A) Zero
B) 2/3
C) \[-\,1\]
D) 1
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, \[{f}'(t)=\frac{2}{1+{{t}^{2}}}\] Þ \[f'(1)=\frac{2}{2}=1\].You need to login to perform this action.
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