MDoubt No 000008 - GATE

GATE [CS, 2015, 2 marks]
Question : If for non-zero $x$, $af(x)+bf\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-25$ where $a\ne b$ then $\displaystyle\int\limits_1^2f(x)dx$ is

(a) $\displaystyle\frac{1}{a^2-b^2}\left[a(\ln 2-25)\frac{47b}{2}\right]$
(b) $\displaystyle\frac{1}{a^2-b^2}\left[a(2\ln 2-25)-\frac{47b}{2}\right]$
(c) $\displaystyle\frac{1}{a^2-b^2}\left[a(2\ln 2-25)+\frac{47b}{2}\right]$
(d) $\displaystyle\frac{1}{a^2-b^2}\left[a(\ln 2-25)-\frac{47b}{2}\right]$

Solution : Given $af(x)+bf\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-25\quad\quad\quad\ldots(1)$
replacing $\dfrac{1}{x}$ by $x$ in $(1)$
$\implies af\left(\dfrac{1}{x}\right)+bf(x)=x-25\quad\quad\quad\ldots(2)$
Solving $(1)$ and $(2)$
$\implies f(x)=\dfrac{1}{a^2-b^2}\left[a\left(\dfrac{1}{x}-25\right)-b(x-25)\right]$
$\displaystyle\int\limits_1^2f(x)dx$
$=\dfrac{1}{a^2-b^2}\left[a(\ln x-25x)-\left(\dfrac{x^2}{2}-25x\right)\right]_1^2$
$=\displaystyle\frac{1}{a^2-b^2}\left[a(\ln 2-25)\frac{47b}{2}\right]$
CORRECT ANSWER : A