Question

If $2f\left(x\right)-3f\left(\frac{1}{x}\right)=x^2(x\ne 0)$, then $f\left(2\right)$ is equal to :

Answer 1

$2f\left(x\right)-3f\left(\frac{1}{x}\right)=x^2...\left(1\right)$

$\text{Replace}x\text{by}\frac{1}{x}\text{in equation 1};$

$2f\left(\frac{1}{x}\right)-3f\left(x\right)=\frac{1}{x^2}...\left(2\right)$

Multiply eqn. 1 by 2 and eqn. 2 by 3 and add them.

On adding, we get

$-5f\left(x\right)=2x^2+\frac{3}{x^2}$

$f\left(x\right)=\frac{-1}{5}(2x^2+\frac{3}{x^2})$

$f\left(2\right)=\frac{-1}{5}(2\times 2^2+\frac{3}{2^2})=\frac{-1}{5}(8+\frac{3}{4})$

$=\frac{-1}{5}\left(\frac{35}{4}\right)=\frac{-7}{4}$

Therefore, Option (A) is correct.