Q41 / 90 · JEE Main 2021 (26 Feb Shift 2) · Mathematics

Let $f (x)$ be a differentiable function at $x = a$ with $f^{'} (a) = 2$ and $f (a) = 4$ . Then $\lim_{x \to a} \frac{x f (a) - a f (x)}{x - a}$ equals:

Solution

$f^{'} (a) = 2, f (a) = 4$

$\lim_{x \to a} \frac{x f (a) - a f (x)}{x - a}$

$\Rightarrow \lim_{x \to a} \frac{f (a) - a f^{'} (x)}{1}$ (L'Hospitals rule)

$= f (a) - a f^{'} (a)$

$= 4 - 2 a$

Hence, the correct answer is option: 'C'.