Question

If $f\left(x\right)$ is function satisfying $f\left(\frac{1}{x}\right)+x^{2}f\left(x\right)=0$ for all non-zero x, then $\underset{\sin \theta }{\overset{\cos ec\theta }{\int }}f\left(x\right)dx$ equals

• A. $\sin \theta +cosec\theta$
• B. ${\sin }^2\theta$
• C. $\cos e{c}^2\theta$
• D. none of these

Answer 1

The correct option is A $\sin \theta +cosec\theta$

Given, $f\left(x\right)$ satisfies :

$f\left(\frac{1}{x}\right)+x^{2}f\left(x\right)=0$ ----- $\left(1\right)$

$\Rightarrow f\left(x\right)=-dfrac1x^{2}f\left(\frac{1}{x}\right)$ ------ $\left(2\right)$

$\Rightarrow f\left(\frac{1}{x}\right)=-x^{2}f\left(x\right)$------- $\left(3\right)$

$\int f(x) dx = \int \left( \dfrac{-1}{x^{2}} \right) f \left( \dfrac{1}{x} \right)

$=$ using integration By part :

$=\int \left(\frac{-1}{x^2}\right)\times \int \left(\frac{1}{x}\right)-\int f\left(\frac{1}{x}\right)\times \int \frac{-1}{x^2}dx$

$=\frac{1}{x}f\left(\frac{1}{x}\right)-\int f^{\prime }\left(\frac{1}{x}\right)\times \frac{1}{x}$

$\Rightarrow f\left(x\right)dx=\frac{1}{x}f\left(\frac{1}{x}\right)-\int \frac{d}{dx}(-x^{2}f(x\left)\right)\times \frac{1}{x}$

$=\frac{1}{x}f\left(\frac{1}{x}\right)-\int \frac{1}{x}\times [-x^2{f}^{\prime }(x)-2xf(x\left)\right]$

$\Rightarrow \int f\left(x\right)dx=\frac{1}{x}f\left(\frac{1}{x}\right)+2\int f\left(x\right)dx+xf\left(x\right)+\int f\left(x\right)dx$

Cancelling $\int f\left(x\right)dx$ on both sides and taking the remaining integral to $LHS$ :

$2\int f\left(x\right)dx=-\left(xf\right(x)+\frac{1}{x}\left(\frac{1}{x}\right))$

$\Rightarrow \int f\left(x\right)dx=\frac{-1}{2}\left(xf\right(x)+\frac{1}{x}\left(\frac{1}{x}\right))$

Changing to definite integral as required from $\sin \theta$ to $\csc \theta$

$\int f\left(x\right)dx=\frac{-1}{2}\left[\csc \theta f\right(\csc \theta )+\sin \theta f(\sin \theta )-(\sin \theta f\left(\sin \theta \right)+\csc \theta f\left(\csc \theta \right)\left)\right]$

$=0$

Hence, anser is option $D$ None of these