Subiecte Bacalaureat rezolvate la Analiza matematica

1. Se considera functia $f:R^{*}\rightarrow R, f\left(x\right)=x^{3}+\frac{3}{x}$

a) Sa se calculeze $f^{'}\left(x\right),x\in R^{*}$

b) Sa se calculeze $\lim\limits_{x\to 1}{\frac{f\left(x\right)-f\left(1\right)}{x-1}}$

c) Sa se determine intervalele de monotonie ale functiei f.

Solutie

Calculam mai inati

a) $f^{'}\left(x\right)=\left(x^{3}+\frac{3}{x}\right)^{'}=\left(x^{3}\right)^{'}+\left(\frac{3}{x}\right)^{'}=3x^{2}+\frac{3^{'}\cdot x-3\cdot x^{'} }{x^{2}}=3x^{2}+\frac{0-3\cdot 1}{x^{2}}=3x^{2}-\frac{3}{x^{2}}$

b) Ca sa calculam limita de la b) trebuie sa stim ca de fapt acea limita este definitia derivatei functiei in punctul x=1, cum am calculat derivata stim ca

$\lim\limits_{x\to 1}{\frac{f\left(x\right)-f\left(1\right)}{x-1}}=f^{'}\left(1\right)$

Deci

$f^{'}\left(1\right)=3\cdot 1^{2}-\frac{3}{1^{2}}=3-3=0$

c) Acum ca sa aflam intervalele de monotonie rezolvam ecuatia:

$f^{'}\left(x\right)=0\Rightarrow 3x^{2}-\frac{3}{x^{2}}=0\Rightarrow \frac{3x^{2}\cdot x^{2}-1\cdot 3}{x^{2}}=0\Rightarrow \frac{3x^{4}-3}{x^{2}}=0$

Cum numitorul este diferit de 0 (acest lucru se observa si din domeniul de definitie), este tot timpul pozitiv, ne ocupam de

numarator. astfel

$3x^{4}-3=0\Rightarrow 3x^{4}=3\Rightarrow x^{4}=1$

Iar solutiile reale sunt $x=\pm 1$

Acum intocmim tabelul de variatie:

Astfel din tabelul de variatie al functiei rezulta ca

– f este crescatoare pe $\left(-\infty, -1\right)\cup\left(1,+\infty\right)$ si

– f este descrescatoare pe $\left(-1,0\right)\cup\left(0,1\right)$

2) Se considera functia $f:\left[0,1\right]\rightarrow R,f\left(x\right)=x\cdot\sqrt{2-x^{2}}$

a) Sa se calculezevolumul corpului obtinut prin rotatie, in jurul axei Ox, a graficului functie f.

b) Sa se calculeze $\int^{1}_{0}f\left(x\right)dx$

c) Sa se calculeze $\lim\limits_{x\to 0}{\frac{\int^{x}_{0}f\left(t\right)dt}{x^{2}}}$

Solutie:

Calculam

$V=\pi\int^{1}_{0}f^{2}\left(x\right) dx=\pi\int^{1}_{0}\left(x\sqrt{2-x^{2}}\right)^{2}=$
$\pi\int^{1}_{0}x^{2}\left(2-x^{2}\right)dx=\pi\int^{1}_{0}\left(2x^{2}-x^{4}\right) dx=$
$\pi\int^{1}_{0}2x^{2}dx-\pi\int^{1}_{0}x^{4}dx=\pi\cdot 2\int^{1}_{0}x^{2}-\pi\cdot \frac{x^{5}}{5}|^{1}_{0}=$
$2\cdot \pi\cdot\frac{x^{3}}{3}|^{1}_{0}-\left(\frac{1^{5}}{5}-\frac{0^{5}}{5}\right)=$
$\pi\left(2\cdot\left(\frac{1^{3}}{3}-\frac{0^{3}}{3}-\frac{1}{5}\right)\right)=$
$\pi\left(2\cdot\frac{1}{3}-\frac{1}{5}\right)=$
$\pi\left(\frac{2}{3}-\frac{1}{5}\right)=\pi\cdot \frac{5\cdot 2-3\cdot 1}{15}=\pi\cdot \frac{10-3}{15}=\pi\ cdot\frac{7}{15}=\frac{7\pi}{15}$

b) Calculam acum integrala

$\int^{1}_{0}d\left(x\right)dx=\int^{1}_{0}x\sqrt{2-x^{2}}dx=$

Ca sa rezolvam integrama de mai sus folosim metoda schimbarii de variabila astfel notam

$2-x^{2}=t\Rightarrow -x^{2}=t-2\Rightarrow x^{2}=2-t\Rightarrow \left(x^{2}\right)^{2}dx=\left(2-t\right)^{'}dx\Rightarrow 2xdx=-1\cdot dt\Rightarrow xdx=-\frac{1}{2}dt$

Acum ne ocupam de captele intervalului, astfel

Pentru

$x=0\Rightarrow 2-0^{2}=t\Rightarrow t=2$

Pentru

$x=1\Rightarrow 2-1^{2}=t\Rightarrow 2-1=t\Rightarrow 1=t$

Astfel integrala devine:

$\int^{1}_{0}x\sqrt{2-x^{2}}dx=\int^{1}_{2}\sqrt{t}\cdot \left(-\frac{1}{2}\right)dt=$
$\int^{1}_{2}t^{\frac{1}{2}\left(-\frac{1}{2}\right)}dt=$
$-\frac{1}{2}\int^{1}_{2}t^{\frac{1}{2}}dt=$
$-\frac{1}{2}\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}|^{1}_{2}=$
$-\frac{1}{2}\cdot\frac{t^{\frac{1+2\cdot 1}{1}}}{\frac{1+2\cdot 1}{2}}|^{1}_{2}=-\frac{1}{2}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}|^{1}_{2}=$
$=-\frac{1}{2}\cdot\frac{2}{3}^{(2}\cdot t^{\frac{3}{2}}|^{1}_{2}=$
$-\frac{1}{3}\cdot \sqrt{t^{3}}^{1}_{2}=-\frac{1}{3}\left(\sqrt{1^{3}}-\sqrt{2^{3}}\right)=$
$-\frac{1}{3}\left(1-2\sqrt{2}\right)=\frac{2\sqrt{2}-1}{3}$

c) Acum sa calculam

$\lim\limits_{x\to 0}{\frac{\int^{x}_{0}f\left(t\right) dt}{x^{2}}}$

Calculam mai intai integrala

$\int|^{x}_{0}f\left(t\right)dt=\int^{x}_{0}t\sqrt{2-t^{2}}dt$

Astfel la fel ca mai sus rezolvam integrala prin metoda schimbarii de variabila, astfel

$2-t^{2}=y\Rightarrow -2t dt=dy\Rightarrow tdt=\frac{-dy}{2}$
Acum calculam capetele intervalului

$t=0\Rightarrow 2-0^{2}=y\Rightarrow 2=y$

Iar pentru

$t=x\Rightarrow 2-x^{2}=y$

Acum trecem la integrala

$\int|^{x}_{0}f\left(t\right)dt=\int^{x}_{0}t\sqrt{2-t^{2}}dt=\int^{2-x^{2}}_{2}\sqrt{y}\left(\frac{-dy}{2}\right)=$
$-\frac{1}{2}\int^{2-x^{2}}_{2}y^{\frac{1}{2}}dt=-\frac{1}{2} \frac{y^{\frac{1}{2}+1}}{\frac{1}{2}+1}|^{2-x^{2}}_{2}=$
$-\frac{1}{2}\frac{y^{\frac{3}{2}}}{\frac{3}{2}}|^{2-x^{2}}_{2}=-\frac{1}{2}\cdot\frac{2}{3}\sqrt{y^{3}}|^{2-x^{2}}_{2}=$
$-\frac{1}{3}\left(\sqrt{\left(2-x^{2}\right)^{3}}-\sqrt{2^{3}}\right)=$
$-\frac{1}{3}\left(2-x^{2}\right)\sqrt{2-x^{2}}+\frac{1}{3}\cdot 2\sqrt{2}$

Astfel acum daca revenim la limita obtinem ca

$\lim\limits_{x\to 0}{\frac{-\frac{1}{3}\left(2-x^{2}\right)\sqrt{2-x^{2}}+\frac{1}{3}\cdot 2\sqrt{2}}{x^{2}}}=$
$-\frac{1}{3}\lim\limits_{x\to 0}{\frac{\left(2-x^{2}\right)\sqrt{2-x^{2}}- 2\sqrt{2}}{x^{2}}}=$
Observam ca suntem in cazul de nedeterminare $\frac{0}{0}$ si daca aplicam regulile lui L’Hospital obtinem

$-\frac{1}{3}\lim\limits_{x\to 0}{\frac{-2x\cdot\sqrt{2-x^{2}}+\left(2-x^{2}\right)\cdot\frac{1}{2\sqrt{2-x^{2}}}\cdot\left(-2x\right)}{2x}}=$
$-\frac{1}{3}\lim\limits_{x\to 0}{\frac{-2x\left(\sqrt{2-x^{2}}+\frac{2-x^{2}}{2\sqrt{2-x^{2}}}\right)}{2x}}$
$-\frac{1}{3}\lim\limits_{x\to 0}{\frac{-1\left(\sqrt{2-x^{2}}+\frac{2-x^{2}}{2\sqrt{2-x^{2}}}\right)}{1}}=$
$\frac{1}{3}\lim\limits_{x\to 0}{\frac{2\sqrt{2-x^{2}}\cdot\sqrt{2-x^{2}}+2-x^{2}}{2\sqrt{2-x^{2}}}}$

$=\frac{1}{3}\lim\limits_{x\to 0}{\frac{2\left(2-x^{2}\right)+2-x^{2}}{2\sqrt{2-x^{2}}}}=$
$\frac{1}{3}\lim\limits_{x\to 0}{\frac{4-2x^{2}+2-x^{2}}{2\sqrt{2-x^{2}}}}=$
$\frac{1}{3}\lim\limits_{x\to 0}{\frac{6-3x^{2}}{2\sqrt{2-x^{2}}}}=$
$\frac{1}{3}\cdot\frac{6-3\cdot 0^{2}}{2\sqrt{2-0^{2}}}=\frac{1}{3}\cdot\frac{6}{2\sqrt{2}}=\frac{6}{6\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

Subiecte Bacalaureat rezolvate la Analiza matematica

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