Question

If $f\left(x\right)$ be such that $f\left(x\right)=$max $\{|2-x|,2-x^3\},$ then

• A. $f\left(x\right)$ is continuous $\forall x\in R$
• B. $f\left(x\right)$ is differentiable $\forall x\in R$
• C. $f\left(x\right)$ is non-differentiable at one point only
• D. $f\left(x\right)$ is non-differentiable at 4 points only

Answer 1

Answer: (A), (D)

The correct options are
A $f\left(x\right)$ is continuous $\forall x\in R$
D $f\left(x\right)$ is non-differentiable at 4 points only

Based on the figure, we find the intersection points of both the curves.
$y=|2-x|=\left\{\begin{array}{cc}2-x& x<2\\ x-2& x>2\end{array}\right\}$
So, if $2-x>2-x^3$ we get $x<x^3$ i.e. $x(1+x)(1-x)<0$
$\Rightarrow x>1$ and $x<-1$
But taking the intersection with the original condition, we get $x<-1$ and $1<x<2$
If $x-2>2-x^3,x^3+x-4>0$, this is always true after $x>2$.
So the function is always continuous but not differential at 4 points due to change in curvature.