Indefinite Integration 3 Question 12

12. Let f:R→R be a differentiable function and f(1)=4. Then, the value of limx→1∫4f(x)2tx−1dt is

(a) 8f′(1)

(b) 4f′(1)

(c) 2f′(1)

(d) f′(1)

(1990, 2M)

Show Answer

Answer:

Correct Answer: 12. (a)

Solution:

  1. limx→1∫4f(x)2tx−1dt=limx→1∫4f(x)2tdtx−1

[using L’ Hospital’s rule]

=limx→12f(x)⋅f′(x)1=2f(1)⋅f′(1)

=8f′(1)

[∵f(1)=4]

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