A) \[0\]
B) \[1\]
C) \[2\]
D) \[3\]
Correct Answer: A
Solution :
Given that, \[f\left( \frac{1}{x} \right)+{{x}^{2}}\,f(x)=0\] ?.(i) Let \[I=\int_{\sin \,\theta }^{\text{cosec }\theta }{f(x)\,dx}\] Put \[x=\frac{1}{t}\Rightarrow dx=-\frac{1}{{{t}^{2}}}dt\] \[\therefore \] \[I=-\int_{\cos ec\,\theta }^{\sin \,\theta }{f\left( \frac{1}{t} \right).\frac{1}{{{t}^{2}}}\,\,dt}\] \[=\int_{\cos ec\,\theta }^{\sin \,\theta }{\frac{1}{{{x}^{2}}}}f\left( \frac{1}{x} \right)\,dx\] \[=-\int_{\cos ec\,\theta }^{\sin \,\theta }{f(x)\,dx}\] [From Eq. (i)] \[\Rightarrow \] \[I=-I\] \[\Rightarrow \] \[I=0\]
You need to login to perform this action.
You will be redirected in
3 sec
