + \dotsb + a_n^x - n}{x}} 129 | \end{align*} 130 | 其中 131 | \begin{align*} 132 | \lim_{x \to 0} \frac{a_1^x + a_2^x + \dotsb + a_n^x - n}{x} 133 | &= \ln (a_1a_2\dotsi a_n)\\ 134 | \mathhint{\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) ^n = e}\\ 135 | \lim_{x \to 0} \left( 1 + \frac{a_1^x + a_2^x + \dotsb + a_n^x - n}{n} 136 | \right) ^{\frac{n}{a_1^x + a_2^x + \dotsb + a_n^x - n}} &=e 137 | \end{align*} 138 | \end{solu} 139 | \end{mathques} 140 | 141 | \begin{mathques}(2){2020-09-08}{极限} 142 | \begin{ques} 143 | 设函数$f(x) = \lim_{n \to \infty} \frac{1 + x}{1 + x^{2n}}$，讨论函数的间断 144 | 点，其结论为(\quad). 145 | \begin{multichoice} 146 | \task 不存在间断点 147 | \task \answer{存在间断点$x = 1$} 148 | \task 存在间断点$x = 0$ 149 | \task 存在间断点$x = -1$ 150 | \end{multichoice} 151 | \end{ques} 152 | \begin{solu} 153 | \begin{mathideabox} 154 | 函数$f(x)$以$x$为自变量，但是在对$n$求极限的时候，$x$被看做常数，因此应根据 155 | $x$的不同取值求出对应的极限 156 | \end{mathideabox} 157 | 当$|x| < 1$时, $\lim_{n \to \infty} x^{2n} = 0$，所以$f(x) = 1 + x$ 158 | 159 | 当$|x| > 1$时，$\lim_{n \to \infty} \frac{1 + x}{1 + x^{2n}} = 0$ 160 | 161 | 又$f(1) = 1, f(-1) = 0$，则 162 | \[ 163 | f(x) = \lim_{n \to \infty} \frac{1 + x}{1 + x^{2n}} = 164 | \begin{dcases} 165 | 0, & x \le -1 \\ 166 | 1 + x, & -1 1 \\ 169 | \end{dcases} 170 | \] 171 | 故$x = 1$为间断点 172 | \end{solu} 173 | \end{mathques} 174 | 175 | \begin{mathques}(2){2020-09-09}{极限} 176 | \begin{ques} 177 | 设$f(\sin^2x) = \frac{x}{\sin x}$，则$f(x) = $\mathblank. 178 | \end{ques} 179 | \begin{solu} 180 | \answer{ 181 | $\frac{\arcsin \sqrt{x}}{\sqrt{x}} \matherror{(x > 0)}$ 182 | }\par 183 | 设$u = \sin^2 x$，则$\sin x = \pm \sqrt{u}$. 184 | 185 | 当$\sin x = \sqrt{u}$时，$x = \arcsin \sqrt{u}$ 186 | 187 | 当$\sin x = - \sqrt{u}$时，$\matherror{\sin(-x) = \sqrt{u}, x = -\arcsin 188 | \sqrt{u}}$.因此 189 | \begin{align*} 190 | f(u) &= \frac{\arcsin \sqrt{u}}{\sqrt{u}} \\ 191 | f(x) &= \frac{\arcsin \sqrt{x}}{\sqrt{x}} (x > 0) 192 | \end{align*} 193 | \end{solu} 194 | \end{mathques} 195 | 196 | \begin{mathques}{2020-09-07}{极限} 197 | \begin{ques} 198 | 设$f\left(x + \frac{1}{x}\right) = \frac{x + x^3}{1 + x^4}$，则$f(x) = 199 | $\mathblank 200 | \end{ques} 201 | \begin{solu} 202 | \answer{$\frac{x}{x^2 - 2}$} 203 | 204 | \[ 205 | f\left(\matherror{ x + \frac{1}{x}}\right) = \frac{x + x^3}{1 + x^4} = 206 | \frac{x^2 \left( \matherror{\textstyle\frac{1}{x} + x} \right) }{x^2 207 | \left( \frac{1}{x^2} + x^2 \right) } = \frac{\frac{1}{x} + x}{\left( 208 | \matherror{\textstyle\frac{1}{x} + x} \right) ^2 - 2} 209 | \] 210 | 因此$f(x) = \frac{x}{x^2 - 2}$ 211 | \end{solu} 212 | \end{mathques} 213 | 214 | \begin{mathques}(2){2020-09-09}{极限} 215 | \begin{ques} 216 | 求极限$\lim_{x \to 0^+} \frac{2e^{\frac{1}{x}} + e^{-\frac{1}{x}}} 217 | {e^{\frac{2}{x}} - e^{- \frac{1}{x}}}$. 218 | \end{ques} 219 | \begin{solu} 220 | \begin{align*} 221 | \lim_{x \to 0^+} \frac{2e^{\frac{1}{x}} + e^{-\frac{1}{x}}} {e^{\frac{2} 222 | {x}} - e^{- \frac{1}{x}}} 223 | &= \lim_{t \to +\infty} \frac{2e^t + e^{-t}}{e^{2t} - e^{-t}} 224 | \mathhint[r]{\text{令}t = \frac{1}{x}}\\ 225 | &= \lim_{t \to +\infty} \frac{2 + e^{-2t}}{e^t - e^{-2t}} 226 | \mathhint[r]{\text{同除}e^t}\\ 227 | &= 0 228 | \end{align*} 229 | \end{solu} 230 | \end{mathques} 231 | 232 | \begin{mathques}(3){2020-09-11}{极限} 233 | \begin{ques} 234 | 求极限$\lim_{x \to 0} \frac{\sin x + x^2 \sin \frac{1}{x}}{(2 + x^2) \ln(1 + 235 | x)}$. 236 | \end{ques} 237 | \begin{solu} 238 | \begin{mathideabox} 239 | 所给极限为“$\frac{0}{0}$”型，先进行\emph{等价无穷小代换}，再拆分 240 | \end{mathideabox} 241 | \begin{align*} 242 | \lim_{x \to 0} \frac{\sin x + x^2 \sin \frac{1}{x}}{(2 + x^2) \ln(1 + x)} 243 | &= \lim_{x \to 0} \frac{1}{2 + x^2} \cdot \frac{\sin x + x^2 \sin \frac{1} 244 | {x}}{x} \mathhint[r]{\text{等价无穷小代换}}\\ 245 | \mathhint[l]{\text{将非零因子}\frac{1}{2 + x^2}\text{单独求极限}} 246 | &= \frac{1}{2} \lim_{x \to 0} \left( \frac{\sin x}{x} + x \sin \frac{1}{x} 247 | \right) \\ 248 | &= \frac{1}{2} 249 | \end{align*} 250 | \end{solu} 251 | \end{mathques} 252 | 253 | \begin{mathques}{2020-09-11}{一元微分} 254 | \begin{ques} 255 | 设函数$f(x)$在$x = 0$处连续，且$\lim_{x \to 0} \frac{f(x^2)}{x^2} = 1$，则 256 | (\quad) 257 | \begin{multichoice} 258 | \task $f(0) = 0$且$f'_-(0)$存在 259 | \task $f(0) = 1$且$f'_-(0)$存在 260 | \task \answer{$f(0) = 0$且$f'_+(0)$存在} 261 | \task $f(0) = 1$且$f'_+(0)$存在 262 | \end{multichoice} 263 | \end{ques} 264 | \begin{solu} 265 | 因为$f(x)$在$x = 0$处连续，且$\lim_{x \to 0} \frac{f(x^2)}{x^2} = 1$ 266 | 267 | 所以$\lim_{x \to 0} f(x^2) = 0$，即$f(0) = 0$ 268 | 269 | 从而有 270 | \begin{align*} 271 | \lim_{x \to 0} \frac{f(x^2)}{x^2} 272 | &= \matherror{\lim_{x \to 0} \frac{f(x^2) - f(0)}{x^2 - 0}}\\ 273 | \mathhint[l]{t = x^2} &= \matherror{\lim_{t \to \color{red} 0^+} 274 | \frac{f(t) - f(0)}{t - 0}}\\ 275 | &= f'_+(0) 276 | \end{align*} 277 | \end{solu} 278 | \end{mathques} 279 | 280 | \begin{mathques}{2020-09-11}{一元微分} 281 | \begin{ques} 282 | 设函数$f(x) = \abs{x^3 - 1}\varphi (x)$，其中$\varphi (x) $在$x = 1$处连续， 283 | 则$\varphi (1) = 0$是$f(x)$在$x = 1$处可导的(\quad) 284 | \begin{multichoice} 285 | \task \answer{充分必要条件} 286 | \task 充分但非必要条件 287 | \task 必要但非充分条件 288 | \task 既非充分也非必要条件 289 | \end{multichoice} 290 | \end{ques} 291 | \begin{solu} 292 | \mathidea{$\varphi(1) = 0 \Longrightarrow f(x)$在$x = 1$处可导}\par 293 | 由$\varphi(1) = 0$可得$\mathhint{f(1) = 0}$ 294 | \begin{align*} 295 | f'_+(1) &= \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} 296 | = \lim_{x \to 1^+} \frac{\abs{x^3 - 1}\varphi(x)}{x - 1} 297 | = \lim_{x \to 1^+} (x^2 + x + 1)\varphi(x) = 0\\ 298 | f'_-(1) &= \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} 299 | = \lim_{x \to 1^-} \frac{\abs{x^3 - 1}\varphi(x)}{x - 1} 300 | = \lim_{x \to 1^-} (x^2 + x + 1)\varphi(x) = 0 301 | \end{align*} 302 | 即$f'_+(1) = f'_-(1) = 0$，则$f'(1) = 0$ 303 | 304 | \mathidea{$\varphi(1) = 0 \Longleftarrow f(x)$在$x = 1$处可导}\par 305 | 设$f(x)$在$x = 1$处可导，因为$f(1) = 0$，所以 306 | \begin{align*} 307 | f'_+(1) &= \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} 308 | = \lim_{x \to 1^+} \frac{\abs{x^3 - 1}\varphi(x)}{x - 1} 309 | = \lim_{x \to 1^+} (x^2 + x + 1)\varphi(x) = 3\varphi(1)\\ 310 | f'_-(1) &= \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} 311 | = \lim_{x \to 1^-} \frac{\abs{x^3 - 1}\varphi(x)}{x - 1} 312 | = \lim_{x \to 1^-} (x^2 + x + 1)\varphi(x) = -3\varphi(1) 313 | \end{align*} 314 | 由$f'_+(1) = f'_-(1) = 0$得，$3\varphi(1) = -3\varphi(1)$，即$\varphi(1) = 0$ 315 | \end{solu} 316 | \end{mathques} 317 | 318 | \begin{mathques}{2020-09-11}{一元微分} 319 | \begin{ques} 320 | 设函数$y = f(x)$由方程$y = x\ln y$所确定，求$\dv{y}{x}$. 321 | \end{ques} 322 | \begin{solu} 323 | \mathmethod 方程两边同时对$x$求导，得$y' = \ln y + x\cdot \frac{1}{y} \cdot 324 | y'$，解得$y' = \frac{y \ln y}{y - x}$. 325 | \mathmethod 令$F(x, y) = y - x\ln y$，则$F'_x = - \ln y, F'_y = 1 - \frac{x}{y} 326 | $，故$\dv{y}{x} = - \frac{F'_x}{F'_y} = \frac{y\ln y}{y - x}$. 327 | \end{solu} 328 | \end{mathques} 329 | 330 | \begin{mathques}(2){2020-09-13}{一元微分} 331 | \begin{ques} 332 | 已知$f'(x) = Ae^x$($A$为正常数)，求$f(x)$的反函数的二阶导数. 333 | \end{ques} 334 | \begin{solu} 335 | \begin{mathideabox}[反函数定理] 336 | 如果从$\mathbb{R}^n$的一个开集$\mathrm{U}$到$\mathbb{R}^n$的连续可微函数 337 | $F$的全微分在点$p$可逆（$F$在点p的雅可比行列式不为零），则$F$在点$p$附近具 338 | 有反函数，且 339 | \[ 340 | J_{F^{-1}}(F(p)) = [J_F(p)]^{-1} 341 | \] 342 | 其中$[\cdot]^{-1}$表示逆矩阵，$J_G(q)$是函数$G$在点$q$的雅可比矩阵。 343 | 344 | 也可由链式法则推出：设$G,H$是两个函数，分别在$H(p)$和$p$处有全导数，则 345 | \[ 346 | J_{G\circ H}(p) = J_G(H(p))\cdot J_H(p) 347 | \] 348 | 设$G$为$F$，$H$为$F^{-1}$，$G\circ H$则为恒等函数，其雅可比矩阵也是单位矩 349 | 阵。 350 | 351 | 当$y$为$x$的一元函数时，其雅可比矩阵为$[\dd y / \dd x]$，即 352 | \[ 353 | \dv{x}{y} = \frac{1}{\dv{y}{x}} 354 | \] 355 | \end{mathideabox} 356 | 设$y = f(x)$，则$\dv{x}{y} = \frac{1}{f'(x)}$ 357 | \begin{align*} 358 | \dv[2]{y}{x} &= \dv{y}(\dv{x}{y})\\ 359 | &= \matherror{\dv{x}(\frac{1}{f'(x)}) \cdot \dv{x}{y}}\\ 360 | &= \matherror{-\frac{f''(x)}{\qty(f'(x))^2} \cdot \frac{1}{f'(x)}}\\ 361 | &= -\frac{f''(x)}{\qty(f'(x))^3} = - \frac{Ae^x}{(Ae^x)^3} = -\frac{1} 362 | {A^2e^{2x}} 363 | \end{align*} 364 | \end{solu} 365 | \end{mathques} 366 | 367 | \begin{mathques}(2){2020-09-13}{一元微分} 368 | \begin{ques} 369 | 设可导函数$y = f(x)$有反函数$g(x)$，且$f(a) = 3, f'(a) = 1, f''(a) = 2$，求 370 | $g''(3)$. 371 | \end{ques} 372 | \begin{solu} 373 | \mathmethod 374 | 由链式求导法则得$\qty(g \circ f(x))' = g'\qty(f(x)) \cdot f'(x)$，即 375 | \[ 376 | \matherror{f'(x) \cdot g'(f(x)) = 1} 377 | \] 378 | 两边同时对$x$求导，有 379 | \[ 380 | f''(x)g'(f(x)) + \qty(f'(x))^2 \cdot g''(f(x)) = 0 381 | \] 382 | 两边同乘$f'(x)$ 383 | \[ 384 | f''(x) + \qty(f'(x))^3 \cdot g''(f(x)) = 0 385 | \] 386 | 将$x = a$代入得 387 | \[ 388 | f''(a) + \qty(f'(a))^3 \cdot g''(f(a)) = 0 389 | \] 390 | 即 391 | \[ 392 | 2 + g''(3) = 0 393 | \] 394 | 则$g''(3) = -2$ 395 | \mathmethod 公式法 396 | \[ 397 | g''(3) = - \frac{f''(a)}{\qty(f(a))^3} = -2 398 | \] 399 | \begin{mathideabox}[公式推导过程] 400 | 已知 401 | $\left\{\begin{aligned} 402 | \eval{y}_{x = a} &= 3\\ 403 | \eval{\dv{y}{x}}_{x = a} &= 1\\ 404 | \eval{\dv[2]{y}{x}}_{x = a} &= 2 405 | \end{aligned}\right.$ 406 | , 求$\eval{\dv[2]{x}{y}}_{y = 3}$ 407 | 408 | \begin{align*} 409 | \dv[2]{x}{y} &= \dv{y}(\dv{x}{y}) = \dv{x}(\dv{x}{y}) \cdot \dv{x}{y} \\ 410 | &= \dv{x}(\frac{1}{\dv{y}{x}}) \cdot \frac{1}{\dv{y}{x}} \\ 411 | &= - \frac{\dv{x}(\dv{y}{x})}{\qty(\dv{y}{x})^2} \cdot \frac{1}{\dv{y}{x}}\\ 412 | &= - \frac{\dv[2]{y}{x}}{\qty(\dv{y}{x})^3} 413 | \end{align*} 414 | 即$\eval{\dv[2]{x}{y}}_{y = 3} = \eval{- \frac{\dv[2]{y}{x}}{\qty(\dv{y} 415 | {x})^3}}_{x = a}$ 416 | \end{mathideabox} 417 | \end{solu} 418 | \end{mathques} 419 | 420 | \begin{mathques}{2020-09-13}{一元微分} 421 | \begin{ques} 422 | 设$f(x) = 423 | \begin{dcases} 424 | x^2 \sin \frac{\pi}{x}, & x < 0, \\ 425 | A, & x = 0, \\ 426 | ax^2 + b, & x > 0,\\ 427 | \end{dcases} 428 | $求常数$A, a, b$的值，使$f(x)$在$x = 0$处可导，并求$f'(0)$. 429 | \end{ques} 430 | \begin{solu} 431 | \begin{mathideabox}[可导必然连续（反之不一定）] 432 | 设函数$y = f(x)$上一点$x_0$，函数在这一点可导，即$f'(x_0) = \lim_{\Delta x 433 | \to 0} \frac{\Delta y}{\Delta x}存在$，其中 434 | \[ 435 | \Delta y = f(x_0 + \Delta x) - f(x_0) 436 | \] 437 | 则 438 | \begin{align*} 439 | \lim_{\Delta x \to 0} \Delta y &= \lim_{\Delta x \to 0} \qty(\frac{\Delta 440 | y}{\Delta x}\cdot \Delta x)\\ 441 | &= f'(x_0) \cdot 0 = 0 442 | \end{align*} 443 | 即函数$f(x)$在$x_0$处连续 444 | \end{mathideabox} 445 | \begin{mathideabox}[导数存在必然有左右导数相等] 446 | 由极限的性质可知： 447 | \[ 448 | \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} 449 | =\lim_{\Delta x \to 0^-} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} 450 | =\lim_{\Delta x \to 0^+} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} 451 | \] 452 | \end{mathideabox} 453 | 可导必然连续，即 454 | \[ 455 | \lim_{x \to 0^-} x^2 \frac{\pi}{x} = \lim_{x \to 0^+} (ax^2 + b) = A 456 | \] 457 | 则$A = b = 0$. 458 | 又 459 | \begin{align*} 460 | f'_-(0) &= \lim_{x \to 0^-} \frac{x^2 \sin \frac{\pi}{x} - 0}{x - 0} = 0\\ 461 | f'_+(0) &= \lim_{x \to 0^+} \frac{ax^2 - 0}{x - 0} = 0 462 | \end{align*} 463 | 所以，$a$可以为任意常数，且$f'(0) = 0$ 464 | \end{solu} 465 | \end{mathques} 466 | 467 | \begin{mathques}(2){2020-09-18}{一元微分} 468 | \begin{ques} 469 | 设函数$y = f(x)$由$ 470 | \begin{dcases} 471 | x = \ln(1 + t^2) + 1,\\ 472 | y = 2 \arctan t - (t + 1)^2,\\ 473 | \end{dcases} 474 | $确定，求$\dv{y}{x}, \dv[2]{y}{x}$ 475 | \end{ques} 476 | \begin{solu} 477 | \begin{align*} 478 | \dv{y}{t} &= \frac{2}{\matherror{1 + t^2}} - 2(t + 1)\\ 479 | \dv{x}{t} &= \frac{2t}{1 + t^2} 480 | \end{align*} 481 | 由链式法则，$\dv{y}{t} = \dv{y}{x}\cdot\dv{x}{t}$，则 482 | \begin{align*} 483 | \dv{y}{x} &= \frac{\dv{y}{t}}{\dv{x}{t}} 484 | = \frac{\frac{2}{1 + t^2} - 2(t + 1)}{\frac{2t}{1 + t^2}} 485 | = -(t^2 + t + 1)\\ 486 | \dv[2]{y}{x} &= \dv{x}(\dv{y}{x})\\ 487 | &= \dv{x}(-(t^2 + t + 1))\\ 488 | &= \dv{t}(-(t^2 + t + 1)) \cdot \dv{t}{x}\\ 489 | &= -\frac{1 + 2t}{\dv{x}{t}}\\ 490 | &= -\frac{(1 + 2t)(1 + t^2)}{2t} 491 | \end{align*} 492 | \end{solu} 493 | \end{mathques} 494 | 495 | \begin{mathques}(2){2020-09-15}{一元微分} 496 | \begin{ques} 497 | 设$f(x)$满足$f(0) = 0$，且$f'(0)$存在，求$\lim_{x \to 0} \frac{f(1 - 498 | \sqrt{\cos x})}{\ln(1 - x\sin x)}$ 499 | \end{ques} 500 | \begin{solu} 501 | \begin{align*} 502 | \lim_{x \to 0} \frac{f(1 - \sqrt{\cos x})}{\ln(1 - x\sin x)} 503 | &=\lim_{x \to 0} \frac{f(1 - \sqrt{\cos x}) - f(0)}{(1 - x\sin x) - 0} 504 | \cdot \lim_{x \to 0} \frac{1 - \sqrt{\cos x}}{\ln(1 - x\sin x)}\\ 505 | &=f'(0) \lim_{x \to 0} \frac{1 - \sqrt{\cos x}}{\ln(1 - x\sin x)}\\ 506 | \mathidea{等价无穷小替换}&=f'(0)\lim_{x \to 0} \frac{1 - \cos x}{-x\sin x} 507 | \cdot \frac{1}{1 + \sqrt{\cos x}}\\ 508 | &= -\frac{1}{2}f'(0)\lim_{x \to 0} \frac{1 - \cos x}{x^2}\\ 509 | &= -\frac{1}{4}f'(0). 510 | \end{align*} 511 | \end{solu} 512 | \end{mathques} 513 | 514 | \begin{mathques}{2020-09-12}{一元微分} 515 | \begin{ques} 516 | 设函数$f(x)$在$x = 0$处连续，且$\lim_{x \to 0} \frac{f(x)}{x}$存在，则(\quad) 517 | \begin{multichoice} 518 | \task $f(0) \neq 0$，但$f'(0)$可能不存在 519 | \task $f(0) = 0$，但$f'(0)$可能不存在 520 | \task \answer{$f'(0)$存在，但$f'(0)$不一定等于零} 521 | \task $f'(0)$存在，且必定有$f'(0) = 0$ 522 | \end{multichoice} 523 | \end{ques} 524 | \begin{solu} 525 | \begin{mathideabox}[连续的定义] 526 | 函数$f(x)$在点$p$处连续当且仅当$\lim_{x \to p} f(x) = f(p)$ (隐含着极限必 527 | 须存在) 528 | \end{mathideabox} 529 | \begin{mathideabox}[重要结论] 530 | 若$f(x)$在$x = 0$处连续，且$\lim_{x \to 0} \frac{f(x)}{x} = A$，则 531 | \begin{align*} 532 | f(0) &= 0\\ 533 | f'(0) &= A 534 | \end{align*} 535 | \end{mathideabox} 536 | $f(x)$在$x = 0$处连续，则$\lim_{x \to 0} f(x)$存在且$\lim_{x \to 0} f(x) = 537 | f(0)$ 538 | 539 | 由$\lim_{x \to 0} \frac{f(x)}{x}$存在且分母极限为$0$可知，分子极限也为$0$， 540 | 即$f(0) = 0$ 541 | 542 | 又 543 | \[ 544 | f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} 545 | \frac{f(x)}{x} 546 | \] 547 | 即$f'(0)$存在 548 | \end{solu} 549 | \end{mathques} 550 | 551 | \begin{mathques}{2020-09-12}{一元微分} 552 | \begin{ques} 553 | 设$f(x) = x^a \abs{x}$，$a$为正整数，则函数$f(x)$在点$x = 0$处(\quad) 554 | \begin{multichoice} 555 | \task 不存在极限 556 | \task 存在极限，但不连续 557 | \task 连续但不可导 558 | \task \answer{可导} 559 | \end{multichoice} 560 | \end{ques} 561 | \begin{solu} 562 | $\lim_{x \to 0} f(x) = 0 = f(0)$，故$f(x)$在$x = 0$处连续 563 | \begin{align*} 564 | f'_+(0) &= \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} 565 | \frac{x^a \abs{x}}{x} = \matherror{\lim_{x \to 0^+} \frac{x^{a + 1}}{x} = 566 | 0}\\ 567 | f'_-(0) &= \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} 568 | \frac{x^a \abs{x}}{x} = \lim_{x \to 0^-} \frac{-x^{a + 1}}{x} = 0 569 | \end{align*} 570 | 即$f'_+(0) = f'_-(0) = 0$，$f(x)$在点$x = 0$处可导，且导数为$0$. 571 | \end{solu} 572 | \end{mathques} 573 | 574 | \begin{mathques}{2020-09-12}{一元微分} 575 | \begin{ques} 576 | 若$y = f(x)$可导，则当$\Delta x \to 0$时，$\Delta y - \dd y$为$\Delta x$的 577 | (\quad) 578 | \begin{multichoice} 579 | \task \answer{高阶无穷小} 580 | \task 低阶无穷小 581 | \task 同阶但不等价无穷小 582 | \task 等阶无穷小 583 | \end{multichoice} 584 | \end{ques} 585 | \begin{solu} 586 | \begin{mathideabox}[微分定义] 587 | 设函数$y = f(x)$在区间$\mathcal{I}$有定义。对于$\mathcal{I}$内的一点$x_0$， 588 | 当$x_0$变动到附近的$x_0 + \Delta x$（也在此区间）时，如果函数的增量$\Delta 589 | y = f\qty(x_0 + \Delta x - f(x_0))$可以表示为 590 | \[ 591 | \Delta y = A\Delta x + o(\Delta x) 592 | \] 593 | 其中$A$是不依赖于$\Delta x$的常数，$o(\Delta x)$是比$\Delta x$高阶的无穷小， 594 | 则称函数$f(x)$在点$x_0$是可微的. 595 | 596 | $A\Delta x$称作函数在点$x_0$对应与自变量增量$\Delta x$的微分，记做$\dd 597 | y$，即$\dd y = A\Delta x$ 598 | 599 | 把自变量$x$的增量$\Delta x$称作自变量的微分，记做$\dd x$，即$\dd x = 600 | \Delta x$ 601 | \end{mathideabox} 602 | 由微分的定义可知$\Delta y - \dd y = o(\Delta x)$为$\Delta x$的高阶无穷小 603 | \end{solu} 604 | \end{mathques} 605 | 606 | \begin{mathques}{2020-09-12}{一元微分} 607 | \begin{ques} 608 | 设函数 609 | \[ 610 | f(x) = 611 | \begin{dcases} 612 | x^3\sin \frac{1}{x}, & x \neq 0,\\ 613 | 0, & x= 0,\\ 614 | \end{dcases} 615 | \] 616 | 讨论$f(x)$在$x = 0$处的可导性以及$f'(x)$在$x = 0$处的连续性. 617 | \end{ques} 618 | \begin{solu} 619 | \begin{mathideabox} 620 | 证明$f(x)$在$x_0$处可导，即证明 621 | \begin{align*} 622 | f'(x_0) &= \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} 623 | = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} 624 | \\ 625 | \text{或}f'(x_0) &= \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} 626 | \end{align*} 627 | 存在 628 | \par\vspace{1em} 629 | 证明$f(x)$在$x_0$处连续，即证明 630 | \[ 631 | \lim_{x \to x_0} f(x) = f(x_0) 632 | \] 633 | \end{mathideabox} 634 | 635 | \matherror{\text{当}x = 0\text{时}}，有 636 | \[ 637 | f'(0) = \lim_{x \to 0} \frac{x^3 \sin \frac{1}{x} - 0}{x - 0} = 0, 638 | \] 639 | 函数$f(x)$在$x = 0$处可导. 640 | 641 | \matherror{\text{当}x \neq 0\text{时}}，有 642 | \[ 643 | f'(x) = 3x^2 \sin \frac{1}{x} - x\cos \frac{1}{x} 644 | \] 645 | 因为 646 | \[ 647 | \lim_{x \to 0} f'(x) = \lim_{x \to 0} \qty(3x^2 \sin \frac{1}{x} - x\cos 648 | \frac{1}{x}) = 0 = f'(0), 649 | \] 650 | 故函数f'(x)在$x = 0$处连续 651 | \end{solu} 652 | \end{mathques} 653 | 654 | \begin{mathques}{2020-09-16}{一元微分}[最后忘记给$a$加平方] 655 | \begin{ques} 656 | 已知曲线$y = x^3 - 3a^2x + b$与$x$轴相切，则$b^2$可通过$a$表示为 657 | $b^2=$\mathblank 658 | \end{ques} 659 | \begin{solu} 660 | \answer{$4a^6$} 661 | 662 | 因为曲线$y = x^3 - 3a^2 x + b$与$x$轴相切，所以，在切点处有 663 | \[ 664 | y' = 3x^2 - 3a^2 = 0. 665 | \] 666 | 由此可解出切点横坐标$x = a$或$x = -a$，且在切点处有 667 | \[ 668 | y(a) = a^3 - 3a^3 + b = 0 \qq{或} y(-a) = -a^3 + 3 a^3 + b = 0 669 | \] 670 | 则$b = \pm 2a^3$，故$b^2 = 4a^6$ 671 | \end{solu} 672 | \end{mathques} 673 | 674 | \begin{mathques}{2020-09-16}{一元微分}[没有考虑导数的分母也会跟着改变正负号] 675 | \begin{ques} 676 | 设$y = \frac{x^3 + 4}{x^2}$，求函数的增减区间及极值 677 | \end{ques} 678 | \begin{solu} 679 | $\matherror{y' = 1 - \frac{8}{x^3}}$，故驻点为$x = 2$，又 680 | \[ 681 | \begin{NiceMatrix}[columns-width=2cm] 682 | \toprule 683 | x & (-\infty, 0) & (0, 2) & 2 & (2, +\infty)\\ 684 | \midrule 685 | y' & + & - & 0 & + \\ 686 | y & \nearrow & \searrow & 3 & \nearrow\\ 687 | \bottomrule 688 | \end{NiceMatrix} 689 | \] 690 | 所以，$(-\infty, 0)$以及$(2, +\infty)$为增区间，$(0, 2)$为减区间，$x = 2$为 691 | 极小值点，极小值为$y = 3$. 692 | \end{solu} 693 | \end{mathques} 694 | 695 | \begin{mathques}{2020-09-16}{一元微分}[算到后面忘记开根号] 696 | \begin{ques} 697 | 给定曲线$y = \frac{1}{x^2}$ 698 | \begin{queslist} 699 | \item 求曲线在横坐标为$x_0$的点处的切线方程 700 | \item 求曲线的切线被两坐标轴所截线段的最短长度 701 | \end{queslist} 702 | \end{ques} 703 | \begin{solu} 704 | \begin{solulist} 705 | \item 由$y' = -\frac{2}{x^3}$可知曲线$y = \frac{1}{x^2}$在横坐标$x_0$处的切 706 | 线方程为 707 | \[ 708 | y - \frac{1}{x_0^2} = - \frac{2}{x_0^3}(x - x_0) 709 | \] 710 | \item 将$x = 0, y = 0$代入$x = x_0$处切线方程可得该切线在$x$轴，$y$轴上的截 711 | 距分别为 712 | \[ 713 | X = \frac{3}{2}x_0 \qc Y = \frac{3}{x_0^2}. 714 | \] 715 | 设该切线被两坐标轴所截线段长度为$L$，则$L^2 = X^2 + Y^2 = \frac{9}{4}x_0^2 716 | + \frac{9}{x_0^4}.$ 717 | 718 | 令$\dv{L^2}{x_0} = \frac{9}{2}x_0 - \frac{36}{x_0^5} = 0$，得驻点$x_0 = 719 | \pm \sqrt{2}.$ 720 | 721 | 又$\dv[2]{L^2}{x_0} = \frac{9}{2} + \frac{180}{x_0^6}$，显然$\eval{\dv[2] 722 | {L^2}{x_0}}_{x_0 = \pm \sqrt{2}} > 0$，由此可知，$L^2$在$x_0 = \pm \sqrt{2} 723 | $处取得极小值，即最小值，$L_{\min}^2 = \frac{27}{4}$ 724 | 725 | 故$L_{\min} = \frac{3}{2}\sqrt{3}.$ 726 | \end{solulist} 727 | \end{solu} 728 | \end{mathques} 729 | 730 | \begin{mathques}(2){2020-09-23}{一元微分}[题目问的是最小值不是极小值] 731 | \begin{ques} 732 | 设$a > 1, f(t) = a^t - at$在$(-\infty, +\infty)$内的驻点为$t(a)$，问$a$为何 733 | 值时，$t(a)$最小，并求出最小值. 734 | \end{ques} 735 | \begin{solu} 736 | \begin{mathideabox} 737 | 先求驻点函数$t(a)$，然后再求$t(a)$的最小值和最小值点 738 | \end{mathideabox} 739 | 令$f'(t) = a^t\ln a - a = 0$，得唯一驻点函数$t(a) = 1 - \frac{\ln\ln a}{\ln 740 | a}.$ 741 | 742 | 令$t'(a) = - \frac{\frac{1}{a} - \frac{1}{a}\ln \ln a}{(\ln a)^2} = 0$，得唯 743 | 一驻点$a = e^e.$ 744 | 745 | 当$a < e^e$时，$t'(a) < 0$；当$a > e^e$时，$t'(a) > 0$，所以，$t(e^e) = 1 - 746 | \frac{1}{e}$为极小值，也是最小值. 747 | \end{solu} 748 | \end{mathques} 749 | 750 | \begin{mathques}{2020-09-17}{一元微分}[没有直接两边取对数，最后没有化到最简] 751 | \begin{ques} 752 | 设$y = y(x)$由方程$x = y^y$确定，则$\dd y = $\mathblank 753 | \end{ques} 754 | \begin{solu} 755 | \answer{$\frac{1}{x(1 + \ln y)}\dd x$} 756 | 757 | 两边取对数得，$\ln x = y\ln y,$ 758 | 759 | 两边关于$x$求导，得$\frac{1}{x} = (1 + \ln y) y'$ 760 | 761 | 则$y' = \frac{1}{x (1 + \ln y)}$ 762 | 763 | 即$\dd y = y' \dd x = \frac{1}{x(1 + \ln y)}\dd x$ 764 | \end{solu} 765 | \end{mathques} 766 | 767 | \begin{mathques}{2020-09-17}{一元微分}[漏抄了] 768 | \begin{ques} 769 | 设$y = e^{x^2}$，求$\dv{y}{x}, \dv{y}{(x^2)}, \dv[2]{y}{x}.$ 770 | \end{ques} 771 | \begin{solu} 772 | \begin{mathideabox}[区分微分符号] 773 | $\dd[n]x = \qty(\dd x)^n$是“微分的幂”，$\dd (x^n) = nx^{n - 1}\dd x$是“幂 774 | 的微分” 775 | \end{mathideabox} 776 | \begin{align*} 777 | \dv{y}{x} &= 2xe^{x^2},\\ 778 | \dv{y}{(x^2)} &= \matherror{\dv{y}{x}\cdot \dv{x}{(x^2)}} = 2xe^{x^2}\cdot 779 | \frac{1}{2x} = e^{x^2},\\ 780 | \dv[2]{y}{x} &= 2(1 + 2x^2)e^{x^2}. 781 | \end{align*} 782 | \end{solu} 783 | \end{mathques} 784 | 785 | \begin{mathques}{2020-09-17}{一元微分}[计算错误，没有化简后再求导] 786 | \begin{ques} 787 | 设$y = \ln \sqrt{\frac{1 - x}{1 + x^2}}$，求$\eval{y''}_{x = 0}.$ 788 | \end{ques} 789 | \begin{solu} 790 | \begin{align*} 791 | y &= \matherror{\ln \sqrt{\frac{1 - x}{1 + x^2}} = \frac{1}{2} \qty(\ln (1 792 | - x) - \ln(1 + x^2))},\\ 793 | y' &= \frac{1}{2}\qty(\frac{-1}{1 - x} - \frac{2x}{1 + x^2}) 794 | = - \frac{1}{2}\qty(\frac{1}{1 - x} + \frac{2x}{1 + x^2}),\\ 795 | y'' &= -\frac{1}{2}\qty(\frac{1}{(1 - x)^2} + 2\cdot \frac{1 - x^2}{(1 + 796 | x^2)^2}),\\ 797 | \eval{y''}_{x = 0} &= -\frac{3}{2}. 798 | \end{align*} 799 | \end{solu} 800 | \end{mathques} 801 | 802 | \begin{mathques}(2){2020-09-27}{中值定理}[没有注意被积函数的导数与欲证部分的关系+ 803 | 积分中值定理不熟悉] 804 | \begin{ques} 805 | 设$f(x)$在$[0, 1]$上连续，在$(0, 1)$内可导，且$f(1) = k\int_{0}^{\frac{1}{k}} 806 | x e^{1-x} f(x) \dd x (k > 1).$证明至少存在一点$\xi \in (0, 1)$，使得$f'(\xi) 807 | = (1 - \xi^{-1}) f(\xi)$. 808 | \end{ques} 809 | \begin{solu} 810 | \begin{mathideabox} 811 | 注意到被积函数$x e^{1 - x} f(x)$的导数 812 | \[ 813 | \qty (xe^{1-x} f(x))' = xe^{1-x} \qty(f'(x) - (1 - x^{-1})f(x)) 814 | \] 815 | 中含有$f'(x) - (1 - x^{-1})f(x),$这正是欲证结论中的一部分 816 | \end{mathideabox} 817 | \mathproof 令$F(x) = xe^{1-x} f(x).$由积分中值定理可知 818 | \begin{align*} 819 | f(1) &= k \int_{0}^{\frac{1}{k}} x e^{1-x} f(x) \dd x \\ 820 | &= \eta e^{1 - \eta} f(\eta) \\ 821 | &= \matherror{1} \cdot e^{1 - 1} f(1) 822 | \end{align*} 823 | 其中$\eta \in \qty(0, \frac{1}{k}) \subset \matherror{(0, 1)}$, 824 | 又$F(x) = xe^{1 - x}f(x)$在$[\eta, 1]$上连续，在$(\eta, 1)$内可导，且$F(\eta) 825 | = F(1),$则$F(x) = xe^{1 - x}f(x)$在$[\eta, 1]$上满足罗尔定理的条件 826 | 827 | 即存在$\xi \in (\eta, 1) \subset (0, 1),$使得$F'(\xi) = 0$, 即$f'(\xi) = (1 - 828 | \xi^{-1}) f(\xi).$ 829 | \end{solu} 830 | \end{mathques} 831 | 832 | \begin{mathques}(4){2020-10-31}{中值定理}[费马定理+脱帽性不熟悉] 833 | \begin{ques} 834 | 设$f(x)$在$[a, b]$上可导，若$f'_+(a) \neq f'_-(b)$，证明对于任意的介于 835 | $f'_+(a)$与$f'_-(b)$之间的$\mu$，存在$\xi \in (a, b)$，使得$f'(\xi) = \mu$. 836 | \end{ques} 837 | \begin{solu} 838 | \mathproof 因$f'_+(a) \neq f'_-(b)$，不妨设$f'_+(a) < f'_-(b)$. 并设$F(x) = 839 | f(x) - \mu x$，则函数$F(x)$在$[a, b]$上可导，且$F'_+(a) = f'_+(a) - \mu < 0, 840 | F'_(b) = f'_-(b) - \mu > 0$，于是 841 | \begin{align*} 842 | F'_+(a) &= \lim_{x \to a^+} \frac{F(x) - F(a)}{x - a} < 0,\\ 843 | F'_+(b) &= \lim_{x \to b^-} \frac{F(x) - F(b)}{x - b} > 0, 844 | \end{align*} 845 | 根据\matherror*{极限的保号性}，知： 846 | {\parindent=2\ccwd\par 847 | 在点$x=a$的某个右邻域内，$\frac{F(x) - F(a)}{x - a} < 0$，即$F(x) < F(a)$； 848 | \vspace{1em} 849 | 850 | 在点$x=b$的某个左邻域内，$\frac{F(x) - F(b)}{x - b} > 0$，即$F(x) > F(a)$. 851 | } 852 | 853 | 故$F(a)$和$F(b)$均不是函数$F(x)$在$[a, b]$上的最小值，又因$F(x)$一定可以取得 854 | 最小值，则其最小值必在$(a, b)$内取到，设函数$F(x)$在$(a, b)$内的最小值点是 855 | $\xi$，根据费马定理，得$F'(\xi) = 0$，即$f'(\xi) = \mu$. 856 | \end{solu} 857 | \end{mathques} 858 | 859 | \begin{mathques}(2){2020-09-27}{中值定理}[没有联系第一小问+拉格朗日中值定理不熟悉] 860 | \begin{ques} 861 | 已知$f(x)$在$[0, 1]$上连续，在$(0, 1)$内可导，且$f(0) = 0, f(1) = 1$. 证明： 862 | \begin{queslist} 863 | \item 存在$\xi \in (0, 1)$，使得$f(\xi) = 1 - \xi$; 864 | \item 存在$\eta , \tau \in (0, 1), \eta \neq \tau$，使得$f'(\eta)f'(\tau) 865 | = 1.$ 866 | \end{queslist} 867 | \end{ques} 868 | \begin{solu} 869 | \begin{solulist} 870 | \item 令$F(x) = f(x) - 1 + x$, 可得 871 | \begin{align*} 872 | F(0) &= f(0) - 1 + 0 = -1 < 0,\\ 873 | F(1) &= f(1) - 1 + 1 = \phantom{-}1 > 0, 874 | \end{align*} 875 | 由零点定理可知，存在$\xi \in (0, 1)$，使得$F(\xi) = 0$，即$f(\xi) = 1 - 876 | \xi$. 877 | \item 用$\xi$将$[0, 1]$划分为$[0, \xi], [\xi, 1]$，再用拉格朗日中值定理 878 | \begin{align*} 879 | f(\xi) - f(0) &= f'(\eta)(\xi - 0), & &\eta \in (0, \xi),\\ 880 | f(1) - f(\xi) &= f'(\tau)(1 - \xi), & &\tau \in (\xi, 1), 881 | \end{align*} 882 | 则 883 | \begin{align*} 884 | f'(\eta) &= \frac{f(\xi) - f(0)}{\xi} = \frac{1 - \xi}{\xi}, \\ 885 | f'(\tau) &= \frac{f(1) - f(\xi)}{ 1 - \xi } = \frac{\xi}{1 - \xi}, 886 | \end{align*} 887 | 故$f'(\eta)f'(\tau) = 1$. 888 | \end{solulist} 889 | \end{solu} 890 | \end{mathques} 891 | 892 | \begin{mathques}(2){2020-10-01}{零点问题与微分不等式}[没有想到用取对数后的式子证 893 | 明] 894 | \begin{ques} 895 | 设$0 < x < 1$，求证$xe^{-x} > \frac{1}{x} e^{-\frac{1}{x}}$. 896 | \end{ques} 897 | \begin{solu} 898 | \mathproof 要证明$xe^{-x} > \frac{1}{x} e^{-\frac{1}{x}}$，只需证$\ln x - x 899 | > -\ln x - \frac{1}{x}$，即$2\ln x -x + \frac{1}{x} > 0$. 900 | 901 | 设$f(x) = 1\ln x - x + \frac{1}{x}, x \in (0, 1)$，则$f'(x) = \frac{-(x - 902 | 1)^2}{x^2} < 0$，故$f(x)$在$(0, 1)$上单调递减 903 | 904 | 则$f(x) > f(1) = 0$. 905 | \end{solu} 906 | \end{mathques} 907 | 908 | \begin{mathques}(2){2020-10-01}{零点问题与微分不等式}[计算繁琐] 909 | \begin{ques} 910 | 证明：$\qty(\ln \frac{1+x}{x} - \frac{1}{1+x})^2 < \frac{1}{x(1+x)^2} (x > 0) 911 | .$ 912 | \end{ques} 913 | \begin{solu} 914 | 只需证明当$x > 0$时，$\abs{\ln \frac{1+x}{x} - \frac{1}{1+x}} - \frac{1} 915 | {\sqrt{x}(1 + x)} < 0$. 916 | 917 | 令 918 | \[ 919 | f(x) = \ln \frac{1 + x}{x} - \frac{1}{1 + x} = \ln (1 + x) - \ln x - 920 | \frac{1}{1 + x}, 921 | \] 922 | 923 | 则 924 | \begin{align*} 925 | f'(x) &= \frac{1}{1+x} - \frac{1}{x} + \frac{1}{(1+x)^2} \\ 926 | &= \frac{(1+x)x - (1+x)^2 + x}{x(1 + x)^2} \\ 927 | &= \frac{-1}{x(1+x)^2} < 0, 928 | \end{align*} 929 | 又$\matherror{\lim_{x \to +\infty} f(x) = 0}$，所以$f(x) > 0$. 930 | 令 931 | \[ 932 | g(x) = \ln \frac{x + 1}{x} - \frac{1}{1 + x} - \frac{1}{\sqrt{x}(1 + x)}, 933 | \] 934 | 则 935 | \begin{align*} 936 | g'(x) &= \frac{1}{1 + x} - \frac{1}{x} + \frac{1}{(1+x)^2} + \frac{\frac{1} 937 | {2\sqrt{x}}\cdot (1 + x) + \sqrt{x}}{x(1 + x)^2} \\ 938 | &= \frac{-2\sqrt{x} + 1 + x + 2x}{2x^{\frac{3}{2}} (1 + x)^2} \\ 939 | &= \frac{1 + 3x - 2\sqrt{x}}{2x^{\frac{3}{2}} (1 + x)^2}. 940 | \end{align*} 941 | 再令$h(x) = 1 + 3x - 2\sqrt{x}$，则$h'(x) = 3 - \frac{1}{\sqrt{x}}$，令$h'(x) 942 | = 0$得$x = \frac{1}{9}$为唯一极小值点，也就是最小值点，且$h_{\min} = \frac{2} 943 | {3} > 0$，故$h(x) > 0$，则$g'(x) > 0$，即$g(x)$单调增加 944 | 945 | 又$\lim_{x \to +\infty} f(x) = 0$且$\lim_{x \to +\infty} - \frac{1} 946 | {\sqrt{x}(1 + x)} = 0$故$g(x) < 0$，证毕. 947 | \end{solu} 948 | \end{mathques} 949 | 950 | \begin{mathques}{2020-09-28}{一元函数几何应用}[极坐标系下的曲线需要转化成直角 951 | 坐标系下的参数方程后，再求切线方程] 952 | \begin{ques} 953 | 曲线$r = \cos 2\theta$在$\theta = \frac{\pi}{4}$处的切线方程为\mathblank. 954 | \end{ques} 955 | \begin{solu} 956 | \answer{$y = x$} 957 | 曲线参数方程：$ 958 | \begin{dcases} 959 | x = \matherror{r\cos \theta} = \cos 2\theta\cos \theta, \\ 960 | y = \matherror{r\sin \theta} = \cos 2\theta\sin \theta, \\ 961 | \end{dcases} 962 | \theta = \frac{\pi}{4} 963 | $对应$(x_0, y_0) = (0, 0),$ 964 | \[ 965 | \eval{\dv{y}{x}}_{\theta = \frac{\pi}{4}} = \eval{\frac{-2\sin 2\theta \sin 966 | \theta + \cos 2\theta \cos \theta}{-2 \sin 2\theta \cos \theta - \cos 967 | 2\theta \sin \theta}}_{\theta = \frac{\pi}{4}} = 1, 968 | \] 969 | 切线方程为$y = x.$ 970 | \end{solu} 971 | \end{mathques} 972 | 973 | \begin{mathques}{2020-09-28}{一元函数几何应用}[手残计算错误] 974 | \begin{ques} 975 | 函数$y = (x - 1)^2 (x - 2)^2 (-3 \le x \le 4)$的值域是\mathblank. 976 | \end{ques} 977 | \begin{solu} 978 | \answer{$[0, 400]$} 979 | \begin{align*} 980 | \dv{y}{x} &= 2 (x - 1) (x - 2)^2 + 2(x - 1)^2 (x - 2) \\ 981 | &= 2(x - 1) (x - 2) (2x - 3). 982 | \end{align*} 983 | 令$\dv{y}{x} = 0$，得驻点：$x = 1, 2, \frac{3}{2},$ 984 | \begin{align*} 985 | y_{\max} &= \max{y(1), y(2), y\qty(\frac{3}{2}), y(-3), y(4)} \\ 986 | &= \max{0, 0, \frac{1}{16}, 400, 36} = 400, \\ 987 | y_{\min} &= \min{y(1), y(2), y\qty(\frac{3}{2}), y(-3), y(4)} \\ 988 | &= \max{0, 0, \frac{1}{16}, 400, 36} = 0, 989 | \end{align*} 990 | 值域是$[0, 400]$ 991 | \end{solu} 992 | \end{mathques} 993 | 994 | \begin{mathques}{2020-09-28}{一元函数几何应用}[并不是二阶倒数取不到$0$值就没有 995 | 拐点，只要发生正负变化即可\par 二阶导数无意义点原函数不一定无意义， 拐点的 996 | $y$值通过原函数求出] 997 | \begin{ques} 998 | 曲线$y = x^{\frac{5}{3}} + 3x + 5$的拐点坐标为\mathblank. 999 | \end{ques} 1000 | \begin{solu} 1001 | \answer{$(0, 5)$} 1002 | 1003 | $y = x^{\frac{5}{3}} + 3x + 5$的定义域为$(-\infty, +\infty)$. 1004 | \begin{align*} 1005 | y' &= \frac{5}{3}x^{\frac{2}{3}} + 3 \\ 1006 | y'' &= \frac{10}{9}x^{-\frac{1}{3}}. 1007 | \end{align*} 1008 | 当$x = 0$时，$y''$不存在； 1009 | 1010 | 当$x < 0$时，$y'' < 0$； 1011 | 1012 | 当$x > 0$时，$y'' > 0$； 1013 | 1014 | 当$x = 0$时，$y = 5$，即拐点为$(0, 5)$. 1015 | \end{solu} 1016 | \end{mathques} 1017 | 1018 | \begin{mathques}{2020-09-28}{一元函数几何应用}[问的是极小值点不是极小值] 1019 | \begin{ques} 1020 | 设$y = x^3 - 3x^2 - 9x + 2$，则$y$在$[-2, 2]$上的最小值点为\mathblank. 1021 | \end{ques} 1022 | \begin{solu} 1023 | \answer{$x = 2$} 1024 | 1025 | $y = x^3 - 3x^2 - 9x + 2$的定义域为$(-\infty, +\infty)$. 1026 | 1027 | 令$y' = 0$得$y$的两个驻点$x_1 = -1, x_2 = 3（舍掉）.$又 1028 | \[ 1029 | \eval{y}_{x = -1} = 7\qc \eval{y}_{x = -2} = 0 \qc \eval{y}_{x = 2} = -20, 1030 | \] 1031 | 比较上述三个值可知$y$在$[-2, 2]$上的最小值点为$x = 2.$ 1032 | \end{solu} 1033 | \end{mathques} 1034 | 1035 | \begin{mathques}{2020-09-28}{一元函数几何应用}[$b$计算错误] 1036 | \begin{ques} 1037 | 曲线$y = \frac{x^2}{x + 2}$的斜渐近线为\mathblank. 1038 | \end{ques} 1039 | \begin{solu} 1040 | \answer{$y = x - 2$} 1041 | 1042 | 由于 1043 | \begin{align*} 1044 | a &= \lim_{x \to \infty} \dv{y}{x} = \lim_{x \to \infty} \frac{x^2}{x(x + 2)} = 1045 | 1, \\ 1046 | b &= \lim_{x \to \infty} (y - ax) \\ 1047 | &= \lim_{x \to \infty} \qty(\frac{x^2}{x + 2} - x) \\ 1048 | &= \lim_{x \to \infty} \frac{x^2 - x^2 - 2x}{x + 2} = -2 1049 | \end{align*} 1050 | 因此所求斜渐近线方程为$y = x - 2.$ 1051 | \end{solu} 1052 | \end{mathques} 1053 | 1054 | \begin{mathques}(2){2020-10-28}{一元函数几何应用}[不知道曲率公式] 1055 | \begin{ques} 1056 | 曲线$4x^2 + y^2 = 4$在点$(0, 2)$处的曲率为\mathblank. 1057 | \end{ques} 1058 | \begin{solu} 1059 | \answer{2} 1060 | 1061 | $4x^2 + y^2 = 4$两边对$x$求导可得$8x + 2yy' = 0$，则有$y' = -\frac{4x}{y}$, 1062 | \[ 1063 | y'' = -4\cdot \frac{y - xy'}{y^2} = \frac{-4\qty(y + \frac{4x^2}{y})}{y^2} 1064 | = -\frac{16}{y^3}, 1065 | \] 1066 | 点$(0, 2)$在曲线上，则$\eval{y''}_{x = 0} = -2, \eval{y'}_{x = 0} = 0$ 1067 | 1068 | 由曲率公式$k = \frac{\abs{y''}}{\qty(1 + (y')^2)^{\frac{3}{2}}}$，可得$k = 1069 | 2.$ 1070 | \end{solu} 1071 | \end{mathques} 1072 | 1073 | \begin{mathques}{2020-09-28}{一元函数几何应用}[极限不会求] 1074 | \begin{ques} 1075 | 曲线$y = \frac{2x^2 - 3}{5x^2 + 2\sin x}$的水平渐近线为\mathblank. 1076 | \end{ques} 1077 | \begin{solu} 1078 | \answer{$y = \frac{2}{5}$} 1079 | 1080 | \[ 1081 | \lim_{x \to \infty} y = \lim_{x \to \infty} \frac{2 - \frac{3}{x^2}}{5 + 1082 | \frac{2}{x^2} \sin x} = \frac{2}{5} 1083 | \] 1084 | 因此$y = \frac{2}{5}$为曲线的水平渐近线. 1085 | \end{solu} 1086 | \end{mathques} 1087 | 1088 | \begin{mathques}{2020-09-29}{一元微分应用}[无思路] 1089 | \begin{ques} 1090 | 证明恒等式：$\arcsin x + \arccos x = \frac{\pi}{2} (-1 \le x \le 1)$. 1091 | \end{ques} 1092 | \begin{solu} 1093 | \matherror*{构造函数$f(x) = \arcsin x + \arccos x, \abs{x} \le 1$}，则其导数为 1094 | \[ 1095 | f'(x) = \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - x^2}} = 0, 1096 | \] 1097 | 从而$f(x)$在其定义域上恒为常数，即$f(x) \equiv C$，$C$是常数. 1098 | 1099 | 令$x = 0$得 1100 | \[ 1101 | C = f(x) = \arcsin 0 + \arccos 0 = \frac{\pi}{2}, 1102 | \] 1103 | 即$\arcsin x + \arccos x = \frac{\pi}{2} (-1 \le x \le 1).$ 1104 | \end{solu} 1105 | \end{mathques} 1106 | 1107 | \begin{mathques}{2020-09-29}{一元微分应用}[无思路] 1108 | \begin{ques} 1109 | 若$\frac{a_n}{n + 1} + \frac{a_{n - 1}}{n} + \dotsb + a_0 = 0$，证明方程 1110 | $a_n x^n + a_{n-1} x^{n-1} + \dotsb + a_0 = 0$在$(0, 1)$内至少有一个实根. 1111 | \end{ques} 1112 | \begin{solu} 1113 | \mathidea{零点存在问题，用罗尔定理} 1114 | 1115 | 记 1116 | \[ 1117 | f(x) = \frac{a_n x^{n + 1}}{n + 1} + \frac{a_{n-1} x^n}{n} + \dotsb + a_0 x 1118 | \quad\mathhint[r]{\text{构造函数}} 1119 | \] 1120 | 则 1121 | \[ 1122 | f'(x) = a_n x^n + a_{n-1} x^{n-1} + \dotsb + a_0 = 0.\quad\mathhint[r] 1123 | {\text{欲求函数}} 1124 | \] 1125 | $f(x)$在$[0, 1]$上连续，在$(0, 1)$内可导，又$f(0) = 0, f(1) = \frac{a_n}{n + 1126 | 1} + \frac{a_{n - 1}}{n} + \dotsb + a_0 = 0 $，由罗尔定理知，方程$f'(x) = 0$ 1127 | 即$a_n x^n + a_{n-1} x^{n-1} + \dotsb + a_0 = 0$在$(0, 1)$内至少有一个实根. 1128 | \end{solu} 1129 | \end{mathques} 1130 | 1131 | \begin{mathques}(2){2020-10-03}{一元微分应用}[无法证明零点存在] 1132 | \begin{ques} 1133 | 设在$[1, +\infty)$上，$f''(x) < 0, f(1) = 2, f'(1) = -3.$ 证明：$f(x) = 0$在 1134 | $(1, +\infty)$内只有一个实根. 1135 | \end{ques} 1136 | \begin{solu} 1137 | \mathidea{零点唯一性可通过证明零点存在和函数单调} 1138 | 1139 | 在$[1, +\infty)$上，$f''(x) < 0, f'(x)$单调减少，$f'(x) < f'(1) = -3 < 0$， 1140 | 所以$f(x)$在$[1, +\infty)$上单调减少. 1141 | 1142 | 由拉格朗日中值定理当$x > 1$时，存在$\xi \in (1, \xi)$使得 1143 | \[ 1144 | f'(\xi) = \frac{f(x) - f(1)}{x - 1} 1145 | \] 1146 | 即 1147 | \begin{align*} 1148 | f(x) &= f(1) + f'(\xi)(x - 1) \\ 1149 | &< 2 + f'(1)(x - 1) \\ 1150 | & = 5 - 3x 1151 | \end{align*} 1152 | 取$x = \frac{5}{3}$，则$f\qty(\frac{5}{3}) < 0$，又$f(1) = 2 > 0$，由零点定理得$f(x) = 0$ 1153 | 在$(1, \frac{5}{3})$内至少有一个实根. 1154 | 1155 | 综上所述，$f(x) = 0$在$(1, +\infty)$内有且只有一个实根. 1156 | \end{solu} 1157 | \end{mathques} 1158 | 1159 | \begin{mathques}{2020-10-02}{一元微分物理应用}[变化速度是对$t$的导数] 1160 | \begin{ques} 1161 | 质点$P$沿抛物线$x = y^2(y > 0)$移动，$P$的横坐标$x$的变化速度为5\si{cm/s}. 1162 | 当$x = 9$时，点$P$到原点$O$的距离变化速度为\mathblank. 1163 | \end{ques} 1164 | \begin{solu} 1165 | \answer{$\frac{95}{6\sqrt{10}}$\si{cm/s}} 1166 | 1167 | \mathidea{已知$x = y^2(y > 0), \dv{x}{t} = 5$，求$\eval{\dv{s}{t}}_{x = 9}$.} 1168 | 1169 | $P$到原点$O$的距离$s = \sqrt{x ^2 + y^2}$，则 1170 | \[ 1171 | \dv{s}{t} = \dv{t}\sqrt{x^2 + y^2} = \dv{x}\sqrt{x^2 + y^2} \cdot 1172 | \matherror{\dv{x}{t}} = \frac{5(2x + 1)}{2\sqrt{x^2 + x}}. 1173 | \] 1174 | 当$x = 9$时，$\eval{\dv{x}{t}}_{x = 9} = \frac{95}{6\sqrt{10}}$ (\si{cm/s}). 1175 | \end{solu} 1176 | \end{mathques} 1177 | 1178 | \begin{mathques}(2){2020-11-01}{一元微分物理应用}[计算错误] 1179 | \begin{ques} 1180 | 球的半径以5\si{cm/s}的速度匀速增长，问球的半径为50\si{cm}时，求的表面积和体 1181 | 积的增长速度各是多少？ 1182 | \end{ques} 1183 | \begin{solu} 1184 | 表面积为$S = 4\pi r^2$，则 1185 | \[ 1186 | \dv{S}{t} = 8\pi r\dv{r}{t} = 8\pi \cdot 50 \cdot 5 = 2000\pi (\si{cm^2 / 1187 | s}). 1188 | \] 1189 | 体积为$V = \frac{4}{3}\pi r^3$，则 1190 | \[ 1191 | \dv{V}{t} = 4\pi r^2\dv{r}{t} = 4\pi \cdot (50)^2 \cdot 5 = 50000\pi (cm^3 1192 | / s). 1193 | \] 1194 | \end{solu} 1195 | \end{mathques} 1196 | \begin{mathques}{2020-10-09}{一元积分计算}[没想到可以直接凑微分计算代入计算判 1197 | 断是否收敛] 1198 | \begin{ques} 1199 | 下列反常积分中收敛的是(\quad). 1200 | \begin{multichoice}[4] 1201 | \task $\int_{e}^{+\infty} \frac{\ln x}{x} \dd x$ 1202 | \task $\int_{e}^{+\infty} \frac{\dd x}{x \ln x}$ 1203 | \task \answer{$\int_{e}^{+\infty} \frac{\dd x}{x (\ln x)^2}$} 1204 | \task $\int_{e}^{+\infty} \frac{\dd x}{x(\ln x)^{\frac{1}{2}}}$ 1205 | \end{multichoice} 1206 | \end{ques} 1207 | \begin{solu} 1208 | \begin{multichoice}[1] 1209 | \task $\int_{e}^{+\infty} \frac{\ln x}{x} = \eval{\frac{1}{2} (\ln x)^2}_{e} 1210 | ^{+\infty}$，发散 1211 | \task $\int_{e}^{+\infty} \frac{\dd x}{x \ln x} = \int_{e}^{+\infty} 1212 | \frac{\dd (\ln x)}{\ln x} = \eval{\ln(\ln x)}_{e}^{+\infty}$，发散 1213 | \task $\int_{e}^{+\infty} \frac{\dd x}{x (\ln x)^2} = \int_{e}^{+\infty} 1214 | \frac{\dd (\ln x)}{(\ln x)^2} = \eval{-\frac{1}{\ln x}}_{e}^{+\infty} = - (0 1215 | - 1) = 1$，收敛 1216 | \task $\int_{e}^{+\infty} \frac{\dd x}{x(\ln x)^{\frac{1}{2}}} = \int_{e} 1217 | ^{+\infty} (\ln x)^{-\frac{1}{2}} \dd (\ln x) = \eval{2(\ln x)^{\frac{1}{2}}} 1218 | _{e}^{+\infty}$，发散 1219 | \end{multichoice} 1220 | \end{solu} 1221 | \end{mathques} 1222 | 1223 | \begin{mathques}(2){2020-10-13}{一元积分计算}[对于反三角函数分部积分不熟悉] 1224 | \begin{ques} 1225 | $\int \frac{\arcsin \sqrt{x}}{\sqrt{x}}\dd x =$\mathblank. 1226 | \end{ques} 1227 | \begin{solu} 1228 | \mathidea{去掉根号会使计算变得简单} 1229 | 1230 | 令$\sqrt{x} = t$，则$x = t^2$，则 1231 | \begin{align*} 1232 | \int \frac{\arcsin \sqrt{x}}{\sqrt{x}}\dd x &= 2 \int t \cdot 1233 | \frac{\arcsin t}{t} \dd t\\ 1234 | &= 2 \int \arcsin t \dd t\\ 1235 | &= 2\matherror{(t\arcsin t + \sqrt{1 - t^2})} + C\\ 1236 | &= 2\sqrt{x}\arcsin \sqrt{x} + 2\sqrt{1 - x} + C 1237 | \end{align*} 1238 | \end{solu} 1239 | \end{mathques} 1240 | 1241 | \begin{mathques}(2){2020-10-13}{一元积分计算}[漏了$C$] 1242 | \begin{ques} 1243 | $\int e^{\sqrt{2x + 1}} \dd x = $\mathblank. 1244 | \end{ques} 1245 | \begin{solu} 1246 | \answer{$(\sqrt{2x + 1} - 1)e^{\sqrt{2x + 1}} + C$} 1247 | \mathmethod 令$\sqrt{2x + 1} = t$，则 1248 | \begin{align*} 1249 | \int e^{\sqrt{2x + 1}} &= \int te^t \dd t = \int t \dd (e^t)\\ 1250 | \mathidea{\small 分部积分法} &= te^t - e^t + C\\ 1251 | &= (\sqrt{2x + 1} - 1)e^{\sqrt{2x + 1}} + C 1252 | \end{align*} 1253 | \mathmethod 凑微分法 1254 | \begin{align*} 1255 | \int e^{\sqrt{2x + 1}} &= \int \sqrt{2x + 1}e^{\sqrt{2x + 1}} \dd 1256 | (\sqrt{2x + 1})\\ 1257 | &= \int \sqrt{2x + 1} \dd (e ^{\sqrt{2x + 1}})\\ 1258 | \mathidea{\small 分部积分} &= \sqrt{2x + 1} e^{\sqrt{2x + 1}} - \int 1259 | e^{\sqrt{2x + 1}} \dd (\sqrt{2x + 1})\\ 1260 | &= \sqrt{2x + 1}e^{\sqrt{2x + 1}} - e^{\sqrt{2x + 1}} + C\\ 1261 | &= (\sqrt{2x + 1} - 1)e^{\sqrt{2x + 1}} + C 1262 | \end{align*} 1263 | \end{solu} 1264 | \end{mathques} 1265 | 1266 | \begin{mathques}(5){2020-10-24}{一元积分计算}[无思路，换元法和分部积分不熟悉] 1267 | \begin{ques} 1268 | 设$x > 1$，则$\int \frac{xe^x}{\sqrt{e^x - 2}}\dd x = $\mathblank. 1269 | \end{ques} 1270 | \begin{solu} 1271 | \answer{$2(x - 2)\sqrt{e^x - 2} + 4\sqrt{2}\arctan \sqrt{\frac{e^x}{2} - 1} 1272 | + C$} 1273 | 1274 | 令$\sqrt{e^2 - 2} = t$ 1275 | 1276 | \begin{minipage}{\linewidth} 1277 | \begin{align*} 1278 | \int \frac{xe^x}{\sqrt{e^x - 2}}\dd x &= 2\int \ln(t^2 + 2) \dd t\\ 1279 | \mathidea{分部积分} &= \matherror{2t}\cdot\ln(t^2 + 2) - 1280 | \mathmark[blue!30!cyan](t2-a){ 1281 | 4\int \frac{t^2}{t^2 + 2} \dd t 1282 | }\\ 1283 | &= 2t\ln(t^2 + 2) - 4t + 4\sqrt{2}\arctan \frac{t}{\sqrt{2}} + C\\ 1284 | &= 2(x - 2)\sqrt{e^x - 2} + 4\sqrt{2}\arctan \sqrt{\frac{e^x}{2} - 1} + C 1285 | \end{align*} 1286 | \[ 1287 | \mathmark[blue!30!cyan][ 1288 | \path[draw=blue!30!cyan, arrows={-Implies[]}, thick, opacity=0.5, double distance=2mm] 1289 | (t2-a.east) to[bend left=90, distance=3cm] (frame.east); 1290 | ]{ 1291 | \begin{aligned} 1292 | \int \frac{t^2}{t^2 + 2} \dd t &= \int \qty(1 - \frac{1}{\qty(\frac{t} 1293 | {\sqrt{2}})^2 + 1})\dd t\\ 1294 | &= t - \sqrt{2}\int \frac{\dd (\frac{t}{\sqrt{2}})}{\qty(\frac{t} 1295 | {\sqrt{2}})^2 + 1}\\ 1296 | &= t - \sqrt{2}\arctan \frac{t}{\sqrt{2}} + C 1297 | \end{aligned} 1298 | } 1299 | \] 1300 | \end{minipage} 1301 | \end{solu} 1302 | \end{mathques} 1303 | 1304 | \begin{mathques}(3){2020-10-15}{一元积分计算}[无思路，三角函数积分变换不熟悉] 1305 | \begin{ques} 1306 | $\int \frac{\dd x}{1 + \sin x + \cos x} =$\mathblank. 1307 | \end{ques} 1308 | \begin{solu} 1309 | \answer{$\ln \abs{1 + \tan \frac{x}{2}} + C$} 1310 | \begin{align*} 1311 | \int \frac{\dd x}{1 + \sin x + \cos x} &= \int \frac{\dd x}{2\sin \frac{x} 1312 | {2}\cos \frac{x}{2} + 2\cos[2](\frac{x}{2})}\\ 1313 | &= \frac{1}{2}\int \frac{\dd x}{\cos[2](\frac{x}{2})\cdot \qty(1 + \tan 1314 | \frac{x}{2})}\\ 1315 | &= \int \frac{\dd (1 + \tan \frac{x}{2})}{1 + \tan \frac{x}{2}}\\ 1316 | &= \ln \abs{1 + \tan \frac{x}{2}} + C 1317 | \end{align*} 1318 | \end{solu} 1319 | \end{mathques} 1320 | 1321 | \begin{mathques}{2020-10-11}{一元积分计算}[三角函数变换不熟悉] 1322 | \begin{ques} 1323 | 求不定积分$\int \frac{x\cos[4](\frac{x}{2})}{\sin[3](x)}\dd x$ 1324 | \end{ques} 1325 | \begin{solu} 1326 | \begin{mathideabox} 1327 | 当被积函数为幂函数与三角函数的乘积，且不能用凑微分法积分时，一定要用分部积分 1328 | 法积分。\par 1329 | 因为分子与分母中三角函数的角度分别为$\frac{x}{2}$和$x$，所以首先要用三角恒等 1330 | 式变成同角度的三角函数形式。 1331 | \end{mathideabox} 1332 | \mathmethod 1333 | \begin{align*} 1334 | \int \frac{x\cos[4](\frac{x}{2})}{\sin[3](x)}\dd x &= \int \frac{x\cos[4] 1335 | (\frac{x}{2})}{8\sin[3](\frac{x}{2})\cos[3](\frac{x}{2})}\dd x \\ 1336 | &= \int \frac{x\cos \frac{x}{2}}{8\sin[3](\frac{x}{2})}\dd x\\ 1337 | &= \matherror{\frac{1}{4}\int x \sin[-3](\frac{x}{2}) \dd(\sin \frac{x}{2})} 1338 | \\ 1339 | &= \matherror{-\frac{1}{8}\int x\dd(\sin[-2](\frac{x}{2}))}\\ 1340 | &= \frac{-x}{8\sin[2](\frac{x}{2})} + \matherror{\frac{1}{8}\int \frac{\dd 1341 | x}{\sin[2](\frac{x}{2})}}\\ 1342 | &= \frac{-x}{8\sin[2](\frac{x}{2})} - \matherror{\frac{1}{4}\cot \frac{x} 1343 | {2}} + C 1344 | \end{align*} 1345 | \mathmethod 1346 | \begin{align*} 1347 | \int \frac{x\cos[4](\frac{x}{2})}{\sin[3](x)}\dd x &= \int \frac{x\cos 1348 | \frac{x}{2}}{8\sin[3](\frac{x}{2})}\dd x\\ 1349 | &= -\frac{1}{4}\int x \cot \frac{x}{2} \dd(\cot \frac{x}{2})\\ 1350 | &= -\frac{1}{8}\int x\dd(\cot[2](\frac{x}{2}))\\ 1351 | &= -\frac{1}{8}x\cot[2](\frac{x}{2}) + \frac{1}{8}\int \cot[2](\frac{x}{2}) 1352 | \dd x \\ 1353 | &= -\frac{1}{8}x\cot[2](\frac{x}{2}) + \frac{1}{8} \int \qty(\csc[2] 1354 | (\frac{x}{2}) - 1)\dd x\\ 1355 | &= \frac{-x}{8\sin[2](\frac{x}{2})} - \frac{1}{4}\cot \frac{x}{2} + C 1356 | \end{align*} 1357 | \end{solu} 1358 | \end{mathques} 1359 | 1360 | \begin{mathques}(5){2020-11-04}{一元积分计算}[没有想到把整个式子直接进行换元] 1361 | \begin{ques} 1362 | 计算$\int \arcsin \sqrt{\frac{x}{a + x}}\dd x (a > 0\text{是常数})$. 1363 | \end{ques} 1364 | \begin{solu} 1365 | 令$\arcsin \sqrt{\frac{x}{a + x}} = t, x = \frac{a\sin[2](t)}{1 - \sin[2](t)} 1366 | = a\tan[2](t)$，则 1367 | \begin{align*} 1368 | \int \arcsin \sqrt{\frac{x}{a + x}}\dd x &= \int t\dd(a\tan[2](t))\\ 1369 | &= at\tan[2](t) - a\int \matherror{\tan[2](t)} \dd t\\ 1370 | &= at\tan[2](t) + a\int \matherror{\qty(1 - \sec[2](t))}\dd t\\ 1371 | &= at\tan[2](t) + at - a\tan t + C\\ 1372 | &= (a + x)\arcsin \sqrt{\frac{x}{a + x}} - \sqrt{ax} + C 1373 | \end{align*} 1374 | \end{solu} 1375 | \end{mathques} 1376 | 1377 | \begin{mathques}(2){2020-10-12}{一元积分计算}[不需要已知分部积分到底，只需要分到 1378 | 可以积分即可] 1379 | \begin{ques} 1380 | 求不定积分$\int \frac{\arctan e^x}{e^x} \dd x$ 1381 | \end{ques} 1382 | \begin{solu} 1383 | \mathidea{不满足凑微分法的条件，用分部积分} 1384 | 1385 | \mathmethod 1386 | \begin{align*} 1387 | \int \frac{\arctan e^x}{e^x} \dd x &= -\int \arctan e^x \dd(e^{-x})\\ 1388 | &= -e^{-x}\arctan e^x + \int \frac{e^{-x}\cdot e^x}{1 + e^{2x}} \dd x\\ 1389 | &= -e^{-x}\arctan e^x + \int \qty(1 - \frac{e^{2x}}{1 + e^{2x}}) \dd x\\ 1390 | &= -e^{-x}\arctan e^x + x - \frac{1}{2}\ln(1 + e^{2x}) + C 1391 | \end{align*} 1392 | \mathmethod 令$e^x = t$，则$x = \ln t, \dd x = \frac{1}{t} \dd t$ 1393 | \begin{align*} 1394 | \int \frac{\arctan e^x}{e^x} \dd x &= \int \frac{\arctan t}{t^2} \dd t\\ 1395 | &= - \int \arctan t\dd(\frac{1}{t})\\ 1396 | &= - \frac{1}{t}\arctan t + \matherror{\int \frac{\dd t}{t(1 + t^2)}}\\ 1397 | &= -\frac{1}{t}\arctan t + \matherror{\int \frac{\dd t}{t} - \int 1398 | \frac{t\dd t}{1 + t^2}}\\ 1399 | &= -\frac{1}{t}\arctan t + \ln t - \frac{1}{2}\ln(1 + t^2) + C\\ 1400 | &= -\frac{1}{e^x}\arctan e^x + x - \frac{1}{2} \ln(1 + e^{2x}) + C 1401 | \end{align*} 1402 | \end{solu} 1403 | \end{mathques} 1404 | 1405 | \begin{mathques}(6){2020-10-22}{一元积分计算}[没想到需要分类讨论，对三角函数积分 1406 | 不熟悉] 1407 | \begin{ques} 1408 | 计算不定积分$\int \frac{\tan x}{a^2\sin[2](x) + b^2\cos[2](x)}\dd x(a, 1409 | b\text{不同时为零})$. 1410 | \end{ques} 1411 | \begin{solu} 1412 | \begin{solulist} 1413 | \item 当$a = 0, b \neq 0$时，原式$= \frac{1}{b^2} \int \frac{\sin x} 1414 | {\cos[3](x)}\dd x = \frac{1}{2b^2\cos[2](x)} + C$； 1415 | \item 当$a \neq 0, b = 0$时， 1416 | \begin{align*} 1417 | \text{原式} &= \frac{1}{a^2}\int \frac{1}{\sin x\cos x}\dd x\\ 1418 | &= \frac{2}{a^2}\int \frac{1}{\sin 2x}\dd x\\ 1419 | &= \matherror{\frac{1}{a^2} \int \csc 2x \dd (2x)}\\ 1420 | &= \frac{1}{a^2}\ln\abs{\csc 2x - \cot 2x} + C 1421 | \end{align*} 1422 | \item 当$ab \neq 0$时， 1423 | \begin{align*} 1424 | \text{原式} &= \int \frac{\tan x}{\cos[2](x)\qty{a^2 \tan[2](x) + b^2}} 1425 | \dd x\\ 1426 | &= \matherror{\int \frac{\tan x}{a^2\tan[2](x) + b^2}\dd(\tan x)}\\ 1427 | &= \matherror{\frac{1}{2a^2}\int \frac{\dd(a^2 \tan[2](x) + b^2)}{a^2 1428 | \tan[2](x) + b^2}}\\ 1429 | &= \frac{1}{2a^2}\ln(a^2 \tan[2](x) + b^2) + C 1430 | \end{align*} 1431 | \end{solulist} 1432 | \end{solu} 1433 | \end{mathques} 1434 | 1435 | \begin{mathques}(2){2020-10-12}{一元积分计算}[夹逼准则和凑定积分可以一起使用] 1436 | \begin{ques} 1437 | 求极限$\lim_{n \to \infty} \sum_{i = 1}^n \frac{\sin \frac{i\pi}{n}}{n + 1438 | \frac{1}{i}}$. 1439 | \end{ques} 1440 | \begin{solu} 1441 | 当各项分母相同且均为$n$时， 1442 | \[ 1443 | \lim_{n \to \infty} \sum_{i = 1}^n \frac{\sin \frac{i\pi}{n}}{n + \frac{1} 1444 | {i}} = \lim_{n \to \infty} \frac{1}{n}\sum_{i = 1}^n \sin \frac{i}{n}\pi = 1445 | \int_{0}^{1} \sin \pi x\dd x. 1446 | \] 1447 | 先对$\sum_{i = 1}^n \frac{\sin \frac{i\pi}{n}}{n + \frac{1}{i}}$进行放缩，有 1448 | \[ 1449 | \sum_{i = 1}^n \frac{\sin \frac{i\pi}{n}}{n + 1} \le \sum_{i = 1}^n 1450 | \frac{\sin \frac{i\pi}{n}}{n + \frac{1}{i}} \le \sum_{i = 1}^n \frac{\sin 1451 | \frac{i\pi}{n}}{n}, 1452 | \] 1453 | 又 1454 | \begin{align*} 1455 | \lim_{n \to \infty} \sum_{i = 1}^n \frac{\sin \frac{i\pi}{n}}{n} &= 1456 | \lim_{n \to \infty} \matherror{\frac{1}{n}} \sum_{i = 1}^n \matherror{\sin 1457 | \frac{i} {n}\pi} = \int_{0}^{1}\sin \pi x\dd x = \frac{2}{\pi},\\ 1458 | \lim_{n \to \infty} \sum_{i = 1}^n \frac{\sin \frac{i\pi}{n}}{n + 1} &= 1459 | \lim_{n \to \infty} \matherror{\frac{n}{n + 1} \cdot \frac{1}{n}} \sum_{i 1460 | = 1}^n \sin \frac{i} {n}\pi = \int_{0}^{1}\sin \pi x\dd x = 1461 | \frac{2}{\pi}, 1462 | \end{align*} 1463 | 由\matherror*{夹逼准则}即得 1464 | \[ 1465 | \lim_{n \to \infty} \sum_{i = 1}^n \frac{\sin \frac{i\pi}{n}}{n + \frac{1} 1466 | {i}} = \frac{2}{\pi}. 1467 | \] 1468 | \end{solu} 1469 | \end{mathques} 1470 | 1471 | \begin{mathques}(3){2020-11-11}{一元积分计算}[没能求出正确的通项公式来凑积分] 1472 | \begin{ques} 1473 | $\lim_{n \to \infty} \qty(\frac{1}{n} \cdot \frac{2}{n} \cdot \dotsi \cdot 1474 | \frac{n}{n})^{\frac{1}{n}} =$\mathblank. 1475 | \end{ques} 1476 | \begin{solu} 1477 | \answer{$\frac{1}{e}$} 1478 | 1479 | 令$x_n = \lim_{n \to \infty} \qty(\frac{1}{n} \cdot \frac{2}{n} \cdot \dotsi 1480 | \cdot \frac{n}{n})^{\frac{1}{n}}$，则 1481 | \[ 1482 | \ln x_n = \frac{1}{n}\qty(\ln \frac{1}{n} + \ln\frac{2}{n} + \dotsb + \ln 1483 | \frac{n}{n}) = \sum_{k = 1}^n \frac{1}{n} \ln \frac{k}{n}, 1484 | \] 1485 | 于是 1486 | \begin{align*} 1487 | \lim_{n \to \infty} \ln x_n &= \lim_{n \to \infty} \sum_{k = 1}^n \frac{1} 1488 | {n} \ln \frac{k}{n}\\ 1489 | &= \int_{0}^{1} \ln x\dd x\\ 1490 | &= \eval{x \ln x}_0^1 - \int_{0}^{1}x \cdot \frac{1}{x} \dd x\\ 1491 | &= 0 - 1 = -1. 1492 | \end{align*} 1493 | 故$\lim_{n \to \infty} x_n = e^{-1} = \frac{1}{e}.$ 1494 | \end{solu} 1495 | \end{mathques} 1496 | 1497 | \begin{mathques}(4){2020-10-18}{一元积分计算}[没有想到正确的换元方法] 1498 | \begin{ques} 1499 | 计算定积分$\int_{\frac{1}{2}}^{\frac{3}{2}} \frac{(1 - x)\arcsin(1 - x)} 1500 | {\sqrt{2x - x^2}}\dd x.$ 1501 | \end{ques} 1502 | \begin{solu} 1503 | 令$1 - x = \sin t$，则 1504 | \begin{align*} 1505 | \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{(1 - x)\arcsin(1 - x)} {\sqrt{2x - 1506 | x^2}}\dd x &= \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{t\sin t}{\cos t} 1507 | \cos t\dd t\\ 1508 | &= \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} t\sin t\dd t\\ 1509 | &= -\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}t\dd(\cos t)\\ 1510 | &= \eval{-(t\cos t - \sin t)}_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\\ 1511 | &= 1 - \frac{\sqrt{3}\pi}{6}. 1512 | \end{align*} 1513 | \end{solu} 1514 | \end{mathques} 1515 | 1516 | \begin{mathques}(4){2020-11-18}{一元积分计算}[无思路] 1517 | \begin{ques} 1518 | 计算定积分$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{\frac{x}{2}} \frac{\cos 1519 | x - \sin x}{\sqrt{\cos x}}\dd x.$ 1520 | \end{ques} 1521 | \begin{solu} 1522 | \begin{mathideabox} 1523 | 当被积函数为指函数与三角函数的乘积时，不能使用凑微分法求出原函数，只能使用分 1524 | 部积分法 1525 | \end{mathideabox} 1526 | \begin{align*} 1527 | \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{\frac{x}{2}} \frac{\cos x - \sin 1528 | x}{\sqrt{\cos x}}\dd x &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{\frac{x} 1529 | {2}} \sqrt{\cos x}\dd x - \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{\frac{x} 1530 | {2}} \frac{\sin x}{\sqrt{\cos x}}\dd x\\ 1531 | &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{\frac{x}{2}}\sqrt{\cos x}\dd x 1532 | + 2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{\frac{x}{2}}\dd (\sqrt{\cos x}) 1533 | \\ 1534 | &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{\frac{x}{2}}\sqrt{\cos x}\dd x 1535 | + \eval{2e^{\frac{x}{2}}\sqrt{\cos x}}_{-\frac{\pi}{4}}^{\frac{\pi}{4}} - 1536 | \matherror{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}e^{\frac{x}{2}} \sqrt{\cos 1537 | x} \dd x}\\ 1538 | &= \sqrt[4]{8}\qty(e^{\frac{\pi}{8}} - e^{-\frac{\pi}{8}}). 1539 | \end{align*} 1540 | \end{solu} 1541 | \end{mathques} 1542 | 1543 | \begin{mathques}(3){2020-10-14}{一元积分计算}[没有想到可以通过平移区间来化简] 1544 | \begin{ques} 1545 | 计算定积分$I = \int_{0}^{\pi}\frac{x\sin x}{1 + \cos[2](x)}\dd x.$ 1546 | \end{ques} 1547 | \begin{solu} 1548 | \begin{mathideabox} 1549 | 被积函数中既有幂函数，又有三角函数，是一种较复杂的分式形式，很难直接求出其原 1550 | 函数，但是，注意到被积函数为偶函数，其中$x$及$\sin x$为奇函数，可考虑将其化 1551 | 为对称区间上的定积分，为此，需将积分区间$[0, \pi]$平移成$\qty[-\frac{\pi}{2}, 1552 | \frac{\pi}{2}]$ 1553 | \end{mathideabox} 1554 | 设$x = \frac{\pi}{2} + u$，则有 1555 | \begin{align*} 1556 | I &= \int_{0}^{\pi}\frac{x\sin x}{1 + \cos[2](x)}\dd x\\ 1557 | &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\qty(\frac{\pi}{2} + u) 1558 | \cos u}{1 + \sin[2](u)}\dd u\\ 1559 | &= \frac{\pi}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1560 | \mathmark[red][ 1561 | \node[red, font=\small] (a) at ([shift={(1.5, 1)}]frame.north east) {偶函数}; 1562 | \draw[->, red] (frame.north east) to[bend left] (a); 1563 | ]{ 1564 | \frac{\cos u}{1 + \sin[2](u)} 1565 | } 1566 | \dd u + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1567 | \mathmark[cyan][ 1568 | \node[cyan, font=\small] (b) at ([shift={(1, 1)}]frame.north east) {奇函数}; 1569 | \draw[->, cyan] (frame.north) to[bend left] (b); 1570 | ]{ 1571 | \frac{u\cos u}{1 + \sin[2](u)} 1572 | }\dd u\\ 1573 | &= \pi\int_{0}^{\frac{\pi}{2}} \frac{\dd (\sin u)}{1 + \sin[2](u)} + 0\\ 1574 | &= \eval{\pi \arctan(\sin u)}_{0}^{\frac{\pi}{2}}\\ 1575 | &= \pi \times \frac{\pi}{4} = \frac{1}{4} \pi^2. 1576 | \end{align*} 1577 | \end{solu} 1578 | \end{mathques} 1579 | 1580 | \begin{mathques}(4){2020-10-16}{一元积分计算}[计算错误，三角函数积分不熟悉] 1581 | \begin{ques} 1582 | 设$f(x) = 1583 | \begin{dcases} 1584 | \frac{1}{1 + \sin x}, & x \ge 0,\\ 1585 | \frac{1}{1 + e^x}, & x < 0, 1586 | \end{dcases} 1587 | $求$\int_{-1}^{\frac{\pi}{4}}f(x)\dd x.$ 1588 | \end{ques} 1589 | \begin{solu} 1590 | \[ 1591 | \int_{-1}^{\frac{\pi}{4}}f(x) \dd x = \int_{-1}^{0} f(x) \dd x + \int_{0} 1592 | ^{\frac{\pi}{4}} f(x) \dd x = \int_{-1}^{0} \frac{\dd x}{1 + e^x} + 1593 | \int_{0}^{\frac{\pi}{4}} \frac{\dd x}{1 + \sin x}, 1594 | \] 1595 | 令$e^x = t$，则 1596 | \begin{align*} 1597 | \int_{-1}^{0}\frac{\dd x}{1 + e^x} &= \int_{e^{-1}}^{1} \frac{1}{1 + t} 1598 | \cdot \frac{1}{t} \dd t\\ 1599 | &= \int_{e^{-1}}^{1} \qty(\frac{1}{t} - \frac{1}{1 + t}) \dd t\\ 1600 | &= \eval{\ln \frac{t}{1 + t}}_{e^{-1}}^1\\ 1601 | &= -\ln 2 + \ln (1 + e),\\ 1602 | \int_{0}^{\frac{\pi}{4}}\frac{\dd x}{1 + \sin x} &= \int_{0}^{\frac{\pi} 1603 | {4}} \frac{1 - \sin x}{\cos[2](x)} \dd x\\ 1604 | &= \int_{0}^{\frac{\pi}{4}} \sec[2](x)\dd x - \int_{0}^{\frac{\pi}{4}} 1605 | \frac{\sin x}{\cos[2](x)} \dd x\\ 1606 | &= \eval{\tan x}_{0}^{\frac{\pi}{4}} - \eval{\frac{1}{\cos x}}_{0} 1607 | ^{\frac{\pi}{4}} \\ 1608 | & = 1 - (\sqrt{2} - 1) = 2 - \sqrt{2}. 1609 | \end{align*} 1610 | 从而 1611 | \[ 1612 | \int_{-1}^{\frac{\pi}{4}} f(x)\dd x = - \ln 2 + \ln(1 + e) + 2 - \sqrt{2}. 1613 | \] 1614 | \end{solu} 1615 | \end{mathques} 1616 | 1617 | \begin{mathques}{2020-10-11}{一元积分计算}[当作复合函数求导，里面的也要求导] 1618 | \begin{ques} 1619 | 设下列不定积分存在，则下列命题正确的是(\quad). 1620 | \begin{multichoice} 1621 | \task \answer{$\int f'(2x)\dd x = \frac{1}{2}f(2x) + C$} 1622 | \task $\qty[\int f(2x) \dd x]' = 2f(2x)$ 1623 | \task $\int f'(2x)\dd x = f(2x) + C$ 1624 | \task $\qty[\int f(2x) \dd x]' = \frac{1}{2}f(2x)$ 1625 | \end{multichoice} 1626 | \end{ques} 1627 | \begin{solu} 1628 | 令$t = 2x$，则$\dd x = \frac{1}{2}\dd t$，则 1629 | \begin{align*} 1630 | \int f'(2x)\dd x &= \frac{1}{2}\int f'(t) \dd t\\ 1631 | &= \frac{1}{2}f(t) + C\\ 1632 | &= \frac{1}{2}f(2x) + C 1633 | \end{align*} 1634 | 故(A)选项正确 1635 | \begin{align*} 1636 | \qty[\int f(2x) \dd x]' &= \frac{\dd}{\matherror{\dd x}} \qty(\int f(2x) 1637 | \dd x)\\ 1638 | &= \dv{t}(\frac{1}{2}\int f(t)\dd t)\cdot \dv{t}{x}\\ 1639 | &= \frac{1}{2}f(t) \cdot 2\\ 1640 | &= f(2x) 1641 | \end{align*} 1642 | \end{solu} 1643 | \end{mathques} 1644 | 1645 | \begin{mathques}{2020-10-11}{一元积分计算}[只要函数恒小于，积分也恒小于，就算 1646 | 是负的也是小于] 1647 | \begin{ques} 1648 | 设函数$f(x)$与$g(x)$在$[0, 1]$上连续，且$f(x) \le g(x)$，则对任意$c \in (0, 1649 | 1)$，有(\quad). 1650 | \begin{multichoice} 1651 | \task $\int_{\frac{1}{2}}^{c} f(t) \dd t \ge \int_{\frac{1}{2}}^{c} g(t) 1652 | \dd t$ 1653 | \task $\int_{\frac{1}{2}}^{c} f(t) \dd t \le \int_{\frac{1}{2}}^{c} g(t) 1654 | \dd t$ 1655 | \task $\int_{c}^{1} f(t) \dd t \ge \int_{c}^{1} g(t) \dd t$ 1656 | \task \answer{$\int_{c}^{1} f(t) \dd t \le \int_{c}^{1} g(t) \dd t$} 1657 | \end{multichoice} 1658 | \end{ques} 1659 | \begin{solu} 1660 | \begin{mathideabox}[定积分性质] 1661 | 若连续函数$f(x), g(x)$在$[a, b]$上满足$f(x) \le g(x)$，则当$a < b$时， 1662 | $\int_{a}^{b}f(x) \dd x \le \int_{a}^{b}g(x)\dd x$ 1663 | \end{mathideabox} 1664 | 由于$c \in (0, 1)$，因此$c < 1$恒成立，而$c$可能大于$\frac{1}{2}$，也可能小 1665 | 于$\frac{1}{2}$，可知(A)，(B)不正确。 1666 | 1667 | 由于$f(x) \le g(x)$，可知应有$\int_{c}^{1}f(t) \dd t \le \int_{c}^{1} g(t) 1668 | \dd t$. 1669 | \end{solu} 1670 | \end{mathques} 1671 | 1672 | \begin{mathques}{2020-10-11}{一元积分计算}[$x$和$-x$图画错了orz] 1673 | \begin{ques} 1674 | 设$f(x) = 1675 | \begin{dcases} 1676 | 1, & x > 0,\\ 1677 | 0, & x = 0,\\ 1678 | -1, & x < 0, 1679 | \end{dcases} 1680 | F(x) = \int_{0}^{x}f(t)\dd t 1681 | $，则(\quad). 1682 | \begin{multichoice}[1] 1683 | \task $F(x)$在$x = 0$处不连续 1684 | \task \answer{$F(x)$在$(-\infty, +\infty)$内连续，但在$x = 0$处不可导} 1685 | \task $F(x)$在$(-\infty, +\infty)$内可导，且满足$F'(x) = f(x)$ 1686 | \task $F(x)$在$(-\infty, +\infty)$内可导，但不一定满足$F'(x) = f(x)$ 1687 | \end{multichoice} 1688 | \end{ques} 1689 | \begin{solu} 1690 | 当$f(x)$为连续函数时，$F(x) = \int_{0}^{x}f(t)\dd t$为$f(x)$的原函数，此时有 1691 | $\qty[\int_{0}^{x}f(t)\dd t]' = f(x)'$ 1692 | 1693 | 如果$f(x)$不为连续函数，上述结论不一定成立，因此不能轻易判定(C)成立。 1694 | 1695 | 由于$f(x)$为分段函数，因此$F(x) = \int_{0}^{x}f(t) \dd t$也为分段函数 1696 | 1697 | 当$x < 0$时，$F(x) = \int_{0}^{x}(-1)\dd t = -x$; 1698 | 1699 | 当$x > 0$时，$F(x) = \int_{0}^{x}1\dd t$ = x; 1700 | 1701 | 当$x = 0$时，$F(0) = 0$. 1702 | 1703 | 因此$\matherror{F(x) = \abs{x}}$，可知$F(x)$在$(-\infty, +\infty)$内连续，但在 1704 | $x = 0$处不可导。 1705 | \end{solu} 1706 | \end{mathques} 1707 | 1708 | \begin{mathques}(3){2020-11-16}{一元积分计算}[凑定积分的公式理解错误] 1709 | \begin{ques} 1710 | $\lim_{n \to \infty} \ln \sqrt[n]{\qty(1 + \frac{1}{n})^2 \qty(1 + \frac{2} 1711 | {n})^2 \dotsi \qty(1 + \frac{n}{n})^2 } =$\mathblank. 1712 | \end{ques} 1713 | \begin{solu} 1714 | \answer{$2(2\ln 2 - 1)$} 1715 | \begin{align*} 1716 | \text{原式} &= 2\lim_{n \to \infty} \frac{1}{n}\qty(\ln (1 + \frac{1}{n}) 1717 | + \ln(1 + \frac{2}{n}) + \dotsb + \ln(1 + \frac{n}{n}))\\ 1718 | &= 2\lim_{n \to \infty} \sum_{i = 1}^n \ln(1 + \frac{i}{n})\cdot \frac{1} 1719 | {n}\\ 1720 | &= \matherror{2\int_{0}^{1} \ln(1+x)\dd x}\\ 1721 | &= \eval{2x\ln x}_1^2 - 2\int_{1}^{2} x \cdot \frac{1}{x} \dd x\\ 1722 | &= 2(2\ln 2 - 1). 1723 | \end{align*} 1724 | \end{solu} 1725 | \end{mathques} 1726 | 1727 | \begin{mathques}{2020-10-11}{一元积分计算}[凑定积分的公式理解错误] 1728 | \begin{ques} 1729 | $\lim_{n \to \infty} \qty(\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 1730 | 2n}} + \frac{1}{\sqrt{n^2 + 3n}} + \dotsb + \frac{1}{\sqrt{n^2 + n^2}}) = 1731 | $\mathblank. 1732 | \end{ques} 1733 | \begin{solu} 1734 | \answer{$2\sqrt{2} - 2$} 1735 | \begin{mathideabox}[凑定积分] 1736 | 根据定积分的定义凑定积分：$\int_{0}^{1}f(x)dx = \lim_{n \to \infty} 1737 | \sum_{i = 1}^n f(\frac{i}{n}) \cdot \frac{1}{n}$ 1738 | 1739 | 能凑成$\frac{i}{n}$的用定积分定义；凑不成的，先用放缩法，常见的几种凑定积 1740 | 分定义的式子有 1741 | \begin{flalign*} 1742 | \lim_{n \to \infty} \qty(\frac{1}{n + 1} + \frac{1}{n + 2} + \dotsb 1743 | + \frac{1}{n + n}) &= \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{n 1744 | + i} &\\ 1745 | &= \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{1 + \frac{i}{n}} 1746 | \cdot \frac{1}{n}\\ 1747 | &= \int_{0}^{1} \frac{1}{1 + x}\dd x\\ 1748 | &= \eval{\ln(1 + x)}_0^1 = \ln2. 1749 | \end{flalign*} 1750 | \begin{flalign*} 1751 | \lim_{n \to \infty} \qty(\frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + 1752 | \dotsb + \frac{n}{n^2 + n^2}) &= \lim_{n \to \infty} \sum_{i = 1}^n 1753 | \frac{n}{n^2 + i^2} &\\ 1754 | &= \lim_{n \to \infty} \sum_{i = 1}^n \frac{n^2}{n^2 + i^2} \cdot 1755 | \frac{1}{n}\\ 1756 | &= \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{1 + \qty(\frac{i} 1757 | {n})^2} \cdot \frac{1}{n}\\ 1758 | &= \int_{0}^{1}\frac{1}{1 + x^2}\dd x\\ 1759 | &= \eval{\arctan x}_0^1 = \frac{\pi}{4}. 1760 | \end{flalign*} 1761 | \end{mathideabox} 1762 | \begin{align*} 1763 | \text{原式} &= \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{\sqrt{n^2 + ni}} 1764 | \\ 1765 | &= \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{\sqrt{1 + \frac{i}{n}}} 1766 | \cdot \frac{1}{n}\\ 1767 | &= \int_{0}^{1} \frac{1}{\sqrt{1 + x}}\dd x \\ 1768 | &= \eval{2\sqrt{1 + x}}_0^1 = 2\sqrt{2} - 2. 1769 | \end{align*} 1770 | \end{solu} 1771 | \end{mathques} 1772 | 1773 | \begin{mathques}{2020-10-14}{一元积分计算}[积分的方法不一样] 1774 | \begin{ques} 1775 | 不定积分$\int \frac{1}{\sqrt{x(4-x)}}\dd x =$\mathblank. 1776 | \end{ques} 1777 | \begin{solu} 1778 | \answer{$\arcsin \frac{x - 2}{x} + C$} 1779 | 1780 | 由于$\sqrt{x(4 - x)} = \sqrt{4 - (x - 2)^2} = 2\sqrt{1 - \qty(\frac{x - 2} 1781 | {2})^2}$，所以 1782 | \begin{align*} 1783 | \int \frac{1}{\sqrt{x(4-x)}}\dd x &= \int \frac{\dd x}{2\sqrt{1 - 1784 | \qty(\frac{x - 2}{2})^2}} \\ 1785 | &= \int \frac{\dd(\frac{x - 2}{2})}{\sqrt{1 - \qty(\frac{x - 2}{2})^2}}\\ 1786 | &= \arcsin \frac{x - 2}{2} + C. 1787 | \end{align*} 1788 | \end{solu} 1789 | \end{mathques} 1790 | 1791 | \begin{mathques}(2){2020-10-28}{一元积分计算}[没有想到可以把1变形] 1792 | \begin{ques} 1793 | 不定积分$\int \frac{1}{\sin[2](x)\cos[4](x)}\dd x =$\mathblank. 1794 | \end{ques} 1795 | \begin{solu} 1796 | \answer{$\frac{1}{3}\tan[3](x) + 2\tan x - \cot x + C$} 1797 | \begin{align*} 1798 | \int \frac{1}{\sin[2](x)\cos[4](x)}\dd x &= \int \frac{\matherror{\qty(\sin[2] 1799 | (x) + \cos[2](x))^2}}{\sin[2](x)\cos[4](x)} \dd x\\ 1800 | &= \int \frac{\sin[2](x)}{\cos[4](x)}\dd x + \int \frac{2}{\cos[2](x)}\dd 1801 | x + \int \frac{1}{\sin[2](x)}\dd x\\ 1802 | &= \int \tan[2](x) \dd (\tan x) + 2\tan x - \cot x\\ 1803 | &= \frac{1}{3}\tan[3](x) + 2\tan x - \cot x + C. 1804 | \end{align*} 1805 | \end{solu} 1806 | \end{mathques} 1807 | 1808 | \begin{mathques}(3){2020-10-18}{一元积分计算}[抄错了orz] 1809 | \begin{ques} 1810 | 不定积分$\int \frac{\sin x - 3\cos x}{\sin[3](x)}\dd x =$\mathblank. 1811 | \end{ques} 1812 | \begin{solu} 1813 | \answer{$-\cot x + \frac{3}{2\sin[2](x)} + C$} 1814 | 先将分式变形得 1815 | \begin{align*} 1816 | \int \frac{\sin x - 3\cos x}{\sin[3](x)}\dd x &= \int \frac{1}{\sin[2](x)} 1817 | \dd x - 3\int \frac{\dd(\sin x)}{\sin[3](x)}\\ 1818 | &= -\cot x - 3\cdot \frac{1}{-3 + 1}\qty(\sin x)^{-3 + 1} + C\\ 1819 | &= -\cot x + \frac{3}{2\sin[2](x)} + C 1820 | \end{align*} 1821 | \end{solu} 1822 | \end{mathques} 1823 | 1824 | \begin{mathques}(3){2020-10-21}{一元积分计算}[思路错误] 1825 | \begin{ques} 1826 | 不定积分$\int \frac{x^3}{\sqrt{1 - x^2}}\dd x =$\mathblank. 1827 | \end{ques} 1828 | \begin{solu} 1829 | \answer{$\frac{1}{3}(1 - x^2)^{\frac{3}{2}} - (1 - x^2)^{\frac{1}{2}} + C$} 1830 | 1831 | \mathmethod 令$t = 1 - x^2$，则$\dd t = -2x\dd x$ 1832 | \begin{align*} 1833 | \int \frac{x^3}{\sqrt{1 - x^2}}\dd x &= -\frac{1}{2}\int \frac{1 - t} 1834 | {\sqrt{t}} \dd t\\ 1835 | &= -\frac{1}{2}\int \qty(\frac{1}{\sqrt{t}} - \sqrt{t}) \dd t\\ 1836 | &= \frac{1}{3}t^{\frac{3}{2}} - t^{\frac{1}{2}} + C\\ 1837 | &= \frac{1}{3}(1 - x^2)^{\frac{3}{2}} - (1 - x^2)^{\frac{1}{2}} + C 1838 | \end{align*} 1839 | 1840 | \mathmethod 令$x = \sin t$，则 1841 | \begin{align*} 1842 | \int \frac{x^3}{\sqrt{1 - x^2}}\dd x &= \int \sin[3](t)\dd t\\ 1843 | &= \matherror{\int \sin[2](t)\cdot \sin t\dd t}\\ 1844 | &= \matherror{\int \qty(\cos[2](t) - 1) \dd (\cos t)}\\ 1845 | &= \frac{1}{3} \cos[3](t) - \cos t + C. 1846 | \end{align*} 1847 | 利用$\cos t = \sqrt{1 - x^2}$得 1848 | \[ 1849 | \text{原式}= \frac{1}{3}(1 - x^2)^{\frac{3}{2}} - (1 - x^2)^{\frac{1}{2}} + C 1850 | \] 1851 | \end{solu} 1852 | \end{mathques} 1853 | 1854 | \begin{mathques}{2020-10-15}{一元积分计算}[没有想到洛必达，以及分母也需要求导] 1855 | \begin{ques} 1856 | $\lim_{x \to 0} \frac{\int_{0}^{x} \sin[2](t) \dd t}{x^3} =$\mathblank. 1857 | \end{ques} 1858 | \begin{solu} 1859 | \answer{$\frac{1}{3}$} 1860 | 1861 | 此极限属于“$\frac{0}{0}$”型，用洛必达法则，则 1862 | \[ 1863 | \lim_{x \to 0} \frac{\int_{0}^{x} \sin[2](t) \dd t}{x^3} = \lim_{x \to 0} 1864 | \frac{\sin[2](x)}{3x^2} = \frac{1}{3}. 1865 | \] 1866 | \end{solu} 1867 | \end{mathques} 1868 | 1869 | \begin{mathques}(2){2020-10-17}{一元积分计算}[中途积分计算错误，简单的可以直接求 1870 | 出没必要和复杂的一起变$\dd x$乘来乘去；倍角要化解成已知角进行还原] 1871 | \begin{ques} 1872 | 求不定积分$\int x^2\sqrt{1 - x^2}\dd x$. 1873 | \end{ques} 1874 | \begin{solu} 1875 | 令$x = \sin t$，则 1876 | \begin{align*} 1877 | \int x^2\sqrt{1 - x^2}\dd x &= \int \sin[2](t)\cdot \cos t\dd(\sin t)\\ 1878 | &= \int \sin[2](t) \cdot \cos[2](t) \dd t\\ 1879 | &= \frac{1}{4}\int \sin[2](2t)\dd t\\ 1880 | &= \frac{1}{4}\int \frac{1}{2}(1 - \cos 4t)\dd t\\ 1881 | &= \frac{1}{8}t - \frac{1}{32}\sin 4t + C\\ 1882 | &= \frac{1}{8}\qty(\arcsin x - x\sqrt{1 - x^2}(1 - 2x^2)) + C. 1883 | \end{align*} 1884 | \end{solu} 1885 | \end{mathques} 1886 | 1887 | \begin{mathques}{2020-10-15}{一元积分计算}[没有对根号进行换元；反三角函数基本 1888 | 都需要进行分部积分后求出] 1889 | \begin{ques} 1890 | 求不定积分$\int \frac{\arcsin \sqrt{x}}{\sqrt{x}}\dd x$. 1891 | \end{ques} 1892 | \begin{solu} 1893 | 令$t = \sqrt{x}$，则$x = t^2$，即有 1894 | \begin{align*} 1895 | \int \frac{\arcsin \sqrt{x}}{\sqrt{x}}\dd x &= 2\int \arcsin t\dd t\\ 1896 | &= 2\qty(t\cdot \arcsin t - \int \frac{t}{\sqrt{1 - t^2}}\dd t)\\ 1897 | &= 2\qty(t\cdot \arcsin t + \sqrt{1 - t^2}) + C\\ 1898 | &= 2\sqrt{x}\arcsin\sqrt{x} + 2\sqrt{1 - x} + C. 1899 | \end{align*} 1900 | \end{solu} 1901 | \end{mathques} 1902 | 1903 | \begin{mathques}{2020-10-15}{一元积分计算}[计算过程中符号出错；像答案直接代入 1904 | 计算出错的可能性小些] 1905 | \begin{ques} 1906 | 设$xe^{-x}$为$f(x)$的一个原函数，求$\int xf(x)\dd x$. 1907 | \end{ques} 1908 | \begin{solu} 1909 | 因为$xe^{-x}$为$f(x)$的一个原函数，故 1910 | \[ 1911 | f(x) = (xe^{-x})' = e^{-x}(1 - x). 1912 | \] 1913 | 所以 1914 | \begin{align*} 1915 | \int xf(x)\dd x &= \int e^{-x}(x - x^2) \dd x\\ 1916 | \mathidea{分部积分}&= -e^{-x}(x - x^2) + \int e^{-x}(1 - 2x)\dd x\\ 1917 | \mathidea{再次分部积分}&= e^{-x}(x^2 - x) -e^{-x}(1 - 2x) - \int 2e^{-x} 1918 | \dd x\\ 1919 | &= e^{-x}(x^2 + x + 1) + C. 1920 | \end{align*} 1921 | \end{solu} 1922 | \end{mathques} 1923 | 1924 | \begin{mathques}(3){2020-11-02}{一元积分计算}[凑反三角函数的公式不熟悉] 1925 | \begin{ques} 1926 | 求不定积分$\int \frac{\dd x}{(x^2 + 1)(x^2 + x + 1)}.$ 1927 | \end{ques} 1928 | \begin{solu} 1929 | \begin{mathideabox}[部分分式分解] 1930 | 设$f$和$g$是非零多项式，将$g$写作若干互不相同的不可约多项式的幂的乘积： 1931 | \[ 1932 | g = \prod_{i = 1}^k p_i^{n_i}. 1933 | \] 1934 | 则存在多项式$b$（不唯一）和多项式$a_{ij}$，且$\deg a_{ij} < \deg p_i$使得 1935 | \[ 1936 | \frac{f}{g} = b + \sum_{i = 1}^k\sum_{j = 1}^{n_i} \frac{a_{ij}}{p_i^j}. 1937 | \] 1938 | 如果$\deg f < \deg g$，则$b = 0$. 1939 | \tcblower 1940 | 比如可设$\frac{3x + 5}{(1 - 2x)^2} = \frac{A}{(1 - 2x)^2} + \frac{B}{1 - 2x} 1941 | .$ 1942 | 则$3x + 5 = A + B(1 - 2x)$，解方程 1943 | \[ 1944 | \begin{dcases} 1945 | 5 = A + B\\ 1946 | 3x = -2Bx 1947 | \end{dcases} 1948 | \] 1949 | 得$A = \frac{13}{2}, B = -\frac{3}{2}$ 1950 | \end{mathideabox} 1951 | 由部分分式分解设 1952 | \[ 1953 | \frac{1}{(x^2 + 1)(x^2 + x + 1)} = \frac{ax + b}{x^2 + 1} + \frac{cx + d} 1954 | {x^2 + x + 1} 1955 | \] 1956 | 消去分母得 1957 | \[ 1958 | (a + c)x^3 + (a + b + d)x^2 + (a + b + c)x + (b + d) = 1 1959 | \] 1960 | 解方程组 1961 | \[ 1962 | \begin{dcases} 1963 | a + c = 0,\\ 1964 | a + b + d = 0,\\ 1965 | a + b + c = 0,\\ 1966 | b + d = 1, 1967 | \end{dcases} 1968 | \] 1969 | 得$a = -1, b = 0, c=d = 1$ 1970 | 则 1971 | \begin{align*} 1972 | \int \frac{\dd x}{(x^2 + 1)(x^2 + x + 1)} &= \int \qty(\frac{-x}{x^2 + 1} 1973 | + \matherror{\frac{x + 1}{x^2 + x + 1}})\dd x\\ 1974 | &= -\frac{1}{2}\int \frac{\dd (x^2 + 1)}{x^2 + 1} + \matherror{\frac{1} 1975 | {2}\int \frac{2x + 1 + 1}{x^2 + x + 1}\dd x} \\ 1976 | &= -\frac{1}{2}\ln(x^2 + 1) + \frac{1}{2}\int \frac{\dd (x ^2 + x + 1)} 1977 | {x^2 + x + 1} + \frac{1}{2}\int \frac{\dd x}{x^2 + x + 1}\\ 1978 | &= -\frac{1}{2}\ln(x^2 + 1) + \frac{1}{2}\ln(x^2 + x + 1) + \frac{1}{2} 1979 | \int \frac{\dd(x + \frac{1}{2})}{\qty(x + \frac{1}{2})^2 + \frac{3}{4}}\\ 1980 | &= -\frac{1}{2}\ln \frac{x^2 + 1}{x^2 + x + 1} + 1981 | \matherror{\frac{\sqrt{3}}{3}\arctan \frac{2x + 1}{\sqrt{3}}} + C. 1982 | \end{align*} 1983 | \end{solu} 1984 | \end{mathques} 1985 | 1986 | \begin{mathques}{2020-10-15}{一元积分计算}[忘了有理函数的积分如何分解（见 1987 | 《30》p. 113）] 1988 | \begin{ques} 1989 | 求不定积分$\int \frac{1}{x^3 + 4x^2 + 5x + 2}\dd x.$ 1990 | \end{ques} 1991 | \begin{solu} 1992 | \begin{mathideabox}[试根法] 1993 | 对于复杂多项式，求其因式时，用试根法： 1994 | 1995 | 对于整数多项式方程 1996 | \[ 1997 | a_nx^n + a_{n - 1}x^{n - 1} + \dotsb + a_0 = 0 1998 | \] 1999 | 设其有理数解为$x = \frac{p}{q}$，则 2000 | $p$为$a_0$的因数，$q$为$a_n$的因数，且$p, q$互质（最大公约数为1） 2001 | \tcblower 2002 | 对于$x^3 + 4x^2 + 5x + 2$，若有有理数根，则必定在$\pm1, \pm2$中， 2003 | 易得$-1$为其一个有理跟，则 2004 | \[ 2005 | x^3 + 4x^2 + 5x + 2 = (x + 1)(x^2 + 3x + 2) = (x + 1)^2(x + 2) 2006 | \] 2007 | \end{mathideabox} 2008 | 设$\frac{1}{x^3 + 4x^2 + 5x + 2} = \frac{1}{(x + 1)^2 (x + 2)} = \frac{A}{x + 2009 | 2} + \frac{B}{x + 1} + \frac{C}{(x + 1)2}$ 2010 | 2011 | 消去分母得 2012 | \[ 2013 | (A + B)x^2 + (2A + 3B +C)x + (A + 2B + 2C) = 1, 2014 | \] 2015 | 解方程组 2016 | \[ 2017 | A + B = 0, 2018 | 2A + 3B + C = 0, 2019 | A + 2B + 2C = 1, 2020 | \] 2021 | 得$A = 1, B = -1, C = 1$，则 2022 | \begin{align*} 2023 | \int \frac{1}{x^3 + 4x^2 + 5x + 2}\dd x &= \int \frac{\dd x}{x + 2} - \int 2024 | \frac{\dd x}{x + 1} + \int \frac{\dd x}{(x + 1)^2}\\ 2025 | &= \ln\abs{x + 2} - \ln\abs{x + 1} - \frac{1}{x + 1} + C. 2026 | \end{align*} 2027 | \end{solu} 2028 | \end{mathques} 2029 | 2030 | \begin{mathques}{2020-10-18}{一元积分应用}[计算错误] 2031 | \begin{ques} 2032 | 设曲线$y = x^2 - 2x, y = 0, x = 1, x = 3$围成一平面图形$A$，求： 2033 | \begin{queslist} 2034 | \item $A$的面积 2035 | \item 该平面图形绕$y$轴旋转一周所得旋转体的体积$V$. 2036 | \end{queslist} 2037 | \end{ques} 2038 | \begin{solu} 2039 | \columnratio{0.65} 2040 | \begin{paracol}{2} 2041 | \begin{rightcolumn} 2042 | \centering 2043 | \begin{tikzpicture} 2044 | \begin{axis}[ 2045 | math, 2046 | width=\linewidth, 2047 | xtick=\empty, 2048 | ytick={-1, 3}, 2049 | xmax=4, 2050 | ymin=-1.2 2051 | ] 2052 | \path[name path=xaxis] (\pgfkeysvalueof{/pgfplots/xmin}, 0) -- 2053 | (\pgfkeysvalueof{/pgfplots/xmax},0); 2054 | \addplot[domain=-0.5:3.1, name path=A] {x^2 - 2*x}; 2055 | \addplot[gray, pattern=north east lines] fill between 2056 | [of=A and xaxis, soft clip={domain=1:3}]; 2057 | \node[above left] at (1, 0) {$1$}; 2058 | \node[above left] at (2, 0) {$2$}; 2059 | \node[above right] at (3, 0) {$3$}; 2060 | \node[below left] at (0, 0) {$O$}; 2061 | \node[pin=-45:{$S_1$}, inner sep=0pt] at (1.4, -0.5) {}; 2062 | \node[pin=30:{$S_2$}, inner sep=0pt] at (2.8, 1.2) {}; 2063 | \draw (1, -1.5) -- (1, 3); 2064 | \draw (3, -1.5) -- (3, 3); 2065 | \draw[dashed] (0, 3) -- (3, 3); 2066 | \end{axis} 2067 | \end{tikzpicture} 2068 | \end{rightcolumn} 2069 | \begin{leftcolumn} 2070 | \begin{solulist} 2071 | \item 画出图形，可见$S = S_1 + S_2$，其中 2072 | \begin{align*} 2073 | S_1 &= \int_{1}^{2} (2x - x^2)\dd x = \frac{2}{3},\\ 2074 | S_2 &= \int_{2}^{3} (x^2 - 2x)\dd x = \frac{4}{3}, 2075 | \end{align*} 2076 | 故$S = 2$. 2077 | \item \mathmethod $V = V_1 + V_2$，其中 2078 | \begin{align*} 2079 | V_1 &= \pi\int_{-1}^{0}(1 + \sqrt{1 + y})^2 \dd y - \pi = \frac{11\pi} 2080 | {6},\\ 2081 | V_2 &= 27\pi - \pi\int_{0}^{3}(1 + \sqrt{1 + y})^2 \dd y = \frac{43\pi} 2082 | {6}, 2083 | \end{align*} 2084 | 故$V = 9\pi$ 2085 | \mathmethod 2086 | \begin{align*} 2087 | V_1 &= 2\pi\int_{1}^{2}x\qty(0 - (x^2 - 2x))\dd x = \frac{11\pi}{6},\\ 2088 | V_2 &= 2\pi\int_{2}^{3}x(x^2 - 2x)\dd x = \frac{43\pi}{6}, 2089 | \end{align*} 2090 | 故$V = 9\pi$ 2091 | \end{solulist} 2092 | \end{leftcolumn} 2093 | \end{paracol} 2094 | \end{solu} 2095 | \end{mathques} 2096 | 2097 | \begin{mathques}(3){2020-10-22}{一元积分应用}[$a$的正负大小并不确定，注意$y$的设 2098 | 法] 2099 | \begin{ques} 2100 | 已知一抛物线经过$x$轴上两点$A(1, 0), B(3, 0)$. 2101 | \begin{queslist} 2102 | \item 求证两坐标轴与该抛物线所围面积等于$x$轴与该抛物线所围图形的面积； 2103 | \item 计算(1)中两个平面图形绕$x$轴旋转一周所产生的两个旋转体的体 2104 | 积之比. 2105 | \end{queslist} 2106 | \end{ques} 2107 | \begin{solu} 2108 | \begin{solulist} 2109 | \item \mathproof \mathmethod 2110 | 设过$A(1, 0), B(3, 0)$两点的抛物线方程为 2111 | \[ 2112 | y=\matherror{a}(x - 1)(x - 3), \mathidea[r]{$a$正负号未定} 2113 | \] 2114 | 则两坐标轴与该抛物线所围图形的面积为 2115 | \[ 2116 | S_1 = \int_{0}^{1} \abs{a(x - 1)(x - 3)}\dd x = \abs{a}\int_{0}^{1}(x^2 2117 | - 4x + 3)\dd x = \frac{4}{3}\abs{a}. 2118 | \] 2119 | $x$轴与该抛物线所围图形的面积为 2120 | \[ 2121 | S_1 = \int_{1}^{3} \abs{a(x - 1)(x - 3)}\dd x = -\abs{a}\int_{1}^{3}(x^2 2122 | - 4x + 3)\dd x = \frac{4}{3}\abs{a}. 2123 | \] 2124 | \mathmethod 因为$\int_{0}^{3} a(x - 1)(x-3)\dd x = \eval{a\qty(\frac{1}{3} 2125 | x^3 - 2x^2 + 3x)}_{0}^3 = 0$，所以$S_1 = S_2$. 2126 | \item 两个平面图形绕$x$轴旋转一周所产生的两个旋转体的体积分别为 2127 | \begin{align*} 2128 | V_1 &= \pi\int_{0}^{1}a^2\qty((x - 1)(x - 3))^2\dd x = \frac{38}{15}\pi 2129 | a^2,\\ 2130 | V_2 &= \pi\int_{1}^{3}a^2\qty((x - 1)(x - 3))^2\dd x = \frac{16}{15}\pi 2131 | a^2, 2132 | \end{align*} 2133 | 故$\frac{V_1}{V_2} = \frac{19}{8}.$ 2134 | \end{solulist} 2135 | \end{solu} 2136 | \end{mathques} 2137 | 2138 | \begin{mathques}{2020-10-18}{一元积分应用}[算到后面把$\pi$漏了orz] 2139 | \begin{ques} 2140 | 求曲线$y = \sqrt{x}e^{-x}$与$x$轴所围成的平面图形在$[0, +\infty)$内肉$x$轴旋 2141 | 转一周所成的旋转体的体积. 2142 | \end{ques} 2143 | \begin{solu} 2144 | 由题意得，该平面图形绕$x$轴旋转一周所成的旋转体体积为 2145 | \begin{align*} 2146 | V_x &= \pi \int_{0}^{+\infty}\qty(\sqrt{x}e^{-x})^2\dd x 2147 | = \pi\int_{0}^{+\infty}xe^{-2x}\dd x\\ 2148 | &= -\frac{1}{2}\pi\int_{0}^{+\infty}x \dd(e^{-2x}) 2149 | = -\frac{\pi}{2}\qty(\eval{xe^{-2x}}_0^{+\infty} - \int_{0}^{+\infty} 2150 | e^{-2x}\dd x)\\ 2151 | &= -\frac{\pi}{2}\qty(0 + \eval{\frac{1}{2}e^{-2x}}_0^{+\infty}) = 2152 | \frac{\pi}{4}. 2153 | \end{align*} 2154 | \end{solu} 2155 | \end{mathques} 2156 | 2157 | \begin{mathques}(3){2020-10-24}{一元积分应用}[没有考虑$a$小于$0$的结果；绕$x$轴旋 2158 | 转的体积是$y^2$之间的差不是$y$相减后的平方] 2159 | \begin{ques} 2160 | 设直线$y = ax$与抛物线$y = x^2$所围成的图形的面积为$S_1$，它们与直线$x = 1$ 2161 | 所围成图形的面积为$S_2$，并且$a < 1$. 2162 | \begin{queslist} 2163 | \item 试确定$a$的值，使$S_1 + S_2$达到最小，并求出最小值； 2164 | \item 求该最小值所对应的平面图形绕$x$轴旋转一周所得旋转体的体积. 2165 | \end{queslist} 2166 | \end{ques} 2167 | \begin{solu} 2168 | 因为$a < 1$，所以可分成$\matherror{0 < a < 1, a \le 0}$两种情况，分别画出两 2169 | 种情况下的图形（见图 \ref{fig:int-10-18})，求出$S_1 + S_2$的最小值后，即可确 2170 | 定$a$的值. 2171 | \begin{figure}[ht] 2172 | \centering 2173 | \subcaptionbox{$0 < a < 1$}[0.4\linewidth]{ 2174 | \begin{tikzpicture} 2175 | \begin{axis}[ 2176 | math, 2177 | width=\linewidth, 2178 | xtick={1}, 2179 | ytick=\empty, 2180 | extra x ticks={2 / 3}, 2181 | extra x tick labels={$a$}, 2182 | xticklabel style={text height=1.5ex}, 2183 | ymin=-0.7, 2184 | xmax=1.5, 2185 | ] 2186 | \addplot[domain=0:1.1, name path=A] {x^2} 2187 | node [pos=1, left, font=\small] {$y = x^2$} 2188 | ; 2189 | \addplot[domain=0:1.2, name path=B] {2/3*x} 2190 | node [pos=0.9, below right, font=\small] {$y = ax$} 2191 | ; 2192 | \addplot fill between [ 2193 | of = A and B, 2194 | soft clip={domain=0:1}, 2195 | split, 2196 | every segment no 0/.style={fill=green!10}, 2197 | every segment no 1/.style={fill=red!10}, 2198 | ]; 2199 | 2200 | \node[pin={[pin distance=3mm]120:{$S_1$}}, inner sep=0pt] at (3/8, 0.2) {}; 2201 | \node[pin={[pin distance=3mm]40:{$S_2$}}, inner sep=0pt] at (0.95, 0.8) {}; 2202 | \draw[dashed] (2/3, 0) -- (2/3, 4/9); 2203 | \draw[dashed] (1, 0) -- (1, 2/3); 2204 | \draw (1, 2/3) -- (1, 1); 2205 | \end{axis} 2206 | \end{tikzpicture} 2207 | } 2208 | \subcaptionbox{$a \le 1$}[0.4\linewidth]{ 2209 | \begin{tikzpicture} 2210 | \begin{axis}[ 2211 | math, 2212 | width=\linewidth, 2213 | xtick=\empty, 2214 | ytick=\empty, 2215 | ymin=-0.7, 2216 | ] 2217 | \addplot[domain=-0.9:1.1, name path=A] {x^2} 2218 | node [pos=1, left, font=\small] {$y = x^2$} 2219 | ; 2220 | \addplot[domain=-1:1.1, name path=B] {-0.6*x} 2221 | node [pos=0.85, below left, font=\small] {$y = ax$} 2222 | ; 2223 | \addplot fill between [ 2224 | of = A and B, 2225 | soft clip={domain=-0.6:1}, 2226 | split, 2227 | every segment no 0/.style={fill=green!10}, 2228 | every segment no 1/.style={fill=red!10}, 2229 | ]; 2230 | 2231 | \node[pin={[pin distance=3mm]90:{$S_1$}}, inner sep=0pt] at (-0.3, 0.12) {}; 2232 | \node[font=\small] at (0.8, 0.3) {$S_2$}; 2233 | \node[text height=2ex, below right] at (1, 0) {$1$}; 2234 | \node[text height=2ex, below] at (-0.6, 0) {$a$}; 2235 | \draw (1, -0.6) -- (1, 1); 2236 | \draw[dashed] (-0.6, 0) -- (-0.6, 0.36); 2237 | \end{axis} 2238 | \end{tikzpicture} 2239 | } 2240 | \caption{示意图} 2241 | \label{fig:int-10-18} 2242 | \end{figure} 2243 | \begin{solulist} 2244 | \item 当$0 < a < 1$时， 2245 | \[ 2246 | S = S_1 + S_2 = \int_{0}^{a}(ax - x^2)\dd x + \int_{a}^{1}(x^2 - ax)\dd 2247 | x = \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}. 2248 | \] 2249 | 令$S'(a) = a^2 - \frac{1}{2} = 0$，求得$a = \frac{1}{\sqrt{2}}$. 又 2250 | $S''\qty(\frac{1}{\sqrt{2}}) = \frac{2 - \sqrt{2}}{6}$是极小值，即最小值； 2251 | 2252 | 当$\matherror{a \le 0}$时， 2253 | \[ 2254 | S = S_1 + S_2 = \int_{a}^{0}(ax - x^2)\dd x + \int_{0}^{1}(x^2 - ax)\dd 2255 | x = -\frac{a^3}{6} - \frac{a}{2} + \frac{1}{3}. 2256 | \] 2257 | 因为 2258 | \[ 2259 | S'(a) = -\frac{1}{2}a^2 - \frac{1}{2} < 0, 2260 | \] 2261 | 所以$S$单调递减，故$a = 0$时，$S$取得最小值，此时$S = \frac{1}{3}$. 2262 | 2263 | 比较可知，当$a = \frac{1}{\sqrt{2}}$时，$S\qty(\frac{1}{\sqrt{2}}) = 2264 | \frac{2 - \sqrt{2}}{6}$是最小值. 2265 | \item 由(1)可知旋转体体积为 2266 | \[ 2267 | V_x = \pi\int_{0}^{\frac{1}{\sqrt{2}}} \qty(\frac{1}{2}x^2 - x^4) \dd x + 2268 | \pi\int_{\frac{1}{\sqrt{2}}}^{1} \qty(x^4 - \frac{1}{2}x^2)\dd x = 2269 | \frac{\sqrt{2} + 1}{30}\pi 2270 | \] 2271 | \end{solulist} 2272 | \end{solu} 2273 | \end{mathques} 2274 | 2275 | \begin{mathques}(2){2020-10-21}{一元积分应用}[没能正确将直角坐标系下的方程转化为 2276 | 极坐标系下的方程；对称性的判断不熟悉] 2277 | \begin{ques} 2278 | 双纽线$\qty(x^2 + y^2)^2 = x^2 - y^2$所围成的区域面积用定积分表示为(\quad). 2279 | \begin{multichoice} 2280 | \task \answer{$2\int_{0}^{\frac{\pi}{4}}\cos 2\theta\dd \theta$} 2281 | \task $4\int_{0}^{\frac{\pi}{4}}\cos 2\theta\dd \theta$ 2282 | \task $2\int_{0}^{\frac{\pi}{4}}\sqrt{\cos 2\theta}\dd \theta$ 2283 | \task $\frac{1}{2}\int_{0}^{\frac{\pi}{4}}(\cos 2\theta)^2\dd \theta$ 2284 | \end{multichoice} 2285 | \end{ques} 2286 | \begin{solu} 2287 | \begin{mathideabox}[对称性] 2288 | 对于$f(x, y) = 0$，若 2289 | \begin{solulist} 2290 | \item $f(x, -y) = 0$，则图像关于$x$轴对称 2291 | \item $f(-x, y) = 0$，则图像关于$y$轴对称 2292 | \item $f(-x, -y) = 0$，则图像关于原点对称 2293 | \end{solulist} 2294 | \tcbsubtitle[before skip=\baselineskip]{直角坐标转极坐标} 2295 | \[ 2296 | \begin{dcases} 2297 | x = r\cos\theta\\ 2298 | y = r\sin\theta 2299 | \end{dcases} 2300 | \] 2301 | \end{mathideabox} 2302 | 双纽线的极坐标方程为$r^2 = \cos 2\theta$，根据对称性，所求面积为 2303 | \[ 2304 | S = 4\cdot \frac{1}{2}\int_{0}^{\frac{\pi}{4}} r^2\dd \theta = 2\int_{0} 2305 | ^{\frac{\pi}{4}}\cos 2\theta\dd \theta 2306 | \] 2307 | \end{solu} 2308 | \end{mathques} 2309 | 2310 | \begin{mathques}{2020-10-19}{一元积分应用}[算到后面忘了负号] 2311 | \begin{ques} 2312 | 函数$y = \frac{x}{\sqrt{1 - x^2}}$在区间$\qty[0, \frac{1}{2}]$上的平均值为 2313 | \mathblank. 2314 | \end{ques} 2315 | \begin{solu} 2316 | \answer{$2 - \sqrt{3}$} 2317 | 2318 | 由平均值$\bar{y} = \frac{1}{b - a}\int_{a}^{b}y(x)\dd x$可知 2319 | \begin{align*} 2320 | \bar{y} &= \frac{1}{\frac{1}{2}}\int_{0}^{\frac{1}{2}} \frac{x}{\sqrt{1 - 2321 | x^2}}\dd x\\ 2322 | &= - \int_{0}^{\frac{1}{2}}(1 - x^2)^{-\frac{1}{2}} \dd(1 - x^2)\\ 2323 | &=\eval{-2(1 - x^2)^{\frac{1}{2}}}_0^{\frac{1}{2}} = 2 - \sqrt{3}. 2324 | \end{align*} 2325 | \end{solu} 2326 | \end{mathques} 2327 | 2328 | \begin{mathques}{2020-10-19}{一元积分应用}[计算错误] 2329 | \begin{ques} 2330 | 设$y = x^2$定义在$[0, 2]$上，$t$为$[0, 2]$上任意一点，问当$t$为何值时，能使 2331 | 下图中阴影面积之和最小. 2332 | \par\hfill 2333 | \begin{tikzpicture} 2334 | \begin{axis}[ 2335 | math, 2336 | xtick={2}, 2337 | ytick=\empty, 2338 | width=0.3\linewidth, 2339 | ymin=-0.5, 2340 | xmin=-0.3 2341 | ] 2342 | \addplot[domain=0:2, name path=A] {x^2}; 2343 | \draw[name path=B] (0, 1.44) -- (2, 1.44); 2344 | \addplot fill between [ 2345 | of = A and B, 2346 | soft clip={domain=0:2}, 2347 | split, 2348 | every segment no 0/.style={fill=green!10}, 2349 | every segment no 1/.style={fill=red!10}, 2350 | ]; 2351 | \draw[dashed] (1.2, 0) -- (1.2, 1.44); 2352 | \draw (2, 0) -- (2, 4); 2353 | \node[text height=1.5ex, below] at (1.2, 0) {$t$}; 2354 | \node[left] at (0, 1.44) {$t^2$}; 2355 | \node[font=\small] at (0.5, 1) {$S_1$}; 2356 | \node[font=\small] at (1.7, 2) {$S_2$}; 2357 | \end{axis} 2358 | \end{tikzpicture} 2359 | \end{ques} 2360 | \begin{solu} 2361 | \begin{align*} 2362 | S_1 &= \int_{0}^{t}(t^2 - x^2)\dd x = \frac{2}{3}t^2,\\ 2363 | S_2 &= \int_{t}^{2}(x^2 - t^2)\dd x = \matherror{\frac{2}{3}t^3 - 2t^2 + 2364 | \frac{8}{3}},\\ 2365 | S &= S_1 + S_2 = \frac{4}{3}t^3 - 2t^2 + \frac{8}{3},\\ 2366 | S' &= 4t^2 - 4t 2367 | \end{align*} 2368 | 令$S' = 0$得$t = 0$或$t = 1$, 2369 | \[ 2370 | S'' = 8t - 4\qc \matherror{\eval{S''}_{t = 0} = -4}\qc 2371 | \matherror{\eval{S''}_{t = 1} = 4}. 2372 | \] 2373 | 可知当$t = 1$时，$S$取极小值，\matherror*{由于在$(0, 2)$内驻点唯一，$S(1) = 2374 | 2$也是最小值.} 2375 | \end{solu} 2376 | \end{mathques} 2377 | 2378 | \begin{mathques}{2020-10-19}{一元积分应用}[是外面的体积减去里面的体积，不是减 2379 | 了以后绕圈] 2380 | \begin{ques} 2381 | 设直线$y = ax(0 < a < 1)$与抛物线$y = x^2$所围成图形为$S_1$，它们相交后的部 2382 | 分与直线$x = 1$所围成图形为$S_2$. 确定$a$的值，使平面图形$S_1$与$S_2$绕$x$轴 2383 | 旋转一周所得旋转体体积之和最小. 2384 | \end{ques} 2385 | \begin{solu} 2386 | 当$0 < a < 1$时，直线$y = ax$倾角取值范围为$\qty(0, \frac{\pi}{4})$，它与抛 2387 | 物线下第一象限相交. 2388 | 2389 | 由$ 2390 | \begin{dcases} 2391 | y = x^2,\\ 2392 | y = ax, 2393 | \end{dcases} 2394 | $解得交点$(0, 0), (a, a^2)$.故旋转体体积之和为 2395 | \[ 2396 | V = \int_{0}^{a} \pi(a^2x^2 - x^4)\dd x + \int_{a}^{1}\pi(x^4 - a^2x^2)\dd 2397 | x = \pi\qty(\frac{4}{15}a^5 - \frac{1}{3}a^2 + \frac{1}{5}). 2398 | \] 2399 | 又$V' = \frac{2}{3}\pi a(2a^3 - 1)$，令$V' = 0$得$a = \sqrt[3]{\frac{1}{2}}$， 2400 | $a = 0$（舍），$V'' = \frac{2}{3}\pi(8a^3 - 1)$，$\eval{V''}_{a = \sqrt[3] 2401 | {\frac{1}{2}}} > 0.$ 2402 | 2403 | 故$a = \sqrt[3]{\frac{1}{2}}$时，$V$取得极小值，由于驻点唯一，故$a = \sqrt[3] 2404 | {\frac{1}{2}}$时，$V$取得最小值. 2405 | \end{solu} 2406 | \end{mathques} 2407 | 2408 | \begin{mathques}(2){2020-10-21}{一元积分应用}[是外面的体积减去里面的体积，不是减 2409 | 了以后绕圈] 2410 | \begin{ques} 2411 | 过曲线$y = x^2$上点$(1, 1)$做切线，求切线方程，并求切线与曲线及$x$轴所围图形 2412 | 绕$x$轴旋转一周所得旋转体体积. 2413 | \end{ques} 2414 | \begin{solu} 2415 | $\eval{y'}_{x = 1} = 2$，故过点$(1, 1)$切线方程为$y = 2x - 1$. 2416 | 2417 | 该切线与$x$轴的交点为$\qty(\frac{1}{2}, 0)$，则旋转体体积为 2418 | \[ 2419 | V = \pi\int_{0}^{1}x^4\dd x - \pi\int_{\frac{1}{2}}^{1}(2x - 1)^2\dd x = 2420 | \frac{1}{5}\pi - \frac{1}{6}\pi = \frac{1}{30}\pi. 2421 | \] 2422 | \end{solu} 2423 | \end{mathques} 2424 | 2425 | \begin{mathques}{2020-10-23}{积分等式与不等式}[没有想到用$G(x)$的原函数；罗尔 2426 | 定理的方法] 2427 | \begin{ques} 2428 | 设$f(x), g(x)$在$[a, b]$上连续，证明至少存在一点$\xi \in (a, b)$，使得 2429 | \[ 2430 | f(\xi)\int_{\xi}^{b}g(x)\dd x = g(\xi)\int_{a}^{\xi}f(x) \dd x 2431 | \] 2432 | \end{ques} 2433 | \begin{solu} 2434 | \begin{mathideabox} 2435 | 若令$G(x) = f(x)\int_{x}^{b}g(t)\dd t - g(x) \int_{a}^{x}f(t)\dd t$，无法 2436 | 验证其满足零点定理，所以应求$G(x)$的原函数，使用罗尔定理证明。 2437 | \end{mathideabox} 2438 | \mathproof 设$F(x)$为$G(x) = f(x)\int_{x}^{b}g(t)\dd t - g(x) \int_{a}^{x} 2439 | f(t)\dd t$的原函数，则 2440 | \begin{align*} 2441 | F(x) &= \int \qty(f(x)\int_{x}^{b}g(t)\dd t)\dd x - \int\qty(g(x)\int_{a} 2442 | ^{x}f(t)\dd t)\dd x\\ 2443 | &= \int\qty(\int_{x}^{b}g(t)\dd t)\dd(\int_{a}^{x}f(t)\dd t) + 2444 | \int\qty(\int_{a}^{x}f(t)\dd t)\dd(\int_{x}^{a}g(t)\dd t)\\ 2445 | &= \int_{a}^{x}f(t)\dd t\int_{x}^{b}g(t)\dd t 2446 | \end{align*} 2447 | 因为$f(x), g(x)$在$[a, b]$上连续，所以$F(x)$在$[a, b]$上连续，在$(a, b)$内可 2448 | 导 2449 | 2450 | 又$\matherror{F(a) = F(b) = 0}$，由罗尔定理可得，至少存在一点$\xi \in (a, b) 2451 | $，使得$F'(\xi) = 0$，即 2452 | \[ 2453 | f(\xi)\int_{\xi}^{b}g(x)\dd x = g(\xi)\int_{a}^{\xi}f(x) \dd x 2454 | \] 2455 | \end{solu} 2456 | \end{mathques} 2457 | 2458 | \begin{mathques}(2){2020-10-25}{积分等式与不等式}[卡在将非积分转化为变限积分的地 2459 | 方] 2460 | \begin{ques} 2461 | 设$f(x)$在$[a, b]$上连续且严格单调增加，证明 2462 | \[ 2463 | (a + b)\int_{a}^{b}f(x)\dd x < 2\int_{a}^{b}xf(x)\dd x. 2464 | \] 2465 | \end{ques} 2466 | \begin{solu} 2467 | \mathproof 令$F(t) = (a + t)\int_{a}^{t}f(x)\dd x - 2\int_{a}^{t}xf(x)\dd x, 2468 | t \in [a, b]$，则 2469 | \begin{align*} 2470 | F'(t) &= \int_{a}^{t}f(x)\dd x + (a + t)f(t) - 2tf(t)\\ 2471 | &= \int_{a}^{t}f(x)\dd x - \matherror{(t - a)f(t)}\\ 2472 | &= \int_{a}^{t}f(x)\dd x - \matherror{\int_{a}^{t}f(t)\dd x}\\ 2473 | &= \int_{a}^{t}\qty(f(x) - f(t))\dd x. 2474 | \end{align*} 2475 | 因为$f(x)$在$[a, b]$上严格单调增加，所以$f(x) - f(t) < 0$，则有 2476 | \[ 2477 | F'(t) = \int_{a}^{t}\qty(f(x) - f(t))\dd x < 0, 2478 | \] 2479 | 即$F(t)$严格单调减少，又$F(a) = 0$，故$F(b) < 0$，即 2480 | \[ 2481 | (a + b)\int_{a}^{b}f(x) \dd x < 2\int_{a}^{b}x f(x)\dd x. 2482 | \] 2483 | \end{solu} 2484 | \end{mathques} 2485 | 2486 | \begin{mathques}(2){2020-10-25}{积分等式与不等式}[卡在积分中值定理没有成功转换] 2487 | \begin{ques} 2488 | 设$f(x)$在$[0, 1]$上连续且单调递减，证明：当$0 < \lambda < 1$时，$\int_{0} 2489 | ^{\lambda}f(x)\dd x \ge \lambda \int_{0}^{1}f(x)\dd x.$ 2490 | \end{ques} 2491 | \begin{solu} 2492 | 要证原不等式成立，只需证$\frac{\int_{0}^{\lambda}f(x)\dd x}{\lambda} \ge 2493 | \int_{0}^{1}f(x)\dd x$. 2494 | 2495 | 令$F(t) = \frac{\int_{0}^{t}f(x)\dd x}{t}$，由于$F(\lambda) = \frac{\int_{0} 2496 | ^{\lambda}f(x)\dd x}{\lambda}, F(1) = \int_{0}^{1}f(x)\dd x$，故只需证当 2497 | $\lambda \in (0, 1)$时，有 2498 | \begin{equation} 2499 | F(\lambda) \ge F(1) \tag{$*$} 2500 | \end{equation} 2501 | $F(t)$在$(0, 1]$内连续，在$(0, 1)$内可导，且 2502 | \[ 2503 | F'(t) = \frac{f(t)t - \int_{0}^{t}f(x)\dd x}{t^2} = \frac{f(t)t - f(c)t} 2504 | {t^2} = \frac{f(t) - f(c)}{t}, 2505 | \] 2506 | 其中$0 \le c \le t$，则$f(c) \ge f(t)$，即有$F'(t) \le 0$，故$F(t)$在$(0, 1] 2507 | $内单调递减. 2508 | 2509 | 又$0 < \lambda < 1$，则$F(\lambda) \ge F(1)$，即$(*)$式成立 2510 | \end{solu} 2511 | \end{mathques} 2512 | 2513 | \begin{mathques}(2){2020-10-25}{积分等式与不等式}[没有想到三角变换] 2514 | \begin{ques} 2515 | 设$f(x)$为$[0, 1]$上的连续函数，证明： 2516 | \begin{queslist} 2517 | \item $\int_{0}^{\frac{\pi}{2}}f(\sin x)\dd x = \int_{0}^{\frac{\pi}{2}} 2518 | f(\cos x)\dd x$； 2519 | \item $\int_{0}^{\pi}xf(\sin x)\dd x = \frac{\pi}{2}\int_{0}^{\pi}f(\sin x) 2520 | \dd x$. 2521 | \end{queslist} 2522 | 并计算$\int_{0}^{\pi}\frac{x\sin x}{1 + \cos[2](x)}\dd x$. 2523 | \end{ques} 2524 | \begin{solu} 2525 | \begin{solulist} 2526 | \item 设$\matherror{x = \frac{\pi}{2} - t}$，则$\dd x = -\dd t$，当$x = 0$ 2527 | 时，$t = \frac{\pi}{2}$；当$x = \frac{\pi}{2}$时，$t = 0$. 则 2528 | \[ 2529 | \int_{0}^{\frac{\pi}{2}}f(\sin x)\dd x = -\int_{\frac{\pi}{2}}^{0} 2530 | f\qty(\sin(\frac{\pi}{2} - t))\dd t = \int_{0}^{\frac{\pi}{2}}f(\cos t) 2531 | \dd t = \int_{0}^{\frac{\pi}{2}}f(\cos x)\dd x. 2532 | \] 2533 | \item 设$\matherror{x = \pi - t}$，则$\dd x = -\dd t$，且当$x = 0$时，$t = 2534 | \pi$；当$x = \pi$时，$t = 0$，则 2535 | \begin{align*} 2536 | \int_{0}^{\pi}xf(\sin x)\dd x &= - \int_{\pi}^{0}(\pi - t) 2537 | f\qty(\sin(\pi - t))\dd t\\ 2538 | &= \int_{0}^{\pi}(\pi - t)f(\sin t)\dd t\\ 2539 | &= \pi\int_{0}^{\pi}f(\sin t)\dd t - \int_{0}^{\pi}tf(\sin t)\dd t\\ 2540 | &= \pi\int_{0}^{\pi}f(\sin x)\dd x - \matherror{\int_{0}^{\pi}xf(\sin 2541 | x)\dd x}\\ 2542 | \end{align*} 2543 | 故 2544 | \[ 2545 | \int_{0}^{\pi}xf(\sin x)\dd x = \frac{\pi}{2}\int_{0}^{\pi}f(\sin x) 2546 | \dd x. 2547 | \] 2548 | \end{solulist} 2549 | 由上述结论可得 2550 | \begin{align*} 2551 | \int_{0}^{\pi}\frac{x\sin x}{1 + \cos[2](x)}\dd x &= \frac{\pi}{2}\int_{0} 2552 | ^{\pi}\frac{\sin x}{1 + \cos[2](x)}\dd x\\ 2553 | &= -\frac{\pi}{2}\int_{0}^{\pi}\frac{\dd (\cos x)}{1 + \cos[2](x)}\\ 2554 | &= \eval{-\frac{\pi}{2}\arctan(\cos x)}_0^{\pi} = \frac{\pi^2}{4}. 2555 | \end{align*} 2556 | \end{solu} 2557 | \end{mathques} 2558 | 2559 | \begin{mathques}(2){2020-10-25}{积分等式与不等式}[没有想到定积分可以当做常数看 2560 | 待，然后再凑出这个常数] 2561 | \begin{ques} 2562 | 设$f(x)$在$[-\pi, \pi]$上连续，且有$f(x) = \frac{x}{1 + \cos[2](x)} + 2563 | \int_{-\pi}^{\pi}f(x)\sin x\dd x$，求$f(x)$. 2564 | \end{ques} 2565 | \begin{solu} 2566 | 由于$\int_{-\pi}^{\pi}f(x)\sin x\dd x$存在，记$A = \int_{-\pi}^{\pi}f(x)\sin 2567 | x\dd x$，则$f(x) = \frac{x}{1 + \cos[2](x)} + A$， 2568 | \[ 2569 | \int_{-\pi}^{\pi}f(x)\sin x\dd x = \int_{-\pi}^{\pi}\frac{x\sin x}{1 + 2570 | \cos[2](x)}\dd x + \matherror{\int_{-\pi}^{\pi}A\sin x\dd x}, 2571 | \] 2572 | 令$x = \pi - t$，从而有 2573 | \begin{align*} 2574 | A &= \int_{-\pi}^{\pi}\frac{x\sin x}{1 + \cos[2](x)}\dd x\\ 2575 | &= 2\int_{0}^{\pi} \frac{x\sin x}{1 + \cos[2](x)}\dd x\\ 2576 | &= -2\int_{\pi}^{0}\frac{(\pi - t)\sin(\pi - t)}{1 + \cos[2](\pi - t)}\dd 2577 | t\\ 2578 | &= 2\int_{0}^{\pi}\frac{(\pi - t)\sin t}{1 + \cos[2](t)}\dd t\\ 2579 | &= 2\pi\int_{0}^{\pi}\frac{\sin t}{1 + \cos[2](t)}\dd t - A 2580 | \end{align*} 2581 | 故 2582 | \[ 2583 | A = -\pi\int_{0}^{\pi}\frac{\dd (\cos t)}{1 + \cos[2](t)} 2584 | = \eval{-\pi\arctan(\cos t)}_0^{\pi} = \frac{\pi^2}{2} 2585 | \] 2586 | 即 2587 | \[ 2588 | f(x) = \frac{x}{1 + \cos[2](x)} + \frac{\pi^2}{2} 2589 | \] 2590 | \end{solu} 2591 | \end{mathques} 2592 | 2593 | \begin{mathques}{2020-10-31}{多元函数微分学}[多元的极限不会求] 2594 | \begin{ques} 2595 | 函数$f(x, y) = \sqrt{\abs{xy}}$在点$(0, 0)$处(\quad). 2596 | \begin{multichoice} 2597 | \task 偏导数不存在 2598 | \task \answer{偏导数存在，但不可微} 2599 | \task 可微，但偏导数不连续 2600 | \task 偏导数连续 2601 | \end{multichoice} 2602 | \end{ques} 2603 | \begin{solu} 2604 | \begin{align*} 2605 | f'_x(0, 0) &= \lim_{\Delta x \to 0} \frac{f(0 + \Delta x, 0) - f(0, 0)} 2606 | {\Delta x} = \lim_{\Delta x \to 0} \frac{\sqrt{\abs{\Delta x \cdot 0}} - 0} 2607 | {\Delta x} = 0 = A,\\ 2608 | f'_y(0, 0) &= \lim_{\Delta y \to 0} \frac{f(0 + \Delta y, 0) - f(0, 0)} 2609 | {\Delta y} = 0 = B,\\ 2610 | \end{align*} 2611 | 故$f(x, y)$在$(0, 0)$处偏导数存在，又 2612 | \[ 2613 | \Delta z = f(0 + \Delta x, 0 + \Delta y) - f(0, 0) = \sqrt{\abs{\Delta 2614 | x\Delta y}} - 0= \sqrt{\abs{\Delta x\Delta y}} 2615 | \] 2616 | 故有 2617 | \[ 2618 | \lim_{\substack{\Delta x \to 0\\\Delta y \to 0}} \frac{\Delta z 2619 | - A\Delta x - B\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 2620 | \lim_{\substack{\Delta x \to 0\\\Delta y \to 0}} 2621 | \frac{\sqrt{\abs{\Delta x\cdot \Delta y}}}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} 2622 | \] 2623 | 不存在，故$f(x, y)$在点$(0, 0)$处不可微. 2624 | \end{solu} 2625 | \end{mathques} 2626 | 2627 | \begin{mathques}{2020-10-31}{多元函数微分学}[后缀写错] 2628 | \begin{ques} 2629 | 设函数$f(x)$可微，且$f'(0) = \frac{1}{2}$，则$z = f(4x^2 - y^2)$在点$(1, 2)$ 2630 | 处的全微分$\eval{\dd z}_{(1, 2)}=$\mathblank. 2631 | \end{ques} 2632 | \begin{solu} 2633 | \answer{$4\dd x - 2\dd y$} 2634 | 2635 | 令$u = 4x^2 - y^2$，则$z = f(u)$，则 2636 | \begin{align*} 2637 | \pdv{z}{x} &= f'(u)(4x^2 - y^2)'_x = 8xf'(u), & \eval{\pdv{z}{x}}_{(1, 2)} 2638 | &= 8f'(0) = 4,\\ 2639 | \pdv{z}{y} &= f'(u)(4x^2 - y^2)'_y = -2yf'(u), & \eval{\pdv{z}{y}}_{(1, 2)} 2640 | &= 8f'(0) = -2, 2641 | \end{align*} 2642 | 故$\eval{\dd z}_{(1, 2)} = 4\dd x-2\dd y$ 2643 | \end{solu} 2644 | \end{mathques} 2645 | 2646 | \begin{mathques}(3){2020-11-04}{多元函数微分学}[抄错；需要合并] 2647 | \begin{ques} 2648 | 设$z = f(x^2y^2, e^{xy})$，其中$f(u, v)$有二阶连续偏导数，求$z''_{xx}, 2649 | z''_{yy}, z''_{xy}.$ 2650 | \end{ques} 2651 | \begin{solu} 2652 | 令$u = x^2y^2, v = e^{xy}$，则$\pdv{u}{x} = 2xy^2, \pdv{u}{y} = 2x^2y, 2653 | \pdv{v}{x} = ye^{xy}, \pdv{v}{y} = xe^{xy}$，故 2654 | \begin{align*} 2655 | \pdv{z}{x} &= \pdv{z}{u}\cdot\pdv{u}{x} + \pdv{z}{v}\cdot\pdv{v}{x} 2656 | = \pdv{z}{u}\cdot 2xy^2 + \pdv{z}{v}\cdot ye^{xy},\\ 2657 | \pdv{z}{y} &= \pdv{z}{u}\cdot\pdv{u}{y} + \pdv{z}{v}\cdot\pdv{v}{y} 2658 | = \pdv{z}{u}\cdot 2x^2y + \pdv{z}{v}\cdot xe^{xy},\\ 2659 | \end{align*} 2660 | 故 2661 | \begin{align*} 2662 | \pdv[2]{z}{x} &= \qty(\pdv[2]{z}{u}\cdot\pdv{u}{x} + \pdv{z}{u}{v} 2663 | \cdot\pdv{v}{x})\cdot2xy^2 + \pdv{z}{u}\cdot2y^2 + \qty(\pdv{z}{u}{v} 2664 | \cdot\pdv{u}{x} + \pdv[2]{z}{v}\cdot\pdv{v}{x})\cdot ye^{xy} + \pdv{z}{v} 2665 | \cdot y^2e^{xy}\\ 2666 | &= \qty(\pdv[2]{z}{u}\cdot2xy^2 + \pdv{z}{u}{v}\cdot ye^{xy}) \cdot2xy^2 + 2667 | \pdv{z}{u}\cdot2y^2 + \qty(\pdv{z}{u}{v}\cdot2xy^2 + \pdv[2]{z}{v}\cdot 2668 | ye^{xy})\cdot ye^{xy} + \pdv{z}{v} \cdot y^2e^{xy}\\ 2669 | &= \pdv[2]{z}{u}\cdot 4x^2y^4 + \pdv[2]{z}{v}\cdot y^2e^{2xy} + \pdv{z}{u} 2670 | {v}\cdot 4xy^3e^{xy} + \pdv{z}{u}\cdot 2y^2 + \pdv{z}{v} \cdot y^2e^{xy} 2671 | ,\\ 2672 | \intertext{同理可得} 2673 | \pdv[2]{z}{y} &= \pdv[2]{z}{u} \cdot 4x^4y^2 + \pdv[2]{z}{v}\cdot 2674 | x^2e^{2xy} + \pdv{z}{u} {v}\cdot 4x^3ye^{xy} + \pdv{z}{u}\cdot 2x^2 + 2675 | \pdv{z}{v} \cdot x^2e^{xy} ,\\ 2676 | \pdv{z}{x}{y} &= \qty(\pdv[2]{z}{u}\cdot\pdv{u}{y} + \pdv{z}{u}{v} 2677 | \cdot\pdv{v}{y})\cdot2xy^2 + \pdv{z}{u}\cdot4xy + \qty(\pdv{z}{u}{v} 2678 | \cdot\pdv{u}{y} + \pdv[2]{z}{v}\cdot\pdv{v}{y})\cdot ye^{xy} + \pdv{z}{v} 2679 | \cdot (1 + xy)e^{xy}\\ 2680 | &= \qty(\pdv[2]{z}{u}\cdot2x^2y + \pdv{z}{u}{v}\cdot xe^{xy}) \cdot2xy^2 + 2681 | \pdv{z}{u}\cdot4xy + \qty(\pdv{z}{u}{v}\cdot2x^2y + \pdv[2]{z}{v}\cdot 2682 | xe^{xy})\cdot ye^{xy} + \pdv{z}{v} \cdot (1 + xy)e^{xy}\\ 2683 | &= \pdv[2]{z}{u}\cdot 4x^3y^3 + \pdv[2]{z}{v}\cdot xye^{2xy} + \pdv{z}{u} 2684 | {v}\cdot 4x^2y^2e^{xy} + \pdv{z}{u}\cdot 4xy + \pdv{z}{v} \cdot (1 + xy) 2685 | e^{xy}. 2686 | \end{align*} 2687 | \end{solu} 2688 | \end{mathques} 2689 | 2690 | \begin{mathques}{2020-10-31}{多元函数微分学}[$xy$和$xz$需要分别对$x$进行求导] 2691 | \begin{ques} 2692 | 设函数$u = f(x^2, xy, xz)$具有一阶连续偏导数，又函数$y = y(x), z = z(x)$分别 2693 | 由 2694 | \[ 2695 | \sin xy = y\qc e^{z} = \int_{0}^{xz}\sin t\dd t 2696 | \] 2697 | 确定，求$\dv{u}{x}$. 2698 | \end{ques} 2699 | \begin{solu} 2700 | \begin{equation} 2701 | \dv{u}{x} = f'_1 \cdot 2x + f'_2\cdot\matherror{\qty(y + x\dv{y}{x})} + 2702 | f'_3\cdot\matherror{\qty(z + x\dv{z}{x})}. 2703 | \tag{1} 2704 | \end{equation} 2705 | 对$\sin xy = y$两边求导得 2706 | \[ 2707 | \cos xy \cdot \qty(y + x\dv{y}{x}) = \dv{y}{x}, 2708 | \] 2709 | 故 2710 | \begin{equation} 2711 | \dv{y}{x} = \frac{y\cos xy}{1 - x\cos xy}. 2712 | \tag{2} 2713 | \end{equation} 2714 | 2715 | $e^{z} = \int_{0}^{xz} \sin t\dd t$两边对$x$求导得 2716 | \[ 2717 | e^z\dv{z}{x} = \sin xz \cdot\qty(z + x\dv{z}{x}), 2718 | \] 2719 | 故 2720 | \begin{equation} 2721 | \dv{z}{x} = \frac{z\sin xz}{e^z - x\sin xz}. 2722 | \tag{3} 2723 | \end{equation} 2724 | 将$(2), (3)$代入$(1)$得 2725 | \[ 2726 | \dv{u}{x} = f'_1 \cdot 2x + f'_2 \cdot\qty(y + \frac{xy\cos xy}{1 - x\cos 2727 | xy}) + f'_3\cdot \qty(z + \frac{xz\sin xz}{e^z - x\sin xz}). 2728 | \] 2729 | \end{solu} 2730 | \end{mathques} 2731 | 2732 | \begin{mathques}(2){2020-11-02}{多元函数微分学}[边界值直接代入$y$漏考虑了$x$的边 2733 | 界点，以后这种还是直接用拉格朗日] 2734 | \begin{ques} 2735 | 求$f(x, y) = x^2 - y^2 + 2$在椭圆域$D = \Bqty{(x, y)\bigg|x^2 + \frac{y^2}{4} 2736 | \le 1}$上的最大值和最小值. 2737 | \end{ques} 2738 | \begin{solu} 2739 | \begin{mathideabox} 2740 | 因为只求函数$f(x, y) = x^2 - y^2 + 2$在椭圆域$D$上的最大值和最小值，而不求 2741 | 极值，所以，只需求出$D$的内部以及$D$的边界上的驻点和导数不存在的点（不用判 2742 | 断它们是否为极值点），并求出这些点的函数值，然后比较它们的大小即可。 2743 | \end{mathideabox} 2744 | 令$ 2745 | \begin{dcases} 2746 | f'_x = 2x = 0,\\ 2747 | f'_y = -2y = 0, 2748 | \end{dcases} 2749 | $得唯一驻点$(0, 0), f(x, y)$在$D$内部没有导数不存在的点 2750 | 2751 | 再求$f(x, y)$在$D$的边界上的驻点和导数不存在的点\\ 2752 | 即变成了一个条件极值问题，\matherror*{边界方程$x^2 + \frac{y^2}{4} = 1$为约 2753 | 束条件}，构造拉格朗日函数 2754 | \[ 2755 | F(x, y, \lambda) = x^2 - y^2 + 2 + \lambda(x^2 + \frac{y^2}{4} - 1), 2756 | \] 2757 | 令 2758 | \[ 2759 | \begin{dcases} 2760 | F'_x = 2x + 2\lambda x = 0,\\ 2761 | F'_y = -2y + \frac{\lambda}{2}y = 0,\\ 2762 | F'_{\lambda} = x^2 + \frac{y^2}{4} - 1 = 0, 2763 | \end{dcases} 2764 | \] 2765 | 解上述方程得四个驻点$(0, 2), (0, -2), (1, 0), (-1, 0)$，计算得 2766 | \[ 2767 | f(0, 0) = 2\qc f(0, 2) = f(0, -2) = -2, f(1, 0) = f(-1, 0) = 3, 2768 | \] 2769 | 故函数最大值最小值分别为$f_{\max} = 3, f_{\min} = -2.$ 2770 | \end{solu} 2771 | \end{mathques} 2772 | 2773 | \begin{mathques}{2020-11-03}{多元函数微分学}[不知道代换方式] 2774 | \begin{ques} 2775 | 利用变量代换$u = x, v = \frac{y}{x}$, 可将方程$x\pdv{z}{x} + y\pdv{z}{y} = 2776 | z$化成新方程(\quad). 2777 | \begin{multichoice} 2778 | \task \answer{$u\pdv{z}{u} = z$} 2779 | \task $v\pdv{z}{v} = z$ 2780 | \task $u\pdv{z}{v} = z$ 2781 | \task $v\pdv{z}{u} = z$ 2782 | \end{multichoice} 2783 | \end{ques} 2784 | \begin{solu} 2785 | $z$是$x, y$的函数，由链式法则 2786 | \begin{align*} 2787 | \pdv{z}{x} &= \pdv{z}{u}\cdot \dv{u}{x} + \pdv{z}{v}\cdot \pdv{v}{x} = 2788 | \pdv{z}{u} - \frac{y}{x^2}\cdot\pdv{z}{v}\\ 2789 | \pdv{z}{y} &= \pdv{z}{v}\cdot \pdv{v}{y} = \frac{1}{x}\cdot \pdv{z}{v} 2790 | \end{align*} 2791 | 故$z = x\cdot \pdv{z}{u} - \frac{y}{x}\pdv{z}{v} + \frac{y}{x}\pdv{z}{v} = 2792 | u\pdv{z}{u}$. 2793 | \end{solu} 2794 | \end{mathques} 2795 | 2796 | \begin{mathques}{2020-11-03}{多元函数微分学}[不知道代换方法] 2797 | \begin{ques} 2798 | 设函数$z = xyf(\frac{y}{x})$，其中$f(u)$可导，则$x\pdv{z}{x} + y\pdv{z}{y} 2799 | =$\mathblank. 2800 | \end{ques} 2801 | \begin{solu} 2802 | \answer{2z} 2803 | 2804 | 设$u = \frac{y}{x}$，则$z = xyf(u)$ 2805 | \begin{align*} 2806 | \pdv{z}{x} &= yf(u) + xyf'(u)\pdv{u}{x} = yf(u) - \frac{y^2}{x}f'(u),\\ 2807 | \pdv{z}{y} &= xf(u) + xyf'(u)\pdv{u}{y} = xf(u) + yf'(u). 2808 | \end{align*} 2809 | 故 2810 | \[ 2811 | x\pdv{z}{x} + y\pdv{z}{y} = xyf(u) - y^2f'(u) + xyf(u) + y^2f'(u) = 2xyf(u) 2812 | = 2z. 2813 | \] 2814 | \end{solu} 2815 | \end{mathques} 2816 | 2817 | \begin{mathques}{2020-11-03}{多元函数微分学}[二重极限不会求] 2818 | \begin{ques} 2819 | 求下列极限： 2820 | \begin{queslist} 2821 | \item $\lim_{(x, y) \to (0, 0)}\frac{\sin(x^2y)}{x^2 + y^2}.$ 2822 | \item $\lim_{(x, y) \to (0, 0)}\frac{x^2y}{x^4 + y^2}.$ 2823 | \end{queslist} 2824 | \end{ques} 2825 | \begin{solu} 2826 | \begin{solulist} 2827 | \item $\lim_{(x, y) \to (0, 0)}\frac{\sin(x^2y)}{x^2 + y^2} = \lim_{(x, y) 2828 | \to (0, 0)} \frac{\sin(x^2y)}{x^2y}\cdot \frac{x^2}{x^2 + y^2}\cdot y$，则 2829 | \begin{align*} 2830 | \lim_{(x, y) \to (0, 0)} \frac{\sin(x^2y)}{x^2y} &= 1,\mathidea[r]{重要 2831 | 极限}\\ 2832 | \lim_{(x, y) \to (0, 0)} \frac{x^2}{x^2 + y^2}\cdot y &= 0,\mathidea[r] 2833 | {无穷小乘有界函数} 2834 | \end{align*} 2835 | 故$\lim_{(x, y) \to (0, 0)}\frac{\sin(x^2y)}{x^2 + y^2} = 1\times 0 = 0.$ 2836 | \item 令$y = kx$则 2837 | \[ 2838 | \lim_{(x, y) \to (0, 0)}\frac{x^2y}{x^4 + y^2} = \lim_{x \to 0} 2839 | \frac{kx^3}{x^4 + (kx)^2} = \lim_{x \to 0} \frac{kx}{x^2 + k^2} = 0. 2840 | \] 2841 | 令$y = x^2$则 2842 | \[ 2843 | \lim_{(x, y) \to (0, 0)}\frac{\sin(x^2y)}{x^2 + y^2} = \frac{1}{2}. 2844 | \] 2845 | 故$\lim_{(x, y) \to (0, 0)}\frac{x^2y}{x^4 + y^2}$不存在. 2846 | \end{solulist} 2847 | \end{solu} 2848 | \end{mathques} 2849 | 2850 | \begin{mathques}{2020-11-03}{多元函数微分学}[不知道代换方法，题目中的$G'_z$不 2851 | 为$0$算是提示] 2852 | \begin{ques} 2853 | 设函数$z = z(x, y)$由$G(x, y, z) = F(xy, yz) = 0$确定，其中$F$为可微函数，且 2854 | $G'_z \neq 0$，求$x\pdv{z}{x} - y\pdv{z}{y}.$ 2855 | \end{ques} 2856 | \begin{solu} 2857 | 由$G(x, y, z) = F(xy, yz)$，可得 2858 | \begin{gather*} 2859 | G'_x = F'_1\cdot y\qc G'_y = F'_1\cdot x + F'_2\cdot z\qc G'_z = F'_2\cdot 2860 | y\neq 0.\\ 2861 | \end{gather*} 2862 | $G(x, y, z) = 0$两边分别对$x, y$求偏导得 2863 | \[ 2864 | G'_x + G'_z\pdv{z}{x} = 0\qc G'_y + G'_z\pdv{z}{y} = 0, 2865 | \] 2866 | 故 2867 | \begin{align*} 2868 | \pdv{z}{x} &= -\frac{G'_x}{G'_z} = -\frac{F'_1}{F'_2}\\ 2869 | \pdv{z}{y} &= -\frac{G'_y}{G'_z} = -\frac{xF'_1 + zF'_2}{yF'_2}. 2870 | \end{align*} 2871 | 则 2872 | \[ 2873 | x\pdv{z}{x} - y\pdv{z}{y} = -\frac{xF'_1}{F'_2} + \frac{xF'_1 + zF'_2} 2874 | {F'_2} = z\\ 2875 | \] 2876 | \end{solu} 2877 | \end{mathques} 2878 | 2879 | \begin{mathques}{2020-11-03}{多元函数微分学}[链式法则不熟悉] 2880 | \begin{ques} 2881 | 设$x = x(y, z), y = y(z, x), z = z(x, y)$均为方程$f(x, y, z) = 0$所确定的具 2882 | 有连续偏导数的函数，证明$x'_y\cdot y'_z \cdot z'_x = -1.$ 2883 | \end{ques} 2884 | \begin{solu} 2885 | \mathproof 由$x = x(y, z)$得，$f(x(y, z), y, z) = 0$，两边对$y$求导得$f'_1 2886 | \cdot x'_y + f'_2 = 0$，解得$x'_y = -\frac{f'_2}{f'_1}$. 2887 | \begin{mathideabox} 2888 | $f(x(y, z), y, z)$是看成关于$y, z$的函数，对$y$求偏导时是把$z$看成常数的因 2889 | 此不会出现$\pdv{z}{y}$，可以理解成$g'_x(x, y) = g'_x + g'_x\cdot 2890 | \dv{y(\text{常数})}{x} $ 2891 | \end{mathideabox} 2892 | 同理$y'z = -\frac{f'_3}{f'_2}, z'_x = -\frac{f'_1}{f'_3}$，故$x'_y\cdot y'_z 2893 | \cdot z'_x = -1.$ 2894 | \end{solu} 2895 | \end{mathques} 2896 | 2897 | \begin{mathques}(2){2020-11-05}{多元函数微分学}[求导符漏了] 2898 | \begin{ques} 2899 | 设$f(u)$具有二阶连续导数，且$z = xf(\frac{y}{x}) + yf(\frac{x}{y})$，求 2900 | $\pdv{z}{x}{y}.$ 2901 | \end{ques} 2902 | \begin{solu} 2903 | 设$u = \frac{y}{x}, v = \frac{x}{y}$，则 2904 | \begin{align*} 2905 | z &= xf(u) + yf(v),\\ 2906 | \pdv{z}{x} &= f(u) + xf'(u)\cdot \qty(-\frac{y}{x^2}) + yf'(v)\cdot 2907 | \frac{1}{y}\\ 2908 | &= f(u) - \frac{y}{x}f'(u) + f'(v),\\ 2909 | \pdv{z}{x}{y} &= f'(u) \cdot \frac{1}{x} - \frac{1}{x}f'(u) - \frac{y}{x} 2910 | y''(u)\cdot \frac{1}{x} + f''(v) \cdot \qty(-\frac{x}{y^2})\\ 2911 | &= -\frac{y}{x^2}f''(u) - \frac{x}{y^2}f''(v)\\ 2912 | &= -\frac{y}{x^2}f''(\frac{y}{x}) - \frac{x}{y^2}f''(\frac{x}{y}). 2913 | \end{align*} 2914 | \end{solu} 2915 | \end{mathques} 2916 | 2917 | \begin{mathques}{2020-11-03}{多元函数微分学}[求出驻点后没有判断是极大值还是极 2918 | 小值] 2919 | \begin{ques} 2920 | 求二元函数$f(x, y) = x^2y^2 + x\ln x$的极值. 2921 | \end{ques} 2922 | \begin{solu} 2923 | \[ 2924 | f'_x(x, y) = 2xy^2 + 1 + \ln x\qc f'_y(x, y) = 2x^2y. 2925 | \] 2926 | 令 2927 | \[ 2928 | \begin{dcases} 2929 | f'_x(x, y) = 0,\\ 2930 | f'_y(x, y) = 0, 2931 | \end{dcases} 2932 | \] 2933 | 解得唯一驻点$\qty(\frac{1}{e}, 0).$由于 2934 | \begin{gather*} 2935 | \matherror{ 2936 | f''_{xx} = 2y^2 + \frac{1}{x}\qc f''_{xy} = 4xy, f''_{yy} = 2x^2, 2937 | }\\ 2938 | A = f''_{xx}\qty(\frac{1}{e}, 0) = e\qc B = f''_{xy}\qty(\frac{1}{e}, 0) = 0 2939 | \qc A = f''_{yy}\qty(\frac{1}{e}, 0) = \frac{2}{e^2}, 2940 | \end{gather*} 2941 | 故$\matherror{B^2 - AC = - \frac{2}{e} < 0, A = e > 0}$，则$\qty(\frac{1}{e}, 0) 2942 | $为$f(x, y)$的极小值点，极小值为$-\frac{1}{e}.$ 2943 | \end{solu} 2944 | \end{mathques} 2945 | 2946 | \begin{mathques}{2020-11-3}{多元函数微分学} 2947 | \begin{ques} 2948 | 求函数$u = xyz$在条件$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a} 2949 | (x > 0, y > 0, z > 0, a > 0)下的极值$. 2950 | \end{ques} 2951 | \begin{solu} 2952 | 作拉格朗日函数$L(x, y, z, \lambda) = xyz + \lambda\qty(\frac{1}{x} + \frac{1}{y} 2953 | + \frac{1}{z} - \frac{1}{a})$.令 2954 | \begin{empheq}[left=\empheqlbrace]{align} 2955 | L'_x &= yz - \frac{\lambda}{x^2} = 0,\tag{1}\\ 2956 | L'_y &= xz - \frac{\lambda}{y^2} = 0,\tag{2}\\ 2957 | L'_z &= xy - \frac{\lambda}{z^2} = 0,\tag{3} 2958 | \end{empheq} 2959 | $(1)x + (2)y + (3)z$可得 2960 | \begin{equation} 2961 | 3xyz - \lambda\qty(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 0,\tag{4} 2962 | \end{equation} 2963 | 注意到$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a}$，$(4)$式化为 2964 | \[ 2965 | 3xyz = \frac{\lambda}{a}, 2966 | \] 2967 | 代回$(1), (2), (3)$得到$x = y = z = 3a, \lambda = 81a^4$. 2968 | 2969 | 故$(3a, 3a, 3a)$是函数$u = xyz$在给定约束条件下唯一可能的极值点. 2970 | 2971 | 又 2972 | \begin{align*} 2973 | \dd L &= xy\dd z + xz \dd y + yz \dd x - \lambda\qty(\frac{1}{x^2}\dd x + 2974 | \frac{1}{y^2}\dd y + \frac{1}{z^2}\dd x)\\ 2975 | \dd[2]{L} &= 2(x\dd y\dd z + y\dd x\dd z + z\dd x \dd y) + 2976 | 2\lambda\qty(\frac{1}{x^3}\dd {x^2} + \frac{1}{y^3}\dd {y^2} + \frac{1} 2977 | {z^3}\dd {z^2})\\ 2978 | \end{align*} 2979 | 又由 2980 | \[ 2981 | \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a} 2982 | \] 2983 | 得 2984 | \[ 2985 | \frac{1}{x^2}\dd x + \frac{1}{y^2}\dd y + \frac{1}{z^2} \dd z = 0 2986 | \] 2987 | 即在点$(3a, 3a, 3a)$处有$\dd x + \dd y + \dd z = 0$，则 2988 | \begin{align*} 2989 | \eval{\dd[2]L}_{(3a, 3a, 3a)} &= 6a(\dd x\dd y + \dd y\dd z + \dd z\dd x) 2990 | + \frac{2\lambda}{27a^3}(\dd{x^2} + \dd{y^2} + \dd{z^2})\\ 2991 | &= 6a(\dd x\dd y + \dd y\dd z + \dd z\dd x + \dd{x^2} + \dd{y^2} + \dd{z^2}) 2992 | \end{align*} 2993 | 2994 | \matherror*{要证明$\dd[2]{L} > 0$} 2995 | 2996 | \end{solu} 2997 | \end{mathques} 2998 | 2999 | \begin{mathques}{2020-11-06}{二重积分}[不知道怎么求$y$的范围] 3000 | \begin{ques} 3001 | 累次积分$\int_{0}^{\frac{\pi}{2}}\dd\theta\int_{0}^{\cos\theta}f(r\cos\theta, 3002 | r\sin\theta)r\dd r$可以写成(\quad). 3003 | \begin{multichoice} 3004 | \task $\int_{0}^{1}\dd y\int_{0}^{\sqrt{y - y^2}}f(x, y)\dd x$ 3005 | \task $\int_{0}^{1}\dd y\int_{0}^{\sqrt{1 - y^2}}f(x, y)\dd x$ 3006 | \task $\int_{0}^{1}\dd x\int_{0}^{1}f(x, y)\dd y$ 3007 | \task \answer{$\int_{0}^{1}\dd x\int_{0}^{\sqrt{x - x^2}}f(x, y)\dd y$} 3008 | \end{multichoice} 3009 | \end{ques} 3010 | \begin{solu} 3011 | 积分区域$D = \Bqty{(r, \theta) \bigg| 0 \le \theta \le \frac{\pi}{2}, 0 \le 3012 | r \le \cos \theta}$ 3013 | 3014 | 先对$r$积分，且$0 = r_1(\theta) \le r \le r_2(\theta)$，$r_2(\theta)$为圆的 3015 | 方程$\qty(x - \frac{1}{2})^2 + y^2 = \frac{1}{4}$，然后对$\theta$积分且$0 3016 | \le \theta \le \frac{\pi}{2}$，实际上积分区域$D$就是圆$\qty(x - \frac{1}{2}) 3017 | ^2 + y^2 = \frac{1}{4}$的上半部分。 3018 | \end{solu} 3019 | \end{mathques} 3020 | 3021 | \begin{mathques}(2){2020-11-08}{二重积分}[漏了$r$导致积分算错] 3022 | \begin{ques} 3023 | 求二重积分$\iint\limits_{D}\frac{1 - x^2 - y^2}{1 + x^2 + y^2}\dd x\dd y$， 3024 | 其中$D$是$\{(x, y)|x^2 + y^2 \le 1\}$在第一象限的部分. 3025 | \end{ques} 3026 | \begin{solu} 3027 | 利用极坐标得 3028 | \begin{align*} 3029 | \text{原式} &= \int_{0}^{\frac{\pi}{2}}\dd \theta \int_{0}^{1} \frac{1 - 3030 | r^2}{1 + r^2}\matherror{r}\dd r = \frac{\pi}{2}\int_{0}^{1}\qty(\frac{2}{1 3031 | + r^2} - 1)r\dd r\\ 3032 | &= \eval{\frac{\pi}{2}\qty(\ln(1 + r^2) - \frac{1}{2}r^2)}_{0}^{1} = 3033 | \frac{\pi}{2}\qty(\ln 2 - \frac{1}{2}). 3034 | \end{align*} 3035 | \end{solu} 3036 | \end{mathques} 3037 | 3038 | \begin{mathques}{2020-11-06}{二重积分}[0-1的部分算错了] 3039 | \begin{ques} 3040 | 设$F(t) = \iint\limits_{x + y \le t}f(x, y)\dd x\dd y$，其中$f(x, y) = 3041 | \begin{dcases} 3042 | 1, & 0 \le x \le 1, 0 \le y \le 1,\\ 3043 | 0, & \text{其他},\\ 3044 | \end{dcases} 3045 | $求$F(t)$的表达式. 3046 | \end{ques} 3047 | \begin{solu} 3048 | 由被积函数$f(x, y)$的表达式可知$F(t)$在数值上等于区域$x + y \le t$与正方形$0 3049 | \le x, y \le 1,$的公共部分的面积. 3050 | 3051 | 当$t \le 0$时，$F(t) = 0$; 3052 | 3053 | 当$0 < t \le 1$时，$F(t) = \frac{1}{2}t^2$; 3054 | 3055 | 当$1 < t \le 2$时，$F(t) = 1 - \frac{1}{2}(2 - t)^2$; 3056 | 3057 | 当$t > 2$时，$F(t) = 1.$ 3058 | 3059 | 综上所述， 3060 | \[ 3061 | F(t) = 3062 | \begin{dcases} 3063 | 0, & t \le 0,\\ 3064 | \frac{1}{2}t^2, & 0 < t \le 1,\\ 3065 | -\frac{1}{2}t^2 + 2t - 1, & 1 < t \le 2,\\ 3066 | 1, & t > 2. 3067 | \end{dcases} 3068 | \] 3069 | \end{solu} 3070 | \end{mathques} 3071 | 3072 | \begin{mathques}(2){2020-11-18}{二重积分}[半径忘记开平方] 3073 | \begin{ques} 3074 | 设函数$f(x, y)$连续，交换二次积分次序$\int_{0}^{1}\dd y\int_{\sqrt{y}} 3075 | ^{\sqrt{2 - y^2}}\dd x=$\mathblank. 3076 | \end{ques} 3077 | \begin{solu} 3078 | 积分区域用不等式表示为 3079 | \[ 3080 | 0 \le y \le 1\qc \sqrt{y} \le x \le \sqrt{2 - y^2} 3081 | \] 3082 | 求解 3083 | \[ 3084 | \begin{dcases} 3085 | \sqrt{y} = x,\\ 3086 | x = \sqrt{2 - y^2},\\ 3087 | \end{dcases} 3088 | \] 3089 | 得$x = 1, y = 1$ 3090 | 3091 | 当$0 \le x \le 1$时，$0 \le y \le x^2$；当$1 \le x \le \sqrt{2}$时，$0 \le y 3092 | \le \sqrt{2 - x^2}$. 3093 | 3094 | 故$\int_{0}^{1}\dd y\int_{\sqrt{y}} ^{\sqrt{2 - y^2}}\dd x = \int_{0}^{1}\dd 3095 | x\int_{0}^{x^2}f(x, y)\dd y + \int_{1}^{\sqrt{2}}\dd x\int_{0}^{\sqrt{2 - 3096 | x^2}}f(x, y)\dd y$. 3097 | \end{solu} 3098 | \end{mathques} 3099 | 3100 | \begin{mathques}(3){2020-11-16}{二重积分}[范围画错] 3101 | \begin{ques} 3102 | 设$f(u)$为连续函数，则二次积分$\int_{0}^{\frac{\pi}{2}}\dd 3103 | \theta\int_{2\cos\theta}^{2}f(r^2)r\dd r$在直角坐标系下化为二次积分 3104 | \mathblank. 3105 | \end{ques} 3106 | \begin{solu} 3107 | \answer{$\int_{0}^{2}\dd x\int_{\sqrt{2x - x^2}}^{\sqrt{4 - x^2}}f(x^2 + y^2) 3108 | \dd y$} 3109 | 3110 | 区域$D$在极坐标系下表示为 3111 | \[ 3112 | 0 \le \theta \le \frac{\pi}{2}\qc 2\cos\theta \le r \le 2, 3113 | \] 3114 | 则图像在第一象限. 3115 | 3116 | \matherror*{由$r = 2\cos\theta$可得$r^2 = 2r\cos\theta$，对应$x^2 + y^2 = 3117 | 2x$}. 3118 | 3119 | \matherror*{由$r = 2$有$r^2 = 4$，对应$x^2 + y^2 = 4$}. 3120 | 3121 | 故在直角坐标系下（要画图）可表示为 3122 | \[ 3123 | 0 \le x \le 2, \sqrt{2x - x^2} \le y \le \sqrt{4 - x^2}. 3124 | \] 3125 | 即 3126 | \[ 3127 | \text{原式} = \int_{0}^{2}\dd x\int_{\sqrt{2x - x^2}}^{\sqrt{4 - x^2}} 3128 | f(x^2 + y^2) \dd y 3129 | \] 3130 | \end{solu} 3131 | \end{mathques} 3132 | 3133 | \begin{mathques}{2020-11-08}{二重积分}[画图画错] 3134 | \begin{ques} 3135 | 设$f(x, y)$为连续函数，直角坐标系下的二次积分$\int_{0}^{1}\dd x\int_{0} 3136 | ^{\sqrt{x - x^2}}f(x, y)\dd y$可化为极坐标系下的二次积分\mathblank. 3137 | \end{ques} 3138 | \begin{solu} 3139 | \answer{$\int_{0}^{\frac{\pi}{2}}\dd \theta\int_{0}^{\cos\theta} 3140 | f(r\cos\theta, r\sin\theta)r\dd r$} 3141 | 3142 | 所给积分区域$D$为 3143 | \[ 3144 | 0 \le x \le 1, 0 \le y \le \sqrt{x - x^2}. 3145 | \] 3146 | $y = \sqrt{x - x^2}$可化为$\qty(x - \frac{1}{2})^2 + y^2 = \qty(\frac{1}{2}) 3147 | ^2$ 3148 | 3149 | 画出图形，又$y = \sqrt{x - x^2}$在极坐标系下化为$r = \cos\theta$，故积分区域 3150 | $D$可以表示为 3151 | \[ 3152 | 0 \le \theta \le \frac{\pi}{2}\qc 0 \le r \le \cos\theta 3153 | \] 3154 | 故$\int_{0}^{1}\dd x\int_{0}^{\sqrt{x - x^2}}f(x, y)\dd y = \int_{0} 3155 | ^{\frac{\pi}{2}}\dd \theta\int_{0}^{\cos\theta} f(r\cos\theta, r\sin\theta) 3156 | r\dd r$ 3157 | \end{solu} 3158 | \end{mathques} 3159 | \end{document} 3160 | -------------------------------------------------------------------------------- /preamble.tex: -------------------------------------------------------------------------------- 1 | \usepackage{xcolor} 2 | \usepackage{geometry} 3 | \usepackage{fancyhdr} 4 | \usepackage{etoc} 5 | \usepackage{tasks} 6 | \usepackage{tipa} 7 | \usepackage{xparse} 8 | \usepackage{multicol} 9 | \usepackage{enumitem} 10 | \usepackage{amsmath, amssymb, bm} 11 | \usepackage{mathtools} 12 | \usepackage{physics} 13 | \usepackage{tabularx} 14 | \usepackage{booktabs} 15 | \usepackage{extarrows} 16 | \usepackage{xeCJKfntef} 17 | \usepackage{tikz} 18 | \usepackage[most]{tcolorbox} 19 | \usepackage{stackengine} 20 | \usepackage[calc]{datetime2} 21 | \DTMsetdatestyle{iso} 22 | \usepackage[hidelinks]{hyperref} 23 | \hypersetup{ 24 | bookmarksopen, 25 | CJKbookmarks, 26 | } 27 | \usepackage{bookmark} 28 | \usepackage[xspace]{ellipsis} 29 | 30 | \setCJKmainfont{Source Han Serif SC} 31 | \setCJKsansfont{Source Han Sans SC} 32 | \newCJKfontfamily[kai]\kaiti{Adobe Kaiti Std R} 33 | 34 | %% toc style 35 | \newcommand*{\DotsAndPage}{% 36 | \nobreak\hskip 5pt\leaders\hbox{% 37 | \bfseries\normalsize\hbox to .75ex {\hss.\hss} 38 | }\hfill\nobreak 39 | \makebox{\large\bfseries\normalsize\itshape\etocpage}\par 40 | } 41 | \etocsetstyle{section} 42 | {} 43 | {\leftskip 0pt \rightskip .75cm \parfillskip-\rightskip 44 | \pagebreak[1]\smallskip} 45 | {\Large\rmfamily\bfseries\scshape{\etocname}\DotsAndPage} 46 | {} 47 | \etocsetstyle{subsection} 48 | {} 49 | {\leftskip 1cm \rightskip .75cm \parfillskip-\rightskip 50 | \pagebreak[1]\smallskip} 51 | {\large\rmfamily\bfseries\scshape{\color{teal}\etocname}\DotsAndPage} 52 | {} 53 | \renewcommand{\etoccolumnsep}{4em} 54 | \etocmulticolstyle[3]{ 55 | \noindent\bfseries\Large 56 | \leaders\hrule height1pt\hfill 57 | \MakeUppercase{Table of Contents} 58 | } 59 | 60 | \ctexset{ 61 | section/format=\raggedright\bfseries\huge, 62 | } 63 | 64 | \newcolumntype{L}{>{\raggedright\arraybackslash}X} 65 | \newcolumntype{R}{>{\raggedleft\arraybackslash}X} 66 | 67 | \renewcommand{\ellipsisgap}{0.1em} 68 | \setlength\parindent{0pt} 69 | 70 | \newtcbox{\reviewtoday}{ 71 | colframe=red!20, colback=yellow!10, fontupper=\small\itshape, 72 | size=small, 73 | } 74 | 75 | % \ExplSyntaxOn 76 | % \NewDocumentCommand{\defineclist}{mm} 77 | % { 78 | % \clist_clear_new:c { l__#1_clist } 79 | % \clist_set:cn { l__#1_clist } { #2 } 80 | % } 81 | % \ExplSyntaxOff 82 | 83 | \makeatletter 84 | \ExplSyntaxOn 85 | \cs_new_protected:Npn \custom_label:n #1 86 | { 87 | \iow_shipout:Nx \@auxout { \token_to_str:N \newlabel{#1}{{}{\thepage}{}{#1}{}} } 88 | \hypertarget{#1}{} 89 | } 90 | 91 | \NewDocumentCommand {\customlabel} {m} 92 | { 93 | \custom_label:n {#1} 94 | } 95 | \ExplSyntaxOff 96 | \makeatother 97 | \endinput 98 | -------------------------------------------------------------------------------- /review.code.tex: -------------------------------------------------------------------------------- 1 | \makeatletter 2 | \renewrobustcmd*{\DTMsaveddatediff}[3]{% 3 | \ifcsundef{@dtm@#1@year}% 4 | {% 5 | \PackageError{datetime2-calc}{Unknown date `#1'}{}% 6 | }% 7 | {% 8 | \ifcsundef{@dtm@#2@year}% 9 | {% 10 | \PackageError{datetime2-calc}{Unknown date `#1'}{}% 11 | }% 12 | {% 13 | \pgfcalendardatetojulian 14 | {\csname @dtm@#1@year\endcsname 15 | -\csname @dtm@#1@month\endcsname 16 | -\csname @dtm@#1@day\endcsname} 17 | {#3}% 18 | \pgfcalendardatetojulian 19 | {\csname @dtm@#2@year\endcsname 20 | -\csname @dtm@#2@month\endcsname 21 | -\csname @dtm@#2@day\endcsname} 22 | {\@dtm@julianday}% 23 | \advance#3 by -\@dtm@julianday\relax 24 | }% 25 | }% 26 | } 27 | \makeatother 28 | 29 | \ExplSyntaxOn 30 | 31 | \newcount\datediff 32 | 33 | \bool_new:N \g_review_bool 34 | \bool_new:N \g__review_today_bool 35 | \tl_new:N \g__review_date_tl 36 | 37 | \bool_gset_true:N \g__review_today_bool 38 | \tl_gset:Nn \g__review_date_tl { \today } 39 | 40 | \NewDocumentCommand { \setdate } { m } 41 | { 42 | \tl_gset:Nn \g__review_date_tl { #1 } 43 | % \seq_gclear_new:N \g__review_date_seq 44 | \bool_gset_false:N \g__review_today_bool 45 | } 46 | 47 | % #1 clist #2 date to be compared 48 | \prg_new_protected_conditional:Nnn \review_if:Nn {T, F, TF} 49 | { 50 | \DTMsavedate{recent}{#2} 51 | \bool_if:NTF \g__review_today_bool 52 | { 53 | \DTMsavenow{today} 54 | } 55 | { 56 | \DTMsavedate{today}{\g__review_date_tl} 57 | } 58 | \DTMsaveddatediff{today}{recent}{\datediff}% specious blank 59 | \bool_gset_false:N \g_review_bool 60 | \clist_map_inline:Nn #1 61 | { 62 | \int_compare:nT {\datediff == ##1} 63 | { 64 | \bool_gset_true:N \g_review_bool 65 | \clist_map_break: 66 | } 67 | } 68 | \bool_if:nTF {\g_review_bool} 69 | {\prg_return_true:} 70 | {\prg_return_false:} 71 | } 72 | \ExplSyntaxOff 73 | \endinput 74 | -------------------------------------------------------------------------------- /rv-utils.sty: -------------------------------------------------------------------------------- 1 | \RequirePackage{expl3} 2 | \ProvidesExplPackage 3 | {rv-utils} 4 | {2020/09/26} 5 | {v1.0} 6 | {rv-utils} 7 | 8 | \cs_generate_variant:Nn \rv_do_kv:nn { nx, xn, xx } 9 | 10 | \cs_new_protected:Npn \rv_parse_kv:w #1=#2\rv_nil 11 | { 12 | \tl_clear_new:N \rv_ra 13 | \tl_clear_new:N \rv_rb 14 | \tl_set:Nn \rv_ra { #1 } 15 | \tl_set:Nn \rv_rb { #2 } 16 | } 17 | 18 | % #1 key=val #2 code 19 | \cs_new_protected:Nn \rv_do_kv:nn 20 | { 21 | \cs_set_protected:Npn \__rv_do_kv_code:nn ##1 ##2 { #2 } 22 | \cs_generate_variant:Nn \__rv_do_kv_code:nn { oo } 23 | \rv_parse_kv:w #1\rv_nil 24 | \__rv_do_kv_code:oo { \rv_ra } { \rv_rb } 25 | } 26 | 27 | \cs_new_protected:Nn \rv_do_kv_variable:nNNn 28 | { 29 | \cs_set_protected:Npn \__rv_do_kv_variable_code: { #4 } 30 | \rv_parse_kv:w #1\rv_nil 31 | \tl_clear_new:N #2 32 | \tl_clear_new:N #3 33 | \tl_set_eq:NN #2 \rv_ra 34 | \tl_set_eq:NN #3 \rv_rb 35 | \__rv_do_kv_variable_code: 36 | } 37 | 38 | \file_input_stop: 39 | -------------------------------------------------------------------------------- /sdutil.sty: -------------------------------------------------------------------------------- 1 | %% 2 | %% This is file `sdutil.sty', 3 | %% generated with the docstrip utility. 4 | %% 5 | %% The original source files were: 6 | %% 7 | %% showdata.dtx (with options: `sdutil') 8 | %% 9 | %% Copyright (C) 2020 Li Changkai 10 | %% 11 | %% This file may be distributed and/or modified under the 12 | %% conditions of the LaTeX Project Public License, either version 1.3 13 | %% of this license or (at your option) any later version. 14 | %% The latest version of this license is in: 15 | %% 16 | %% http://www.latex-project.org/lppl.txt 17 | %% 18 | %% and version 1.3 or later is part of all distributions of LaTeX 19 | %% version 2005/12/01 or later. 20 | %% 21 | \def\myfileversion{1.0} 22 | \def\myfiledate{2020/10/12} 23 | \RequirePackage[calc]{datetime2} 24 | \RequirePackage{expl3} 25 | \ProvidesExplPackage 26 | { sdutil } 27 | { \myfiledate } 28 | { \myfileversion } 29 | { Provide common functions for showdata } 30 | \newcount\l__sdutil_date_diff 31 | \bool_new:N \l__sdutil_review_bool 32 | \cs_generate_variant:Nn \seq_set_split:Nnn { Nno } 33 | \prg_generate_conditional_variant:Nnn \seq_if_in:Nn { Nx, cx } { T, F, TF, p } 34 | \prg_generate_conditional_variant:Nnn \clist_if_in:Nn { Nx, cx } { T, F, TF, p } 35 | \cs_new_protected:Npn \sdutil_parse_kv:w #1=#2\sd_nil 36 | { 37 | \tl_clear_new:N \l__sdutil_tmpa_tl 38 | \tl_clear_new:N \l__sdutil_tmpb_tl 39 | \tl_set:Nn \l__sdutil_tmpa_tl { #1 } 40 | \tl_set:Nn \l__sdutil_tmpb_tl { #2 } 41 | } 42 | \cs_new_protected:Nn \sdutil_get_kv:nNN 43 | { 44 | \sdutil_parse_kv:w #1\sd_nil 45 | \tl_set_eq:NN #2 \l__sdutil_tmpa_tl 46 | \tl_set_eq:NN #3 \l__sdutil_tmpb_tl 47 | } 48 | \cs_new_protected:Nn \sdutil_do_kv:nn 49 | { 50 | \cs_set_protected:Npn \__sdutil_do_kv_code:nn ##1 ##2 { #2 } 51 | \cs_generate_variant:Nn \__sdutil_do_kv_code:nn { VV } 52 | \sdutil_parse_kv:w #1\rv_nil 53 | \__sdutil_do_kv_code:VV \l__sdutil_tmpa_tl \l__sdutil_tmpb_tl 54 | } 55 | \cs_generate_variant:Nn \sdutil_do_kv:nn { nx, xn, xx } 56 | \cs_new_protected:Nn \sdutil_do_kv_variable:nNNn 57 | { 58 | \cs_set_protected:Npn \__sdutil_do_kv_variable_code: { #4 } 59 | \sdutil_parse_kv:w #1\rv_nil 60 | \tl_clear_new:N #2 61 | \tl_clear_new:N #3 62 | \tl_set_eq:NN #2 \l__sdutil_tmpa_tl 63 | \tl_set_eq:NN #3 \l__sdutil_tmpb_tl 64 | \__sdutil_do_kv_variable_code: 65 | } 66 | \cs_new_protected:Nn \sdutil_seq_append_clist:Nn 67 | { 68 | \seq_clear_new:N \l__sdutil_tmpa_seq 69 | \seq_set_from_clist:Nn \l__sdutil_tmpa_seq { #2 } 70 | \seq_concat:NNN #1 #1 \l__sdutil_tmpa_seq 71 | } 72 | \cs_generate_variant:Nn \sdutil_seq_append_clist:Nn { cn } 73 | \cs_new_protected:Nn \sdutil_seq_gappend_clist:Nn 74 | { 75 | \seq_clear_new:N \l__sdutil_tmpa_seq 76 | \seq_set_from_clist:Nn \l__sdutil_tmpa_seq { #2 } 77 | \seq_gconcat:NNN #1 #1 \l__sdutil_tmpa_seq 78 | } 79 | \cs_generate_variant:Nn \sdutil_seq_gappend_clist:Nn { cn } 80 | \prg_new_conditional:Nnn \sdutil_if_review:Nnn { T, F, TF, p } 81 | { 82 | \DTMsavedate{recent}{#2} 83 | \DTMsavedate{today}{#3} 84 | \DTMsaveddatediff{today}{recent}{\l__sdutil_date_diff}% specious blank 85 | \bool_set_false:N \l__sdutil_review_bool 86 | \seq_if_in:NxTF #1 { \the\l__sdutil_date_diff } 87 | { \prg_return_true: } 88 | { \prg_return_false: } 89 | } 90 | \prg_generate_conditional_variant:Nnn \sdutil_if_review:Nnn { cnn } { T, F, TF, p } 91 | \cs_new_protected:Nn \sdutil_regex_replace_all:nnnN 92 | { 93 | \tl_set:Nn #4 { #3 } 94 | \regex_replace_all:nnN { #1 } { #2 } #4 95 | } 96 | \cs_new_protected:Nn \sdutil_set_msg_cs:n 97 | { 98 | \str_if_in:nnTF { #1 } { = } 99 | { \sdutil_get_kv:nNN { #1 } } 100 | { \sdutil_get_kv:nNN { #1=#1 } } 101 | \l_tmpa_tl \l_tmpb_tl 102 | \regex_replace_all:nnN { - } { _ } \l_tmpb_tl 103 | \cs_gset_protected:cn { __\l_tmpb_tl _msg_new:nn } 104 | { 105 | \exp_args:NV \msg_new:nnn \l_tmpa_tl { ##1 } { ##2 } 106 | } 107 | \cs_gset_protected:cn { __\l_tmpb_tl _msg_new:nnn } 108 | { 109 | \exp_args:NV \msg_new:nnnn \l_tmpa_tl { ##1 } { ##2 } { ##3 } 110 | } 111 | \clist_map_inline:nn { fatal, critical, error, warning, info, log, none } 112 | { 113 | \__sdutil_set_msg_issue_cs:VVn \l_tmpa_tl \l_tmpb_tl { ##1 } 114 | } 115 | \cs_gset:Nn \msg_fatal_text:n { Package~ ##1~ Error } 116 | \cs_gset:Nn \msg_critical_text:n { Package~ ##1~ Error } 117 | } 118 | \cs_new_protected:Nn \__sdutil_set_msg_issue_cs:nnn 119 | { 120 | \cs_gset_protected:cn { __#2_msg_#3:n } 121 | { 122 | \cs_if_exist_use:c { msg_#3:nn } { #1 } { ##1 } 123 | } 124 | \cs_gset_protected:cn { __#2_msg_#3:nn } 125 | { 126 | \cs_if_exist_use:c { msg_#3:nnn } { #1 } { ##1 } { ##2 } 127 | } 128 | \cs_gset_protected:cn { __#2_msg_#3:nnn } 129 | { 130 | \cs_if_exist_use:c { msg_#3:nnnn } { #1 } { ##1 } { ##2 } { ##3 } 131 | } 132 | \cs_gset_protected:cn { __#2_msg_#3:nnnn } 133 | { 134 | \cs_if_exist_use:c { msg_#3:nnnnn } { #1 } { ##1 } 135 | { ##2 } { ##3 } { ##4 } 136 | } 137 | \cs_gset_protected:cn { __#2_msg_#3:nnnnn } 138 | { 139 | \cs_if_exist_use:c { msg_#3:nnnnnn } { #1 } { ##1 } 140 | { ##2 } { ##3 } { ##4 } { ##5 } 141 | } 142 | \cs_generate_variant:cn { __#2_msg_#3:nn } { nx } 143 | \cs_generate_variant:cn { __#2_msg_#3:nnn } { nxx } 144 | \cs_generate_variant:cn { __#2_msg_#3:nnnn } { nxxx } 145 | \cs_generate_variant:cn { __#2_msg_#3:nnnnn } { nxxxx } 146 | } 147 | \cs_generate_variant:Nn \__sdutil_set_msg_issue_cs:nnn { VVn } 148 | \cs_set_eq:NN \IfBlankTF \tl_if_blank:nTF 149 | \cs_set_eq:NN \IfBlankT \tl_if_blank:nT 150 | \cs_set_eq:NN \IfBlankF \tl_if_blank:nF 151 | \endinput 152 | %% 153 | %% End of file `sdutil.sty'. 154 | -------------------------------------------------------------------------------- /showdata.sty: -------------------------------------------------------------------------------- 1 | %% 2 | %% This is file `showdata.sty', 3 | %% generated with the docstrip utility. 4 | %% 5 | %% The original source files were: 6 | %% 7 | %% showdata.dtx (with options: `package') 8 | %% 9 | %% Copyright (C) 2020 Li Changkai 10 | %% 11 | %% This file may be distributed and/or modified under the 12 | %% conditions of the LaTeX Project Public License, either version 1.3 13 | %% of this license or (at your option) any later version. 14 | %% The latest version of this license is in: 15 | %% 16 | %% http://www.latex-project.org/lppl.txt 17 | %% 18 | %% and version 1.3 or later is part of all distributions of LaTeX 19 | %% version 2005/12/01 or later. 20 | %% 21 | \def\myfileversion{1.0} 22 | \def\myfiledate{2020/10/12} 23 | \RequirePackage{expl3, xparse, sdutil} 24 | \ProvidesExplPackage 25 | { showdata } 26 | { \myfiledate } 27 | { \myfileversion } 28 | { Store and show your data } 29 | \seq_new:N \g__showdata_domain_seq 30 | \seq_set_from_clist:Nn \g__showdata_reserved_counter_seq 31 | { sdenv, sdgroup, sditem } 32 | \prop_gset_from_keyval:Nn \g__showdata_reserved_counter_prop 33 | { sdenv=, sdgroup=, sditem= } 34 | \sdutil_set_msg_cs:n { showdata } 35 | \__showdata_msg_new:nn { domain-not-set } 36 | { 37 | Domains~ are~ not~ set~ or~ empty.~ Please~ set~ domains~ by~ command~ 38 | \sdsetdomain~ first. 39 | } 40 | \__showdata_msg_new:nn { sd-not-define } 41 | { 42 | #4~ uses~ an~ undefined~ #2~ `#3'~ of~ domain~ `#1'~ \msg_line_context:. 43 | } 44 | \__showdata_msg_new:nn { domain-not-define } 45 | { 46 | domain `#1'~ not~ defined! 47 | } 48 | \__showdata_msg_new:nn { reserved-counter } 49 | { 50 | counter~ name~ `#1'~ is~ reserved~ for~ package. 51 | } 52 | \__showdata_msg_new:nn { counter-read-only } 53 | { 54 | reserved counter~ `#1'~ is~ read-only. 55 | } 56 | \__showdata_msg_new:nn { counter-use-only } 57 | { 58 | counter~ `#1'~ is~ used~ only~ in~ #2. 59 | } 60 | \__showdata_msg_new:nn { addto-multi-times } 61 | { 62 | You~ have~ used~ \c_backslash_str sdaddtogroup~ to~ append~ subgroup~ to~ 63 | the~ local~ group~ `#2'~ more~ than~ once~ in~ batch~ environment~ of~ 64 | domain~ `#1'.~ This~ will~ cause~ the~ same~ subgroup~ to~ be~ printed~ 65 | multiple~ times. 66 | } 67 | \__showdata_msg_new:nn { step-set-not-allowed } 68 | { 69 | \c_backslash_str #1~ is~ not~ allowed~ in~ \c_backslash_str sdsetshowstyle. 70 | } 71 | \__showdata_msg_new:nn { show-counter-use-only } 72 | { 73 | Only~ counters~ `sditem'~ and~ `sdgroup'~ are~ available~ in~ \c_backslash_str 74 | sdsetshowstle. 75 | } 76 | \__showdata_msg_new:nn { show-global-ignore-optional } 77 | { 78 | optional~ argument~ of~ \c_backslash_str sduse~ in~ \c_backslash_str 79 | sdsetshowstyle*~ is~ ignored. 80 | } 81 | \__showdata_msg_new:nn { item-index-out-of-range } 82 | { 83 | index~ out~ of~ range~ when~ requiring~ the~ item~ of~ property~ `#4'~ of~ 84 | #2~ group~ `#3'~ in~ domain~ `#1',~ nothing~ returned. 85 | } 86 | \__showdata_msg_new:nn { global-item-index-out-of-range-detail } 87 | { 88 | property~ `#2'~ of~ global~ group~ `#1'~ has~ #3~ items~ but~ you~ are~ 89 | requiring~ the~ item~ of~ index~ #4. 90 | \\ \\ 91 | Please~ make~ sure~ there~ are~ an~ equal~ number~ of~ items~ of~ 92 | properties~ in~ global~ group~ `#1'. 93 | } 94 | 95 | \__showdata_msg_new:nn { local-item-index-out-of-range-detail } 96 | { 97 | subgroup of property~ `#2'~ of~ local~ group~ `#1'~ has~ #3~ items~ but~ you~ are~ 98 | requiring~ the~ item~ of~ index~ #4. 99 | \\ \\ 100 | Please~ make~ sure~ there~ are~ an~ equal~ number~ of~ items~ of~ 101 | properties~ in~ local~ group~ `#1'. 102 | } 103 | \cs_new_protected:Nn \__showdata_check_undefined:nnnn 104 | { 105 | \str_if_in:nnTF { #2 } { = } 106 | { \sdutil_get_kv:nNN { #2 } } 107 | { \sdutil_get_kv:nNN { #2=#2 } } 108 | \l_tmpa_tl \l_tmpb_tl 109 | \regex_replace_all:nnN { \s } { _ } \l_tmpa_tl 110 | \seq_if_in:cnF { g__showdata_#1_\l_tmpa_tl _seq } { #3 } 111 | { 112 | \sdutil_regex_replace_all:nnnN 113 | { . } { \c{string} \0 } { #4 } \l_tmpc_tl 114 | \__showdata_msg_error:nxxxx { sd-not-define } 115 | { #1 } { \l_tmpb_tl } { #3 } { \l_tmpc_tl } 116 | } 117 | } 118 | \cs_new_protected:Nn \__showdata_check_domain_not_set: 119 | { 120 | \bool_if:nT 121 | { 122 | ! \seq_if_exist_p:N \g__showdata_domain_seq || 123 | \seq_if_empty_p:N \g__showdata_domain_seq 124 | } 125 | { \__showdata_msg_critical:n { domain-not-set } } 126 | } 127 | \cs_new_protected:Nn \__showdata_check_domain_not_define:nn 128 | { 129 | \seq_if_in:NnF \g__showdata_domain_seq { #1 } 130 | { 131 | \cs_if_exist_use:c { __showdata_msg_#2:nn } 132 | { domain-not-define } { #1 } 133 | } 134 | } 135 | \cs_new_protected:Nn \__showdata_check_reserved_counter:n 136 | { 137 | \prop_if_in:NnT \g__showdata_reserved_counter_prop { #1 } 138 | { 139 | \__showdata_msg_error:nn { reserved-counter } { #1 } 140 | } 141 | } 142 | \cs_new_protected:Nn \__showdata_check_counter_use:nn 143 | { 144 | \str_if_eq:nnF { #1 } { #2 } 145 | { 146 | \__showdata_check_counter_use_only:n { #1 } 147 | } 148 | } 149 | \cs_new_protected:Nn \__showdata_check_counter_use_only:n 150 | { 151 | \str_case:nn { #1 } 152 | { 153 | { sdenv } 154 | { 155 | \__showdata_msg_error:nnn { counter-use-only } 156 | { #1 } { group~environment } 157 | } 158 | { sdgroup } 159 | { 160 | \__showdata_msg_error:nxx { counter-use-only } 161 | { #1 } { \c_backslash_str sdsetshowstyle } 162 | } 163 | { sditem } 164 | { 165 | \__showdata_msg_error:nxx { counter-use-only } { #1 } 166 | { 167 | \c_backslash_str sdsetshowstyle~ or~ 168 | \c_backslash_str sdsetshowstyle* 169 | } 170 | } 171 | } 172 | } 173 | \cs_new_protected:Nn \__showdata_check_couter_not_define:nnnn 174 | { 175 | \prop_if_in:cnF { g__showdata_#1_#2_counter_prop } { #3 } 176 | { 177 | \str_if_eq:nnTF { #2 } { global } 178 | { 179 | \__showdata_msg_error:nxxxx { sd-not-define } { #1 } 180 | { #2~counter } { #3 } { \c_backslash_str#4* } 181 | } 182 | { 183 | \__showdata_msg_error:nxxxx { sd-not-define } { #1 } 184 | { #2~counter } { #3 } { \c_backslash_str#4 } 185 | } 186 | } 187 | } 188 | \cs_new_protected:Nn \__showdata_check_group_add_once:nn 189 | { 190 | \int_compare:nT 191 | { 192 | \int_use:c 193 | { g__showdata_#1_addto_local_group_#2_counter_int } 194 | > 1 195 | } 196 | { 197 | \__showdata_msg_warning:nnn { addto-multi-times } { #1 } { #2 } 198 | } 199 | } 200 | \cs_new_protected:Nn \__showdata_check_counter_not_allowed:nn 201 | { 202 | \str_if_eq:nnT { #1 } { show } 203 | { 204 | \__showdata_msg_fatal:nn { step-set-not-allowed } { #2 } 205 | } 206 | } 207 | \cs_new_protected:Nn \__showdata_check_show_counter_use:nn 208 | { 209 | \bool_if:nT 210 | { 211 | \str_if_eq_p:nn { #1 } { show } && 212 | ! \str_if_eq_p:nn { #2 } { sditem } && 213 | ! \str_if_eq_p:nn { #2 } { sdgroup } 214 | } 215 | { 216 | \__showdata_msg_fatal:n { show-counter-use-only } 217 | } 218 | } 219 | \cs_new_protected:Nn \__showdata_check_invalid_index:nnnnnn 220 | { 221 | \int_compare:nT 222 | { \int_use:c { #6 } > \seq_count:c { #5 } } 223 | { 224 | \__showdata_msg_warning:nnnnn { item-index-out-of-range } 225 | { #1 } { #2 } { #3 } { #4 } 226 | 227 | \__showdata_msg_warning:nxxxx { #2-item-index-out-of-range-detail } 228 | { #3 } { #4 } { \seq_count:c { #5 } } { \int_use:c { #6 } } 229 | } 230 | } 231 | \NewDocumentCommand { \sdsetdomain } { m } 232 | { 233 | \__showdata_set_domain:n { #1 } 234 | } 235 | \cs_new_protected:Nn \__showdata_set_domain:n 236 | { 237 | \seq_set_from_clist:Nn \g__showdata_domain_seq { #1 } 238 | \clist_map_inline:nn { #1 } 239 | { \__showdata_set_single_domain:n { ##1 } } 240 | } 241 | \cs_new_protected:Nn \__showdata_set_single_domain:n 242 | { 243 | \__showdata_init_single_domain_attr:n { #1 } 244 | \__showdata_def_single_domain_opt:n { #1 } 245 | } 246 | \cs_new_protected:Nn \__showdata_init_single_domain_attr:n 247 | { 248 | \seq_gset_from_clist:cn { g__showdata_#1_global_group_seq } 249 | { star, date, all } 250 | \seq_gset_from_clist:cn { g__showdata_#1_local_group_seq } 251 | { date, all } 252 | \seq_gclear_new:c { g__showdata_#1_global_counter_seq } 253 | \seq_gclear_new:c { g__showdata_#1_local_counter_seq } 254 | \prop_gclear_new:c { g__showdata_#1_global_counter_prop } 255 | \prop_gclear_new:c { g__showdata_#1_local_counter_prop } 256 | \seq_gclear_new:c { g__showdata_#1_prop_seq } 257 | } 258 | \cs_new_protected:Nn \__showdata_def_single_domain_opt:n 259 | { 260 | \keys_define:nn { showdata / #1 } 261 | { 262 | prop .code:n = 263 | { 264 | \sdutil_seq_gappend_clist:cn 265 | { g__showdata_#1_prop_seq } { ##1 } 266 | }, 267 | local-group .code:n = 268 | { 269 | \sdutil_seq_gappend_clist:cn 270 | { g__showdata_#1_local_group_seq } { ##1 } 271 | }, 272 | global-group .code:n = 273 | { 274 | \sdutil_seq_gappend_clist:cn 275 | { g__showdata_#1_global_group_seq } { ##1 } 276 | }, 277 | group .code:n = 278 | { 279 | \sdutil_seq_gappend_clist:cn 280 | { g__showdata_#1_global_group_seq } { ##1 } 281 | \sdutil_seq_gappend_clist:cn 282 | { g__showdata_#1_local_group_seq } { ##1 } 283 | }, 284 | local-counter .code:n = 285 | { 286 | \__showdata_counter_seq_append:nnn { #1 } { local } { ##1 } 287 | }, 288 | global-counter .code:n = 289 | { 290 | \__showdata_counter_seq_append:nnn { #1 } { global } { ##1 } 291 | }, 292 | counter .code:n = 293 | { 294 | \__showdata_counter_seq_append:nnn { #1 } { local } { ##1 } 295 | \__showdata_counter_seq_append:nnn { #1 } { global } { ##1 } 296 | }, 297 | } 298 | } 299 | \cs_new_protected:Nn \__showdata_counter_seq_append:nnn 300 | { 301 | \clist_map_inline:nn { #3 } 302 | { 303 | \seq_gput_right:cn { g__showdata_#1_#2_counter_seq } { ##1 } 304 | \prop_gput:cnn { g__showdata_#1_#2_counter_prop } { ##1 } { } 305 | } 306 | } 307 | \NewDocumentCommand { \sdsetup } { m } 308 | { 309 | \__showdata_check_domain_not_set: 310 | \keys_set:nn { showdata } { #1 } 311 | } 312 | \NewDocumentCommand { \sdsetupsub } { m m } 313 | { 314 | \__showdata_check_domain_not_set: 315 | \clist_map_inline:nn { #1 } 316 | { \__showdata_setup_single:nn { ##1 } { #2 } } 317 | } 318 | \NewDocumentCommand { \sdsetupall } { m } 319 | { 320 | \__showdata_check_domain_not_set: 321 | \seq_map_inline:Nn \g__showdata_domain_seq 322 | { \__showdata_setup_single:nn { ##1 } { #1 } } 323 | } 324 | \cs_new_protected:Nn \__showdata_setup_single:nn 325 | { 326 | \__showdata_check_domain_not_define:nn { #1 } { critical } 327 | \keys_set:nn { showdata / #1 } { #2 } 328 | } 329 | \clist_map_inline:nn { New, Renew, Declare, Provide } 330 | { 331 | \cs_new_protected:cpn { #1sdBatchEnvironment } ##1##2##3##4 332 | { 333 | \__showdata_new_batch_env:nnnnn { #1 } { ##1 } { ##2 } { ##3} { ##4 } 334 | } 335 | } 336 | \cs_new_protected:Nn \__showdata_new_batch_env:nnnnn 337 | { 338 | \__showdata_check_domain_not_set: 339 | \tl_set:Nn \l__showdata_env_tmpa_tl { #4 } 340 | \tl_set:Nn \l__showdata_env_tmpb_tl { #5 } 341 | \clist_map_inline:nn { #2 } 342 | { 343 | \__showdata_parse_single_batch_env:nnnVV { #1 } { ##1 } { #3 } 344 | \l__showdata_env_tmpa_tl 345 | \l__showdata_env_tmpb_tl 346 | } 347 | } 348 | \cs_new_protected:Nn \__showdata_parse_single_batch_env:nnnnn 349 | { 350 | \str_if_in:nnTF { #2 } { = } 351 | { \sdutil_get_kv:nNN { #2 } } 352 | { \sdutil_get_kv:nNN { #2=sd#2 } } 353 | \l__showdata_domain_tl \l__showdata_env_tl 354 | \__showdata_init_before_env:V \l__showdata_domain_tl 355 | \__showdata_set_single_batch_env:nVVnnn { #1 } 356 | \l__showdata_domain_tl \l__showdata_env_tl 357 | { #3 } { #4 } { #5 } 358 | } 359 | \cs_generate_variant:Nn \__showdata_parse_single_batch_env:nnnnn { nnnVV } 360 | \cs_new_protected:Nn \__showdata_init_before_env:n 361 | { 362 | \__showdata_check_domain_not_define:nn { #1 } { critical } 363 | \__showdata_init_global_group:n { #1 } 364 | \__showdata_add_default_counter:n { #1 } 365 | \__showdata_init_counter:nnn { #1 } { env } { global } 366 | \int_gzero_new:c { g__showdata_#1_reserved_counter_sdenv_int } 367 | \__showdata_init_index_seq:n { #1 } 368 | } 369 | \cs_generate_variant:Nn \__showdata_init_before_env:n { V } 370 | \cs_new_protected:Nn \__showdata_init_index_seq:n 371 | { 372 | \seq_map_inline:cn { g__showdata_#1_local_group_seq } 373 | { 374 | \seq_gclear_new:c { g__showdata_#1_subgroup_##1_index_seq } 375 | } 376 | } 377 | \cs_new_protected:Nn \__showdata_init_global_group:n 378 | { 379 | \seq_map_inline:cn { g__showdata_#1_global_group_seq } 380 | { 381 | \seq_map_inline:cn { g__showdata_#1_prop_seq } 382 | { 383 | \seq_gclear_new:c { g__showdata_#1_global_group_##1_prop_####1_seq } 384 | } 385 | } 386 | } 387 | \cs_new_protected:Nn \__showdata_add_default_counter:n 388 | { 389 | \__showdata_set_group_default_counter:nnn { #1 } { global } { global } 390 | \__showdata_set_group_default_counter:nnn { #1 } { global } { local } 391 | \__showdata_set_group_default_counter:nnn { #1 } { local } { local } 392 | } 393 | \cs_new_protected:Nn \__showdata_set_group_default_counter:nnn 394 | { 395 | \seq_map_inline:cn { g__showdata_#1_#3_group_seq } 396 | { 397 | \seq_gput_left:cn { g__showdata_#1_#2_counter_seq } { ##1 } 398 | \prop_gput:cnn { g__showdata_#1_#2_counter_prop } { ##1 } { } 399 | } 400 | } 401 | \cs_new_protected:Nn \__showdata_init_counter:nnn 402 | { 403 | \seq_map_inline:cn { g__showdata_#1_#3_counter_seq } 404 | { 405 | \__showdata_check_reserved_counter:n { ##1 } 406 | \int_gzero_new:c { g__showdata_#1_#2_#3_counter_##1_int } 407 | } 408 | } 409 | \cs_new_protected:Nn \__showdata_set_single_batch_env:nnnnnn 410 | { 411 | \cs_if_exist_use:c { #1DocumentEnvironment } { #3 } { #4 } 412 | { 413 | \__showdata_env_init:n { #2 } 414 | #5 415 | } { #6 } 416 | } 417 | \cs_generate_variant:Nn \__showdata_set_single_batch_env:nnnnnn { nVVnnn } 418 | \cs_new_protected:Nn \__showdata_env_init:n 419 | { 420 | % step counter first 421 | \int_gincr:c { g__showdata_#1_reserved_counter_sdenv_int } 422 | \__showdata_env_init_local_group:n { #1 } 423 | \__showdata_init_counter:nnn { #1 } { env } { local } 424 | \__showdata_env_init_cs:n { #1 } 425 | } 426 | \cs_new_protected:Nn \__showdata_env_init_local_group:n 427 | { 428 | \seq_map_inline:cn { g__showdata_#1_local_group_seq } 429 | { 430 | \seq_map_inline:cn { g__showdata_#1_prop_seq } 431 | { 432 | \seq_gclear_new:c 433 | { 434 | g__showdata_#1_local_group_##1_prop_####1_ 435 | \int_use:c { g__showdata_#1_reserved_counter_sdenv_int } 436 | _seq 437 | } 438 | } 439 | } 440 | } 441 | \cs_new_protected:Nn \__showdata_env_init_cs:n 442 | { 443 | \__showdata_env_init_cs_save:n { #1 } 444 | \__showdata_init_cs_addto_group:n { #1 } 445 | \__showdata_init_cs_counter:nn { #1 } { env } 446 | } 447 | \cs_new_protected:Nn \__showdata_env_init_cs_save:n 448 | { 449 | \DeclareDocumentCommand { \sdsave } { s m m +m } 450 | { 451 | \IfBooleanTF { ##1 } 452 | { 453 | \__showdata_save_global_group_prop:nnnn 454 | { #1 } { ##2 } { ##3 } { ##4 } 455 | } 456 | { 457 | \__showdata_save_local_group_prop:nnnn 458 | { #1 } { ##2 } { ##3 } { ##4 } 459 | } 460 | } 461 | \DeclareDocumentCommand { \sdesave } { s m m +m } 462 | { 463 | \IfBooleanTF { ##1 } 464 | { 465 | \__showdata_save_global_group_prop:nnnx 466 | { #1 } { ##2 } { ##3 } { ##4 } 467 | } 468 | { 469 | \__showdata_save_local_group_prop:nnnx 470 | { #1 } { ##2 } { ##3 } { ##4 } 471 | } 472 | } 473 | } 474 | \cs_new_protected:Nn \__showdata_save_global_group_prop:nnnn 475 | { 476 | \__showdata_check_undefined:nnnn 477 | { #1 } { global~group } { #2 } { \sdsave*~or~\sdesave* } 478 | \__showdata_check_undefined:nnnn 479 | { #1 } { prop=property } { #3 } { \sdsave*~or~\sdesave* } 480 | \seq_gput_right:cn { g__showdata_#1_global_group_#2_prop_#3_seq } { #4 } 481 | } 482 | \cs_generate_variant:Nn \__showdata_save_global_group_prop:nnnn { nnnx } 483 | \cs_new_protected:Nn \__showdata_save_local_group_prop:nnnn 484 | { 485 | \__showdata_check_undefined:nnnn 486 | { #1 } { local~group } { #2 } { \sdsave~or~\sdesave } 487 | \__showdata_check_undefined:nnnn 488 | { #1 } { prop=property } { #3 } { \sdsave~or~\sdesave } 489 | \seq_gput_right:cn 490 | { 491 | g__showdata_#1_local_group_#2_prop_#3_ 492 | \int_use:c { g__showdata_#1_reserved_counter_sdenv_int } 493 | _seq 494 | } { #4 } 495 | } 496 | \cs_generate_variant:Nn \__showdata_save_local_group_prop:nnnn { nnnx } 497 | \cs_new_protected:Nn \__showdata_init_cs_addto_group:n 498 | { 499 | \seq_map_inline:cn { g__showdata_#1_local_group_seq } 500 | { 501 | \int_gzero_new:c { g__showdata_#1_addto_local_group_##1_counter_int } 502 | } 503 | \DeclareDocumentCommand { \sdaddtogroup } { m } 504 | { \__showdata_addto_group:nn { #1 } { ##1 } } 505 | } 506 | \cs_new_protected:Nn \__showdata_addto_group:nn 507 | { 508 | \clist_map_inline:nn { #2 } 509 | { 510 | \int_gincr:c 511 | { g__showdata_#1_addto_local_group_##1_counter_int } 512 | \__showdata_check_undefined:nnnn 513 | { #1 } { local~group } { ##1 } { \sdaddtogroup } 514 | \__showdata_check_group_add_once:nn { #1 } { ##1 } 515 | \seq_gput_right:cx { g__showdata_#1_subgroup_##1_index_seq } 516 | { 517 | \int_use:c { g__showdata_#1_reserved_counter_sdenv_int } 518 | } 519 | } 520 | } 521 | \cs_new_protected:Nn \__showdata_init_cs_counter:nn 522 | { 523 | \clist_map_inline:nn { arabic, alph, Alph, roman, Roman } 524 | { 525 | \exp_args:Nc \DeclareExpandableDocumentCommand { sdc##1 } { s m } 526 | { 527 | \IfBooleanTF { ####1 } 528 | { \__showdata_use_counter:nnnnn { global } } 529 | { \__showdata_use_counter:nnnnn { local } } 530 | { #1 } { #2 } { ####2 } { ##1 } 531 | } 532 | } 533 | \DeclareDocumentCommand { \sdcstep } { s m } 534 | { 535 | \IfBooleanTF { ##1 } 536 | { \__showdata_step_counter:nnnn { global } } 537 | { \__showdata_step_counter:nnnn { local } } 538 | { #1 } { #2 } { ##2 } 539 | } 540 | \DeclareExpandableDocumentCommand { \sdcvalue } { s m } 541 | { 542 | \IfBooleanTF { ##1 } 543 | { \__showdata_get_counter_value:nnnn { global } } 544 | { \__showdata_get_counter_value:nnnn { local } } 545 | { #1 } { #2 } { ##2 } 546 | } 547 | \DeclareDocumentCommand { \sdcset } { s m m } 548 | { 549 | \IfBooleanTF { ##1 } 550 | { \__showdata_set_counter:nnnnx { global } } 551 | { \__showdata_set_counter:nnnnx { local } } 552 | { #1 } { #2 } { ##2 } { ##3 } 553 | } 554 | } 555 | \cs_new:Nn \__showdata_use_counter:nnnnn 556 | { 557 | \prop_if_in:NnTF \g__showdata_reserved_counter_prop { #4 } 558 | { 559 | \str_if_eq:nnT { #3 } { env } 560 | { 561 | \__showdata_check_counter_use:nn { #4 } { sdenv } 562 | } 563 | \__showdata_check_show_counter_use:nn { #3 } { #4 } 564 | \cs_if_exist_use:c { int_to_#5:n } 565 | { 566 | \int_use:c { g__showdata_#2_reserved_counter_#4_int } 567 | } 568 | } 569 | { 570 | \__showdata_check_couter_not_define:nnnn { #2 } { #1 } { #4 } { sdc#5 } 571 | \cs_if_exist_use:c { int_to_#5:n } 572 | { 573 | \int_use:c { g__showdata_#2_#3_#1_counter_#4_int } 574 | } 575 | } 576 | } 577 | \cs_new_protected:Nn \__showdata_step_counter:nnnn 578 | { 579 | \__showdata_check_counter_not_allowed:nn { #3 } { sdcstep } 580 | \seq_if_in:NnTF \g__showdata_reserved_counter_seq { #4 } 581 | { \__showdata_msg_error:nn { counter-read-only } { #4 } } 582 | { 583 | \__showdata_check_couter_not_define:nnnn { #2 } { #1 } { #4 } { sdcstep } 584 | } 585 | \int_gincr:c { g__showdata_#2_#3_#1_counter_#4_int } 586 | } 587 | \cs_new:Nn \__showdata_get_counter_value:nnnn 588 | { 589 | \prop_if_in:NnTF \g__showdata_reserved_counter_prop { #4 } 590 | { 591 | \str_if_eq:nnT { #3 } { env } 592 | { 593 | \__showdata_check_counter_use:nn { #4 } { sdenv } 594 | } 595 | \__showdata_check_show_counter_use:nn { #3 } { #4 } 596 | \int_use:c { g__showdata_#2_reserved_counter_#4_int } 597 | } 598 | { 599 | \__showdata_check_couter_not_define:nnnn { #2 } { #1 } { #4 } { sdcvalue } 600 | \int_use:c { g__showdata_#2_#3_#1_counter_#4_int } 601 | } 602 | } 603 | \cs_new_protected:Nn \__showdata_set_counter:nnnnn 604 | { 605 | \__showdata_check_counter_not_allowed:nn { #3 } { sdcset } 606 | \seq_if_in:NnT \g__showdata_reserved_counter_seq { #4 } 607 | { \__showdata_msg_error:nn { counter-read-only } { #4 } } 608 | { 609 | \__showdata_check_couter_not_define:nnnn { #2 } { #1 } { #4 } { sdcset } 610 | } 611 | \int_gset:cn { g__showdata_#2_#3_#1_counter_#4_int } { #5 } 612 | } 613 | \cs_generate_variant:Nn \__showdata_set_counter:nnnnn { nnnnx } 614 | \NewDocumentCommand { \sdshow } { s m m } 615 | { 616 | \IfBooleanTF { #1 } 617 | { \__showdata_show:nnn { global } } 618 | { \__showdata_show:nnn { local } } 619 | { #2 } { #3 } 620 | } 621 | \cs_new_protected:Nn \__showdata_show:nnn 622 | { 623 | \cs_if_exist_use:c { __showdata_#2_show_cs_#1_#3 } 624 | } 625 | \NewDocumentCommand { \sdsetshowstyle } { s } 626 | { 627 | \IfBooleanTF { #1 } 628 | { \__showdata_set_show_global_style:nnnnnn } 629 | { \__showdata_set_show_local_style:nnnnnnnn } 630 | } 631 | \cs_new_protected:Nn \__showdata_set_show_global_style:nnnnnn 632 | { 633 | \__showdata_check_domain_not_set: 634 | \clist_map_inline:nn { #1 } 635 | { 636 | \__showdata_check_domain_not_define:nn { ##1 } { error } 637 | \__showdata_check_undefined:nnnn 638 | { ##1 } { prop=property } { #3 } { \sdsetshowstyle* } 639 | \clist_map_inline:nn { #2 } 640 | { 641 | \__showdata_check_undefined:nnnn 642 | { ##1 } { global~group } { ####1 } { \sdsetshowstyle* } 643 | \cs_gset_protected:cpn { __showdata_##1_show_cs_global_####1 } 644 | { 645 | \__showdata_show_global:nnnnnn 646 | { #1 } { #2 } { #3 } { #4 } { #5 } { #6 } 647 | } 648 | } 649 | } 650 | } 651 | \cs_new_protected:Nn \__showdata_show_global:nnnnnn 652 | { 653 | \int_zero_new:c { g__showdata_#1_reserved_counter_sditem_int } 654 | \__showdata_init_cs_global_empty_prop:nnn { #1 } { #2 } { #3 } 655 | \__showdata_init_cs_counter:nn { #1 } { show } 656 | \__showdata_show_global_init_cs_use:nn { #1 } { #2 } 657 | #4 658 | \seq_map_inline:cn { g__showdata_#1_global_group_#2_prop_#3_seq } 659 | { 660 | \int_incr:c { g__showdata_#1_reserved_counter_sditem_int } 661 | #6 662 | } 663 | #5 664 | } 665 | \cs_new_protected:Nn \__showdata_init_cs_global_empty_prop:nnn 666 | { 667 | \prg_set_conditional:Nnn \__showdata_if_empty_prop: { T, F, TF, p} 668 | { 669 | \seq_if_empty:cTF { g__showdata_#1_global_group_#2_prop_#3_seq } 670 | { \prg_return_true: } 671 | { \prg_return_false: } 672 | } 673 | \cs_set_eq:NN \sdIfEmptyPropTF \__showdata_if_empty_prop:TF 674 | \cs_set_eq:NN \sdIfEmptyPropT \__showdata_if_empty_prop:T 675 | \cs_set_eq:NN \sdIfEmptyPropF \__showdata_if_empty_prop:F 676 | } 677 | \cs_new_protected:Nn \__showdata_show_global_init_cs_use:nn 678 | { 679 | \DeclareDocumentCommand { \sduse } { o m } 680 | { 681 | \IfValueT { ##1 } 682 | { 683 | \__showdata_msg_warning:n { show-global-ignore-optional } 684 | } 685 | \__showdata_check_undefined:nnnn 686 | { #1 } { prop=property } { ##2 } { \sduse } 687 | \__showdata_check_invalid_index:nnnnnn { #1 } { global } { #2 } { ##2 } 688 | { g__showdata_#1_global_group_#2_prop_##2_seq } 689 | { g__showdata_#1_reserved_counter_sditem_int } 690 | \seq_item:cn { g__showdata_#1_global_group_#2_prop_##2_seq } 691 | { \int_use:c { g__showdata_#1_reserved_counter_sditem_int } } 692 | } 693 | } 694 | \cs_new_protected:Nn \__showdata_set_show_local_style:nnnnnnnn 695 | { 696 | \__showdata_check_domain_not_set: 697 | \clist_map_inline:nn { #1 } 698 | { 699 | \__showdata_check_domain_not_define:nn { ##1 } { error } 700 | \__showdata_check_undefined:nnnn 701 | { ##1 } { prop=property } { #3 } { \setshowstyle } 702 | \clist_map_inline:nn { #2 } 703 | { 704 | \__showdata_check_undefined:nnnn 705 | { ##1 } { local~group } { ####1 } { \setshowstyle } 706 | \cs_gset_protected:cpn { __showdata_##1_show_cs_local_####1 } 707 | { 708 | \__showdata_show_local:nnnnnnnn 709 | { #1 } { #2 } { #3 } { #4 } 710 | { #5 } { #6 } { #7 } { #8 } 711 | } 712 | } 713 | } 714 | } 715 | \cs_new_protected:Nn \__showdata_show_local:nnnnnnnn 716 | { 717 | \int_zero_new:c { g__showdata_#1_reserved_counter_sdgroup_int } 718 | \__showdata_init_cs_no_subgroup:nn { #1 } { #2 } 719 | \__showdata_init_cs_counter:nn { #1 } { show } 720 | #4 721 | \seq_map_inline:cn { g__showdata_#1_subgroup_#2_index_seq } 722 | { 723 | \int_gincr:c { g__showdata_#1_reserved_counter_sdgroup_int } 724 | \__showdata_show_local_init_cs_use:nnn { #1 } { #2 } { ##1 } 725 | \__showdata_show_local_sub:nnnnnnn { #1 } { #2 } { #3 } 726 | { ##1 } { #6 } { #7 } { #8 } 727 | } 728 | #5 729 | } 730 | \cs_new_protected:Nn \__showdata_init_cs_no_subgroup:nn 731 | { 732 | \prg_set_conditional:Nnn \__showdata_if_no_subgroup: { T, F, TF, p} 733 | { 734 | \seq_if_empty:cTF { g__showdata_#1_subgroup_#2_index_seq } 735 | { \prg_return_true: } 736 | { \prg_return_false: } 737 | } 738 | \cs_set_eq:NN \sdIfNoSubgroupTF \__showdata_if_no_subgroup:TF 739 | \cs_set_eq:NN \sdIfNoSubgroupT \__showdata_if_no_subgroup:T 740 | \cs_set_eq:NN \sdIfNoSubgroupF \__showdata_if_no_subgroup:F 741 | } 742 | \cs_new_protected:Nn \__showdata_show_local_sub:nnnnnnn 743 | { 744 | \int_zero_new:c { g__showdata_#1_reserved_counter_sditem_int } 745 | \__showdata_init_cs_local_empty_prop:nnnn { #1 } { #2 } { #3 } { #4 } 746 | #5 747 | \seq_map_inline:cn { g__showdata_#1_local_group_#2_prop_#3_#4_seq } 748 | { 749 | \int_incr:c { g__showdata_#1_reserved_counter_sditem_int } 750 | #7 751 | } 752 | #6 753 | } 754 | \cs_new_protected:Nn \__showdata_init_cs_local_empty_prop:nnnn 755 | { 756 | \prg_set_conditional:Nnn \__showdata_if_empty_prop: { T, F, TF, p} 757 | { 758 | \seq_if_empty:cTF { g__showdata_#1_local_group_#2_prop_#3_#4_seq } 759 | { \prg_return_true: } 760 | { \prg_return_false: } 761 | } 762 | \cs_set_eq:NN \sdIfEmptyPropTF \__showdata_if_empty_prop:TF 763 | \cs_set_eq:NN \sdIfEmptyPropT \__showdata_if_empty_prop:T 764 | \cs_set_eq:NN \sdIfEmptyPropF \__showdata_if_empty_prop:F 765 | } 766 | \cs_new_protected:Nn \__showdata_show_local_init_cs_use:nnn 767 | { 768 | \DeclareDocumentCommand { \sduse } { o m } 769 | { 770 | \IfValueTF { ##1 } 771 | { 772 | \__showdata_show_local_use_global:nnn { #1 } { ##1 } { ##2 } 773 | } 774 | { 775 | \__showdata_show_local_use_local:nnnn { #1 } { #2 } { ##2 } { #3 } 776 | } 777 | } 778 | } 779 | \cs_new_protected:Nn \__showdata_show_local_use_global:nnn 780 | { 781 | \__showdata_check_undefined:nnnn 782 | { #1 } { prop=property } { #3 } { \sduse } 783 | \__showdata_check_invalid_index:nnnnnn { #1 } { global } { #2 } { #3 } 784 | { g__showdata_#1_global_group_#2_prop_#3_seq } 785 | { g__showdata_#1_reserved_counter_sdgroup_int } 786 | \seq_item:cn { g__showdata_#1_global_group_#2_prop_#3_seq } 787 | { \int_use:c { g__showdata_#1_reserved_counter_sdgroup_int } } 788 | } 789 | \cs_new_protected:Nn \__showdata_show_local_use_local:nnnn 790 | { 791 | \__showdata_check_undefined:nnnn 792 | { #1 } { prop=property } { #3 } { \sduse } 793 | \__showdata_check_invalid_index:nnnnnn { #1 } { local } { #2 } { #3 } 794 | { g__showdata_#1_local_group_#2_prop_#3_#4_seq } 795 | { g__showdata_#1_reserved_counter_sditem_int } 796 | \seq_item:cn { g__showdata_#1_local_group_#2_prop_#3_#4_seq } 797 | { \int_use:c { g__showdata_#1_reserved_counter_sditem_int } } 798 | } 799 | \NewDocumentCommand { \NewsdReviewPoint } { m m } 800 | { 801 | \seq_gset_from_clist:cn { g__showdata_review_point_#1_seq } { #2 } 802 | } 803 | \prg_new_conditional:Nnn \__showdata_if_review:nnn { T, F, TF } 804 | { 805 | \sdutil_if_review:cnnTF { g__showdata_review_point_#1_seq } { #2 } { #3 } 806 | { \prg_return_true: } 807 | { \prg_return_false: } 808 | } 809 | \cs_set_eq:NN \sdIfReviewTF \__showdata_if_review:nnnTF 810 | \cs_set_eq:NN \sdIfReviewT \__showdata_if_review:nnnT 811 | \cs_set_eq:NN \sdIfReviewF \__showdata_if_review:nnnF 812 | \endinput 813 | %% 814 | %% End of file `showdata.sty'. 815 | -------------------------------------------------------------------------------- /update.py: -------------------------------------------------------------------------------- 1 | from pathlib import Path 2 | import pendulum 3 | import argparse 4 | 5 | def rename(name: str, args) -> None: 6 | # print(path.) 7 | p = Path('.') 8 | old = p / f'{name}.pdf' 9 | dt = pendulum.today().add(days=args.offset[0]) 10 | new = p / f'{name}{dt.isoformat()[0:10]}.pdf' 11 | if not old.exists(): 12 | print(f'file {old} does not exist!') 13 | return 14 | if new.exists(): 15 | print(f'file {new} has existed!') 16 | return 17 | old.rename(new) 18 | 19 | if __name__ == "__main__": 20 | parser = argparse.ArgumentParser(description = 'update file') 21 | parser.add_argument('offset', metavar='N', type=int, nargs=1, help='offset of the date') 22 | args = parser.parse_args() 23 | files = ['math', 'english', 'general'] 24 | for f in files: 25 | rename(f, args) 26 | --------------------------------------------------------------------------------