MCQ
Let $f:R \to R$ be a differentiable function having $f(2) = 6,f'(2) = \left( {\frac{1}{{48}}} \right).$ Then $\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f(x)} {\frac{{4{t^3}}}{{x - 2}}} dt$ equals
  • A
    $12$
  • $18$
  • C
    $24$
  • D
    $36$

Answer

Correct option: B.
$18$
b
(b) $\mathop {\lim }\limits_{x \to 2} \frac{{\int\limits_6^{f(x)} {4{t^3}dt} }}{{x - 2}}\,\,(0/0\,{\rm{\,\, form}}) = \mathop {\lim }\limits_{x \to 2} \frac{{4{{(f(x))}^3} \times f'(x)}}{1}$

$ = 4{(f(2))^3} \times f'(2) = 18$.

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