Question

A curve passes through $(2,0)$ and the slope of tangent at a point $P(x,y)$ is equal to $\frac{\left(\right(x+1{)}^2+y-3)}{(x+1)}$. Then equation of the curve is:

• A. $y=x^2+2x$
• B. $y=x^2-2x$
• C. $y=2x^2-x$
• D. None of these

Answer 1

The correct option is B $y=x^2-2x$

Let $y=f\left(x\right)$ be the required curve.

Now the slope of $f\left(x\right)$ at $(x,y)$ is $\frac{dy}{dx}$.

According to the problem,

$\frac{dy}{dx}=$$\frac{\left(\right(x+1{)}^2+y-3)}{(x+1)}$

or, $\frac{dy}{dx}-\frac{y}{(x+1)}=(x+1)-\frac{3}{(x+1)}$

or, $\frac{1}{(x+1)}\frac{dy}{dx}-\frac{y}{(x+1{)}^2}=1-\frac{3}{(x+1{)}^2}$

or, $\frac{d}{dx}\left(\frac{y}{x+1}\right)=1-\frac{3}{(x+1{)}^2}$

Now integrating both sides we get,

$\frac{y}{(x+1)}=x+\frac{3}{x+1}+c$ [$c$ being integrating constant]

Given this curve passes through $(2,0)$ then

$0=2+1+c$

or, $c=-3$

The required curve is

$y=3+(x+1)(x-3)$

or, $y=x^2-2x$