The function given by y = ||x| - 1| is differentiable for all real numbers except the points (a) {0,

Q: Wait, why do we check $x=0$ if the outer modulus is $\vert x \vert - 1$?

Great question! The function is nested: $y = \vert \vert x \vert - 1 \vert$. The non-differentiability of a modulus function occurs wherever the expression *inside* the bars equals zero. Since $\vert x \vert$ is part of the function, it creates a sharp corner at $x=0$. Even though the outer modulus is $\vert \vert x \vert - 1 \vert$, the inner $\vert x \vert$ still dictates the behavior at the origin.

Q: Can we just use the derivative formula? Why is graphing better?

You *could* use the derivative, but it would be tedious because you'd have to define the function piecewise across different intervals. Graphing is much faster and more intuitive for JEE problems involving absolute values. It helps you 'see' the sharp corners, which are the visual representation of non-differentiability.

Q: Is a function always non-differentiable at the roots of the modulus?

Almost always, yes! A modulus function $\vert u(x) \vert$ creates a 'bounce' or a sharp corner at the roots of $u(x) = 0$ because the slope changes sign abruptly. As long as the function is continuous at that point, the sharp corner guarantees non-differentiability.

Question

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

Options:

(A) $\{0, 1, -1\}$

(B) $\pm 1$

(C) 1

(D) $-1$

Sigma Percentile · Accepted Answer

### The Geometry of the Absolute Value

Welcome, future engineers! Today, we are going to dissect a beautiful problem involving the absolute value function. The function $y = \vert \vert x \vert - 1 \vert$ might look intimidating at first, but it is actually a masterpiece of geometric transformations.

Let's peel it back like an onion, layer by layer.

### Phase 1: The Foundation ($y = \vert x \vert$)

Every journey begins with a single step. Our base function is $y = \vert x \vert$. This is the classic V-shape that every JEE aspirant knows by heart.

It is perfectly symmetric, opening upwards, and it has a very special property: at the origin $(0, 0)$, it has a sharp corner.

The slope to the left is $-1$ and the slope to the right is $+1$. Since the left-hand derivative does not equal the right-hand derivative, the function is not differentiable at $x = 0$. This is our first critical point.

### Phase 2: The Vertical Shift ($y = \vert x \vert - 1$)

Now, let's apply the next operation: subtracting $1$. When we transform our function to $y = \vert x \vert - 1$, we are simply shifting the entire V-shaped graph downwards by one unit.

The sharp corner, which was previously at $(0, 0)$, now slides down to $(0, -1)$.

Notice what happens to the x-intercepts. The graph now crosses the x-axis at two points. To find them, we set $y = 0$, which gives us:

$$\vert x \vert - 1 = 0 \implies \vert x \vert = 1$$

This tells us the graph hits the x-axis at $x = 1$ and $x = -1$.

### Phase 3: The Final Reflection ($y = \vert \vert x \vert - 1 \vert$)

This is where the magic happens. We apply the outermost modulus. The rule for an outer modulus is simple: it takes any part of the graph that is below the x-axis and reflects it upwards.

Think of the x-axis as a mirror. The V-shaped portion that dipped below the axis between $x = -1$ and $x = 1$ is now flipped.

The sharp corner at $(0, -1)$ is reflected up to $(0, 1)$. The final graph looks like a 'W'.

### Phase 4: Locating the Sharp Corners

Now, let's scan our 'W' graph for sharp corners. We have one at $x = -1$ (where the graph bounces off the x-axis), another at $x = 0$ (the peak of the 'W'), and a third at $x = 1$ (where it bounces again).

These are the three points where the function is continuous but not differentiable. Algebraically, we can verify this by setting the inner expressions to zero:

$$\vert x \vert = 0 \implies x = 0$$

$$\vert x \vert - 1 = 0 \implies x = \pm 1$$

The geometry and the algebra are in perfect harmony! You have successfully navigated the transformation of this function. Keep this visual intuition in your toolkit—it will serve you well in the exam hall. **The points of non-differentiability are $x \in \{-1, 0, 1\}$.**

Animated Solution for Mathematics - Differentiation: The function given by $y = ∣∣ x ∣ - 1∣$ is differentiable for all real numbers except the points

Select Answer:

Visualized Solution

Understanding Differentiability

The Base Function: $y = ∣ x ∣$

Corner of $y = ∣ x ∣$

Vertical Shift: $y = ∣ x ∣ - 1$

x-intercepts of $y = ∣ x ∣ - 1$

The Final Fold: $y = ∣∣ x ∣ - 1∣$

The 'W' Shaped Graph

Locating Non-Differentiable Points

The Three Points

Algebraic Check: Inner Modulus

Algebraic Check: Outer Modulus

Final Answer

The Sigma Insight: Differentiability of a Function

The Geometry of the Absolute Value

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function given by y=∣∣x∣−1∣ is differentiable for all real numbers except the points

Select Answer:

Visualized Solution

Understanding Differentiability

The Base Function: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=∣x∣

Corner of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=∣x∣

Vertical Shift: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=∣x∣−1

x-intercepts of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=∣x∣−1

The Final Fold: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=∣∣x∣−1∣