Is every continuous function differentiable at all points?

No. A classic counterexample is $f(x) = |x|$, which is continuous everywhere on $\mathbb{R}$, but non-differentiable at $x = 0$ due to a sharp corner where $LHD = -1 \neq 1 = RHD$.

How do I quickly find points of non-differentiability for $f(x) = [x]$ or $f(x) = \{x\}$?

Greatest integer $[x]$ and fractional part $\{x\}$ functions are discontinuous at all integer points $x = n \in \mathbb{Z}$. Since discontinuity implies non-differentiability, these functions are automatically non-differentiable at all integer values.

What is the strategic approach to solve functional equation differentiability problems in JEE Advanced?

Always apply the First Principle of Differentiation: express $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ and utilize the given functional relation to isolate $f'(0)$ or $f(0)$ before integrating.

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Solved High-Yield Sample Questions

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Question #1JEE Main 2024

If the function $f(x) = \begin{cases} a x^2 + 1, & x \le 1 \\ 2x + b, & x > 1 \end{cases}$ is differentiable at $x = 1$ , then the value of $a + b$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Enforce continuity at $x = 1$ by equating limits: $\lim_{x \to 1^-} (a x^2 + 1) = \lim_{x \to 1^+} (2x + b) \implies a + 1 = 2 + b \implies a - b = 1$ .

Step 2

Enforce differentiability at $x = 1$ by equating derivatives: $LHD = \left.\frac{d}{dx}(a x^2 + 1)\right|_{x=1} = 2a(1) = 2a$ and $RHD = \left.\frac{d}{dx}(2x + b)\right|_{x=1} = 2$ . Thus $2a = 2 \implies a = 1$ .

Step 3

Substitute $a = 1$ into $a - b = 1$ to obtain $1 - b = 1 \implies b = 0$ . Hence $a + b = 1 + 0 = 1$ .

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Question #2JEE Main 2023

The number of points where the function $f(x) = |x^2 - 5x + 6|$ is not differentiable is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Factorize the inner function $g(x) = x^2 - 5x + 6 = (x - 2)(x - 3)$ . Modulus functions $|g(x)|$ are potentially non-differentiable at roots $x = 2$ and $x = 3$ .

Step 2

Check the derivative of inner function $g'(x) = 2x - 5$ . At $x = 2$ , $g'(2) = -1 \neq 0$ , and at $x = 3$ , $g'(3) = 1 \neq 0$ .

Step 3

Since $g'(x) \neq 0$ at both simple roots $x = 2$ and $x = 3$ , sharp corners form at both points, giving exactly $2$ points of non-differentiability.

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Mastering Continuity and Differentiability: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Definition of Continuity at a Point

First Principle of Differentiability (LHD & RHD)

Intermediate Value Theorem (IVT)

Differentiability of Modulus Functions

Chain Rule for Composite Differentiability

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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