Question

If for a non-zero $x,$ the function $f\left(x\right)$ satisfies the equation $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5(a\ne b)$ then $f^{\prime }\left(x\right)$ is equal to

• A. $\frac{1}{b^2-a^2}(\frac{a}{x^2}+b)$
• B. $\frac{1}{a^2-b^2}(\frac{a}{x^2}+b)$
• C. $\frac{1}{a^2-b^2}(\frac{a}{x^2}-b)$
• D. none of these

Answer 1

The correct option is A $\frac{1}{b^2-a^2}(\frac{a}{x^2}+b)$

We have, $af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5$

Substituting $\frac{1}{x}$ in place of $x,$ we have

$af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5$ ...(2)

Eliminating $f\left(\frac{1}{x}\right)$ from equation (1) and (2), we have

$(a^2-b^2)f\left(x\right)=a(\frac{1}{x}-5)-b(x-5)$

$\Rightarrow f\left(x\right)=\frac{1}{a^2-b^2}[\frac{a}{x}-bx+5(b-a)]$

$\therefore f^{\prime }\left(x\right)=\frac{1}{b^2-a^2}[\frac{a}{x^2}+b]$