Question

Subiectul III
Pct 1 c
2c
Multumesc!

Answer 1

1c. Folosim punctul b, la care am aflat punctul de extrem, x=3, care este punct de minim global:

$x= 3\text{ punct de minim global }\Leftrightarrow f(3)\leq f(x), \forall x \in (1, \infty)\\\left. \begin{aligned} &a>1\Rightarrow a\in(1,\infty)\Rightarrow f(a)\geq f(3)\\ &b>1\Rightarrow b\in(1,\infty)\Rightarrow f(b)\geq f(3) \end{aligned}\right\}\Rightarrow f(a)+f(b)\geq2f(3) = 2\displaystyle\cdot\frac{3^2+3}{3-1}=12$

2c. F primitiva lui f, $F'(x) = f(x)$.

$\displaystyle\int f(x)F^{2014}(x)dx=\int F'(x)F^{2014}(x)dx\\\text{Substitu\c tie: }\\ \begin{aligned} &F(x) = t\\ &F'(x)dx=dt \end{aligned}\\\text{Integrala devine:} \int t^{2014}dt=\frac{1}{2015}t^{2015}+C = \frac1{2015}F^{2015}(x)+C$

$\displaystyle\int\limits_1^5 f(x)F^{2014}(x)dx=\frac1{2015}F^{2015}(x)\big|_1^5=\frac1{2015}\left(F^{2015}(5)-F^{2015}(1)\right)=\\\\=\frac1{2015}\left[\left(\frac{2\cdot5-1}{3}\sqrt{2\cdot5-1}\right)^{2015}-\left(\frac{2\cdot1-1}{3}\sqrt{2\cdot1-1}\right)^{2015}\right]=\\\\\frac1{2015}\left[(3^2)^{2015}-\left(\frac13\right)^{2015}\right]=\frac1{2015}\left(3^{2\cdot2015}-\frac1{3^{2015}}\right)=\\\\=\frac1{2015}\frac{3^{2\cdot2015+2015}-1}{3^{2015}} = \frac{3^{6045}-1}{2015\cdot3^{2015}}$