Let f: R → R be a function such that f(2) = 4 and f ^ ' (2) = 1 . Then, the value of _ x → 2 x ^ 2 f(2) - 4f(x) x - 2 is equal to:

Correct answer: D. 12 Given, f(2) = 4 So, the given limit is in (0)/(0) form. So, _ x → 2 x ^ 2 f(2) - 4f(x) x - 2 Apply L 'Hospital Rule _ x → 2 ( 2xf(2) - 4 f ^ ' (x) 1 ) = (4(4) - 4)/(1) = 12

JEE Main 202127 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2021 · 27 July · Morning Shift · Mathematics · Limits

Let f: R → R be a function such that f(2) = 4 and f ^ ' (2) = 1 ....

Let

f: R \to R

be a function such that

f\left(2\right) = 4

and

{f}^{'}\left(2\right) = 1 .

Then, the value of

\lim_{x \to 2}\dfrac{{x}^{2}f\left(2\right) - 4f\left(x\right)}{x - 2}

is equal to:

Options

A $4$
B $8$
C $16$
D $12$

Correct answer

D. $12$

Step-by-step solution

Given, $f\left(2\right) = 4$

So, the given limit is in $\dfrac{0}{0}$ form.

So, $\lim_{x \to 2}\dfrac{{x}^{2}f\left(2\right) - 4f\left(x\right)}{x - 2}$

Apply $L$ 'Hospital Rule

$\lim_{x \to 2}\left(\dfrac{2xf\left(2\right) - 4{f}^{'}\left(x\right)}{1}\right)$

$= \dfrac{4\left(4\right) - 4}{1} = 12$

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