If $f(x)$ is a function satisfying $f\left(\frac{1}{x}\right)+x^2 f(x)=0$ for all non-zero $x$, then $\int\limits_{\sin \theta}^{~cosec \theta} f(x) d x$ equals
Answer & explanation
Correct answer: option 4
We have,
$f\left(\frac{1}{x}\right)+x^2 f(x)=0 \Rightarrow f(x)=-\frac{1}{x^2} f\left(\frac{1}{x}\right)$
∴ $I=\int\limits_{\sin \theta}^{~cosec \theta} f(x) d x=\int\limits_{\sin \theta}^{~cosec \theta}-\frac{1}{x^2} f\left(\frac{1}{x}\right) d x=\int\limits_{~cosec \theta}^{\sin \theta} f(t) d t$, where $t=\frac{1}{x}$
$\Rightarrow I=-\int\limits_{\sin \theta}^{~cosec \theta} f(t) d t=-I \Rightarrow 2 I=0 \Rightarrow I=0$