If $f(x)=\left\{\begin{array}{ll} e^x, & 0 \leq x \leq 1 \\ 2-e^{x-1}, & 1<x \leq 2 \\ x-e, & 2<x \leq 3 \end{array}\right.$ and $g(x)=\int\limits_0^x f(t) d t, x \in[1,3]$, then which one of the following is incorrect?
Answer & explanation
Correct answer: option 3
It is given that
$f(x)=\left\{\begin{array}{ll} e^x & , 0 \leq x \leq 1 \\ 2-e^{x-1}, & 1<x \leq 2 \\ x-e & , 2<x \leq 3 \end{array}\right]$ and $g(x)=\int\limits_0^x f(t) d t, x \in[1,3]$
∴ $g'(x)=f(x)= \begin{cases}2-e^{x-1} ~~~1<x \leq 2 \\ x-e ~~~~~~~2<x \leq 3\end{cases}$
So, $g'(x)=0 \Rightarrow x=1+\log _e 2$ and $x=e$
Also, $g'(x)>0$ for $x \rightarrow\left(1,1+\log _e 2\right)$ and $g'(x)<0$ for $x \in\left(1+\log _e 2,2\right)$.
So, g(x) attains a local maximum at $x=1+\log _e 2$.
Similarly,
$g'(x)<0$ for $2<x<e$ and, $g'(x)>0$ for $e<x<3$
$\Rightarrow g(x)$ attains local minimum at $x=e$.
We have,
$f'(x)= \begin{cases}e^x &,~ 0<x<1 \\ -e^{x-1} &,~1<x<2 \\ 1& ,~2<x<3\end{cases}$
Clearly, $f'(x)>0$ for $x \in(0,1)<0$ for $x \in(1,2)>0$ for $x \in(2,3)$. So, f(x) attains local maximum at $x=1$ and local minimum at $x=2$.
Hence, option (c) is incorrect.