MCQ
Let $f:R \to R$ be a continuously differentiable function such that $f\left( 2 \right) = 6$ and $f'\left( 2 \right) = \frac{1}{{48}}$. If $\int_6^{f\left( x \right)} {4{t^3}} \,dt = \left( {x - 2} \right)\,g\left( x \right)$, then $\mathop {\lim }\limits_{x \to 2} \,g\left( x \right)$ is equal to
  • A
    $24$
  • $18$
  • C
    $12$
  • D
    $36$

Answer

Correct option: B.
$18$
b
$\mathop {\lim }\limits_{x \to 2} \frac{{\int\limits_6^{f\left( x \right)} {4{t^2}dt} }}{{x - 2}}$

$ = \mathop {\lim }\limits_{x \to 2} \frac{{4.{f^3}\left( x \right).f'\left( x \right)}}{1}$

$ = 4{f^3}\left( 2 \right)f'\left( 2 \right) = 18$

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