For the curve $y(1+x^2)=2-x$, if $\frac{dy}{dx}=\frac{1}{A}$ at the point where the curve crosses the x-axis, then the value of A is:

-5

The correct answer is Option (2) → -5

$y(1+x^2)=2-x$

$\frac{d}{dx}[y(1+x^2)]=\frac{d}{dx}(2-x)$

$y\frac{d}{dx}(1+x^2)+(1+x^2)\frac{dy}{dx}=-1$

$y(2x)+(1+x^2)\frac{dy}{dx}=-1$

$(1+x^2)\frac{dy}{dx}=-1-2xy$

$\frac{dy}{dx}=\frac{-1-2xy}{1+x}$

At the x-axis, $y=0$

∴ Substituting in $y.(1+x^2)=2-x$

$0(1+x^2)=2-x$

$⇒x=2$

∴ Curve crosses x-axis at (2, 0)

Finding $\frac{dy}{dx}$ at (2, 0)

$⇒\frac{dy}{dx}=\frac{-1-2(2)(0)}{1+2^2}=-\frac{1}{5}$

$⇒\frac{dy}{dx}=\frac{1}{A}$

$⇒A=-5$