Question

Let $f\left(x\right)=x^3-x^2+x^{}+1$ and
$g\left(x\right)=\{\begin{array}{c}max\left(f\right(t\left)\right)for0\le t\le xfor0\le x\le 1\\ 3-x+x^{2}for1<x\le 2\end{array}$ then

• A. $g\left(x\right)$ is continuous and derivable at $x=1$
• B. $g\left(x\right)$ is continuous but not derivable at $x=1$
• C. $g\left(x\right)$ is neither continuous nor derivable at $x=1$
• D. $g\left(x\right)$ is derivable but not continuous at $x=1$

Answer 1

Answer: (C)

$g\left(x\right)$ is neither continuous nor derivable at $x=1$

Given:\\

$g\left(x\right)=max\left(f\right(t\left)\right)for0\le t\le xfor0\le x\le 1$

and $f^{\prime }\left(x\right)=3x^2-2x^{}+1>0$

$\Rightarrow$ f(x) is always increasing, $\therefore g\left(x\right)=x^3-x^2+x^{}+1for0\le x\le 1$

Since the function changes definition at $x=1$, we should check continuity at $x=1$

${lim}_{x\to 1^{-}}g\left(x\right)=1^3-1^2+1^{}+1=2$

also, $g\left(x\right)=3-x+x^{2}for1<x\le 2$

${lim}_{x\to 1^{+}}g\left(x\right)=3-1+1^2=3$

${lim}_{x\to 1^{-}}\ne {lim}_{x\to 1^{+}}$

Thererfore, g(x) is not continous and hence not differentiable at $x=1$

Correct answer is C