Question

The area bounded by $y=f(x)$ and the curve $y=\frac{2}{1+x^2}$ where $f$ is a continuous function satisfying the conditions $f(x)\cdot f(y)=f(xy).\forall x,y,\in R$ and $f^{\prime }(1)=2,f(1)=1$ is then

• A. $\left(\pi +\frac{a}{3}\right)$
• B. $\left(\pi -\frac{a}{3}\right)$
• C. $\left(\pi +\frac{2a}{3}\right)$
• D. None of these

Answer 1

Answer: (B)

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$
$\underset{h\to 0}{\lim }\frac{f(x(1+h/x))-f(x)}{h}=\frac{f(x)}{x}\underset{h\to 0}{\lim }\frac{f(1+h/x)-f(1)}{h/x}$
$=\frac{f(x)}{x}\cdot f^{\prime }(1)$
$\therefore f^{\prime }(x)=\frac{2f(x)}{x}$ or $\frac{f^{\prime }(x)}{f(x)}=\frac{2}{x}$
Integrating both sides, we get, $f(x)=c{x}^2$ Since $f(1)=1,$
$\therefore c=1$

So, $f(x)=x^2.$ Now, $\frac{2}{1+x^2}=x^2$
$\Rightarrow x^4+x^2-2=0$
$\Rightarrow x^2=1$
$\Rightarrow x=\pm 1$
Required area $=2\left[{\int }_0^1\left(\frac{2}{1+x^2}-x^2\right)dx\right]$
$=2{\left[2{\tan }^{-1}x-\frac{x^3}{3}\right]}_0^1=2\left[\frac{\pi }{2}-\frac{1}{3}\right]=\left(\pi -\frac{2}{3}\right)$ sq. units.