<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper]{amsart}

\frenchspacing % aktivuje pouďż˝itďż˝ nďż˝kterďż˝ch ďż˝eskďż˝ch typografickďż˝ch pravidel

\usepackage{a4wide}
\usepackage[T1]{fontenc}
%\usepackage[cp1250]{inputenc}
\usepackage[utf8]{inputenc}
\usepackage[czech]{babel}
\usepackage{amsthm,amsmath,amssymb,color}
\theoremstyle{definition}
\newtheorem{example}{PĹ™Ă­klad}

\def\er{\mathbb R}
\def \arctg {\operatorname{arctg}}
\def \tg {\operatorname{tg}}
\def \cotg {\operatorname{cotg}}

\title{Kalkulus 1, 2019--2020\\ zkou\v{s}kov\'e p\'isemky}  % tyto dvďż˝ poloďż˝ky jsou zde v podstatďż˝ formďż˝lnďż˝, ve skuteďż˝nosti nejsou nikde 
\author{Marek C\'uth}

\begin{document}\noindent\begin{center}\textbf{Kalkulus 1, LS 2019-2020\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 26.5.}\end{center}

 \bigskip
 
\begin{example}[13 bod\r{u}]Spo\v{c}t\v{e}te
\[\lim_{x\to 0}\frac{\sin(\exp(2x)-1) + 2\log(2\cos(x)-1)-2x}{x^4}.\]
\end{example}

\begin{example}[14 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[ f(x) = \frac{1}{\cos x + 7}\]
na maxim\'aln\'ich intervalech existence.
\end{example}

\begin{example}[11 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[\int_3^\infty \frac{\arctg(x+7)}{3^x+2}(x^3+4x+2) \, \mathrm{d} x.\]
\end{example}

\begin{example}[12 bod\r{u}] Nalezn\v{e}te obecn\'e \v{r}e\v{s}en\'i rovnice
	\[
	y'' - \frac{2x+2}{x^2+2x+2}y' + \frac{2}{x^2+2x+2}y = x^2+2x+2.
	\]
	(Hint: nejprve doka\v{z}te, \v{z}e $\{x^2+5x+3, x+1\}$ tvo\v{r}\'i fundament\'aln\'i syst\'em \v{r}e\v{s}en\'i homogenn\'i rovnice).
\end{example}
%%%%%%%
\newpage

\noindent\begin{center}\textbf{Kalkulus 1, LS 2019-2020\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 8.6.}\end{center}

 \bigskip
 
\begin{example}[13 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[ f(x) = \frac{3e^{3x} + 5e^{2x} + 11e^x}{(e^x - 2)(e^{2x}+e^x+3)}\]
na maxim\'aln\'ich intervalech existence.
\end{example}

\begin{example}[13 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[\int_{\tg 1}^\infty \frac{\arctg x - 1}{(x-\tg 1)^{5/4}} \, \mathrm{d} x\]
\end{example}

\begin{example}[16 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice 
	\[
	y' = \frac{y^2-y-2}{9x^2+6x+2},\quad y(-\tfrac{1}{3})=\tfrac{3}{2}.
	\]
	\end{example}

\begin{example}[8 bod\r{u}] Ur\v{c}ete a nakreslete defini\v{c}n\'i obor funkce 
	\[f(x,y) = \sqrt{x^2 + \sqrt[3]{y}}\]
	a vy\v{s}et\v{r}ete jejĂ­ parci\'aln\'i derivace.
\end{example}
\newpage

\noindent\begin{center}\textbf{Kalkulus 1, LS 2019-2020\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 22.6.}\end{center}

 \bigskip
 
\begin{example}[12 bod\r{u}]Spo\v{c}t\v{e}te
\[\lim_{x\to 0}\frac{1 - \cos(\log(1+\sin x)) + \sin(\log(1+x)) - x}{x^3}\]
\end{example}

\begin{example}[13 bod\r{u}]Spo\v{c}t\v{e}te integr\'al
\[\int_3^4 \frac{1}{(x-2)\sqrt{-x^2 + 9x - 14}} \, \mathrm{d} x\]
\end{example}

\begin{example}[10 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[\int_{10}^\infty \frac{e^{2/x}-1}{\sin(3/x)}\log(3+\tfrac{1}{x}) \, \mathrm{d} x\]
\end{example}

\begin{example}[15 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice 
\[
	y'' - 4y' + 5y = 3\cos 2x + 24\sin 2x + 5x^2 + 7x - 5,\quad y(0)=5, y'(0)=4.
	\]
\end{example}
\newpage
\begin{document}\noindent\begin{center}\textbf{Kalkulus 1, LS 2019-2020\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 8.7.}\end{center}

 \bigskip
 
\begin{example}[12 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[\int \frac{1}{\sin^2 x + 9\cos^2 x} \, \mathrm{d} x\]
na maxim\'aln\'ich intervalech existence.
\end{example}

\begin{example}[12 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[\int_3^\infty \frac{\log(x^2-5x+7)}{x^2-2x-3} \, \mathrm{d} x\]
\end{example}

\begin{example}[14 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[
	y'  + \frac{2x+1}{x^2+x-6}y = e^x,\quad y(0)=0.
	\]
	\end{example}

\begin{example}[12 bod\r{u}] Ur\v{c}ete a nakreslete defini\v{c}n\'i obor funkce
	\[f(x,y) = \arcsin\Big(\frac{1}{|x| + |y|}\Big)\]
	a vy\v{s}et\v{r}ete jejĂ­ parci\'aln\'i derivace.
\end{example}
\newpage

\begin{document}\noindent\begin{center}\textbf{Kalkulus 1, LS 2019-2020\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 27. 7.}\end{center}

 \bigskip

\begin{example}[12 bod\r{u}]Spo\v{c}t\v{e}te
\[\lim_{x\to 0}\frac{\frac{2+x^2}{e^x} - \log(1-2x+3x^2) - 2}{1-\cos x}\]
\end{example}
 
\begin{example}[12 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[f(x) = \frac{13\sin x + 46}{(\tg x - \frac{4}{\cos x})(\sin x + 3)^2}\]
na maxim\'aln\'ich intervalech existence.
\end{example}

\begin{example}[12 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[\int_1^\infty \Bigg(\sin\Big(\frac{\pi(x+3)}{4(x+2)}\Big) - \frac{\sqrt{2}}{2}\Bigg) \, \mathrm{d} x\]
\end{example}

\begin{example}[14 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[
	y'' - \frac{2x}{x^2-1}y' + \frac{2}{x^2-1}y = x^2-1,\quad y(0)=2, y'(0)=3.
	\]
	(Hint: nejprve doka\v{z}te, \v{z}e $\{x^2+1, x\}$ tvo\v{r}\'i fundament\'aln\'i syst\'em \v{r}e\v{s}en\'i homogenn\'i rovnice).
	\end{example}
	\newpage

\begin{document}\noindent\begin{center}\textbf{Kalkulus 1, LS 2019-2020\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 7. 9.}\end{center}

 \bigskip

\begin{example}[12 bod\r{u}]Spo\v{c}t\v{e}te integr\'al
\[\int_1^\infty \frac{1}{x\sqrt{x^2+4}} \, \mathrm{d} x.\]
\end{example}

\begin{example}[13 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[\int_5^7 \frac{\arctg(x^2+1)}{1-\cos(5-x)}(x^2-9x+20)  \, \mathrm{d} x\]
\end{example}

\begin{example}[15 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[
	y''  - y' - 6y = e^x(-12x-4) - 6x^2 - 2x - 4,\quad y(0)=4, y'(0)=4.
	\]
	\end{example}
	
	\begin{example}[10 bod\r{u}] Ur\v{c}ete a nakreslete defini\v{c}n\'i obor funkce
	\[f(x,y) = (\sqrt{x}+1)\log(y^2 + x - 4)\]
	a vy\v{s}et\v{r}ete jejĂ­ parci\'aln\'i derivace.
\end{example}
\newpage


\begin{document}\noindent\begin{center}\textbf{Kalkulus 1, LS 2019-2020\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 14. 9.}\end{center}

\begin{example}[11 bod\r{u}]Spo\v{c}t\v{e}te
\[\lim_{x\to 0}\frac{(1+x)^{\sin x} + 2\cos(x) - 3}{e^{(x^3)}-1}\]
\end{example}

\begin{example}[13 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[\int \frac{2\log x + 5}{x(\log^2 x + \log x + 3)} \, \mathrm{d} x\]
na maxim\'aln\'ich intervalech existence.
\end{example}

\begin{example}[12 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[\int_0^9 \frac{x^8-7x+6}{\sqrt{9-x}}\log(9-x) \, \mathrm{d} x\]
\end{example}

\begin{example}[14 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[
	y'  + \cotg(x+1) y = x^2 + 3x +2,\quad y(0)=4.
	\]
\end{example}

\end{document}
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