Subiectul III

Se consideră funcția $f:(0,+\infty)\to\mathbb{R}, \ f(x)=\ln{(x+1)}-\ln{x}.$

Calculați $f'(x), \ x\in\(0,+\infty).$
Arătați că funcția este descrescătoare.
Calculați $\displaystyle\lim_{x\to+\infty} {x f(x)}.$

Calculăm

$\begin{align*} f'(x) &= [\ln{(x+1)}-\ln{x}]'\\\\ &=[\ln{(x+1)}]'-(\ln{x})'\\\\ &=\displaystyle\frac{1}{x+1}\cdot(x+1)'-\frac{1}{x}\\\\ &=\displaystyle\frac{x'+1'}{x+1}-\frac{1}{x}\\\\ &=\displaystyle\frac{1+0}{x+1}-\frac{1}{x}\\\\ &=\displaystyle\frac{1}{x+1}-\frac{1}{x}\\ \\&=\displaystyle\frac{x-(x+1)}{x(x+1)}\\\\ &=\displaystyle\frac{x-x-1}{x(x+1)}\\\\ &=-\displaystyle\frac{1}{x(x+1)} \end{align*}$

$\begin{align*}\Leftrightarrow f'(x) =-\displaystyle\frac{1}{x(x+1)} ,\forall\ x\in(0,+\infty). \end{align*}$

Arătăm că funcția este descrescătoare.

$\begin{align*} f'(x) =-\displaystyle\frac{1}{x(x+1)} \end{align*}$ , pentru orice $x\in(0,+\infty).$

f'(x)<0, pentru orice $x\in(0,+\infty) ,$ de unde ne rezultă că funcția este descrescătoare.

Calculăm $\displaystyle\lim_{x\to+\infty} {x f(x)}.$

$\begin{align*} \displaystyle\lim_{x\to+\infty} {x f(x)} &= \displaystyle\lim_{x\to+\infty} \displaystyle\frac{f(x)}{\displaystyle\frac{1}{x}}\\\\ &= \displaystyle\lim_{x\to+\infty}\displaystyle\frac{\ln{(x+1)}-\ln{x}}{\displaystyle\frac{1}{x}} \end{align*}$

Fracția $\displaystyle\frac{\ln{(x+1)}-\ln{x}}{\displaystyle\frac{1}{x}}$ este un caz de $\displaystyle\frac{\infty}{\infty},$ rezultă că aplicăm regula lui L'Hôpital astfel:

$\begin{align*} \displaystyle\lim_{x\to+\infty} \displaystyle\frac{f(x)}{\displaystyle\frac{1}{x}}&=\displaystyle\lim_{x\to+\infty} \displaystyle\frac{f'(x)}{\displaystyle\Big(\frac{1}{x}\Big)'}\\\\ &= \displaystyle\lim_{x\to+\infty}\displaystyle\frac{[\ln{(x+1)}-\ln{x}]'}{\displaystyle -\frac{1}{x^2}} \\\\ &=\displaystyle\lim_{x\to+\infty}\displaystyle\frac{[\ln{(x+1)}]'-[\ln{x}]'}{\displaystyle -\frac{1}{x^2}} \\\\ &=\displaystyle\lim_{x\to+\infty}\displaystyle\frac{\displaystyle\frac{1}{x+1}-\frac{1}{x}}{\displaystyle -\frac{1}{x^2}} \\\\ &= \displaystyle\lim_{x\to+\infty}\displaystyle\frac{\displaystyle-\frac{1}{x(x+1)}}{\displaystyle -\frac{1}{x^2}}\\\\ &=\displaystyle\lim_{x\to+\infty}\Big[\displaystyle-\frac{1}{x(x+1)} : \Big({\displaystyle -\frac{1}{x^2}}\Big)\Big]\\ \\&=\displaystyle\lim_{x\to+\infty}\Big[\displaystyle\frac{1}{x(x+1)} \cdot \Big({x^2}\Big)\Big]\\ \\&=\displaystyle\lim_{x\to+\infty}\displaystyle\frac{x}{x+1}\\ \end{align*}$

Avem din nou o fracție de tipul $\displaystyle\frac{\infty}{\infty},$ rezultă că aplicăm din nou regula lui L'Hôpital:

$\begin{align*} \displaystyle\lim_{x\to+\infty}\displaystyle\frac{x}{x+1}&=\displaystyle\lim_{x\to+\infty}\displaystyle\frac{x'}{(x+1)'}\\\\ &= \displaystyle\lim_{x\to+\infty}\displaystyle\frac{1}{x'+1'}\\ \\&= \displaystyle\lim_{x\to+\infty}\displaystyle\frac{1}{1+0}\\\\ &=\displaystyle\frac{1}{1}\\ \\&=1 \end{align*}$

$\Rightarrow \displaystyle\lim_{x\to+\infty} {x f(x)}=1.$

Se consideră funcția $f:(-2,+\infty)\to\mathbb{R}, \ f(x)=\displaystyle\frac{x}{x+2}.$

Calculați $\displaystyle\int_{0}^{1} (x+2) f(x) dx.$
Arătați că $\displaystyle\int_{2013}^{2014} \Big(f(x)+(x+2)f'(x)\Big) dx=1.$ .
Determinați volumul corpului obținut prin rotația în jurul axei a graficului funcției $g:[1,2]\to\mathbb{R}, \ g(x)=\displaystyle\frac{x}{f(x)}.$

Calculăm $\displaystyle\int_{0}^{1} (x+2) f(x) dx.$

$\begin{align*} \displaystyle\int_{0}^{1} (x+2) f(x) dx &= \displaystyle\int_{0}^{1} (x+2)\cdot \displaystyle\frac{x}{x+2}\ dx\\\\ &= \displaystyle\int_{0}^{1} xdx\\\\ &=\displaystyle\frac{x^2}{2}\Big|_0^1\\\\ &=\displaystyle\frac{1^2}{2}-\displaystyle\frac{0^2}{2}\\\\ &=\displaystyle\frac{1}{2} \end{align*}$

$\Leftrightarrow \displaystyle\int_{0}^{1} (x+2) f(x) dx=\frac{1}{2}.$

Calculăm $f'(x), \ f(x)+(x+2)f'(x)$ și apoi calculăm $\displaystyle\int_{2013}^{2014} \Big(f(x)+(x+2)f'(x)\Big) dx.$

$\begin{align*} f'(x) &= \Big(\displaystyle\frac{x}{x+2}\Big)'\\\\ &= \displaystyle\frac{x'\cdot(x+2)-x(x+2)'}{(x+2)^2}\\ \\&=\displaystyle\frac{1'\cdot(x+2)-x(x'+2')}{(x+2)^2}\\ \\&=\displaystyle\frac{x+2-x(1+0)}{(x+2)^2}\\ \\&=\displaystyle\frac{x+2-x}{(x+2)^2}\\ \\&=\displaystyle\frac{2}{(x+2)^2} \end{align*}$

$\begin{align*}\Leftrightarrow f'(x) &=\displaystyle\frac{2}{(x+2)^2} .\end{align*}$

Atunci:

$\begin{align*} f(x)+(x+2)f'(x) &= \frac{x}{x+2}+(x+2)\frac{2}{(x+2)^2}\\\\ &=\frac{x}{x+2}+\frac{2}{x+2}\\\\ &=\frac{x+2}{x+2}\\\\ &=1 \end{align*}$

Avem că f(x)+(x+2)f'(x)=1, pentru orice $x\in(-2,+\infty).$

Revenind la integrală, obținem:

$\begin{align*} \displaystyle\int_{2013}^{2014} \Big(f(x)+(x+2)f'(x)\Big) dx &=\int_{2013}^{2014} 1\ dx\\\\ &= x\Big|_{2013}^{2014}\\\\ &= 2014-2013\\\\ &=1 \end{align*}$

$\Rightarrow \displaystyle\int_{2013}^{2014} \Big(f(x)+(x+2)f'(x)\Big) dx=1.$

Calculăm .

$\begin{align*} g(x)&=\displaystyle\frac{x}{f(x)}\\\\ &=\displaystyle\frac{x}{\displaystyle\frac{x}{x+2}}\\\\ &=x: \displaystyle\frac{x}{x+2}\\\\ &=x\cdot\displaystyle\frac{x+2}{x}\\\\ &=x+2 \end{align*}$

$\begin{align*} \Leftrightarrow g(x)=x+2. \end{align*}$

Calculăm volumul corpului obținut prin rotația în jurul axei a graficului funcției $g:[1,2]\to\mathbb{R}, \ g(x)=\displaystyle\frac{x}{f(x)}.$

$\begin{align*} V &= \int_{1}^{2} \pi \cdot g^2(x)\ dx \\ \\ &= \pi \int_{1}^{2} (x+2)^2\ dx \\\\ &=\pi \displaystyle\frac{(x+2)^3}{3}\Big|_1^2\\ \\&=\pi \Big[\displaystyle\frac{(2+2)^3}{3}-\displaystyle\frac{(1+2)^3}{3}\Big]\\\\ &=\pi \Big(\displaystyle\frac{4^3}{3}-\displaystyle\frac{3^3}{3}\Big)\\\\ &=\pi \Big(\displaystyle\frac{64}{3}-\displaystyle\frac{27}{3}\Big)\\ \\&=\pi \cdot \displaystyle\frac{64-27}{3}\\\\ &=\pi \cdot \displaystyle\frac{37}{3} \end{align*}$

$\begin{align*} \Leftrightarrow V = \displaystyle\frac{37\pi}{3}. \end{align*}$