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JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Differentiation

Let

f : R \to R

be a function such that

f (2) = 4

and

f^{'} (2) = 1

. Then the value of

lim_{x \to 2} \frac{x ^{2} f ( 2 ) - 4 f ( x )}{x - 2}

is equal to :

(A)12✓ Correct
(B)4
(C)16
(D)8

Answer: (A)

Solution

Using

f (2) = 4

,

x \to 2 lim \frac{x ^{2} f ( 2 ) - 4 f ( x )}{x - 2} = 4 x \to 2 lim \frac{x ^{2} - f ( x )}{x - 2} = 4 x \to 2 lim [\frac{x ^{2} - 4}{x - 2} - \frac{f ( x ) - f ( 2 )}{x - 2}]

The first quotient tends to

4

and the second to

f^{'} (2) = 1

, so the limit is

4 (4 - 1) = 12

.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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