Q51 / 155 · WBJEE 2024 · Mathematics

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function and $f(1)=4$. Then the value of $\lim _{x \rightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} d t$, if $f^{\prime}(1)=2$ is

  • Correct answer
    A
    16
  • B
    8
  • C
    4
  • D
    2

Solution

Hint : $\operatorname{Lim}_{x \rightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} d t,=\operatorname{Lim}_{x \rightarrow 1}\left[\frac{t^2}{x-1}\right]_4^{f(x)}$ $\begin{aligned} & ==\operatorname{Lim}_{x \rightarrow 1} \frac{[f(x)]^2-4^2}{x-1} \quad\left(\frac{0}{0} \text { form }\right)[\because f(1)=4] \\ & =\operatorname{Lim}_{x \rightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}=2 f(1) f^{\prime}(1)=2(4)(2)=16 \end{aligned}$

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

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