Question

The point of inflection of the function $y={\int }_0^x(t^2-3t+2)dt$ is :

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

Answer 1

The correct option is A $(\frac{3}{2},\frac{3}{4})$

Given
$y={\int }_0^x(t^2-3t+2)dt....\left(i\right)$
On differentiating w.r.t $x$, we get
$\frac{dy}{dx}=x^2-3x+\mathrm{2.....}\left(ii\right)$
Again on differentiating w.r.t $x$ we get
$\frac{d^{2}y}{d{x}^2}=2x-\mathrm{3....}\left(iii\right)$
We know that, at point of inflection
$\frac{d^{2}y}{d{x}^2}=0$
$\therefore$ From Eq(iii) we get
$2x-3=0\Rightarrow x=\frac{3}{2}$
Now, we have to check behaviour of $\frac{d^{2}y}{d{x}^2}$ at point $x=\frac{3}{2}$
Clearly at $x=\frac{3}{2}$ sign at $\frac{d^{2}y}{d{x}^2}$ changes
$\therefore (\frac{3}{2},\frac{3}{4})$ is point of inflection.