Question

A curve passes through the point $(2,0)$ and the slope of the tangent at any point $(x,y)$ is $x^2-2x$ for all value of $x$. The point of maxima of the curve is

• A. $(2,\frac{4}{3})$
• B. $(1,\frac{2}{3})$
• C. $(0,\frac{2}{3})$
• D. $(0,\frac{4}{3})$

Answer 1

The correct option is D

$(0,\frac{4}{3})$

$\frac{dy}{dx}=x^2-2x$

$\Rightarrow dy=(x^2-2x)dx$

Integrating, we get

$y=\frac{x^3}{3}-x^2+c$

Since, the curve passes through $(2,0)$ we get

$0=\frac{8}{3}-4+c$ i.e., $c=\frac{4}{3}$

Hence, the equation of the curve is

$y=\frac{x^3}{3}-x^2+\frac{4}{3}$

Now, from $\frac{d^{2}y}{d{x}^2}=2x-2\Rightarrow {\begin{array}{c}\frac{d^{2}y}{d{x}^2}\end{array}|}_{x=0}=-2$ and ${\begin{array}{c}\frac{d^{2}y}{d{x}^2}\end{array}|}_{x=2}=2$

Hence, at $x=0,y$ has a maxima. Thus, the required point is $(0,\frac{4}{3})$