Question

The function $f\left(x\right)={\int }_1^{x}t(e^t-1)(t-1)(t-2{)}^3(t-3{)}^5$ dt has local minimum at $x=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B), (D)

The correct options are
B $1$
D $3$

$f\left(x\right)={\int }_1^{x}t(e^t-1)(t-1)(t-2{)}^3(t-3{)}^{5}dt$

$f^{\prime }\left(x\right)=x(e^x-1)(x-1)(x-2{)}^3(x-3{)}^5=0$ for extrema.

$\Rightarrow x=0,1,2,3$
$\underset{–}{(-)0.(-)1.(+)2.(-)3.(+)}$

For local minima sign change of $f^{\prime }\left(x\right)$ should be $-ve$ to $+ve$

Hence $x=1,3$ are point of local minima.