Question

Let us consider a curve, $y=f\left(x\right)$ passing through the point $(-2,2)$ and slope of the tangent to the curve at any point $(x,f(x\left)\right)$ is given by $f\left(x\right)+x{f}^{\prime }\left(x\right)=x^2$. Then

• A. $x^3+xf\left(x\right)+12=0$
• B. $x^2+2xf\left(x\right)+4=0$
• C. $x^2+2xf\left(x\right)-12=0$
• D. $x^3-3xf\left(x\right)-4=0$

Answer 1

The correct option is D $x^3-3xf\left(x\right)-4=0$

Let $f\left(x\right)=y$
$\therefore y+x\frac{dy}{dx}=x^2$
$\Rightarrow x\cdot dy+y\cdot dx=x^2\cdot dx$
$\Rightarrow \int d\left(xy\right)=\int x^{2}dx\Rightarrow xy=\frac{x^3}{3}+C$
Since curve is passing through $(-2,2)$
$\therefore -4=-\frac{8}{3}+C$
$\Rightarrow C=-\frac{4}{3}$
Hence, the curve is $3xy=x^3-4$
$\therefore x^3-3xf\left(x\right)-4=0$