Indefinite Integration 3 Question 12

12. Let $f: R \rightarrow R$ be a differentiable function and $f(1)=4$. Then, the value of $\lim _{x \rightarrow 1} \int _4^{f(x)} \frac{2 t}{x-1} d t$ is

(a) $8 f^{\prime}(1)$

(b) $4 f^{\prime}(1)$

(c) $2 f^{\prime}(1)$

(d) $f^{\prime}(1)$

(1990, 2M)

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Answer:

Correct Answer: 12. (a)

Solution:

  1. $\lim _{x \rightarrow 1} \int _4^{f(x)} \frac{2 t}{x-1} d t=\lim _{x \rightarrow 1} \frac{\int _4^{f(x)} 2 t d t}{x-1}$

[using L’ Hospital’s rule]

$$ =\lim _{x \rightarrow 1} \frac{2 f(x) \cdot f^{\prime}(x)}{1}=2 f(1) \cdot f^{\prime}(1) $$

$$ =8 f^{\prime}(1) $$

$[\because f(1)=4]$



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