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-\pdfoutput=1
-
-\documentclass{article}
-
-\usepackage{color}
-
-\usepackage{amsmath}
-\DeclareMathOperator{\arctg}{arctg}
-\everymath{\displaystyle}
-
-\usepackage[pdftex,designi]{web}
-\usepackage[noxcolor,pdftex]{exerquiz}
-\usepackage[ImplMulti]{dljslib}
-
-\parindent 0pt
-\def\correctColor{color.green}
-\def\wrongColor{color.red}
-
-\usepackage[picture,finetune]{jeopardy}
-\def\JeopardyPictureFile{joke.jpg}
-
-\def\ChampionMsg{Have a fun with math!}
-\GrandPoohbahtoks{\textColor{1 1 0 rg}\textSize{0}}
-
-\everyCategoryHead{\color{darkgreen}\tiny}
-\pagestyle{empty}
-
-\def\logoB{\rotatebox{90}{%
- \vbox{\normalsize\hbox{\color{red}{Robert Ma\v{r}\'{i}k}}
- \hbox{\color{green}Jeopardy game}}}
- \vfill}
-
-\let\rmdefault\sfdefault
-\hypersetup{pdfpagemode=Window, pdfnewwindow=true, pdfmenubar=true,%
- pdftoolbar=true,colorlinks, pdfwindowui=false,
- pdfpagemode=Window}
-
-
-\AditionalShift=5pt
-\begin{document}
-%\CellHeight=24bp
-
-\Celltoks{\BG{1 1 0.8}\textSize{0}\BC{1 1 0.8}\W{0}}
-\SetGameWidth{0.7\linewidth}
-\everyPushButton{\S{I}\BC{}}
-
-\MakeGameBoard
-
-\begin{category}{Precalculus}
-
-\begin{question}
-$\ln \frac xy=$
-\Ans0 $\ln x+\ln y$
-\Ans1 $\ln x-\ln y$
-\Ans0 $x\ln y$
-\Ans0 $y\ln x$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-The function $y=x^2\cdot \sin x$ is
-\Ans1 odd
-\Ans0 even
-\Ans0 neither odd nor even
-\end{question}
-
-\begin{question}
-$\arctan 1=$
-\Ans0 $\infty$
-\Ans0 $\frac \pi3$
-\Ans1 $\frac \pi4$
-\Ans0 $\frac \pi6$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-The equivalence "$a<b$ if and only if $f(a)<f(b)$" is the property of
-\Ans0 even functions
-\Ans0 one-to-one functions
-\Ans0 continuous functions
-\Ans1 increasing functions
-\Ans0 none of them
-\end{question}
-
-\end{category}
-
-
-\begin{category}{Functions}
-
-\begin{question}
-How many points of inflection is on the graph of the function $y=\sin x$ in
-the open interval $(0,2\pi)$
-\Ans0 none
-\Ans1 one
-\Ans0 two
-\Ans0 three
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-Find points of discontinuity of the function $y=\frac {x-4}{(x-2)\ln x}$
-\Ans0 none
-\Ans0 $0$
-\Ans0 $0$, $1$
-\Ans1 $0$, $1$, $2$
-\Ans0 $0$, $2$
-\Ans0 $0$, $1$, $4$
-\Ans0 $0$, $4$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-Let $f$ be a function and $f^{-1}$ be its inverse. Then $f^{-1}\bigl(f(x)\bigr)=$
-\Ans0 $0$
-\Ans0 $1$
-\Ans1 $x$
-\Ans0 $f(x)$
-\Ans0 $f^{-1}(x)$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-$\arcsin(\sin x)=x$ for every $x\in\mathbf{R}$
-\Ans0 Yes
-\Ans1 No
-\end{question}
-
-\end{category}
-
-
-\begin{category}{Limits}
-
-\begin{question}
-$\lim_{x\to-\infty} \arctg x=$
-\Ans0 $0$
-\Ans0 $\frac\pi2$
-\Ans1 $-\frac\pi2$
-\Ans0 $\infty$
-\Ans0 $-\infty$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-$\lim_{x\to\infty}\sin x=$
-\Ans0 $1$
-\Ans0 $-1$
-\Ans1 does not exist
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-$\lim_{x\to\infty}\frac{2x^3+x^2+4}{x^2-x+2}=$
-\Ans1 $\infty$
-\Ans0 $2$
-\Ans0 $0$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-$\lim_{x\to 0^+}\frac{e^{1/x}(x-1)}{x}$
-\Ans0 $0$
-\Ans0 $1$
-\Ans0 $e$
-\Ans0 $\infty$
-\Ans0 $-1$
-\Ans0 $-e$
-\Ans1 $-\infty$
-\Ans0 none of them
-\end{question}
-
-\end{category}
-
-\begin{category}{Derivative}
-
-
-\begin{question}
-$\left(\frac 1{\sqrt[3]x}\right)'=$
-\Ans0 $\frac 13 x^{-2/3}$
-\Ans0 $-\frac 13 x^{-2/3}$
-%\Ans0 $\frac 13 x^{1/3}$
-\Ans0 $-\frac 13 x^{1/3}$
-\Ans0 $\frac 13 x^{-4/3}$
-\Ans1 $-\frac 13 x^{-4/3}$
-\Ans0 none of them
-\end{question}
-
-
-\begin{question}
-$(x-x\ln x)'=$
-\Ans0 $\ln x$
-\Ans1 $-\ln x$
-\Ans0 $1+\ln x$
-\Ans0 $1-\ln x$
-\Ans0 $0$
-\Ans0 $1-\frac 1x$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-$\left(x^2e^{x^2}\right)'$
-\Ans0 $2xe^{2x}$
-\Ans0 $2xe^{x^2}2x$
-\Ans0 $2xe^{x^2}+x^2e^{x^2}$
-\Ans1 $2xe^{x^2}+x^2e^{x^2}2x$
-\Ans0 $2xe^{x^2}2x+x^2e^{x^2}2x$
-\Ans0 none of them
-\end{question}
-
-\begin{question}
-The definition of the derivative of the function $f$ at the point $a$ is
-\Ans0 $\lim_{h\to 0}\frac{f(x+h)+f(x)}{h}$
-\Ans0 $\lim_{h\to 0}\frac{f(x+h)}{h}$
-\Ans1 $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
-\Ans0 $\lim_{h\to 0}\frac{f(x)-f(x+h)}{h}$
-\Ans0 $\lim_{h\to 0}\frac{f(x-h)-f(x)}{h}$
-\Ans0 none of them
-\end{question}
-
-\end{category}
-
-\everyRespBoxMath{\BG{1 1 1}}
-\begin{category}{Evaluation of derivatives}
- \begin{question}$(x^2+1)'=$
- \RespBoxMath{2x}{3}{0.001}{[1,5]}
- \end{question}
- \begin{question}$(xe^x)'=$
- \RespBoxMath{(x+1)e^x}{3}{0.001}{[1,5]}
- \end{question}
- \begin{question}$\ln(\sin x)=$
- \RespBoxMath{cos(x)/sin(x)}{3}{0.001}{[1,2]}
- \end{question}
- \begin{question}$(xe^{-x})'=$
- \RespBoxMath{(1-x)e^{-x}}{3}{0.001}{[1,5]}
- \end{question}
-\end{category}
-
-
-
-\begin{category}{Theory}
-
-\begin{question} By theorem of Bolzano, the polynomial $y=x^3+2x+4$ has zero on
-\Ans0 $(0,1)$
-\Ans0 $(1,2)$
-\Ans0 $(2,3)$
-\Ans0 $(-1,0)$
-\Ans1 $(-2,-1)$
-\Ans0 $(-3,-2)$
-\Ans0 none of them
-\end{question}
-
-
-\begin{question}
-Let $a\in Im(f)$. Then the solution of the equation $f(x)=a$ exists. This
-solution is unique if and only if
-\Ans1 $f$ is one-to-one
-\Ans0 $f$ is increasing
-\Ans0 $f$ continuous
-\Ans0 $f$ differentiable
-\Ans0 none of them
-\end{question}
-
-
-\begin{question}
-If the function has a derivative at the point $x=a$, then it is
-\Ans0 increasing at $a$.
-\Ans0 decreasing at $a$.
-\Ans0 one-to-one at $a$.
-\Ans1 continuous at $a$.
-\Ans0 undefined at $a$.
-\end{question}
-
-\begin{question}
-If both $y(a)=y'(a)=y''(a)=0$, then the function
-\Ans0 has local maximum at $a$.
-\Ans0 has local minimum at $a$.
-\Ans0 has point of inflection at $a$.
-\Ans1 any of these possibilites may be true, we need more informations.
-\end{question}
-
-\end{category}
-
-
-\end{document}
-%%% Local Variables:
-%%% mode: latex
-%%% TeX-master: t
-%%% End:
-
-