Last Updated: April 2026
CBSE Class 12 Mathematics Chapter 5 — Continuity and Differentiability — is one of the highest-weightage chapters in the CBSE Class 12 Maths board exam 2027. Carrying approximately 8–10 marks directly and forming the foundation for Differentiation (Chapter 6) and Integration (Chapter 7), this chapter requires thorough conceptual understanding. This page provides complete NCERT solutions, important theorems, formulas, and board exam practice questions.
Chapter 5 Overview — Continuity and Differentiability
| Topic | Marks Weightage | Difficulty |
|---|---|---|
| Continuity of Functions | 3–4 marks | Easy–Moderate |
| Differentiability | 2–3 marks | Moderate |
| Differentiation of Composite Functions (Chain Rule) | 4–6 marks | Moderate |
| Implicit Differentiation | 3–4 marks | Moderate |
| Derivatives of Inverse Trigonometric Functions | 3–4 marks | Moderate–High |
| Exponential and Logarithmic Differentiation | 4–5 marks | Moderate–High |
| Parametric Differentiation | 3–4 marks | High |
| Rolle’s Theorem and Mean Value Theorem | 3–4 marks | Moderate |
| Second Order Derivatives | 3–4 marks | Moderate |
Key Formulas — Continuity and Differentiability
Continuity — Definition and Conditions
A function f(x) is continuous at x = a if:
- f(a) is defined
- lim(x→a) f(x) exists
- lim(x→a) f(x) = f(a)
Left-Hand Limit = Right-Hand Limit = f(a)
Standard Derivatives — Must Memorise
| Function f(x) | Derivative f'(x) |
|---|---|
| sin x | cos x |
| cos x | –sin x |
| tan x | sec² x |
| cot x | –cosec² x |
| sec x | sec x · tan x |
| cosec x | –cosec x · cot x |
| eˣ | eˣ |
| ln x (log x) | 1/x |
| aˣ | aˣ · ln a |
| xⁿ | n · x^(n–1) |
Inverse Trigonometric Derivatives
| Function f(x) | Derivative f'(x) |
|---|---|
| sin⁻¹ x | 1/√(1–x²) |
| cos⁻¹ x | –1/√(1–x²) |
| tan⁻¹ x | 1/(1+x²) |
| cot⁻¹ x | –1/(1+x²) |
| sec⁻¹ x | 1/(|x|√(x²–1)) |
| cosec⁻¹ x | –1/(|x|√(x²–1)) |
NCERT Exercise 5.1 — Continuity Solutions (Key Questions)
Question 1: Prove that f(x) = 5x – 3 is continuous at x = 0, x = –3, and x = 5.
Solution:
f(x) = 5x – 3 is a polynomial function. All polynomial functions are continuous everywhere on ℝ.
- At x = 0: f(0) = 5(0) – 3 = –3. lim(x→0) f(x) = –3 = f(0). Continuous. ✓
- At x = –3: f(–3) = 5(–3) – 3 = –18. lim(x→–3) f(x) = –18 = f(–3). Continuous. ✓
- At x = 5: f(5) = 5(5) – 3 = 22. lim(x→5) f(x) = 22 = f(5). Continuous. ✓
Question 10: Find all points of discontinuity for f(x) = |x| + |x – 1|
Solution:
f(x) = |x| + |x – 1| is the sum of two absolute value functions. Both |x| and |x – 1| are continuous everywhere on ℝ. Therefore their sum f(x) is also continuous everywhere — there are no points of discontinuity.
Question 19: For what value of k is f(x) continuous at x = 2?
f(x) = kx + 1 if x ≤ 5, and f(x) = 3x – 5 if x > 5
Solution:
For continuity at x = 5: LHL = RHL = f(5)
- LHL = lim(x→5⁻) (kx + 1) = 5k + 1
- RHL = lim(x→5⁺) (3x – 5) = 15 – 5 = 10
- f(5) = k(5) + 1 = 5k + 1
For continuity: 5k + 1 = 10 → 5k = 9 → k = 9/5
NCERT Exercise 5.2 — Differentiability Solutions
Understanding Differentiability
A function f(x) is differentiable at x = a if:
LHD = RHD
LHD = lim(h→0⁻) [f(a+h) – f(a)] / h
RHD = lim(h→0⁺) [f(a+h) – f(a)] / h
Key Fact: Differentiability implies continuity. But continuity does NOT imply differentiability. f(x) = |x| is continuous at x = 0 but not differentiable there (corner point).
NCERT Exercise 5.4 — Implicit Differentiation
Method for Implicit Differentiation
- Differentiate both sides with respect to x
- Apply chain rule: d/dx [f(y)] = f'(y) · dy/dx
- Collect all dy/dx terms on one side
- Solve for dy/dx
Example: Find dy/dx if x² + y² = 25
Solution:
Differentiating both sides: 2x + 2y(dy/dx) = 0
→ 2y(dy/dx) = –2x
→ dy/dx = –x/y
NCERT Exercise 5.5 — Logarithmic Differentiation
Used when function is of the form: y = [f(x)]^g(x) or products/quotients of complex functions.
Method:
- Take log on both sides: ln y = g(x) · ln[f(x)]
- Differentiate both sides w.r.t. x
- (1/y)(dy/dx) = g'(x)·ln[f(x)] + g(x)·f'(x)/f(x)
- Multiply both sides by y to get dy/dx
Example: Find dy/dx if y = xˢⁱⁿˣ
Solution:
ln y = sin x · ln x
Differentiating: (1/y)(dy/dx) = cos x · ln x + sin x · (1/x)
dy/dx = y[cos x · ln x + sin x/x]
dy/dx = xˢⁱⁿˣ[cos x · ln x + sin x/x]
Rolle’s Theorem and Mean Value Theorem
Rolle’s Theorem
If f(x) is: (1) continuous on [a, b], (2) differentiable on (a, b), (3) f(a) = f(b), then there exists at least one c ∈ (a, b) such that f'(c) = 0.
Lagrange’s Mean Value Theorem (LMVT)
If f(x) is continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that:
f'(c) = [f(b) – f(a)] / (b – a)
Example: Verify Rolle’s Theorem for f(x) = x² – 4x + 3 on [1, 3]
Solution:
- f(1) = 1 – 4 + 3 = 0
- f(3) = 9 – 12 + 3 = 0
- f(1) = f(3) ✓; f is polynomial (continuous and differentiable everywhere)
- f'(x) = 2x – 4
- f'(c) = 0 → 2c – 4 = 0 → c = 2
- c = 2 ∈ (1, 3) ✓ — Rolle’s Theorem verified
Important Board Exam Questions — Chapter 5
3-Mark Questions (Commonly Asked)
- Differentiate sin(cos(x²)) with respect to x
- If y = √(sin x + √(sin x + √(sin x + …))), find dy/dx
- Find dy/dx if y = sin⁻¹(2x/(1+x²))
- If y = (sin x)^(cos x) + (cos x)^(sin x), find dy/dx
- Verify LMVT for f(x) = 2x² – 3x + 1 on [1, 3]
5-Mark Questions (Board Exam Pattern)
- If y = (tan⁻¹ x)², prove that (1+x²)² y₂ + 2x(1+x²) y₁ = 2
- If y = Acos(log x) + Bsin(log x), prove that x²y₂ + xy₁ + y = 0
- Find the second derivative of e^(sin⁻¹ x) and verify given identity
Common Mistakes to Avoid — Chapter 5
- Forgetting to check all 3 conditions for continuity (defined, limit exists, limit = value)
- Confusing LHD and RHD calculation direction
- Forgetting chain rule when differentiating composite functions
- Incorrect sign in cos⁻¹ x derivative (it’s –1/√(1–x²), not +1)
- In implicit differentiation: forgetting the chain rule factor dy/dx when differentiating y-terms
- Logarithmic differentiation: forgetting to multiply back by y at the end
Practice Quiz — Continuity and Differentiability
Practice Quiz — 10 CLAT-Style Questions
Click an option to reveal the answer and explanation.
Frequently Asked Questions — CBSE Class 12 Maths Chapter 5
Q1: How many marks does Chapter 5 carry in CBSE 2027?
Chapter 5 carries 8–10 marks directly. It also underpins Chapter 6 (Applications of Derivatives, 10–12 marks) and Chapter 7 (Integrals). Combined, mastery of Chapter 5 impacts 25–30 marks of the paper.
Q2: What is the difference between continuity and differentiability?
Continuity means no break in the function at a point. Differentiability means a unique derivative exists at that point. Every differentiable function is continuous, but not vice versa — f(x) = |x| is continuous but not differentiable at x = 0 due to the corner point.
Q3: When should I use logarithmic differentiation?
Use it when: (a) function has variable in both base and exponent (y = x^(sin x)), (b) function is a complicated product/quotient, or (c) direct differentiation would be excessively complex. Always take natural log, differentiate, then multiply back by y.
Q4: What is Rolle’s Theorem?
Rolle’s Theorem: if f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then ∃ c ∈ (a,b) where f'(c) = 0. Board exams regularly ask “verify Rolle’s Theorem” problems for 3–4 marks — always verify all three conditions explicitly.
Q5: Most important topics in Chapter 5 for CBSE 2027?
Priority topics: chain rule, logarithmic differentiation, implicit differentiation, inverse trig derivatives, second order derivatives, and Rolle’s/MVT theorem. These cover 90% of actual board exam questions from this chapter.
Conclusion
CBSE Class 12 Mathematics Chapter 5 — Continuity and Differentiability — is both conceptually important and heavily tested in the board exam 2027. Build your foundation with NCERT examples and exercises, then move to previous year board questions. Ready For Boards’ AI answer checking helps you verify your working step-by-step and get personalised feedback — so you score full marks, not partial marks, on these derivation questions.
