Question

Let $f:R\to R$ and $g:R\to R$ be respectively given by $f\left(x\right)=|x|+1$ and $g\left(x\right)=x^2+1$. Define $h:R\to R$ by
$h\left(x\right)=\{\begin{array}{cccc}max& \left\{f\right(x),g(x\left)\right\}& if& x\le 0\\ min& \left\{f\right(x),g(x\left)\right\}& if& x>0\end{array}$.
Then number of points at which $h\left(x\right)$ is not differentiable is

Answer 1

Case 1: When $x<0$

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

At the meeting point $f\left(x\right)=g\left(x\right)$

$x^2+1=-x+1⟹x(x+1)=0$

$x=-1⟹y=2$

$g\left(x\right)$ and $f\left(x\right)$ meet at point $(-1,2)$

Similarly in Case 2: when $x\ge 0$

$f\left(x\right)=g\left(x\right)⟹x+1=x^2+1⟹x^2-x=0⟹x=0,1$

Meeting points are $(0,1)$ and $(1,2)$

From the figure

If $x<-1\Rightarrow max\left\{f\right(x),g(x\left)\right\}=g\left(x\right)$

If $-1<x<0\Rightarrow max\left\{f\right(x),g(x\left)\right\}=f\left(x\right)$

If $0<x<1\Rightarrow min\left\{f\right(x),g(x\left)\right\}=g\left(x\right)$

If $x>1\Rightarrow min\left\{f\right(x),g(x\left)\right\}=f\left(x\right)$

Therefore, Non-differential points are $(0,1),(-1,2)$ and $(1,2)$

Correct answer is $3$