Question

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

has a local minimum at $x=$

• A. 4
• B. 1
• C. 0
• D. 2

Answer 1

The correct option is B 1

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

$\Rightarrow f^{\prime }\left(x\right)=x(e^x–1)(x–1)(x–2{)}^3(x–3{)}^5$

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 $x=1,f^{\prime }\left(x\right)>0or<0$ according as $x>1\text{or}x<1.$

$\Rightarrow f\left(x\right)$ has local minimum at $x=1.$

At $x=2,f^{\prime }\left(x\right)>0\text{or}<0$ according as $x<2\text{or}x>2.$

$\Rightarrow f\left(x\right)$ has local maximum at $x=2.$

At $x=3,f^{\prime }\left(x\right)>0\text{or}<0$ according as $x>3\text{or}x<3.$

$\Rightarrow f\left(x\right)$ has local minima at $x=3.$