Question:
If $2 f(x)-3 f\left(\frac{1}{x}\right)=x^{2}(x \neq 0)$, then $f(2)$ is equal to
(a) $-\frac{7}{4}$
(b) $\frac{5}{2}$
(c) −1
(d) None of these
Solution:
(a) $-\frac{7}{4}$
$2 f(x)-3 f\left(\frac{1}{x}\right)=x^{2} \quad(x \neq 0)$
Replacing $x$ by $\frac{1}{x}$ :
$2 f\left(\frac{1}{x}\right)-3 f(x)=\frac{1}{x^{2}}$ ....(2)
Solving equations $(1) \&(2)$
$-5 f(x)=\frac{3}{x^{2}}+2 x^{2}$
$\Rightarrow f(x)=\frac{-1}{5}\left(\frac{3}{x^{2}}+2 x^{2}\right)$
Thus, $f(2)=\frac{-1}{5}\left(\frac{3}{4}+2 \times 4\right)$
$=\frac{-1}{5}\left(\frac{3+32}{4}\right)$
$=\frac{-7}{4}$
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