A) \[\pi \]
B) \[-\pi \]
C) \[\frac{1}{\pi }\]
D) \[1\]
Correct Answer: C
Solution :
\[\because \] \[\int_{-\infty }^{\infty }{f(x)\,dx=1}\] \[\therefore \] \[\int_{-\infty }^{\infty }{\frac{k}{1+{{x}^{2}}}}\,dx=1\] \[\Rightarrow \] \[\int_{0}^{\infty }{\frac{k}{1+{{x}^{2}}}}\,\,dx=1\] \[\Rightarrow \] \[2\int_{0}^{\infty }{\frac{k}{1+{{x}^{2}}}}\,\,dx=1\] \[\Rightarrow \] \[2k\,({{\tan }^{-1}}x)_{0}^{\infty }=1\] \[\Rightarrow \] \[2k({{\tan }^{-1}}\infty -{{\tan }^{-1}}0)=1\] \[\Rightarrow \] \[2k\left( \frac{\pi }{2}-0 \right)=1\] \[\Rightarrow \] \[k=\frac{1}{\pi }\]
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