Question

Let f be a function satisfying $2f\left(x\right)-3f\left(\frac{1}{x}\right)=x^2$ for any $x\ne 0$, then the value of f(2) is

• A. -2
• B. $\frac{-7}{4}$
• C. $-\frac{7}{8}$
• D. 4

Answer 1

The correct option is B $\frac{-7}{4}$

$2f\left(x\right)-3f\left(\frac{1}{x}\right)=x^2\dots \dots \left(i\right)Replacingxby\frac{1}{x}2f\left(\frac{1}{x}\right)-3f\left(x\right)=\frac{1}{x^2}\dots \dots \left(ii\right)$
Solving (i) and (ii) we get
$-5f\left(x\right)=2x^2+\frac{3}{x^2}f\left(x\right)=\frac{-1}{5}(2x^2+\frac{3}{x^2})\therefore f\left(2\right)=\frac{-1}{5}(8+\frac{3}{4})=\frac{-7}{4}$