Question

Find the sum of the terms of the $G.P$, $a+ar+a{r}^2+...\infty$, where $a$ is the value of $x$ for which the function $7-5^{x-1}-5^{2-x}+2x{\log }_{e}25$ has the greatest value and $r$ is the limit, ${lim}_{x\to 0}{\int }_0^x\frac{t^{2}dt}{\left(x^2\tan \right(\pi +x\left)\right)}$.

Answer 1

Let $f\left(x\right)=7+2x{\log }_{e}25-\frac{5^x}{5}-\frac{25}{5^x}\Rightarrow f^{\prime }\left(x\right)=4{\log }_{e}5-\frac{5^x}{5}\ln 5+25\times 5^{-x}\ln 5f^{\prime }\left(x\right)=0\Rightarrow \ln 5\{4-\frac{5^x}{5}+25\times 5^{-x}\}=0\Rightarrow \frac{5^{2x}-20{\times 5}^x-125}{5^{x+1}}=0\Rightarrow (5^x-25)(5^x+5)=0\Rightarrow (5^x-25)=0or(5^x+5)=0$
but $(5^x+5)=0$ is not possible
Hence $(5^x-25)=0$
$\Rightarrow x=2$
So for this value of $x$ we will get the maximum.
For clarification one can check that for this value of x $f^{\prime \prime }\left(x\right)<0$
Hence $a=2$
Now we calculate the value of $r$.
${lim}_{x\to 0}{\int }_0^x\frac{t^{2}dt}{x^2\tan (\pi +x)}={lim}_{x\to 0}\frac{{\int }_0^x{t}^{2}dt}{x^2\tan (\pi +x)}=\frac{1}{3}{lim}_{x\to 0}\frac{x^3}{x^2\tan x}=\frac{1}{3}$
Hence $r=\frac{1}{3}$
So the sum of infinite G.P is
$\frac{a}{1-r}=\frac{2\times 3}{2}=3$