Differentiability at a Point: Modulus, GIF, and Min/Max Functions

Introduction

After continuity, the next most asked question type in JEE Main (35–40 questions over the past six years) involves checking differentiability of piecewise, modulus, greatest integer, and min/max functions at specific points. The key distinction: a function can be continuous at a point without being differentiable there, but differentiability always implies continuity. This article gives you the complete method—from definition to JEE-level execution.

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1. Fundamental Principle

Definition of Differentiability

A function $f$ is differentiable at $x = a$ if the following limit exists and is finite: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}.$

This requires: $\boxed{\text{LHD} = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h} = \text{RHD}}$

Key Relationships

• Differentiable $\Rightarrow$ Continuous (always true).
• Continuous $\not\Rightarrow$ Differentiable (e.g., $|x|$ at 0).
• Not continuous $\Rightarrow$ Not differentiable (contrapositive of the above).

Sharp Corner Test

If $f$ is continuous at $x = a$ but the graph has a "sharp corner" (abrupt change in slope), then $f$ is not differentiable at $x = a$. The LHD and RHD exist but are unequal.

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2. Non-Differentiable Functions: The Big Three

Type 1: Modulus Functions

$f(x) = |g(x)|$ is not differentiable at points where $g(x) = 0$ (provided $g'(x) \neq 0$ at that point).

Standard example: $f(x) = |x - 3|$

• LHD at $x = 3$: $-1$
• RHD at $x = 3$: $+1$
• Not differentiable.

Counting points of non-differentiability: For $f(x) = |x^2 - 5x + 6| = |(x-2)(x-3)|$, the function is not differentiable at $x = 2$ and $x = 3$ (where the expression inside changes sign).

Type 2: Greatest Integer Function (GIF)

$f(x) = \lfloor g(x) \rfloor$ is discontinuous (and hence not differentiable) at every point where $g(x)$ is an integer and $g$ is strictly monotone.

Example: $f(x) = \lfloor x^2 \rfloor$ on $[0, 2]$ is not differentiable at $x = 1, \sqrt{2}, \sqrt{3}, 2$ (where $x^2$ passes through integers).

Type 3: Min/Max Functions

$f(x) = \min(g(x), h(x))$ or $f(x) = \max(g(x), h(x))$ is not differentiable at points where $g(x) = h(x)$ (the "crossover" points), unless $g$ and $h$ are tangent there.

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3. Worked Examples

Example 1 (Basic Modulus)

Find the number of points where $f(x) = |x^3 - 3x^2 + 2x|$ is not differentiable.

Solution:

Factor: $x^3 - 3x^2 + 2x = x(x-1)(x-2)$.

The expression changes sign at $x = 0, 1, 2$. At each of these, $f$ has a sharp corner (since $g'(x) \neq 0$ at these points).

Answer: 3 points.

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Example 2 (Piecewise Differentiability)

$f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x = 0 \end{cases}$ Is $f$ differentiable at $x = 0$?

Solution:

Check using the definition: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h) = 0.$ (Since $|h \sin(1/h)| \leq |h| \to 0$.)

Yes, $f$ is differentiable at 0, with $f'(0) = 0$.

Important contrast: $g(x) = x \sin(1/x)$ (without the $x^2$) is continuous but not differentiable at 0, because $\lim_{h \to 0} \sin(1/h)$ does not exist.

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Example 3 (Max Function)

Let $f(x) = \max(x, x^2)$ on $[-1, 2]$. Find the points of non-differentiability.

Solution:

Solve $x = x^2$: $x(x-1) = 0$, so $x = 0, 1$.

• For $x < 0$: $x^2 > x$ (since $x$ is negative, $x^2$ is positive), so $f(x) = x^2$.
• For $0 < x < 1$: $x > x^2$, so $f(x) = x$.
• For $x > 1$: $x^2 > x$, so $f(x) = x^2$.

At $x = 0$: LHD $= 2(0) = 0$, RHD $= 1$. Not differentiable. At $x = 1$: LHD $= 1$, RHD $= 2(1) = 2$. Not differentiable.

Answer: 2 points ($x = 0, 1$).

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Example 4 (Modulus + Parameter)

If $f(x) = |x - a| + |x - b|$ where $a < b$, at how many points is $f$ not differentiable?

Solution:

Sharp corners occur at $x = a$ and $x = b$.

At $x = a$: LHD $= -1 + (-1) = -2$, RHD $= 1 + (-1) = 0$. Not differentiable. At $x = b$: LHD $= 1 + (-1) = 0$, RHD $= 1 + 1 = 2$. Not differentiable.

Answer: 2 points.

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Example 5 (Differentiability implies Continuity — Parameter Finding)

$f(x) = \begin{cases} ax + b, & x \leq 1 \\ x^2 + 3x + 2, & x > 1 \end{cases}$ Find $a$ and $b$ so that $f$ is differentiable at $x = 1$.

Solution:

Continuity at $x = 1$: $a + b = 1 + 3 + 2 = 6$.

Differentiability at $x = 1$:

• LHD $= a$
• RHD $= 2(1) + 3 = 5$

So $a = 5$, and $b = 6 - 5 = 1$.

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Example 6 (Counting Non-Differentiable Points)

Find the total number of points of non-differentiability of $f(x) = \max(\sin x, \cos x)$ on $[0, 2\pi]$.

Solution:

$\sin x = \cos x$ when $\tan x = 1$, i.e., $x = \frac{\pi}{4}, \frac{5\pi}{4}$.

At these points, the function switches between $\sin x$ and $\cos x$, creating sharp corners.

Answer: 2 points.

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4. Previous Year JEE Problems

Problem 1 (JEE Main 2023)

Let $f(x) = |2x^2 + 5|x| - 3|$. The number of points of non-differentiability of $f(x)$ in $\mathbb{R}$ is:

Solution:

First solve $2x^2 + 5|x| - 3 = 0$. Let $t = |x| \geq 0$: $2t^2 + 5t - 3 = 0 \Rightarrow t = \frac{-5 + \sqrt{49}}{4} = \frac{1}{2} \quad (\text{taking positive root}).$

So $|x| = \frac{1}{2}$, giving $x = \pm\frac{1}{2}$.

Now $g(x) = 2x^2 + 5|x| - 3$ changes sign at $x = \pm\frac{1}{2}$, creating corners in $f = |g|$.

Also, $g(x) = 2x^2 + 5|x| - 3$ itself has a corner at $x = 0$ due to $|x|$. But since $g(0) = -3 < 0$, and $|g(0)| = 3$, we need to check if $f$ has a corner at 0. Since $g$ is negative in a neighborhood of 0, $f(x) = -(2x^2 + 5|x| - 3) = -2x^2 - 5|x| + 3$, which has a corner at 0.

Points of non-differentiability: $x = -\frac{1}{2}, 0, \frac{1}{2}$. Answer: 3.

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Problem 2 (JEE Main 2021)

The number of points at which $f(x) = |x^2 - 1| + |x^2 - 4|$ is not differentiable is:

Solution:

$x^2 - 1 = 0 \Rightarrow x = \pm 1$. $x^2 - 4 = 0 \Rightarrow x = \pm 2$.

Check each:

• At $x = \pm 1$: $|x^2-1|$ has a corner, but $|x^2-4| = |1-4| = 3$ is smooth. So $f$ is not differentiable.
• At $x = \pm 2$: $|x^2-4|$ has a corner, but $|x^2-1| = |4-1| = 3$ is smooth. So $f$ is not differentiable.

Answer: 4 points ($x = -2, -1, 1, 2$).

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Problem 3 (JEE Main 2024)

Let $f(x) = \min(|x|, 1-|x|)$. The number of points where $f$ is not differentiable in $[-2, 2]$ is:

Solution:

Find where the functions cross: $|x| = 1 - |x| \Rightarrow 2|x| = 1 \Rightarrow x = \pm\frac{1}{2}$.

Analyzing the graph in segments:

1. For $x \in [-0.5, 0.5]$: $|x| < 1-|x|$, so $f(x) = |x|$. This has a sharp corner at $x = 0$.
2. For $|x| > 0.5$: $1-|x| < |x|$, so $f(x) = 1-|x|$.

At the transition points $x = \pm 0.5$:

• At $x = 0.5$: LHD comes from $|x|$ (slope 1), RHD comes from $1-|x|$ (slope $-1$). Sharp corner.
• At $x = -0.5$: LHD comes from $1-|x|$ (slope 1), RHD comes from $|x|$ (slope $-1$). Sharp corner.

Note: At $x = \pm 1$, the function is $f(x) = 1-|x|$, which is linear locally—it is differentiable there.

Answer: 3 points ($x = -0.5, 0, 0.5$).

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5. Key Formulas & Results

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6. Tips for JEE Aspirants

1. Always check continuity first. If $f$ is not continuous at a point, it's automatically not differentiable—no further work needed.
2. For modulus functions, find the zeros of the inner expression. Each zero (where the function crosses the x-axis) is a potential non-differentiable point.
3. For piecewise + parameter problems, you always get 2 equations: one from continuity (LHL = RHL = f(a)) and one from differentiability (LHD = RHD).
4. Draw a rough sketch. Even a quick mental picture of the graph immediately reveals corners and cusps.
5. $x^n \sin(1/x)$ rule of thumb: differentiable at 0 if $n \geq 2$, not differentiable if $n = 1$, not continuous if $n < 1$.
6. Count systematically. For "how many points" questions, list all candidate points first, then verify each one.

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7. Quick-Reference Summary

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Differentiability questions are a goldmine for JEE scorers—they follow predictable patterns. Learn to spot the three types (modulus, GIF, min/max) and you'll handle them effortlessly.