Question

Let the first term $a$ of an infinite G.P. is the value of $x$, where the function $f\left(x\right)=7+2x{\log }_{e}25-5^{x-1}-5^{2-x}$ has the greatest value and the common ratio $r$ is equal to $\underset{x\to 0}{lim}\underset{0}{\overset{x}{\int }}\frac{t^2}{x^2\tan (\pi +x)}dt$. Also, let $S$ be the sum of infinite terms of G.P.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& a& \text{(P)}& 4\\ \text{(B)}& \frac{1}{r}& \text{(Q)}& 3\\ \text{(C)}& S& \text{(R)}& 2\\ \text{(D)}& a-rS& \text{(S)}& 1\\ & & \text{(T)}& 5\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(Q)}$
• B. $\text{(A)}\to \text{(P)},\text{(B)}\to \text{(R)}$
• C. $\text{(A)}\to \text{(S)},\text{(B)}\to \text{(P)}$
• D. $\text{(A)}\to \text{(Q)},\text{(B)}\to \text{(T)}$

Answer 1

The correct option is A $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(Q)}$

$f\left(x\right)=7+2x\ln 25-5^{x-1}-5^{2-x}$
$\Rightarrow f^{\prime }\left(x\right)=2\ln 25-5^{x-1}\ln 5+5^{2-x}\ln 5$
$=\ln 5(4-\frac{5^x}{5}+\frac{25}{5^x})$
$=-\frac{\ln 5}{5\cdot 5^x}(5^{2x}-20\cdot 5^x-125)$
$=-\frac{\ln 5}{5^{x+1}}(5^x-25)(5^x+5)$

For maximum value of $f\left(x\right)$, we equate $f^{\prime }\left(x\right)$ to $0$
$(5^x-25)(5^x+5)=0$
$\Rightarrow 5^x-25=0$
$\Rightarrow x=2$
At $x=2$, $f^{\prime }\left(x\right)$ changes sign from $+ve$ to $-ve$.
Hence, $x=2$ is point of maximum.
$\therefore a=2$

Now, $r=\underset{x\to 0}{lim}\underset{0}{\overset{x}{\int }}\frac{t^2}{x^2\tan (\pi +x)}dt$
$\Rightarrow r=\underset{x\to 0}{lim}\frac{\underset{0}{\overset{x}{\int }}{t}^{2}dt}{x^2\tan x}$
$\Rightarrow r=\underset{x\to 0}{lim}\frac{x^2}{3x^2}=\frac{1}{3}$ (Using L Hospital rule)

$\therefore S=\frac{a}{1-r}=\frac{2}{1-\frac{1}{3}}=3$
$\therefore a-Sr=2-\frac{3}{3}=1$