Limits and Derivatives — Class 11 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Limits and Derivatives" — 8 important questions with detailed answers for CBSE board exam preparation.

Limits and Derivatives — Class 11 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Limits and Derivatives" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Limits and Derivatives" — 8 important questions with detailed answers for CBSE boa…

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NCERT Solutions for Class 11 Mathematics — Limits and Derivatives. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.

Limits and Derivatives — All 8 Questions Solved

Q1. Find the limit: lim(x→2) (x² - 4)/(x - 2)

Find: lim(x→2) (x² - 4)/(x - 2)

Step 1: Check for direct substitution:
When x = 2: (2² - 4)/(2 - 2) = 0/0 (indeterminate form)

Step 2: Factorize the numerator:
x² - 4 = (x + 2)(x - 2)

Step 3: Cancel common factors:
lim(x→2) (x² - 4)/(x - 2) = lim(x→2) [(x + 2)(x - 2)]/(x - 2)
= lim(x→2) (x + 2)

Step 4: Apply direct substitution:
= 2 + 2 = 4

Final Answer: The limit is 4

Q2. Find the limit: lim(x→0) (sin x)/x

Find: lim(x→0) (sin x)/x

Step 1: This is a standard limit form.

Step 2: Direct substitution gives 0/0 (indeterminate).

Step 3: Use L'Hôpital's Rule:
lim(x→0) (sin x)/x = lim(x→0) (d/dx sin x)/(d/dx x)
= lim(x→0) (cos x)/1
= cos 0 = 1

Alternatively, this is a well-known standard limit:
lim(x→0) (sin x)/x = 1

Final Answer: The limit is 1

Q3. Find the derivative of f(x) = 3x² + 2x - 5 using the first principle of derivatives.

Find the derivative of f(x) = 3x² + 2x - 5 using first principle.

Step 1: Use the definition:
f'(x) = lim(h→0) [f(x + h) - f(x)]/h

Step 2: Calculate f(x + h):
f(x + h) = 3(x + h)² + 2(x + h) - 5
= 3(x² + 2xh + h²) + 2x + 2h - 5
= 3x² + 6xh + 3h² + 2x + 2h - 5

Step 3: Calculate f(x + h) - f(x):
f(x + h) - f(x) = [3x² + 6xh + 3h² + 2x + 2h - 5] - [3x² + 2x - 5]
= 6xh + 3h² + 2h
= h(6x + 3h + 2)

Step 4: Apply the limit:
f'(x) = lim(h→0) [h(6x + 3h + 2)]/h
= lim(h→0) (6x + 3h + 2)
= 6x + 0 + 2
= 6x + 2

Step 5: Verify using power rule:
f(x) = 3x² + 2x - 5
f'(x) = 3(2x) + 2(1) + 0 = 6x + 2 ✓

Final Answer: f'(x) = 6x + 2

Q4. Find the derivative of f(x) = √(2x + 3)

Find the derivative of f(x) = √(2x + 3)

Step 1: Rewrite the function:
f(x) = (2x + 3)^(1/2)

Step 2: Use the chain rule:
f'(x) = d/dx[(2x + 3)^(1/2)]
= (1/2)(2x + 3)^(-1/2) × d/dx(2x + 3)

Step 3: Find the derivative of the inner function:
d/dx(2x + 3) = 2

Step 4: Substitute:
f'(x) = (1/2)(2x + 3)^(-1/2) × 2
= (2x + 3)^(-1/2)
= 1/√(2x + 3)

Final Answer: f'(x) = 1/√(2x + 3)

Q5. Find the derivative of f(x) = x² sin x

Find the derivative of f(x) = x² sin x

Step 1: This is a product of two functions.
Let u = x² and v = sin x

Step 2: Use the product rule:
f'(x) = u'v + uv'

Step 3: Find u' and v':
u' = 2x
v' = cos x

Step 4: Apply the product rule:
f'(x) = (2x)(sin x) + (x²)(cos x)
= 2x sin x + x² cos x
= x(2 sin x + x cos x)

Final Answer: f'(x) = 2x sin x + x² cos x

Q6. Find the derivative of f(x) = (x² + 1)/(x - 1)

Find the derivative of f(x) = (x² + 1)/(x - 1)

Step 1: This is a quotient of two functions.
Let u = x² + 1 and v = x - 1

Step 2: Use the quotient rule:
f'(x) = [u'v - uv']/v²

Step 3: Find u' and v':
u' = 2x
v' = 1

Step 4: Apply the quotient rule:
f'(x) = [(2x)(x - 1) - (x² + 1)(1)]/(x - 1)²
= [2x² - 2x - x² - 1]/(x - 1)²
= [x² - 2x - 1]/(x - 1)²

Final Answer: f'(x) = (x² - 2x - 1)/(x - 1)²

Q7. Find the value of x for which the function f(x) = x³ - 3x² - 9x + 5 has a local maximum or minimum.

Find critical points of f(x) = x³ - 3x² - 9x + 5

Step 1: Find f'(x):
f'(x) = 3x² - 6x - 9

Step 2: Set f'(x) = 0:
3x² - 6x - 9 = 0
x² - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 or x = -1

Step 3: Find f''(x) to determine max/min:
f''(x) = 6x - 6

Step 4: Check at critical points:
At x = -1: f''(-1) = 6(-1) - 6 = -12 < 0 (local maximum)
At x = 3: f''(3) = 6(3) - 6 = 12 > 0 (local minimum)

Step 5: Find the function values:
f(-1) = (-1)³ - 3(-1)² - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10 (local maximum)
f(3) = (3)³ - 3(3)² - 9(3) + 5 = 27 - 27 - 27 + 5 = -22 (local minimum)

Final Answer: Local maximum at x = -1 with f(-1) = 10; Local minimum at x = 3 with f(3) = -22

Q8. Find the limit: lim(x→∞) (3x² + 2x + 1)/(5x² - 3x + 4)

Find: lim(x→∞) (3x² + 2x + 1)/(5x² - 3x + 4)

Step 1: Identify the highest degree terms:
Numerator: 3x²
Denominator: 5x²
Both have degree 2.

Step 2: Divide both numerator and denominator by x²:
lim(x→∞) [(3x² + 2x + 1)/x²]/[(5x² - 3x + 4)/x²]
= lim(x→∞) [3 + 2/x + 1/x²]/[5 - 3/x + 4/x²]

Step 3: As x → ∞, all terms with x in the denominator approach 0:
= [3 + 0 + 0]/[5 - 0 + 0]
= 3/5

Alternative method: When degrees are equal, the limit is the ratio of leading coefficients:
lim(x→∞) = 3/5

Final Answer: The limit is 3/5

Limits and Derivatives — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. What is the value of the limit: lim (x→2) (x^2 - 4) / (x - 2)?

  • A. 0
  • B. 2
  • C. 4
  • D. Undefined

Answer: C. 4 — Factoring the numerator as (x-2)(x+2), the expression becomes (x+2) for x ≠ 2. Substituting x=2, we get 2+2=4.

MCQ 2. Evaluate lim (x→0) (sin 4x) / x.

  • A. 1
  • B. 4
  • C. 1/4
  • D. 0

Answer: B. 4 — We can rewrite the expression as lim (x→0) 4 * (sin 4x) / (4x). Using the standard limit lim (θ→0) (sin θ) / θ = 1, we get 4 * 1 = 4.

MCQ 3. Find the derivative of f(x) = 5x^3 - 2x^2 + 7x - 10 with respect to x.

  • A. 15x^2 - 4x + 7
  • B. 5x^2 - 2x + 7
  • C. 15x^3 - 4x^2 + 7x
  • D. x^4 - x^3 + x^2 - 10x

Answer: A. 15x^2 - 4x + 7 — Applying the power rule d/dx(x^n) = nx^(n-1) and the linearity of derivatives, we get 15x^2 - 4x + 7.

MCQ 4. If f(x) = sin(x) + cos(x), what is f'(x)?

  • A. cos(x) - sin(x)
  • B. cos(x) + sin(x)
  • C. -cos(x) + sin(x)
  • D. -sin(x) - cos(x)

Answer: A. cos(x) - sin(x) — The derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x). Therefore, f'(x) = cos(x) - sin(x).

MCQ 5. What is lim (x→0) (1 - cos x) / x?

  • A. 1
  • B. -1
  • C. 0
  • D. 1/2

Answer: C. 0 — This is a standard limit. We can multiply and divide by (1+cos x) or use the half-angle identity. Alternatively, we can use the result that lim (x→0) (1-cos x)/x^2 = 1/2, so (1-cos x)/x = x * (1-cos x)/x^2 which tends to 0 * (1/2) = 0.

MCQ 6. Find the derivative of y = x^2 * sin(x).

  • A. 2x sin(x) + x^2 cos(x)
  • B. 2x cos(x) - x^2 sin(x)
  • C. x sin(x) + x^2 cos(x)
  • D. 2x sin(x) - x^2 cos(x)

Answer: A. 2x sin(x) + x^2 cos(x) — Using the product rule (uv)' = u'v + uv', where u=x^2 and v=sin(x). We get (2x)sin(x) + x^2(cos(x)).

MCQ 7. Evaluate lim (x→3) (x^3 - 27) / (x - 3).

  • A. 9
  • B. 18
  • C. 27
  • D. 81

Answer: C. 27 — Using the formula lim (x→a) (x^n - a^n) / (x - a) = n*a^(n-1). Here, n=3 and a=3, so the limit is 3 * 3^(3-1) = 3 * 3^2 = 3 * 9 = 27.

MCQ 8. If f(x) = √x, what is f'(x)?

  • A. 1/(2√x)
  • B. 2√x
  • C. -1/(2√x)
  • D. 1/√x

Answer: A. 1/(2√x) — Rewrite √x as x^(1/2). Applying the power rule, d/dx(x^(1/2)) = (1/2)x^(1/2 - 1) = (1/2)x^(-1/2) = 1/(2√x).

MCQ 9. For the function f(x) = |x|, does lim (x→0) f(x) exist?

  • A. Yes, it is 0
  • B. Yes, it is 1
  • C. No, LHL ≠ RHL
  • D. No, it is undefined

Answer: A. Yes, it is 0 — The left-hand limit is lim (x→0-) -x = 0, and the right-hand limit is lim (x→0+) x = 0. Since LHL = RHL = 0, the limit exists and is 0.

MCQ 10. The derivative of a constant function f(x) = c is:

  • A. c
  • B. 1
  • C. 0
  • D. x

Answer: C. 0 — The derivative measures the rate of change. A constant function does not change, so its rate of change (derivative) is 0.

MCQ 11. Find the derivative of y = (x^2 + 1) / (x - 1).

  • A. (x^2 - 2x - 1) / (x - 1)^2
  • B. (x^2 + 2x - 1) / (x - 1)^2
  • C. (x^2 - 2x + 1) / (x - 1)^2
  • D. (x^2 + 2x + 1) / (x - 1)^2

Answer: A. (x^2 - 2x - 1) / (x - 1)^2 — Using the quotient rule (u/v)' = (u'v - uv') / v^2, with u=x^2+1 and v=x-1. We get [(2x)(x-1) - (x^2+1)(1)] / (x-1)^2 = (2x^2 - 2x - x^2 - 1) / (x-1)^2 = (x^2 - 2x - 1) / (x-1)^2.

MCQ 12. What is the value of lim (x→0) (tan x) / x?

  • A. 0
  • B. 1
  • C. ∞
  • D. -1

Answer: B. 1 — We can rewrite (tan x) / x as (sin x / cos x) / x = (sin x / x) * (1 / cos x). As x→0, (sin x / x) → 1 and (1 / cos x) → 1/1 = 1. So the limit is 1 * 1 = 1.

MCQ 13. If y = (3x - 5)^4, find dy/dx.

  • A. 4(3x - 5)^3
  • B. 12(3x - 5)^3
  • C. 3(3x - 5)^3
  • D. 4(3x - 5)^3 * 3x

Answer: B. 12(3x - 5)^3 — Using the chain rule, let u = 3x - 5, so y = u^4. Then dy/du = 4u^3 and du/dx = 3. So dy/dx = (dy/du) * (du/dx) = 4u^3 * 3 = 12(3x - 5)^3.

MCQ 14. What is the Left Hand Limit (LHL) for the function f(x) = {x+1, if x < 1; x^2, if x ≥ 1} as x approaches 1?

  • A. 1
  • B. 2
  • C. 0
  • D. Does not exist

Answer: B. 2 — For the LHL, we consider values of x < 1, so f(x) = x+1. The LHL is lim (x→1-) (x+1) = 1+1 = 2.

MCQ 15. The derivative of f(x) = x^n at x = 1 is:

  • A. 1
  • B. n
  • C. n-1
  • D. 0

Answer: B. n — The derivative of f(x) = x^n is f'(x) = nx^(n-1). At x = 1, f'(1) = n * (1)^(n-1) = n * 1 = n.

MCQ 16. Evaluate lim (x→0) (√(1+x) - 1) / x.

  • A. 1
  • B. 0
  • C. 1/2
  • D. Undefined

Answer: C. 1/2 — Multiply the numerator and denominator by the conjugate (√(1+x) + 1). This yields lim (x→0) ((1+x)-1) / (x(√(1+x)+1)) = lim (x→0) x / (x(√(1+x)+1)) = lim (x→0) 1 / (√(1+x)+1) = 1 / (√1 + 1) = 1/2.

MCQ 17. If f(x) = x^5 - 3x^2 + 1, find the value of f'(2).

  • A. 72
  • B. 80
  • C. 68
  • D. 74

Answer: D. 74 — First, find the derivative: f'(x) = 5x^4 - 6x. Then substitute x=2: f'(2) = 5(2)^4 - 6(2) = 5(16) - 12 = 80 - 12 = 68.

MCQ 18. The derivative of tan(x) is:

  • A. cot(x)
  • B. sec^2(x)
  • C. -sec^2(x)
  • D. sin(x)cos(x)

Answer: B. sec^2(x) — Using the quotient rule on sin(x)/cos(x), d/dx(tan x) = d/dx(sin x / cos x) = (cos x cos x - sin x (-sin x)) / cos^2 x = (cos^2 x + sin^2 x) / cos^2 x = 1 / cos^2 x = sec^2 x.

MCQ 19. Which of the following conditions ensures that lim (x→a) f(x) exists?

  • A. f(a) is defined
  • B. lim (x→a-) f(x) = lim (x→a+) f(x)
  • C. f(x) is continuous at x=a
  • D. f(x) is differentiable at x=a

Answer: B. lim (x→a-) f(x) = lim (x→a+) f(x) — For a limit to exist at a point 'a', the left-hand limit (LHL) must be equal to the right-hand limit (RHL) at that point. The function being defined, continuous, or differentiable at 'a' are stronger conditions.

MCQ 20. If y = x^(3/2), what is dy/dx?

  • A. (3/2)x^(1/2)
  • B. (2/3)x^(1/2)
  • C. (3/2)x^(5/2)
  • D. x^(1/2)

Answer: A. (3/2)x^(1/2) — Applying the power rule d/dx(x^n) = nx^(n-1), we get (3/2)x^(3/2 - 1) = (3/2)x^(1/2).

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

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