Definite Integration Question 12

Question 12

  1. Let f:RR be a differentiable function and f(1)=4. Then, the value of $\lim {x \rightarrow 1} \int{4}^{f(x)} \frac{2 t}{x-1} d tis(a)8 f^{\prime}(1)(b)4 f^{\prime}(1)(c)2 f^{\prime}(1)(d)f^{\prime}(1)$

(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]

=limx12f(x)f(x)1=2f(1)f(1)

=8f(1)

[f(1)=4]



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