Explanation

(b)
Given, $3f(x) - 2f\left( \frac{1}{x} \right) = x\ldots(i)$
Let $\frac{1}{x} = y,then\ 3f\left( \frac{1}{y} \right) - 2f(y) = \frac{1}{y}$
$\Rightarrow - 2f(y) + 3f\left( \frac{1}{y} \right) = \frac{1}{y}$
$\Rightarrow - 2f(x) + 3f\left( \frac{1}{x} \right) = \frac{1}{x}\ldots(ii)$
Multiply Eq. (i) by 3 and Eq. (ii) by 2 and then adding, we get
$5f(x) = 3x + \frac{2}{x}$
$\Rightarrow f(x) = \frac{1}{5}\left\lbrack 3x + \frac{2}{x} \right\rbrack$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \frac{1}{5}\left\lbrack 3 - \frac{2}{x^{2}} \right\rbrack$
$\Rightarrow f^{'}(2) = \frac{1}{5}\left\lbrack 3 - \frac{2}{4} \right\rbrack = \frac{1}{2}$