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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {x}^{3} - {x}^{2} + x + 1$

and

$\begin{gathered}\begin{gathered}\bf\: g(x) = \begin{cases} &\sf{max \{f(t) \},0 \leqslant t \leqslant x \: for \: 0 \leqslant x \leqslant 1} \\ &\sf{3 - x + {x}^{2} , \: 1 < x \leqslant 2} \end{cases}\end{gathered}\end{gathered}$

Now, Consider

$\rm :\longmapsto\:f(x) = {x}^{3} - {x}^{2} + x + 1$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}[ \: {x}^{3} - {x}^{2} + x + 1 \: ]$

$\rm :\longmapsto\:f'(x) = {3x}^{2} - 2x + 1$

Now, its a quadratic equation and

$\red{\rm :\longmapsto\:Discriminant = {( - 2)}^{2} - 4(3)(1) = 4 - 12 = - 8 < 0}$

We know, in a quadratic equation, if coefficient of x^2 is positive and Discriminant < 0, it implies quadratic equation is always > 0.

$\bf\implies \:f'(x) > 0$

$\bf\implies \:f(x) \: is \: increasing \: function \: in \: (0,2)$

Now, g(x) is defined as

$\begin{gathered}\begin{gathered}\bf\: g(x) = \begin{cases} &\sf{f(x) , \: \: \: 0 \leqslant x \leqslant 1} \\ &\sf{3 - x + {x}^{2} , \: 1 < x \leqslant 2} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: g(x) = \begin{cases} &\sf{ {x}^{3} - {x}^{2} + x + 1 , \: \: \: 0 \leqslant x \leqslant 1} \\ &\sf{3 - x + {x}^{2} , \: 1 < x \leqslant 2} \end{cases}\end{gathered}\end{gathered}$

To check the continuity, at x = 1.

$\rm :\longmapsto\:g(1) = {1}^{3} - {1}^{2} + 1 + 1 = 1 - 1 + 1 + 1 = 2$

RHL

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 1^+} \: (3 - x + {x}^{2}) \: }$

To evaluate this limit, we use method of Substitution

So, Substitute,

$\red{\rm :\longmapsto\:x = 1 + h, \: \: as \: x \to \: 1, \: \: so \: h \to \: 0}$

$\rm \: = \:\displaystyle\lim_{h \to 0}[3 - (1 + h) + {(1 + h)}^{2}]$

$\rm \: = \:3 - 1 + 1$

$\rm \: = \:3$

Thus, we have

$\bf\implies \:\boxed{ \tt{ \: g(1) \: \ne \: \displaystyle\lim_{x \to 1^+}g(x) \: }}$

So, function is not continuous at x = 1.

Now, We check differentiability at x = 1

$\begin{gathered}\begin{gathered}\bf\: g(x) = \begin{cases} &\sf{ {x}^{3} - {x}^{2} + x + 1 , \: \: \: 0 \leqslant x \leqslant 1} \\ &\sf{3 - x + {x}^{2} , \: 1 < x \leqslant 2} \end{cases}\end{gathered}\end{gathered}$

LHD

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^ - } \frac{g(x) - g(1)}{x - 1}$

$\rm \: = \:\displaystyle\lim_{x \to 1^ - } \frac{ {x}^{3} - {x}^{2} + x + 1 - 2}{x - 1}$

$\rm \: = \:\displaystyle\lim_{x \to 1^ - } \frac{ {x}^{3} - {x}^{2} + x - 1}{x - 1}$

$\rm \: = \:\displaystyle\lim_{x \to 1^ - } \frac{ {x}^{2} (x- 1) +1( x - 1)}{x - 1}$

$\rm \: = \:\displaystyle\lim_{x \to 1^ - } \frac{ ({x}^{2} + 1)( x - 1)}{x - 1}$

$\rm \: = \:\displaystyle\lim_{x \to 1^ - } ( {x}^{2} + 1)$

$\rm \: = \:1 + 1$

$\rm \: = \:2$

Now, RHD

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^ + } \frac{g(x) - g(1)}{x - 1}$

$\rm \: = \:\displaystyle\lim_{x \to 1^+} \frac{3 - x + {x}^{2} - 2 }{x - 1}$

$\rm \: = \:\displaystyle\lim_{x \to 1^+} \frac{1 - x + {x}^{2}}{x - 1}$

$\rm \: = \: \infty$

$\boxed{ \tt{ \: \:\displaystyle\lim_{x \to 1^ - } \frac{g(x) - g(1)}{x - 1} \: \ne \: \displaystyle\lim_{x \to 1^ + } \frac{g(x) - g(1)}{x - 1}}}$

So, it implies g(x) is not differentiable at x = 1.

Hence,

$\red{\boxed{ \sf{ \: g(x) \: is \: neither \: is \: continuous \: nor \: differentiable \: at \: x = 1}} \: }$