Question

A curve passes through $(2,0)$ and the slope of the tangent at any point $(x,y)$ is $x^2-2x$ for all values of $x$. The point of minimum ordinate on the curve where $x>0$ is $(a,b)$'

Then find the value of $a+6b$.

• A. $2$
• B. $4$
• C. $-2$
• D. $-4$

Answer 1

The correct option is A $2$

$\frac{dy}{dx}=x^2-2x$
$y=\frac{x^3}{3}-x^2+C$
Passing through (2, 0)
$\Rightarrow C=\frac{4}{3}$
$\therefore y=\frac{x^3}{3}-x^2+\frac{4}{3}$
$y^{\prime }=x^2-2x=x(x-2)$
For maxima or minima, $y^{\prime }=0$
$\Rightarrow x=0,2$
$y^{\prime \prime }>0$ at $x=2$
At x = 2, y takes the minimum value.
$\therefore$ minimum value of y is $=\frac{8}{3}-4+\frac{4}{3}=0$
$\therefore a=2,b=0$
$a+6b=2$