NCERT Solutions for Class 11 Mathematics Chapter 12 Limits and Derivatives

This worksheet provides complete NCERT Solutions for Class 11 Mathematics Chapter 12 Limits and Derivatives. This chapter introduces students to the fundamental ideas of limits and derivatives, which are essential concepts in higher mathematics. Students learn how to evaluate limits, understand the behaviour of functions, and calculate derivatives using standard mathematical rules. This chapter is important because it builds a strong base for advanced topics such as calculus, continuity, and applications of derivatives in real-life problem solving. This worksheet includes complete and accurate NCERT Solutions presented in a clear and structured manner to support effective learning and exam preparation. The solutions strictly follow the NCERT pattern and help students understand mathematical concepts step by step.

Chapter summary: stories, poems & themes

This chapter focuses on mathematical concepts related to limits and derivatives. It does not include stories or poems, but it involves problem-solving activities and mathematical expressions that help students understand how functions change and how to calculate rates of change. Students learn to apply formulas and rules to solve numerical and algebraic problems. The chapter is concept-based and practice-oriented, requiring logical thinking and careful calculation. The main learning focus of this chapter is to understand mathematical relationships, evaluate limits, and differentiate various algebraic and trigonometric functions.

What this NCERT chapter covers?

• Understanding the concept of limits and their evaluation 
• Learning how to determine whether limits exist or do not exist 
• Applying rules of differentiation to algebraic expressions 
• Differentiating polynomial, rational, and trigonometric functions 
• Using derivative formulas for solving mathematical problems 
• Developing problem-solving and analytical thinking skills 
• Strengthening calculation accuracy and mathematical reasoning 

How to use these NCERT solutions?

Students should first attempt each question from the exercise on their own before checking the answers provided in this worksheet. This helps in building confidence and improving understanding of mathematical concepts. Parents and teachers can use these NCERT Solutions to guide students in solving problems correctly and understanding the step-by-step process of finding limits and derivatives. The solutions follow the exact NCERT order and structure, making revision easier and more systematic. Regular practice using these solutions helps students prepare effectively for school assessments and examinations.

Student tips & learning tricks

• Carefully read each mathematical expression before solving the problem 
• Always apply the correct differentiation rule for each type of function 
• Check whether the limit exists before writing the final answer 
• Avoid calculation errors while simplifying algebraic expressions 
• Practice solving similar problems regularly to improve speed and accuracy 
• Revise standard derivative formulas to strengthen understanding 

Why NCERT solutions are important?

NCERT Solutions are important because they provide accurate and reliable answers based on the official curriculum. These solutions help students develop a strong foundation in mathematics and improve their problem-solving skills. Following NCERT-aligned solutions ensures that students learn the correct method for solving questions and understand the logic behind each step. Regular practice using structured solutions builds confidence, improves performance in exams, and supports long-term academic success.

Complete answer key – NCERT solutions

Exercise 12.1

1. 6 
2. π − 22/7 
3. π 
4. 19/2 
5. 11/4 
6. −1/2 
7. 108/7 
8. 5 
9. b 
10. 2 
11. 1 
12. −1/4 
13. a/b 
14. a/b 
15. 1/π 
16. 1/π 
17. 44 
18. (a + 1)/b 
19. 0 
20. 1 
21. 0 
22. 2 
23. (i) lim f(x) as x → 0 = 3 
(ii) lim f(x) as x → 1 = 6 
24. Does not exist 
25. Does not exist 
26. Does not exist 
27. 0 
28. a + b = 4 
b − a = 4 
Therefore, 
a = 0, b = 4 
29. lim f(x) as x → a₁ = 0 
For a ≠ a₁, a₂, ..., aₙ, 
lim f(x) as x → a = (a − a₁)(a − a₂)...(a − aₙ) 
30. lim f(x) exists for all a ≠ 0 
31. 2 
32. At x = 0: 
n = m 
At x = 1: 
n + m = n + m, so automatically satisfied 
Therefore, m = n, where m and n may be any integers 

Exercise 12.2

1. 20 
2. 1 
3. 99 
4. (i) For f(x) = x³ − 27, 
f′(x) = 3x² 

(ii) For f(x) = (x − 1)(x − 2), 
f′(x) = 2x − 3 

(iii) For f(x) = 1/x², 
f′(x) = −2/x³ 

(iv) For f(x) = (x + 1)/(x − 1), 
f′(x) = −2/(x − 1)² 

5. f(x) = x¹⁰⁰/100 + x⁹⁹/99 + ... + x²/2 + x + 1 
f′(x) = x⁹⁹ + x⁹⁸ + ... + x + 1 
So, 
f′(1) = 100 
f′(0) = 1 
Hence, 
f′(1) = 100f′(0) 

6. d/dx [xⁿ + axⁿ⁻¹ + a²xⁿ⁻² + ... + aⁿ⁻¹x + aⁿ] 
= nxⁿ⁻¹ + (n − 1)axⁿ⁻² + (n − 2)a²xⁿ⁻³ + ... + 2aⁿ⁻²x + aⁿ⁻¹ 

7. (i) d/dx [(x − a)(x − b)] = 2x − a − b 
(ii) d/dx [(ax² + b)²] = 4ax(ax² + b) 
(iii) d/dx [(x − a)/(x − b)] = (a − b)/(x − b)² 

8. (xⁿ − aⁿ)/(x − a) = xⁿ⁻¹ + axⁿ⁻² + a²xⁿ⁻³ + ... + aⁿ⁻¹ 
Therefore, d/dx [(xⁿ − aⁿ)/(x − a)] 

9. (i) d/dx (2x − 3/4) = 2 
(ii) d/dx [(5x³ + 3x − 1)(x − 1)] 
= (15x² + 3)(x − 1) + (5x³ + 3x − 1) 
(iii) d/dx [x⁻³(5 + 3x)] 
= −15x⁻⁴ − 6x⁻³ 
(iv) d/dx [x⁵(3 − 6x⁻⁹)] 
= 15x⁴ + 24x⁻⁵ 
(v) d/dx [x⁻⁴(3 − 4x⁻⁵)] 
= −12x⁻⁵ + 36x⁻¹⁰ 
(vi) d/dx [2/(x + 1) − x²/(3x − 1)] 
= −2/(x + 1)² − x(3x − 2)/(3x − 1)² 

10. d/dx (cos x) = −sin x 

11. (i) d/dx (sin x cos x) = cos²x − sin²x = cos 2x 
(ii) d/dx (sec x) = sec x tan x 
(iii) d/dx (5sec x + 4cos x) = 5sec x tan x − 4sin x 
(iv) d/dx (cosec x) = −cosec x cot x 
(v) d/dx (3cot x + 5cosec x) = −3cosec²x − 5cosec x cot x 
(vi) d/dx (5sin x − 6cos x + 7) = 5cos x + 6sin x 
(vii) d/dx (2tan x − 7sec x) = 2sec²x − 7sec x tan x 

Miscellaneous exercise

1. (i) d/dx (−x) = −1 
(ii) d/dx [(-x)⁻¹] = 1/x² 
(iii) d/dx [sin(x + 1)] = cos(x + 1) 
(iv) d/dx [cos(x − π/8)] = −sin(x − π/8) 

2. d/dx (x + a) = 1 

3. d/dx [(px + q)(r/x + s)] = ps − qr/x² 

4. d/dx [(ax + b)(cx + d)²] 
= a(cx + d)² + 2c(ax + b)(cx + d) 

5. d/dx [(ax + b)/(cx + d)] = (ad − bc)/(cx + d)² 

6. d/dx [(1 + 1/x)/(1 − 1/x)] = −2/(x − 1)² 

7. d/dx [1/(ax² + bx + c)] = −(2ax + b)/(ax² + bx + c)² 

8. d/dx [(ax + b)/(px² + qx + r)] 
= [a(px² + qx + r) − (ax + b)(2px + q)]/(px² + qx + r)² 

9. d/dx [(px² + qx + r)/(ax + b)] 
= [(2px + q)(ax + b) − a(px² + qx + r)]/(ax + b)² 

10. d/dx [a/x⁴ − b/x² + cos x] 
= −4a/x⁵ + 2b/x³ − sin x 

11. d/dx (4√x − 2) = 2/√x 

12. d/dx [(ax + b)ⁿ] = na(ax + b)ⁿ⁻¹ 

13. d/dx [(ax + b)ⁿ(cx + d)ᵐ] 
= na(ax + b)ⁿ⁻¹(cx + d)ᵐ + mc(ax + b)ⁿ(cx + d)ᵐ⁻¹ 

14. d/dx [sin(x + a)] = cos(x + a) 

15. d/dx [cosec x cot x] 
= −cosec x cot²x − cosec³x 

16. d/dx [cos x/(1 + sin x)] = −1/(1 + sin x) 

17. d/dx [(sin x + cos x)/(sin x − cos x)] 
= −2/(sin x − cos x)² 

18. d/dx [(sec x − 1)/(sec x + 1)] 
= 2sec x tan x/(sec x + 1)² 

19. d/dx (sinⁿx) = nsinⁿ⁻¹x cos x 

20. d/dx [(a + bsin x)/(c + dcos x)] 
= [bc cos x + ad sin x + bd]/(c + dcos x)² 

21. d/dx [sin(x + a)/cos x] = cos a · sec²x 

22. d/dx [x⁴(5sin x − 3cos x)] 
= 4x³(5sin x − 3cos x) + x⁴(5cos x + 3sin x) 

23. d/dx [(x² + 1)cos x] 
= 2x cos x − (x² + 1)sin x 

24. d/dx [(ax² + sin x)(p + qcos x)] 
= (2ax + cos x)(p + qcos x) − q(ax² + sin x)sin x 

25. d/dx [(x + cos x)(x − tan x)] 
= (1 − sin x)(x − tan x) + (x + cos x)(1 − sec²x) 

26. d/dx [(4x + 5sin x)/(3x + 7cos x)] 
= [(4 + 5cos x)(3x + 7cos x) − (4x + 5sin x)(3 − 7sin x)]/(3x + 7cos x)² 

27. d/dx [x²cos(π/4)/sin x] 
= cos(π/4) · (2x sin x − x² cos x)/sin²x 

28. d/dx [x/(1 + tan x)] 
= (1 + tan x − xsec²x)/(1 + tan x)² 

29. d/dx [(x + sec x)(x − tan x)] 
= (1 + sec x tan x)(x − tan x) + (x + sec x)(1 − sec²x) 

30. d/dx [x/sinⁿx] 
= [1 − nx cot x]/sinⁿx 

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