Subiecte posibile Simulare Bacalaureat 2014

Prezentam subiecte posibile simulare Bacalaureat 2014 pentru subiectul III
Se considera functia $f:\left(0, +\infty\right)\rightarrow R$ definita prin $f\left(x\right)=\frac{1}{x^{2}}+\frac{1}{\left(x+1\right)^{2}}$
a) Sa se calculeze $f^{'}\left(x\right), x\in\left(0,\infty\right)$
b) Sa se demonstreze ca functia f este descrescatoare pe intervalul $\left(0,+\infty\right)$
c) Sa se calculeze $\lim\limits_{x\to +\infty}{x^{3}f^{'}\left(x\right)}$
Solutie:
a) $f^{'}\left(x\right)=\frac{-1\cdot\left(x^{2}\right)^{'}}{\left(x^{2}\right)^{2}}+\frac{-1\cdot \left(\left(x+1\right)^{2}\right)^{'}}{\left[\left(x+1\right)^{2}\right]^{2}}=\frac{-1\cdot 2x}{x^{4}}+\frac{-1\cdot 2\left(x+1\right)\cdot \left(x+1\right)^{'}}{\left(x+1\right)^{4}}=\frac{-2x}{x^{4}}+\frac{-2\left(x+1\right)\cdot 1}{\left(x+1\right)^{4}}=\frac{-2}{x^{3}}-\frac{2}{\left(x+1\right)^{3}}=\frac{-2\left(x+1\right)^{3}-2\cdot x^{3}}{x^{3}\left(x+1\right)^{3}}=\frac{-2\left[\left(x+1\right)^{3}+x^{3}\right]}{x^{3}\left(x+1\right)^{3}}$ .
b) Observam ca $f^{'}\left(x\right)<0$ , deci f este strict descrescatoare pe $\left(0,+\infty\right)$
c) Acum sa calculam limita
$\lim\limits_{x\to +\infty}{x^{3}\cdot \left(\frac{-2\left[\left(x+1\right)^{3}+x^{3}\right]}{x^{3}\left(x+1\right)^{3}}\right)}=$
$\lim\limits_{x\to +\infty}{\frac{-2\left[\left(x+1\right)^{3}+x^{3}\right]}{\left(x+1\right)^{3}}}=$
$\lim\limits_{x\to +\infty}{-2\left[\cdot \frac{\left(x+1\right)^{3}}{\left(x+1\right)^{3}}+\frac{x^{3}}{\left(x+1\right)^{3}}\right]}=$
$-2\lim\limits_{x\to +\infty}{1+\frac{x^{3}}{\left(x+1\right)^{3}}}=-2\left(1+1\right)=-2\cdot 2=-4$ .
2) Se considera functia $f:\left(0, +\infty\right)\rightarrow R, f\left(x\right)=\frac{\ln x}{x}+x$
a) Sa se calculeze $\int^{e}_{1}\left(f\left(x\right)-\frac{\ln x}{x}\right)dx$
b) Sa se calculeze $\int^{e}_{1} f\left(x\right) dx$
c) Sa se determine ratia progresiei aritmetice avand termenul general $I_{n}=\int^{e^{n+1}}_{e^{n}}\left(f\left(x\right)-x\right) dx, n\geq 1$
Solutie:
Incepem prin a calcula integrala
a) $\int^{e}_{1}\left(\frac{\ln x}{x}+x-\frac{\ln x}{x}\right)dx=\int^{e}_{2} xdx=\frac{x^{2}}{2}|^{e}_{1}=\frac{e^{2}}{2}-\frac{1}{2}=\frac{e^{2}-1}{2}$ .
b) $\int^{e}_{1]}f\left(x\right) dx=\int^{e}_{1}\left(\frac{\ln x}{x}+x\right)dx=$
$\frac{\ln^{2} x}{2}|^{e}_{1}+\frac{x^{2}}{2}|^{e}_{1}=$
$\frac{\ln^{2} e}{2}-\frac{\ln^{2} 1}{2}+\frac{e^{2}}{2}-\frac{1}{2}=$
$\frac{1}{2}-0+\frac{e^{2}-1}{2}=\frac{1}{2}+\frac{e^{2}-1}{2}=$
$\frac{1+e^{2}-1}{2}=\frac{e^{2}}{2}$
Deci avem ca
$\int^{e}_{1}\left(\frac{\ln x}{x}\right)dx=\ln x\cdot\ln x|^{e}_{1}-\int^{e}_{1}\frac{\ln x}{x}dx\Rightarrow \int^{e}_{1}\frac{\ln x}{x}+\int^{e}_{1}\frac{\ln x}{x}=\ln^{2} x|^{e}_{1}\Rightarrow 2\int^{e}_{1}\frac{\ln x}{x}dx =\ln^{2} x|^{e}_{1}\Rightarrow \int^{e}_{1}\frac{\ln x}{x}dx=\frac{\ln^{2} x}{2}|^{e}_{1}$
Observam ca am luat separat si am calculat integrala $\int^{e}_{1}\frac{\ln x}{x} dx$ , am folosit metoda integrarii prin parti, adica am luat $f^{'}\left(x\right)=\ln x$ , iar $g\left(x\right)=\ln x$ , iar prin aplicarea si partea celei de-a doua a integrarii prin parti obtinem integrala de la care am plecat si astfel integrala care am gasit-o am trecut-o cu semn schimbat in partea stanga si astfel am obtinut de doua ori integrala de mai sus si astfel daca impartim prin 2 obtinem integrala de mai sus.
c) $I_{n}=\int^{e^{n+1}}_{e^{n}}\left(f\left(x\right)-x\right)dx=\int^{e^{n+1}}_{e^{n}}\left(\frac{\ln x}{x}+x-x\right)dx=\int^{e^{n+1}}_{e^{n}}\frac{\ln x}{x}dx$
Notam $\ln x=t \\ \frac{1}{x}dx=1\cdot dt \\ x=e^{n+1}\Rightarrow \ln e^{n+1}=t\Rightarrow n+1\cdot \ln e=t\Rightarrow n+1=t \\x=e^{n}\Rightarrow \ln e^{n}=t\Rightarrow n\cdot \ln e=t\Rightarrow t=n$
Deci integrala devine:
$\int^{e^{n+1}}_{e^{n}}\frac{\ln x}{x}dx=\int^{n+1}_{n}t dt=\frac{t^{2}}{2}|^{n+1}_{n}=\frac{\left(n+1\right)^{2}}{2}-\frac{n^{2}}{2}=\frac{\left(n+1\right)^{2}-n^{2}}{2}=\frac{n^{2}+2n+1-n^{2}}{2}=\frac{2n+1}{2}$
Am gasit ca $I_{n}=\frac{2n+1}{2}$
Acum daca calculam
$I_{1}=\frac{2\cdot 1+1}{2}=\frac{2+1}{2}=\frac{3}{2} \\I_{2}=\frac{2\cdot 2+1}{2}=\frac{5}{2}$
I_{2}-I_{1}=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1$
Deci am gasit ca r=1.
Mai calculam
$I_{n+1}-I_{n}=\frac{2\left(n+1\right)+1}{2}-\frac{2n+1}{2}=\frac{2n+2+1}{2}-\frac{2n+1}{2}=\frac{2n+3-2n-1}{2}=\frac{2}{2}=1$ constant.
Si astfel gasim ca $I_{n}$ este progresie aritmetica cu ratia 1.

Subiecte posibile Simulare Bacalaureat 2014

Nota !Acestea au fost subiecte posibile simulare bacalaureat 2014 deci nimic nu este sigur.

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