Question

A curve passes through the point $(2,0)$ and its gradient at the point $(x,y)$ is $x^2-2x$ for all Values of $x$,then the point of maximum ordinate on the curve is

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

Answer 1

Answer: (D)

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

Given $\frac{dy}{dx}=x^2-2x.$
Integrating we get ,
$y=\frac{1}{3}{x}^3-x^2+c.$
Since the curve passes through $(2,0)$,we get
$0=\frac{1}{3}\times 2^3-2^2+c$
$\Rightarrow c=\frac{4}{3}.$
Hence the equation of the curve is $y=\frac{1}{3}{x}^3{x}^2+\frac{4}{3}.$
Now for maximum or minimum ,we have
$\frac{dy}{dx}=0,i.e.x^2-2x=0$
$\therefore x=0,2.$
Now, $\frac{d^{2}y}{d{x}^2}=2x-2=-2$ at $x=0$.
Hence y is maximum at $x=0$
When $x=0$,we get from (1),
$y=\frac{4}{3}$
Hence the point of maximum ordinate on the curve is $(0,\frac{4}{3})$.