Question

If $y=\frac{x+c}{1+x^2}$, then the value of $xy$, where $\frac{dy}{dx}=0$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{5}{4}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

Given, $y=\frac{x+c}{1+x^2}$
$\Rightarrow y+x^{2}y=x+c$
On differntiation we have
$\Rightarrow \frac{dy}{dx}+x^2\left(\frac{dy}{dx}\right)+\left(2x\right)y=1$
$\Rightarrow \frac{dy}{dx}=0$
$\Rightarrow 0+0+2xy=1$
$\Rightarrow xy=\frac{1}{2}$