Question

The points of extremum of ${\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}$ are

• A. $x=-2$
• B. $x=1$
• C. $x=0$
• D. $x=-1$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $x=1$
B $x=-1$
C $x=-2$
D $x=0$

Let $F\left(x\right)={\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt$
$\Rightarrow F^{\prime }\left(x\right)=\frac{x^4-5x^2+4}{2+e^{x^2}}2x$
So for extremum putting $F^{\prime }\left(x\right)=0$, we get $x=0$ or
$x^2=\frac{5\pm \sqrt{25-16}}{2}=\frac{5\pm 3}{2}=4,1$
Hence, $x=0,\pm 2,\pm 1.$