Subiectul III

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

Arătați că $\frac{{f}'(x)}{f(x)}=-\frac{x}{x+1}$ pentru orice $x\in(0,+\infty)$ .
Arătați că funcția este descrescătoare pe $(0,+\infty)$ .
Determinați ecuația asimptotei oblice la graficul funcției $g:(0,+\infty)\to\mathbb{R}$ , $g(x)=\frac{e^{2x}\cdot f^2(x)}{x}$ .

Calculăm $f'(x), \ \frac{{f}'(x)}{f(x)}.$

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

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

Atunci

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

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

Studiem semnul derivatei I:

$\begin{align*} f'(x)=-\displaystyle\frac{x}{e^x} <0,\forall x\in (0,+\infty) \end{align*}$

$\begin{align*} \Rightarrow f \end{align*}$ este descrescătoare pe $(0,+\infty)$ .

Calculăm

$\begin{align*} g(x)&=\displaystyle\frac{e^{2x}\cdot \left(\displaystyle\frac{x+1}{e^x}\right)^2}{x} \\ \\&=\displaystyle\frac{e^{2x}\cdot \displaystyle\frac{(x+1)^2}{e^{2x}}}{x} \\\\&= \displaystyle \frac{(x+1)^2}{x} \\\\&= \displaystyle\frac{x^2+2x+1}{x} \end{align*}$

$\begin{align*} \Leftrightarrow g(x)&= \displaystyle\frac{x^2+2x+1}{x} ,\forall\ x\in(0,+\infty). \end{align*}$

Ecuația asimptotei oblice la graficul funcției $\begin{align*} g \end{align*}$ este $\begin{align*} y=mx+n, \end{align*}$ unde

$\begin{align*} m=\lim_{x\to+\infty}\displaystyle\frac{g(x)}{x}; \end{align*}$

$\begin{align*} n=\lim_{x\to+\infty}[g(x)-mx]. \end{align*}$

$\begin{align*} m&= \lim\limits_{x\to+\infty}\displaystyle\frac{g(x)}{x}\\\\ &= \lim\limits_{x\to+\infty}\displaystyle\frac{\displaystyle\frac{x^2+2x+1}{x}}{x} \\\\ &= \lim\limits_{x\to+\infty }\displaystyle\frac{x^2+2x+1}{x^2} \\\\&= \lim\limits_{x\to+\infty }\displaystyle\frac{x^2\Big(1+\displaystyle\frac{2}{x}+\frac{1}{x^2}\Big)}{x^2} \\ \end{align*}$

Cum

$\begin{align*} \displaystyle\frac{2}{x}\underset{x\to+\infty}{\longrightarrow} 0 \end{align*}$

și

$\begin{align*} \displaystyle\frac{1}{x^2}\underset{x\to+\infty}{\longrightarrow} 0 \end{align*}$

$\begin{align*} \Rightarrow \lim\limits_{x\to+\infty }\displaystyle\frac{x^2\Big(1+\displaystyle\frac{2}{x}+\frac{1}{x^2}\Big)}{x^2}&= \lim\limits_{x\to+\infty }\Big(1+\displaystyle\frac{2}{x}+\frac{1}{x^2}\Big)\\ &=1+0+0\\&=1 \end{align*}$

$\begin{align*} \Leftrightarrow m&=1. \end{align*}$

$\begin{align*} n&=\lim_{x\to+\infty}[g(x)-mx]\\\\ &=\lim_{x\to+\infty}\Big(\displaystyle\frac{x^2+2x+1}{x}-1\cdot x\Big)\\\\ &=\lim_{x\to+\infty}\displaystyle\frac{x^2+2x+1-x\cdot x}{x}\\\\ &=\lim_{x\to+\infty}\displaystyle\frac{x^2+2x+1-x^2}{x}\\\\ &=\lim_{x\to+\infty}\displaystyle\frac{2x+1}{x}\\\\ &=\lim_{x\to+\infty}\displaystyle\frac{x\Big(2+\displaystyle\frac{1}{x}\Big)}{x}\\\\ &=\lim_{x\to+\infty}\Big(2+\displaystyle\frac{1}{x}\Big) \end{align*}$

$\begin{align*} \displaystyle\frac{1}{x}\underset{x\to+\infty}{\longrightarrow} 0 \end{align*}$

$\begin{align*} \Rightarrow \lim_{x\to+\infty}\Big(2+\displaystyle\frac{1}{x}\Big)&=2+0\\&=2 \end{align*}$

$\begin{align*} \Leftrightarrow n=2. \end{align*}$

Rezultă că ecuația asimptotei oblice la graficul funcției $\begin{align*} g \end{align*}$ este y=x+2.

Se consideră funcția $f:\mathbb{R}\to\mathbb{R}$ , $f(x)=x^{2012}+x^{2011}+x^2+x$ .

Determinați primitiva $F:\mathbb{R}\to\mathbb{R}$ a funcției , care verifică relația .
Calculați $\int_{0}^{1}\frac{f(x)}{x+1}\mathrm{d}x$ .
Calculați volumul corpului obținut prin rotația, în jurul axei , a graficului funcției $g:[1,2]\to\mathbb{R}$ , $g(x)=f(x)-x^{2012}-x^{2011}$ .

Aflăm primitiva funcției astfel:

$\begin{align*} F(x)&=\int f(x) \mathrm{d}x\\ \\ &=\int (x^{2012}+x^{2011}+x^2+x)\mathrm{d}x \\\\&= \displaystyle\frac{x^{2013}}{2013}+\displaystyle\frac{x^{2012}}{2012}+\displaystyle\frac{x^3}{3}+\displaystyle\frac{x^2}{2}+C \end{align*}$

$\begin{align*} \Leftrightarrow F(x)= \displaystyle\frac{x^{2013}}{2013}+\displaystyle\frac{x^{2012}}{2012}+\displaystyle\frac{x^3}{3}+\displaystyle\frac{x^2}{2}+C. \end{align*}$

Avem F(0)=1 . Calculăm F(0).

$\begin{align*} F(0)&=\displaystyle\frac{0^{2013}}{2013}+\displaystyle\frac{0^{2012}}{2012}+\displaystyle\frac{0^3}{3}+\displaystyle\frac{0^2}{2}+C\\\\ &=\displaystyle\frac{0}{2013}+\displaystyle\frac{0}{2012}+\displaystyle\frac{0}{3}+\displaystyle\frac{0}{2}+C\\\\ &=0+0+0+0+C\\\\ &=C \end{align*}$

$\begin{align*} \begin{cases} & F(0)=1\\ &F(0)=C \end{cases} \Rightarrow C=1.\end{align*}$

Avem primitiva $F:\mathbb{R}\to\mathbb{R}$ , $\begin{align*} F(x)&= \displaystyle\frac{x^{2013}}{2013}+\displaystyle\frac{x^{2012}}{2012}+\displaystyle\frac{x^3}{3}+\displaystyle\frac{x^2}{2}+1. \end{align*}$

Calculăm $\int_{0}^{1}\frac{f(x)}{x+1}\mathrm{d}x$ .

$\begin{align*} \int_{0}^{1}\displaystyle\frac{f(x)}{x+1}&=\int_{0}^{1}\displaystyle\frac{x^{2012}+x^{2011}+x^2+x}{x+1}\mathrm{d}x\\\\&= \int_{0}^{1} \displaystyle\frac{x^{2011}(x+1)+x(x+1)}{x+1}\mathrm{d}x \\\\&= \int_{0}^{1}\displaystyle\frac{(x^{2011}+x)(x+1)}{x+1}\mathrm{d}x \\\\&=\int_{0}^{1}(x^{2011}+x)\mathrm{d}x\\\\&= \left(\displaystyle\frac{x^{2012}}{2012}+\displaystyle\frac{x^2}{2}\right)\Big|_0^1 \\\\&= \displaystyle\frac{1}{2012}+\displaystyle\frac{1}{2}\\\\&= \displaystyle\frac{1+1006\cdot1}{2012} \\\\&= \displaystyle\frac{1007}{2012} \end{align*}$

$\begin{align*} \Leftrightarrow \int_{0}^{1}\displaystyle\frac{f(x)}{x+1}&=\displaystyle\frac{1007}{2012}. \end{align*}$

Calculăm $\begin{align*} g(x) \end{align*}$ .

$\begin{align*} g(x)&=f(x)-x^{2012}-x^{2011}\\&=x^{2012}+x^{2011}+x^2+x-x^{2012}-x^{2011} \\&= x^2+x \end{align*}$

$\begin{align*} \Leftrightarrow g(x)&= x^2+x, \ \forall\ x\in [1,2]. \end{align*}$

Calculăm volumul corpului astfel:

$\begin{align*} &V=\pi \int_{1}^{2}g^2(x)\mathrm{d}x\\\\ &= \pi \int_{1}^{2}(x^2+x)^2\mathrm{d}x\\\\ &=\pi \int_{1}^{2}\Big((x^2)^2+2\cdot x^2\cdot x+x^2\Big)^2\mathrm{d}x\\\\ &=\pi \int_{1}^{2}(x^4+2x^3+x^2)\mathrm{d}x \\\\ &= \pi \left(\displaystyle\frac{x^5}{5}+2\cdot\displaystyle\frac{x^4}{4}+\displaystyle\frac{x^3}{3}\right)\Big|_1^2\\\\ &=\pi \left(\displaystyle\frac{2^5}{5}+2\cdot\displaystyle\frac{2^4}{4}+\displaystyle\frac{2^3}{3}\right)-\pi \left(\displaystyle\frac{1^5}{5}+2\cdot\displaystyle\frac{1^4}{4}+\displaystyle\frac{1^3}{3}\right)\\\\ &= \pi \left(\displaystyle\frac{32}{5}+\displaystyle\frac{16}{2}+\displaystyle\frac{8}{3}\right)-\pi \left(\displaystyle\frac{1}{5}+\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}\right)\\\\ &=\pi\Big[\displaystyle\frac{32}{5}+\displaystyle\frac{16}{2}+\displaystyle\frac{8}{3}-\Big(\displaystyle\frac{1}{5}+\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}\Big)\Big] \end{align*}$

$\begin{align*} &=\pi \Big(\displaystyle\frac{32}{5}+\displaystyle\frac{16}{2}+\displaystyle\frac{8}{3}-\displaystyle\frac{1}{5}-\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}\Big) \\\\ &= \pi\Big(\displaystyle\frac{32-1}{5}+\displaystyle\frac{16-1}{2}+\displaystyle\frac{8-1}{3}\Big)\\\\ &= \pi\Big(\displaystyle\frac{31}{5}+\displaystyle\frac{15}{2}+\displaystyle\frac{7}{3}\Big)\\\\ &= \pi \cdot \displaystyle\frac{31\cdot2\cdot3+15\cdot5\cdot3+7\cdot5\cdot2}{5\cdot2\cdot3}\\\\ &= \pi \cdot \displaystyle\frac{186+225+70}{30}\\\\ &= \pi \cdot \displaystyle\frac{481}{30}\\\\ &= \displaystyle\frac{481\pi}{30}\\\\ \end{align*}$

$\begin{align*} \Leftrightarrow V=\displaystyle\frac{481\pi}{30}. \end{align*}$