Explanation

(b)
Since, $3f(x) + 2f\left( \frac{x + 59}{x - 1} \right) = 10x + 30\ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Replacing $x$ by $\frac{x + 59}{x - 1}$ in Eq. (i), we get
$\therefore\ \ \ \ 3\left( \frac{x + 59}{x - 1} \right) + 2f(x) = \frac{40x + 560}{x - 1}$ …(ii)
On solving Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(x) = \frac{6x^{2} - 4x - 242}{x - 1}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ f(7) = \frac{6 \times 49 - 28 - 242}{6} = 4$