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\def\er{\mathbb R}
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\begin{document}

\begin{center}\textbf{Kalkulus 1, LS 2023-2024\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 28.5.}\end{center}

 \bigskip
 
\begin{example}[12 bod\r{u}]Spo\v{c}t\v{e}te
\[\lim_{x\to 0}\frac{\sqrt[3]{1+\sin (6x)} - \exp(-x)\log(1+2x)-1}{x(1-\cos x)}\]
\end{example}

\begin{example}[13 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[ f(x) = \frac{1}{6\sin x + \cos x + 11}
\]
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
\[
\int_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}} \frac{\sin x - \tfrac{1}{2} - \tfrac{\sqrt{3}}{2}(x-\tfrac{\pi}{6})}{(x-\tfrac{\pi}{6})^3}\arctan(1+x^2) \mathrm{d} x
\]
    (Můžete bez důkazu použít, že pro $x\in (\tfrac{\pi}{6},\tfrac{\pi}{2})$ je $\sin x - \tfrac{1}{2} - \tfrac{\sqrt{3}}{2}(x-\tfrac{\pi}{6})\neq 0$)
\end{example}

\begin{example}[13 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[
	y'' - 4y' + 4y = e^{2x} + 25 \sin x,\quad y(0)=1,y'(0)=0.
	\]
\end{example}
\newpage
\newpage

\noindent\begin{center}\textbf{Kalkulus 1, LS 2023-2024\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 4.6.}\end{center}

 \bigskip

\begin{example}[13 bod\r{u}]Spo\v{c}t\v{e}te integr\'al
\[
    \int_0^{\tfrac{7}{2}\pi} \frac{7-\cos^2 x}{(2+\sin^2 x)(1+\sin^2 x)} \mathrm{d} x.
    \]
\end{example}

\begin{example}[12 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[
\int_3^\infty \frac{\sqrt{|\sin(x-3)|}}{x-3}\cdot\frac{x^2+5x+2}{4^x} \mathrm{d} x.
\]
\end{example}

\begin{example}[12 bod\r{u}] Nalezn\v{e}te na intervalu $(-2,\infty)$ obecn\'e \v{r}e\v{s}en\'i rovnice
	\[
	y'\cdot (x+3) + \frac{1}{\log(x+3)} y = x(x+3)
	\]
a nalezněte řešení na intervalu $(-2,\infty)$ splňující počáteční podmínku $y(0) = \frac{9}{2}$.
\end{example}

\begin{example}[13 bod\r{u}] Ur\v{c}ete a nakreslete defini\v{c}n\'i obor funkce
	\[f(x,y) = \sqrt{\frac{|x|-\sqrt[3]{y}}{2+\sqrt[3]{y}}}\]
	a vy\v{s}et\v{r}ete její parci\'aln\'i derivace podle proměnné $x$.
\end{example}
%%%%%%%
\newpage

\begin{center}\textbf{Kalkulus 1, LS 2023-2024\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 18.6.}\end{center}

 \bigskip
 
\begin{example}[11 bod\r{u}]Spo\v{c}t\v{e}te
\[\lim_{x\to 0}\frac{(1+3x)^x + (\tfrac{2}{3}-x)\cos(3x) - \tfrac{5}{3} +x}{x^3\log(1+x)}\]
\end{example}

\begin{example}[13 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[ f(x) = \frac{1}{(\sqrt{4x^2+4x+7} + 2x + 2)\sqrt{4x^2+4x+7}}
\]
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
\[
\int_{\log 4}^\infty \frac{\sqrt{x-\log 4}}{e^x-4}\sin x \; \mathrm{d} x.
\]
\end{example}

\begin{example}[13 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[y'' + \frac{1}{x(1-\log x)}y' - \frac{1}{x^2(1-\log x)}y = 1-\log x,\quad y(1)=2,\;y'(1)=\frac{3}{2}\]%5
    (Hint: nejprve doka\v{z}te, \v{z}e $\{x,\log x\}$ 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 2023-2024\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 25.6.}\end{center}

 \bigskip

\begin{example}[13 bod\r{u}]Vyj\'ad\v{r}ete primitivn\'i funkci k 
\[ \scaleto{f(x):=\begin{cases}
        (x^2+1)\cos x & \text{ pro }x\leq 0,\\
        \frac{3e^{2x}-22e^x + 39}{e^{2x} + 6e^x + 13} & \text{ pro }x\geq 0.
    \end{cases}
    }{40pt}
\]
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
\[
\int_5^\infty \frac{1-e^{5-x}}{\sqrt{(x-5)^3}}\log(x^2+3x+1) \mathrm{d} x.
\]
\end{example}

\begin{example}[12 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[
	y'' - 7y' + 12y = \frac{e^{4x}}{e^x + 1},\qquad y(0)=0,\;y'(0)=0.
	\]
\end{example}

\begin{example}[12 bod\r{u}] Uva\v{z}ujme funkci
	\[f(x,y) = y^{|y-\sqrt{x^2+1}|}.\]
Vy\v{s}et\v{r}ete jej\'i parci\'aln\'i derivace podle proměnné $x$.
\end{example}
%%%%%%%
\newpage

\begin{center}\textbf{Kalkulus 1, LS 2023-2024\\Zad\'an\'i p\'isemn\'e \v{c}\'asti zkou\v{s}ky - term\'in 3.9.}\end{center}

 \bigskip
 
\begin{example}[13 bod\r{u}]Spo\v{c}t\v{e}te
\[\lim_{x\to 0}\frac{\exp(1-\cos 2x) - 1 + 6x -2x^3 - x(6-x^2)\cdot \sqrt[4]{1+\frac{4}{3}x}}{x^4}\]
\end{example}

\begin{example}[12 bod\r{u}]Spo\v{c}t\v{e}te integr\'al
\[ \int_2^\infty \frac{\sqrt[3]{\tfrac{x+1}{x-2}}}{(x-2)(x+1)\left(\sqrt[3]{\tfrac{x+1}{x-2}} + 1\right)\left(\sqrt[3]{\tfrac{x+1}{x-2}} + 2\right)} \mathrm{d} x.
\]
\end{example}

\begin{example}[12 bod\r{u}]Vy\v{s}et\v{r}ete konvergenci n\'asleduj\'ic\'iho integr\'alu
\[
\int_{1}^\infty \Big(e^{1/x} - 1 - \frac{1}{x}\Big)\cos\Big(\frac{x^2-2x+1}{x-1}\Big) \mathrm{d} x.
\]
\end{example}

\begin{example}[13 bod\r{u}] Nalezn\v{e}te maxim\'aln\'i \v{r}e\v{s}en\'i rovnice
	\[
	y' = \frac{\sqrt{y^2-4}}{y}\cdot \frac{1}{x^2+3},\quad y(\sqrt{3}) = \sqrt{4 + (\tfrac{\pi}{2\sqrt{3}})^2}.
	\]
\end{example}
\end{document}

