Question

Let $f\left(x\right)=x^3-x^2+x+1$ and $g\left(x\right)=\{\begin{array}{c}max\{f\left(t\right);0\le t\le x\};0\le x\le 1\\ 3-x;1<x\le 2\end{array}$ then

• A. $g\left(x\right)$ is continuous and differentiable in $(0,2)$
• B. $g\left(x\right)$ is discontinuous at finite number of points in $(0,2)$
• C. $g\left(x\right)$ is non-drivable at $2$ points
• D. $g\left(x\right)$ is continuous but non-differentiable at one points

Answer 1

The correct option is A $g\left(x\right)$ is continuous and differentiable in $(0,2)$

$f\left(x\right)=x^3-x^2+x+1$

$g\left(x\right)=\{\begin{array}{c}max.\left(F\right(t):0\le t\le x)0\le x\le 1\\ 3-x;1\le x\le 2\end{array}$

here $F\left(0\right)=1,F\left(1\right)=2$

If $F\left(x\right)$ increasing in $0$ to $1$ then max will be $f\left(t\right)$

$F\left(t\right)=3x^2-2x+1$

$0=9x^2-2x+1$

$x=\frac{2\pm \sqrt{8}i}{6}$

clearly $f{x}^n$ is always increasing

So $g\left(x\right)=x^3-x^2+x+1,0\le x\le 1$

$g-x,1\le x\le 1$

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

$\underset{x\to 1}{lim}g\left(x\right)=2$

So clearly $g\left(x\right)\to$ continuous

$g\left(1\right)=2$ at $x=1$

clearly $F{x}^n$ is deriable