Definite Integration Question 267
Question: If $ f(t)=\int _{-t}^{t}{\frac{dx}{1+x^{2}},} $ then $ {f}’(1) $ is
[Roorkee 2000]
Options:
A) Zero
B) 2/3
C) $ -1 $
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
Given $ f(t)=\int _{-t}^{t}{\frac{dx}{1+x^{2}}} $
$ =[{{\tan }^{-1}}x] _{-t}^{t} $
$ =2{{\tan }^{-1}}t $
Differentiating with respect to t, $ {f}’(t)=\frac{2}{1+t^{2}} $
therefore $ f’(1)=\frac{2}{2}=1 $ .
BETA
