A function f is defined on [-3,3] as f(x) = { min {|x|, 2-x²}, -2 ≤ x ≤ 2 ; [|x|], 2

Q: Wait, why is $x=0$ a point of non-differentiability? It looks like a simple V-shape.

Excellent observation! The function $f(x) = |x|$ at the origin has a left-hand derivative of $-1$ and a right-hand derivative of $+1$. Since these are not equal, the slope is undefined at the vertex. Even though the function is continuous there, the sharp turn makes it non-differentiable.

Q: Why do we reject the negative root when solving $|x| = 2 - x^2$ for $x \ge 0$?

Great catch! When we assume $x \ge 0$, we are working in the first quadrant. The equation $|x| = 2 - x^2$ simplifies to $x = 2 - x^2$. Solving this quadratic gives $x=1$ and $x=-2$. Since we restricted our domain to $x \ge 0$, we must discard $x=-2$. Symmetry handles the negative side for us!

Q: Do I really need to check the boundary points $x = \pm 2$?

Absolutely! In JEE Advanced, boundary points are often where the 'hidden' traps lie. A function must be continuous to be differentiable. If you calculate the limit from the left and the limit from the right at $x=2$, you will find a jump discontinuity. A jump means the function is not continuous, and therefore, it cannot be differentiable.

Question

A function $f$ is defined on $[-3,3]$ as $f(x) = \begin{cases} \min \{|x|, 2-x^2\}, & -2 \le x \le 2 \ [|x|], & 2 < |x| \le 3 \end{cases}$ where $[x]$ denotes the greatest integer $\le x$. The number of points, where $f$ is not differentiable in $(-3,3)$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are going to dissect a beautiful piecewise function. In the world of JEE Advanced, piecewise functions are not just equations; they are stories of different behaviors in different regions.

Our goal is to find the number of points where the function $f(x)$ is not differentiable in the open interval $(-3, 3)$. Remember, non-differentiability is the 'enemy' of smoothness. It occurs either at a sharp corner or at a point of discontinuity.

### The Central Arena

We start in the region $x \in [-2, 2]$. Here, $f(x) = \min(|x|, 2-x^2)$.

Imagine two dancers on a stage: the V-shaped absolute value function $y = |x|$ and the inverted parabola $y = 2-x^2$. The function $f(x)$ is the 'lower envelope'—it always follows the dancer who is closer to the x-axis.

To see where they switch roles, we must find their intersection points. We set $|x| = 2-x^2$. For $x \ge 0$, this becomes:

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

Factoring this, we get $(x+2)(x-1) = 0$. Since we are in the positive domain, we find $x = 1$. By symmetry, the intersection on the negative side is $x = -1$. Thus, the function switches its definition at $x = -1$ and $x = 1$.

### Internal Non-Differentiable Points

Now, let us trace the path. From $x = -2$ to $x = -1$, the parabola $2-x^2$ is lower. At $x = -1$, the V-shape takes over.

At $x = 0$, the V-shape has a sharp corner (a cusp). Then, at $x = 1$, the parabola takes over again.

We have identified three points of non-differentiability so far: $x = -1$ (intersection), $x = 0$ (the cusp of $|x|$), and $x = 1$ (intersection). These are all sharp corners where the slope changes abruptly.

### Outer Regions and Boundary Traps

Now, we look at the outer regions: $2 < |x| \le 3$. This corresponds to $x \in (2, 3)$ and $x \in (-3, -2)$.

In these regions, the function is defined by the behavior of the outer components. We must check the boundaries at $x = 2$ and $x = -2$.

At $x = 2$, the value of our function from the central region is:

$$f(2) = 2 - (2)^2 = -2$$

However, the limit from the right is $2$. A jump from $-2$ to $2$ creates a massive discontinuity. The same happens at $x = -2$. Since the function is discontinuous at these points, it is automatically non-differentiable there.

### The Final Count

Let us tally our findings. We have two points of discontinuity at $x = -2$ and $x = 2$.

We have three points of sharp corners at $x = -1$, $x = 0$, and $x = 1$.

That gives us a total of **$5$ points of non-differentiability**. This problem teaches us that we must always be vigilant—check the intersections, check the corners, and never, ever ignore the boundaries.

Enter Numerical Value:

Visualized Solution

Understanding the Piecewise Function $f (x)$

Analyzing the Region $x \in [- 2, 2]$

Finding Intersection Points

Constructing $f (x)$ in $[- 2, 2]$

Identifying Internal Non-Differentiable Points

Analyzing the Region $2 < ∣ x ∣ \leq 3$

Checking Continuity at $x = 2$ and $x = - 2$

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Central Arena

Internal Non-Differentiable Points

Outer Regions and Boundary Traps

The Final Count

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