NCERT Solutions for Class 11 Maths Chapter 12
NCERT solutions for Class 11 Mathematics Chapter 12 Limits and Derivatives – complete answers & explanations
Class 11 Mathematics Chapter 12 Limits and Derivatives is one of the most important chapters that introduces students to the basic ideas of calculus. In this chapter, students learn how to evaluate limits, understand how functions behave near a point, and calculate derivatives using different mathematical rules. These concepts form the foundation for advanced topics in higher classes and are widely used in science, engineering, and real-life problem solving. A strong understanding of limits and derivatives helps students improve their logical thinking and calculation skills. This blog provides clear and reliable NCERT solutions that help students follow the correct method for solving questions step by step. Parents and teachers can also use these answers to guide students effectively during revision and homework practice. Download the worksheet and practice alongside solutions for better clarity. Book a free trial now to get expert guidance.
What this NCERT chapter covers?
1. Understanding the concept of limits
and how functions behave as values approach a specific point.
2. Learning to evaluate limits using algebraic simplification and substitution methods.
3. Identifying situations where limits exist and where limits do not exist.
4. Introduction to derivatives and their meaning as the rate of change of a function.
5. Applying differentiation rules to polynomial, rational, and algebraic expressions.
6. Differentiating trigonometric functions such as sin x, cos x, tan x, and sec x.
7. Using derivative formulas to solve mathematical problems accurately and efficiently.
8. Developing step-by-step problem-solving skills required for school exams and competitive tests.
9. Strengthening calculation accuracy and improving mathematical reasoning abilities.
10. Building a strong base for advanced mathematics topics like continuity, applications of derivatives, and calculus.
How to use these NCERT solutions?
1. First read each question carefully
and try solving it on your own before looking at the answers.
2. Use the solutions to check whether your method and final answer are correct.
3. If you make a mistake, review the steps and understand where the error occurred.
4. Practice similar questions again to strengthen your understanding of limits and derivatives.
5. Parents and teachers can use these answers to explain correct mathematical procedures to students.
6. Follow the same order of questions while revising so that learning remains structured and systematic.
7. Use these solutions regularly during homework, revision, and exam preparation.
8. Focus on understanding formulas and rules rather than memorising answers.
Important tips & tricks for students
1. Always write each step clearly while solving limit and derivative problems.
2. Apply the correct formula or rule before simplifying the expression.
3. Carefully check signs such as positive and negative values during calculations.
4. Revise standard differentiation formulas regularly to improve speed and accuracy.
5. Avoid skipping steps, as examiners often award marks for correct working.
6. Practice solving a variety of problems to gain confidence in applying different rules.
7. Double-check answers to ensure calculations are correct before submitting your work.
8. Read questions carefully to understand whether the limit exists or does not exist.
9. Maintain neat presentation to make your answers easy to read and evaluate.
NCERT solutions – complete answer key
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
Why NCERT solutions help students?
NCERT solutions help students prepare confidently for exams by providing the correct approach to solving mathematical problems. They improve concept clarity, strengthen problem-solving skills, and ensure that students follow the expected NCERT answering pattern. Regular practice with reliable solutions builds accuracy, boosts confidence, and supports better academic performance.
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