Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}\max \{|x|,x^2\},& |x|\le 2\\ 8-2|x|,& 2<|x|\le 4\end{array}$. Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

Welcome, student. Today, we are not just solving a math problem; we are performing a surgical dissection of a function. When you see a piecewise function like $f(x)=\max \{|x|,x^2\}$ for $|x|\le 2$ and $f(x)=8-2|x|$ for $2<|x|\le 4$, do not panic.

Do not rush to differentiate blindly. Instead, pause and visualize. Imagine this function as a path you are walking along. Where might you stumble? Where might the path have a sharp, jagged edge?

The Inner Core ($|x|\le 2$)

Let's look at the heart of the function: $f(x)=\max \{|x|,x^2\}$. This is a battle between two functions: the linear $|x|$ and the parabolic $x^2$.

For $x$ values very close to zero, say $x=0.5$, we have $|x|=0.5$ and $x^2=0.25$. Clearly, $|x|$ is larger. In fact, for any $x$ in the interval $(-1,1)$, $|x|>x^2$.

So, in this region, our function is simply $f(x)=|x|$. This gives us that classic V-shape centered at the origin.

But as we move further out, towards $x=1$ or $x=-1$, the parabola $x^2$ starts to grow faster. Once $|x|>1$, the square of the number becomes larger than the number itself.

Thus, for $x\in [-2,-1)\cup (1,2]$, the function 'switches' to $f(x)=x^2$. We have effectively stitched a parabola onto the arms of our V-shape.

The Outer Wings ($2<|x|\le 4$)

Now, let's look at the outer regions. For $2<|x|\le 4$, the function is defined as $f(x)=8-2|x|$. This is a linear function.

If $x$ is positive, $f(x)=8-2x$. If $x$ is negative, $f(x)=8+2x$.

These are straight lines with slopes of $-2$ and $+2$, respectively. They act as the 'wings' of our graph, bringing the function down to the x-axis at $x=4$ and $x=-4$.

The Differentiability Hunt

Now, the hunt begins. A function is non-differentiable where it is discontinuous or where it has a 'kink' (a sharp corner). We must check the transition points.

1. At $x=0$: The function is $f(x)=|x|$. The Left-Hand Derivative (LHD) is: The Right-Hand Derivative (RHD) is: Since $-1 eq 1$, $x=0$ is a point of non-differentiability.

2. At $x=1$: To the left, $f(x)=x$, so the LHD is $1$. To the right, $f(x)=x^2$, so the RHD is: Since $1 eq 2$, $x=1$ is a point of non-differentiability.

3. At $x=-1$: By symmetry, the LHD (from $x^2$) is $2(-1)=-2$, and the RHD (from $-x$) is $-1$. Since $-2 eq -1$, $x=-1$ is a point of non-differentiability.

4. At $x=2$: To the left, $f(x)=x^2$, so the LHD is $2(2)=4$. To the right, $f(x)=8-2x$, so the RHD is $-2$. Since $4 eq -2$, $x=2$ is a point of non-differentiability.

5. At $x=-2$: To the left, $f(x)=8+2x$, so the LHD is $2$. To the right, $f(x)=x^2$, so the RHD is $2(-2)=-4$. Since $2 eq -4$, $x=-2$ is a point of non-differentiability.

We have found five points where the slope changes abruptly: $\{-2,-1,0,1,2\}$. Each of these points represents a sharp corner in our graph.

By systematically checking the LHD and RHD at every transition point, we have successfully identified the set $S$. Remember, in JEE Advanced, visualization is your greatest weapon. Trust your graph, verify with calculus, and you will never go wrong. The final set of non-differentiable points is $\{-2,-1,0,1,2\}$.