Q117 / 225 · JEE Main 2005 · Mathematics

Let $f: R \rightarrow R$ be a differentiable function having $f(2)=6, f^{\prime}(2)=\left(\frac{1}{48}\right)$. Then $\lim _{x \rightarrow 2} \int_6^{f(x)} \frac{4 t^3}{x-2} d t$ equals

  • A
    24
  • B
    36
  • C
    12
  • Correct answer
    D
    18

Solution

$$ \lim _{x \rightarrow 2} \int_0^{f(x)} \frac{4 t^3}{x-2} d t $$ Applying $L$ Hospital rule $$ \begin{aligned} & \lim _{x \rightarrow 2}\left[4 f(x)^2 f^{\prime}(x)\right]=4 f(2)^3 f^{\prime}(2) \\ & =4 \times 6^3 \times \frac{1}{48}=18 . \end{aligned} $$

Hence, the correct answer is option: 'D'.

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