Question

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

• A. $2$
• B. $4$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

We have,
$f\left(x\right)=\underset{-1}{\overset{x}{\int }}t(e^t-1)(t-1)(t-2{)}^3(t-3{)}^{5}dx$
$\Rightarrow f^{\prime }\left(x\right)=x(e^x-1)(x-1)(x-2{)}^3(x-3{)}^5$
For extreme points $f^{\prime }\left(x\right)=0\Rightarrow x=0,1,2,3$
At $x=0,f^{\prime }\left(x\right)$ does not change sign as $x$ crosses $0$.
At $x=1,f^{\prime }\left(x\right)>0$ or $<0$ according as $x>1$ or $x<1$.
$\Rightarrow f\left(x\right)$ has local minimum at $x=1$.
At $x=2,f^{\prime }\left(x\right)>0$ or $<0$ according as $x<2$ or $x>2$.
$\Rightarrow f\left(x\right)$ has local maximum at $x=2$.
At $x=3,f^{\prime }\left(x\right)>0$ or $<0$ according as $x>3$ or $x<3$.
$\Rightarrow f\left(x\right)$ has local minima at $x=3$.