Let $f:R \to R$ be a differentiable function having $f\left( 2 \right) = 6,\;f'\left( 2 \right) = \left( {\frac{1}{{48}}} \right)$. Then, $\mathop {\lim }\limits_{x \to 2} \frac{{\mathop \smallint \nolimits_6^{f\left( x \right)} 4\;{t^3}\;dt}}{{x - 2}}$ is equals

JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

Let

f:R \to R

be a differentiable function having

f\left( 2 \right) = 6,\;f'\left( 2 \right) = \left( {\frac{1}{{48}}} \right)

. Then,

\lim\limits_{x \to 2} \frac{{\int \nolimits_6^{f\left( x \right)} 4\;{t^3}\;dt}}{{x - 2}}

is equals

View answer and explanation

Correct option: A

Option A (18) is correct because it is the result of applying L'Hopital's rule and the Fundamental Theorem of Calculus to the given limit.