Question

The functions $f\left(x\right)=\left\{\begin{array}{cc}\begin{array}{rcl}& & e^x\\ & & 2-e^{x-1}\\ & & x-e\end{array}& \begin{array}{rcl}0& \le & x\le 1\\ 1& <& x\le 2\\ 2& <& x\le 3\end{array}\end{array}\right.$and $g\left(x\right)={\int }_0^{x}f\left(t\right)\mathrm{d}t$, $x\in [1,3]$, then$g\left(x\right)$has

• A. Local maxima at $x=1+\ln \left(2\right)$ and minima at $x=e$
• B. Local maxima at $x=1$ and local minima at $x=2$
• C. No local maxima
• D. No local minima
• E. both A and B

Answer 1

The correct option is E

both A and B

Explanation for the correct option:

Given,

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{cc}{e}^x \\ 2-e^{x-1} \\ x-e& \begin{array}{rcl}0& \le & x\le 1\\ 1& <& x\le 2\\ 2& <& x\le 3\end{array}\end{array}\right. \end{gathered}$$

and $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$ for all $x\in [1,3]$

Thus, $g\text{'}\left(x\right)=f\left(x\right)$

To find maxima and minima, set $g\text{'}\left(x\right)=f\left(x\right)=0$

In the interval $0\le x\le 1$, $f\left(x\right)=e^x$

But the function is discontinuous at $f\left(1\right)$ because the left hand limit is $e$ and the right hand limit is $2$ here.

We know that at minima, $g\text{'}\text{'}\left(x\right)>0$ and at maxima, $g\text{'}\text{'}\left(x\right)<0$

Since $g\text{'}\text{'}\left(1\right)=e^1>0$, $g\left(x\right)$ has a local maxima at $x=1$

In the interval $1<x\le 2$, $f\left(x\right)=2-e^{x-1}$

$\Rightarrow g\text{'}\left(x\right)=2-e^{x-1}$

Setting this equal to $0$,

$\begin{array}{ccc}\Rightarrow & 2-e^{x-1}=0& \\ \Rightarrow & e^{x-1}=2& \\ \Rightarrow & \ln \left(e^{x-1}\right)=\ln \left(2\right)& \\ \Rightarrow & \left(x-1\right)·\ln \left(\mathrm{e}\right)=\ln \left(2\right)& \left[\because ln{a}^b=b·ln\left(a\right)\right]\\ \Rightarrow & x-1=\ln \left(2\right)& \left[\because lne=1\right]\\ \Rightarrow & x=1+\ln \left(2\right)& \end{array}$

Here,
$\begin{array}{rcl}g\text{'}\text{'}\left(x\right)& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(2-e^{x-1}\right)\\ & =& -e^{x-1}\end{array}$

Setting, $x=1+\ln \left(2\right)$,

$\begin{array}{rcl}g\text{'}\text{'}\left(1+\ln \left(2\right)\right)& =& -e^{\left[1+\ln \left(2\right)\right]-1}\\ & =& -e^{\ln \left(2\right)}\\ & =& -2\end{array}$

$\Rightarrow g\text{'}\text{'}(1+\ln \left(2\right))<0$

$\therefore$There is a maxima at $x=1+\ln 2$

Evaluating the second derivative at the boundary, (i.e., at $x=2$) because the function is not differentiable at this point,

$$\begin{gathered} g\text{'}\text{'}\left(2\right)=2-e^{2-1} \\ \Rightarrow g\text{'}\text{'}\left(2\right)=2-e \\ \Rightarrow g\text{'}\text{'}\left(2\right)<0 \end{gathered}$$

Thus, at $x=2$, there is a minima

In the interval $2<x\le 3$, $f\left(x\right)=x-e$

$\Rightarrow g\text{'}\left(x\right)=x-e$

Setting this equal to $0$,

$\begin{array}{cc}\Rightarrow & x-e=0\\ \Rightarrow & x=e\end{array}$

Here, $g\text{'}\text{'}\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(x-e\right)=1>0$

$\therefore$There is a minima at $x=e$

Hence, option E is correct.