Question

The function $f\left(x\right)={\int }_{-1}^{x}t\left(e^t-1\right)\left(t-1\right){\left(t-2\right)}^3{\left(t-3\right)}^{5}dt$ for $t\in \left[-1,x\right]$ has a local minimum at,

• A. $0$, $4$
• B. $1$, $3$
• C. $0$, $2$
• D. $2$, $4$

Answer 1

The correct option is B

$1$, $3$

Explanation for the correct option

Step 1: Solve for the critical points

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

Then,
$f\text{'}\left(x\right)=x\left(e^x-1\right)\left(x-1\right){\left(x-2\right)}^3{\left(x-3\right)}^5\left[\because \frac{d}{dx}{\int }_a^{x}f\left(t\right)dt=f\left(x\right)\right]$

The critical points of the functions can be obtained by setting $f\text{'}\left(x\right)=0$
$$\begin{gathered} \Rightarrow x\left(e^x-1\right)\left(x-1\right){\left(x-2\right)}^3{\left(x-3\right)}^5=0 \\ \begin{array}{cccccc}\Rightarrow & x=0& e^x-1=0& x-1=0& {\left(x-2\right)}^3=0& {\left(x-3\right)}^5=0\\ \Rightarrow & & x=0& x=1& x=2& x=3\end{array} \end{gathered}$$

Step 2: Find the local minima

A function has local minima at $a$ if $f\text{'}\left(x\right)<0$ to its left and $f\text{'}\left(x\right)>0$ to its right.

We see that,
$\begin{array}{rcl}f\text{'}\left(-1\right)& =& -1\left(\frac{1}{e}-1\right)\left(-1-1\right){\left(-1-2\right)}^3{\left(-1-3\right)}^5\\ & =& -1\left(0.3676-1\right)\left(-2\right){\left(-3\right)}^3{\left(-4\right)}^5<0\left[\because e=2.72\right]\end{array}$
And,
$\begin{array}{rcl}f\text{'}\left(\frac{1}{2}\right)& =& \frac{1}{2}\left(\sqrt{e}-1\right)\left(\frac{1}{2}-1\right){\left(\frac{1}{2}-2\right)}^3{\left(\frac{1}{2}-3\right)}^5\\ & =& \frac{1}{2}\left(\sqrt{e}-1\right)\left(-\frac{1}{2}\right){\left(-\frac{3}{2}\right)}^3{\left(-\frac{5}{2}\right)}^5<0\end{array}$

Thus, $f\left(x\right)$ does not have local minimum at $x=0$

Also,
$\begin{array}{rcl}f\text{'}\left(\frac{3}{2}\right)& =& \frac{3}{2}\left(e^{\frac{3}{2}}-1\right)\left(\frac{3}{2}-1\right){\left(\frac{3}{2}-2\right)}^3{\left(\frac{3}{2}-3\right)}^5\\ & =& \frac{3}{2}\left(e^{\frac{3}{2}}-1\right)\left(\frac{1}{2}\right){\left(-\frac{1}{2}\right)}^3{\left(-\frac{3}{2}\right)}^5>0\end{array}$

Thus, $f\left(x\right)$ has local minimum at $x=1$

$\begin{array}{rcl}f\text{'}\left(\frac{5}{2}\right)& =& \frac{5}{2}\left(e^{\frac{5}{2}}-1\right)\left(\frac{5}{2}-1\right){\left(\frac{5}{2}-2\right)}^3{\left(\frac{5}{2}-3\right)}^5\\ & =& \frac{5}{2}\left(e^{\frac{5}{2}}-1\right)\left(\frac{3}{2}\right){\left(\frac{1}{2}\right)}^3{\left(-\frac{1}{2}\right)}^5<0\end{array}$
And,
$\begin{array}{rcl}f\text{'}\left(\frac{7}{2}\right)& =& \frac{7}{2}\left(e^{\frac{7}{2}}-1\right)\left(\frac{7}{2}-1\right){\left(\frac{7}{2}-2\right)}^3{\left(\frac{7}{2}-3\right)}^5\\ & =& \frac{7}{2}\left(e^{\frac{7}{2}}-1\right)\left(\frac{5}{2}\right){\left(\frac{3}{2}\right)}^3{\left(\frac{1}{2}\right)}^5>0\end{array}$

Thus, $f\left(x\right)$ has local minimum at $x=3$

So, $f\left(x\right)$has local minima at $x=1$ and $x=3$

Hence option(B) i.e. $1$, $3$ is correct.