Let <img alt="\mathrm{f(x)=x^3-x^2+x+1, g(x)=\left\{\begin{array}{cc}\max \cdot\{f(t), 0 \leq t \leq x\}, & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2\end{array}\right.}" src="https://entrance

Let $\mathrm{f(x)=x^3-x^2+x+1, g(x)=\left\{\begin{array}{cc}\max \cdot\{f(t), 0 \leq t \leq x\}, & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2\end{array}\right.}$
Then in $[0,2]$ the points where $\mathrm{g(x)}$ is not differentiable is/are
Option: 1
1

Option: 2
2

Option: 3
1 and 2

Option: 4
none of these

Answers (1)

$\mathrm{f(t)=t^3-t^2+t+1}$

$\mathrm{\therefore \quad f^{\prime}(t)=3 t^2-2 t+1>0}$

$\mathrm{\therefore \quad f(t) }$ is an increasing function.
Since $\mathrm{0 \leq t \leq x \therefore \quad \max . f(t)=f(x)=x^3-x^2+x+1 }$

Thus $\mathrm{g(x)=x^3-x^2+x+1,0 \leq x \leq 1\: and \: g(x)=3-x, 1<x \leq 2 }$
The only doubtful point for differentiability of $\mathrm{g(x) }$ in $\mathrm{[0,2]}$ is $\mathrm{x=1}$ .

Clearly, $\mathrm{\lim _{x \rightarrow 1^{+} 0} g(x)=\lim _{x \rightarrow 1^{+}}(3-x)=2\: and \: g(1)=2}$

$\mathrm{\therefore \quad g(x)}$ is continuous at $\mathrm{x=1}$

Also $g^{\prime}(x)=3 x^2-2 x+1,0 \leq x<1\: and\: g^{\prime}(x)=-1,1<x \leq 2$
$\therefore \quad g^{\prime}(1-0)=3 \cdot 1^2-2 \cdot 1+1=2$ and $g^{\prime}(1+0)=-1$

Hence $g(x)$ is not differentiable at $x=1$ .