Question

$f\left(x\right)={\int }_0^x(t-1)(t-2{)}^2(t-3{)}^3(t-4{)}^{5}dt(x>0)$ then number of points of extremum of $f\left(x\right)$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is A $4$

To find number of points of extremum of $f\left(x\right)$ we need to find $f^{\prime }\left(x\right)$

We differentiate the above equation w.r.t $x$ by Leibniz integral rule,

$f^{\prime }\left(x\right)=(x-1)(x-2{)}^2(x-3{)}^3(x-4{)}^5$

$f\left(x\right)$ has extremum at points where $f^{\prime }\left(x\right)=0$.

$\therefore f^{\prime }\left(x\right)=0\Rightarrow x=1,x=2,x=3,x=4$

So, we get $4$ points where $f^{\prime }\left(x\right)=0$, means there are $4$ points of extremum of $f\left(x\right)$