Question

If, $f\left(x\right)=\{\begin{array}{cc}{e}^x,& 0\le x\le 1\\ 2-e^{x-1},& 1<x\le 2andg\left(x\right)={\int }_0^{x}f\left(t\right)dt,x\in [1,3]theng\left(x\right)has\\ x-e,& 2<x\le 3\end{array}$

• A. local maxima at $x=1+\ln 2$ and local minima at $x=e$
• B. local maxima at $x=1$ and local minima at $x=2$
• C. no local maxima
• D. no local minima

Answer 1

Answer: (A), (B)

The correct options are
A local maxima at $x=1+\ln 2$ and local minima at $x=e$
B local maxima at $x=1$ and local minima at $x=2$

Given, $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$

$\Rightarrow$ $g^{\prime }\left(x\right)=f\left(x\right)=\{\begin{array}{c}{e}^x;0\le x\le 1\\ 2-e^{x-1};1<x\le 2\\ x-e;2<x\le 3\end{array}$

$\Rightarrow$ $g^{\prime }\left(x\right)=0$, when $x=1+\ln 2$ and $x=e$

$\Rightarrow$ $g^{\prime \prime }\left(x\right)=\left\{\begin{array}{ccc}-e^{x-1}& 1<& x\le 2\\ 1& 2<& x\le 3\end{array}\right\}$

$\Rightarrow$ $g^{\prime \prime }(1+\ln 2)=-e^{\ln 2}<0$ hence at $x=1+\ln 2,g\left(x\right)$ has a local maximum

$\Rightarrow$ $g^{\prime \prime }\left(e\right)=1>0$ hence at $x=e,g\left(x\right)$ has local minimum.

$\Rightarrow$ $\cdot f\left(x\right)$ is discontinuous at $x=1$, then we get local maxima at $x=1$ and local minima at $x=2$.