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:
- $\lim _{x \rightarrow 1} \int _4^{f(x)} \frac{2 t}{x-1} d t=\lim _{x \rightarrow 1} \frac{\int _4^{f(x)} 2 t d t}{x-1}$
[using L’ Hospital’s rule]
$$ =\lim _{x \rightarrow 1} \frac{2 f(x) \cdot f^{\prime}(x)}{1}=2 f(1) \cdot f^{\prime}(1) $$
$$ =8 f^{\prime}(1) $$
$[\because f(1)=4]$
