Question

The point of inflection of the function $y - \int^{x}_{0} \left(t^{2} - 3t + 2 \right) dt$ is

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

Answer 1

Answer: (B)

$\left( \frac{3}{2} , \frac{3}{4}\right)$

Answer 2

Given,
$y=\int_{0}^{x}\left(t^{2}-3 t+2\right) dt\,...(i)$
On differentiating w.r.t. ' $x,$, we get
$\frac{d y}{d x}=x^{2}-3 x+2\,...(ii)$
Again, on differentiating w.r.t. ' $x'$, we get
$\frac{d^{2} y}{d x^{2}}=2 x-3\,...(iii)$
We know that, at point of inflection
$\frac{d^{2} y}{d x^{2}}=0$
$\therefore$ From E (iii), we get
$2 x-3=0$
$\Rightarrow x=\frac{3}{2}$
Now, we have to check behaviour of $\frac{d^{2} y}{d x^{2}}$ at point $x=\frac{3}{2}$
$x=\frac{3}{2}$
Clearly, at $x=\frac{3}{2}$ sign at $\frac{d^{2} y}{d x^{2}}$ changes
$\therefore\left(\frac{3}{2}, \frac{3}{4}\right)$ is point of inflection