Question

For the curve $f\left(x\right)=\frac{1}{1+x^2}$, let two points on it be $A(\alpha ,f(\alpha \left)\right),B(-\frac{1}{\alpha },f(-\frac{1}{\alpha }))(\alpha >0)$. Find the minimum area bounded by the line segments OA, OB and $f\left(x\right)$, where 'O' is the origin.

• A. $\frac{(\pi -1)}{2}$
• B. $\frac{\pi }{2}$
• C. $\frac{(\pi -2)}{2}$
• D. Maximum area is always infinite

Answer 1

The correct option is A $\frac{(\pi -1)}{2}$

The shaded area in the graph is the required area that has to be maximised

Point A $\equiv (\alpha ,\frac{1}{1+{\alpha }^2})$

Point B $\equiv (-\frac{1}{\alpha },\frac{{\alpha }^2}{1+{\alpha }^2})$

Thus the required area to be maximised is,

$A={\int }_{-1/\alpha }^{\alpha }f\left(x\right)dx-$(Area of triangle under segment OB) $-$ (Area of triangle under segment OA)

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

$\therefore A={{\tan }^{-1}x|}_{-1/\alpha }^{\alpha }-(\frac{1}{2}\times \frac{1}{\alpha }\times \frac{{\alpha }^2}{1+{\alpha }^2})-(\frac{1}{2}\times \alpha \times \frac{1}{1+{\alpha }^2})$

$\therefore A={\tan }^{-1}\alpha +{\tan }^{-1}\frac{1}{\alpha }-\frac{\alpha }{1+{\alpha }^2}$

${\tan }^{-1}\alpha +{\tan }^{-1}\frac{1}{\alpha }=\frac{\pi }{2}$

$\therefore A=\frac{\pi }{2}-\frac{\alpha }{1+{\alpha }^2}$

This is maximum when $\frac{\alpha }{1+{\alpha }^2}$ is minimum.

$\therefore \frac{d}{d\alpha }\left(\frac{\alpha }{1+{\alpha }^2}\right)=0$

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

$\therefore \alpha =\pm 1$

Maximum value occurs at $\alpha =1$

Therefore minimum value of bounded area $=\frac{\pi }{2}-\frac{1}{2}$