Question

Let $f\left(x\right)=x^3-x^2+x+1$ and $g\left(x\right)=\{\begin{array}{c}max\text{{}f\left(t\right)\text{}},0\le t\le x,0\le x\le 1\\ 3-x,1<x\le 2\end{array},$
Then which among the following options is/are correct for $g\left(x\right)$ in $[0,2]$

• A. $g\left(x\right)$ is continuous for all $x$
• B. $g\left(x\right)$ is differentiable for all $x$
• C. $g\left(x\right)$ is discontinuous at $x=1$
• D. $g\left(x\right)$ is not differentiable at $x=1$

Answer 1

Answer: (A), (D)

$g\left(x\right)$ is not differentiable at $x=1$

Here, $f\left(x\right)=x^3-x^2+x+1$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-2x+1,$ which is positive $\forall x\in R$.
Hence, $f\left(x\right)$ is stricly increasing in $(0,2).$
$g\left(x\right)=\{\begin{array}{c}max\text{{}f\left(t\right)\text{}},0\le t\le x,0\le x\le 1\\ 3-x,1<x\le 2\end{array}$

As $f\left(x\right)$ is increasing function.
So, $max\text{{}f\left(t\right)\text{}},0\le t\le x,0\le x\le 1=f\left(x\right)$

$\therefore g\left(x\right)=\{\begin{array}{c}{x}^3-x^2+x+1,0\le x\le 1\\ 3-x,1<x\le 2\end{array}$
Clearly, $g\left(1\right)=g\left(1^{+}\right)=g\left(1^{-}\right)=2$
Hence, $g\left(x\right)$ is continuous for all $x\in [0,2]$
Also,
$g^{\prime }\left(x\right)=\{\begin{array}{c}3x^2-2x+1,0<x<1\\ -1,1<x<2\end{array}$
At $x=1,L.H.D=2$ but $R.H.D=-1$
Clearly, $L.H.D\ne R.H.D$
Hence, $g\left(x\right)$ is not differentiable at $x=1.$