Question

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

(a) $-\frac{7}{4}$
(b) $\frac{5}{2}$
(c) −1
(d) None of these

Answer 1

(a) $-\frac{7}{4}$

2f (x) − $3f\left(\frac{1}{x}\right)=x^2$ (x ≠ 0) ....(1)
$$\begin{gathered} \mathrm{Replacing}x\text{by}\frac{1}{x}: \\ 2f\left(\frac{1}{x}\right)-3f\left(x\right)=\frac{1}{x^2}...\left(2\right) \\ \mathrm{Solving}\mathrm{equations}\left(1\right)\&\left(2\right) \\ -5f\left(x\right)=\frac{3}{x^2}+2x^2 \\ \Rightarrow f\left(x\right)=\frac{-1}{5}\left(\frac{3}{x^2}+2x^2\right) \\ \mathrm{Thus},f\left(2\right)=\frac{-1}{5}\left(\frac{3}{4}+2\times 4\right) \\ =\frac{-1}{5}\left(\frac{3+32}{4}\right) \\ =\frac{-7}{4} \end{gathered}$$