Let $f(x)$ be a differentiable function at $x=a$ with $f^{\prime}(a)=2$ and $f(a)=4$. Then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$ equals :

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Differentiation

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 :

Answer: (B)

Add and subtract

a f (a)

in the numerator:

\frac{x f ( a ) - a f ( x )}{x - a} = \frac{f ( a ) ( x - a ) - a [ f ( x ) - f ( a ) ]}{x - a} = f (a) - a \cdot \frac{f ( x ) - f ( a )}{x - a}

x \to a

the difference quotient tends to

f^{'} (a)

, so the limit equals

f (a) - a f^{'} (a) = 4 - 2 a

The limit is

4 - 2 a

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option