Question

Let S be the set of points in the interval $\left( -4,4 \right)$ at which function f $f\left( x \right)=\left\{ \begin{matrix}
   \max \left\{ \left| x \right|,{{x}^{2}} \right\},\text{ }\left| x \right|<2 \\
   8-2\left| x \right|,\text{ }2<\left| x \right|\le 4 \\
\end{matrix} \right.$ is not differentiable. Then S is
$\begin{align}
  & \left( A \right)\text{ is an Empty set} \\
 & \left( B \right)equals\text{ }\left\{ -2,-1,1,2 \right\} \\
 & \left( C \right)equals\text{ }\left\{ -2,-1,0,1,2 \right\} \\
 & \left( D \right)equals\text{ }\left\{ -2,2 \right\} \\
\end{align}$

Answer 1

Hint: The major idea behind solving this question is dividing the given interval into several subparts and finding the function values for each subpart and then we differentiate the function and check both ends of the interval values after differentiation. If they are the same then the function is differentiable else not at that point.

Complete step-by-step solution
We are given that the function f is $f\left( x \right)=\left\{ \begin{matrix}
   \max \left\{ \left| x \right|,{{x}^{2}} \right\},\text{ }\left| x \right|\text{ }<2 \\
   8-2\left| x \right|,\text{ }2\text{ }<\left| x \right|\text{ }\ \le 4 \\
\end{matrix} \right.$.
Let us consider the region $\left| x \right|<2$, that is $-2<$ $x<2$. In that region the function is $f\left( x \right)=\max \left\{ \left| x \right|,{{x}^{2}} \right\}$.
And dividing the region further into two parts we get
When $-2<$ $x<0$, function is $f\left( x \right)=\max \left\{ -x,{{x}^{2}} \right\}$.
When $0<$$x<2$, function is $f\left( x \right)=\max \left\{ x,{{x}^{2}} \right\}$.
Now let us simplify the functions further by finding the region in which $-x$ is maximum and in which it is minimum.
$\begin{align}
  & -x>{{x}^{2}} \\
 &\Rightarrow {{x}^{2}}+x<0 \\
 &\Rightarrow x\left( x+1 \right)<0 \\
 &\Rightarrow x\in \left( -1,0 \right) \\
\end{align}$
So, we get
When $-2<$$x<-1$, function is $f\left( x \right)={{x}^{2}}$
When $-1<$ $x<0$, function is $f\left( x \right)=-x$.
Similarly, let us find the function values when $0<$$x<2$.
$\begin{align}
  & x>{{x}^{2}} \\
 &\Rightarrow {{x}^{2}}-x<0 \\
 &\Rightarrow x\left( x-1 \right)<0 \\
 &\Rightarrow x\in \left( 0,1 \right) \\
\end{align}$
So, we get
When $1<$$x<2$, function is $f\left( x \right)={{x}^{2}}$
When $0<$$x<1$, function is $f\left( x \right)=x$
Now let us consider the other interval $2<$$\left| x \right|\le 4$, It can be divide into two parts as
$-4\le $$x<-2$ and $2<$$x\le 4$. Then in those intervals the function values will be
When $-4\le $$x<-2$, $\left| x \right|=-x$. So, $f\left( x \right)=8-2\left( -x \right)=8+2x$
When $2<$$x\le 4$, $\left| x \right|=x$. So, $f\left( x \right)=8-2x$.
So, overall our function will be
$f\left( x \right)=\left\{ \begin{matrix}
   8+2x\text{ }; -4\le x<-2 \\
  {{x}^{2}} \text{ }; -2\lt x <-1 \\
  -x\text{ }; -1\lt x<0 \\
   x\text{ }; 0\lt x<1 \\
 {{x}^{2}}\text{ }; 1\lt x < 2 \\
   8-2x\text{ }; 2\lt x\le 4 \\
\end{matrix} \right.$
The possible points where the function is not differentiable are -4, -2, -1, 0, 1, 2, 4, as they are the points where the curve is changing.
Now let us differentiate the function with respective to x. Then we get
${f}'\left( x \right)=\left\{ \begin{matrix}
   2\text{ }; -4\le x<-2 \\
   2x\text{ }; -2< x<-1\\
   -1\text{ }; -1< x<0 \\
   \text{1}; 0< x<1 \\
   2x\text{ }; 1< x<2 \\
   -2\text{ }; 2< x\le 4 \\
\end{matrix} \right.$
Now let us verify if their values at both ends are equal.
From above we can see that except at x=-4, 4 all the points have different values of ${f}'\left( x \right)$ on their right-hand side and left-hand side. So, at all those points function is not differentiable.
So, we set $S=\left\{ -2,-1,0,1,2 \right\}$.
Hence answer is Option C.

Note: The possibility of making a mistake is one might miss writing the interval $2<\left| x \right|\le 4$ into two parts by thinking that it is between two positive values. But should check all the possibilities while dividing the intervals.