Continuity and Differentiability — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Continuity and Differentiability" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Continuity and Differentiability" — 7 important questions with detailed answers fo…
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NCERT Solutions for Class 12 Mathematics — Continuity and Differentiability. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Continuity and Differentiability — All 7 Questions Solved
Q1. Check whether the function f(x) = |x| is continuous at x = 0.
To check continuity at x = 0, we verify: lim[x→0⁻] f(x) = lim[x→0⁺] f(x) = f(0)
Given: f(x) = |x|
Step 1: Find f(0)
f(0) = |0| = 0
Step 2: Find left-hand limit as x → 0⁻
For x < 0: f(x) = |x| = -x
lim[x→0⁻] f(x) = lim[x→0⁻] (-x) = 0
Step 3: Find right-hand limit as x → 0⁺
For x > 0: f(x) = |x| = x
lim[x→0⁺] f(x) = lim[x→0⁺] x = 0
Step 4: Check continuity
Since lim[x→0⁻] f(x) = 0 = lim[x→0⁺] f(x) = f(0)
CONCLUSION: f(x) = |x| is continuous at x = 0.
Q2. Determine whether f(x) = 1/(x-2) is continuous at x = 2.
To check continuity at x = 2, we need to verify if lim[x→2] f(x) exists and equals f(2).
Given: f(x) = 1/(x-2)
Step 1: Check if f(2) is defined
f(2) = 1/(2-2) = 1/0, which is undefined.
Since f(2) is not defined, the function cannot be continuous at x = 2.
Alternatively, examining the limits:
lim[x→2⁻] f(x) = 1/(2⁻ - 2) = 1/0⁻ = -∞
lim[x→2⁺] f(x) = 1/(2⁺ - 2) = 1/0⁺ = +∞
The left and right limits are not equal and both are infinite.
CONCLUSION: f(x) = 1/(x-2) is NOT continuous at x = 2. In fact, x = 2 is a point of discontinuity (vertical asymptote).
Q3. Find the derivative of f(x) = x² at x = 3 using first principles.
Given: f(x) = x²
We find f'(3) using the definition of derivative.
f'(x) = lim[h→0] (f(x+h) - f(x))/h
At x = 3:
f'(3) = lim[h→0] (f(3+h) - f(3))/h
= lim[h→0] ((3+h)² - 3²)/h
= lim[h→0] ((9 + 6h + h²) - 9)/h
= lim[h→0] (6h + h²)/h
= lim[h→0] (6 + h)
= 6 + 0
= 6
CONCLUSION: f'(3) = 6
Q4. Find the derivative of f(x) = sin(x).
Given: f(x) = sin(x)
We find f'(x) using the definition of derivative.
f'(x) = lim[h→0] (f(x+h) - f(x))/h
= lim[h→0] (sin(x+h) - sin(x))/h
Using the formula: sin(A) - sin(B) = 2 cos((A+B)/2) sin((A-B)/2)
sin(x+h) - sin(x) = 2 cos((x+h+x)/2) sin((x+h-x)/2)
= 2 cos((2x+h)/2) sin(h/2)
= 2 cos(x + h/2) sin(h/2)
f'(x) = lim[h→0] (2 cos(x + h/2) sin(h/2))/h
= lim[h→0] cos(x + h/2) × (2sin(h/2)/h)
= lim[h→0] cos(x + h/2) × (sin(h/2)/(h/2))
= cos(x) × 1
= cos(x)
where we used lim[u→0] sin(u)/u = 1
CONCLUSION: d/dx[sin(x)] = cos(x)
Q5. Check if f(x) = x² is differentiable at x = 1.
To check if f(x) = x² is differentiable at x = 1, we check if the left and right derivatives are equal.
Left derivative at x = 1:
f'(1⁻) = lim[h→0⁻] (f(1+h) - f(1))/h
= lim[h→0⁻] ((1+h)² - 1)/h
= lim[h→0⁻] (1 + 2h + h² - 1)/h
= lim[h→0⁻] (2h + h²)/h
= lim[h→0⁻] (2 + h)
= 2
Right derivative at x = 1:
f'(1⁺) = lim[h→0⁺] (f(1+h) - f(1))/h
= lim[h→0⁺] ((1+h)² - 1)/h
= lim[h→0⁺] (2h + h²)/h
= lim[h→0⁺] (2 + h)
= 2
Since f'(1⁻) = f'(1⁺) = 2, the left and right derivatives are equal.
CONCLUSION: f(x) = x² is differentiable at x = 1, and f'(1) = 2.
Q6. Find the derivative of f(x) = e^x.
Given: f(x) = e^x
We find f'(x) using the definition of derivative.
f'(x) = lim[h→0] (f(x+h) - f(x))/h
= lim[h→0] (e^(x+h) - e^x)/h
= lim[h→0] (e^x × e^h - e^x)/h
= lim[h→0] e^x(e^h - 1)/h
= e^x × lim[h→0] (e^h - 1)/h
We know that lim[h→0] (e^h - 1)/h = 1 (a standard limit)
f'(x) = e^x × 1 = e^x
CONCLUSION: d/dx[e^x] = e^x
This shows that the exponential function is its own derivative.
Q7. Show that f(x) = |x - 1| is continuous but not differentiable at x = 1.
Given: f(x) = |x - 1|
PART 1: CHECKING CONTINUITY AT x = 1
f(1) = |1 - 1| = 0
lim[x→1⁻] f(x) = lim[x→1⁻] |x - 1| = lim[x→1⁻] (-(x-1)) = 0
lim[x→1⁺] f(x) = lim[x→1⁺] |x - 1| = lim[x→1⁺] (x-1) = 0
Since the left limit = right limit = f(1) = 0, f is continuous at x = 1. ✓
PART 2: CHECKING DIFFERENTIABILITY AT x = 1
Left derivative at x = 1:
f'(1⁻) = lim[h→0⁻] (f(1+h) - f(1))/h
= lim[h→0⁻] (|h| - 0)/h
= lim[h→0⁻] (-h)/h (since h < 0, |h| = -h)
= lim[h→0⁻] (-1)
= -1
Right derivative at x = 1:
f'(1⁺) = lim[h→0⁺] (f(1+h) - f(1))/h
= lim[h→0⁺] (|h| - 0)/h
= lim[h→0⁺] h/h (since h > 0, |h| = h)
= lim[h→0⁺] 1
= 1
Since f'(1⁻) = -1 ≠ 1 = f'(1⁺), the left and right derivatives are not equal.
CONCLUSION: f(x) = |x - 1| is continuous at x = 1 but NOT differentiable at x = 1. The function has a sharp corner at x = 1.
Continuity and Differentiability — 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. Which of the following functions is continuous everywhere?
- A. f(x) = 1/x
- B. f(x) = tan(x)
- C. f(x) = |x|
- D. f(x) = [x] (greatest integer function)
Answer: C. f(x) = |x| — The modulus function f(x) = |x| is continuous for all real numbers. 1/x is discontinuous at x=0, tan(x) is discontinuous at odd multiples of π/2, and [x] is discontinuous at all integer points.
MCQ 2. If a function f(x) is differentiable at a point x = a, then it must be:
- A. Discontinuous at x = a
- B. Continuous at x = a
- C. Not defined at x = a
- D. Both continuous and discontinuous at x = a
Answer: B. Continuous at x = a — A fundamental theorem states that if a function is differentiable at a point, it must also be continuous at that point. However, the converse is not always true.
MCQ 3. The derivative of sin(x²) with respect to x is:
- A. cos(x²)
- B. 2x cos(x²)
- C. -cos(x²)
- D. 2x sin(x²)
Answer: B. 2x cos(x²) — Using the chain rule, if y = sin(u) where u = x², then dy/dx = (dy/du) * (du/dx) = cos(x²) * 2x = 2x cos(x²).
MCQ 4. For what value of k is the function f(x) = { (kx+1, if x <= π), (sin x + 1, if x > π) } continuous at x = π?
- A. 0
- B. 1
- C. -1
- D. π
Answer: A. 0 — For continuity at x=π, LHL = RHL = f(π). LHL = lim (x→π-) (kx+1) = kπ+1. RHL = lim (x→π+) (sin x + 1) = sin(π)+1 = 0+1 = 1. f(π) = kπ+1. Equating LHL and RHL: kπ+1 = 1, which implies kπ = 0. Since π ≠ 0, k must be 0.
MCQ 5. If y = log(log x), then dy/dx is:
- A. 1/log x
- B. 1/(x log x)
- C. x log x
- D. 1/x
Answer: B. 1/(x log x) — Using the chain rule, if y = log(u) where u = log x, then dy/dx = (dy/du) * (du/dx) = (1/log x) * (1/x) = 1/(x log x).
MCQ 6. The derivative of tan⁻¹(x) with respect to x is:
- A. 1/(1+x²)
- B. 1/(1-x²)
- C. √(1-x²)
- D. -1/(1+x²)
Answer: A. 1/(1+x²) — The standard derivative of tan⁻¹(x) with respect to x is 1/(1+x²).
MCQ 7. If y = x^x, then dy/dx is:
- A. x^x (1 + log x)
- B. x^x log x
- C. x^(x-1)
- D. x^x
Answer: A. x^x (1 + log x) — To differentiate y = x^x, take natural logarithm on both sides: log y = x log x. Differentiate implicitly: (1/y) dy/dx = 1 * log x + x * (1/x) = log x + 1. So, dy/dx = y(1 + log x) = x^x (1 + log x).
MCQ 8. If x = a cos³θ, y = a sin³θ, then dy/dx is:
- A. -tan θ
- B. tan θ
- C. -cot θ
- D. cot θ
Answer: A. -tan θ — First, find dx/dθ = a * 3 cos²θ * (-sin θ) = -3a cos²θ sin θ. Then, find dy/dθ = a * 3 sin²θ * (cos θ) = 3a sin²θ cos θ. Finally, dy/dx = (dy/dθ) / (dx/dθ) = (3a sin²θ cos θ) / (-3a cos²θ sin θ) = -sin θ / cos θ = -tan θ.
MCQ 9. The second derivative of e^(2x) with respect to x is:
- A. e^(2x)
- B. 2e^(2x)
- C. 4e^(2x)
- D. e^x
Answer: C. 4e^(2x) — Let y = e^(2x). First derivative: dy/dx = 2e^(2x). Second derivative: d²y/dx² = d/dx (2e^(2x)) = 2 * (2e^(2x)) = 4e^(2x).
MCQ 10. Which of the following functions is not differentiable at x = 0?
- A. f(x) = x²
- B. f(x) = sin x
- C. f(x) = |x|
- D. f(x) = e^x
Answer: C. f(x) = |x| — The function f(x) = |x| has a sharp corner at x = 0, meaning its left-hand derivative and right-hand derivative at x=0 are not equal. Therefore, it is not differentiable at x=0.
MCQ 11. If y = x sin x, then dy/dx is:
- A. sin x + x cos x
- B. sin x - x cos x
- C. cos x
- D. x cos x
Answer: A. sin x + x cos x — Using the product rule, d/dx(uv) = u'v + uv'. Here u=x and v=sin x. So, dy/dx = (1)(sin x) + (x)(cos x) = sin x + x cos x.
MCQ 12. For the function f(x) = { (x²/2, if x <= 0), (2x, if x > 0) }, is it continuous at x = 0?
- A. Yes
- B. No
- C. Cannot be determined
- D. Only left continuous
Answer: A. Yes — LHL = lim (x→0-) x²/2 = 0. RHL = lim (x→0+) 2x = 0. f(0) = 0²/2 = 0. Since LHL = RHL = f(0), the function is continuous at x=0.
MCQ 13. The derivative of √(sin x) with respect to x is:
- A. cos x / (2√(sin x))
- B. 1 / (2√(sin x))
- C. cos x
- D. -cos x / (2√(sin x))
Answer: A. cos x / (2√(sin x)) — Using the chain rule, if y = √u where u = sin x, then dy/dx = (dy/du) * (du/dx) = (1/(2√u)) * (cos x) = cos x / (2√(sin x)).
MCQ 14. If x² + y² = 25, then dy/dx is:
- A. x/y
- B. -x/y
- C. y/x
- D. -y/x
Answer: B. -x/y — Differentiate implicitly with respect to x: d/dx(x²) + d/dx(y²) = d/dx(25). This gives 2x + 2y (dy/dx) = 0. So, 2y (dy/dx) = -2x, and dy/dx = -x/y.
MCQ 15. The derivative of (x^5 + 3x - 1)^(10) with respect to x is:
- A. 10(x^5 + 3x - 1)^9
- B. 10(x^5 + 3x - 1)^9 * (5x^4 + 3)
- C. 5x^4 + 3
- D. 10(5x^4 + 3)
Answer: B. 10(x^5 + 3x - 1)^9 * (5x^4 + 3) — Using the chain rule, let u = x^5 + 3x - 1. Then y = u^10. dy/dx = (dy/du) * (du/dx) = 10u^9 * (5x^4 + 3) = 10(x^5 + 3x - 1)^9 * (5x^4 + 3).
MCQ 16. For what values of a and b is the function f(x) = { (5ax-2b, if x < 1), (11, if x = 1), (3ax+b, if x > 1) } continuous at x = 1?
- A. a=3, b=2
- B. a=1, b=8
- C. a=2, b=5
- D. a=5, b=2
Answer: A. a=3, b=2 — For continuity at x=1, LHL = RHL = f(1). LHL = lim (x→1-) (5ax-2b) = 5a-2b. RHL = lim (x→1+) (3ax+b) = 3a+b. f(1) = 11. So we have two equations: 5a-2b = 11 and 3a+b = 11. Solving these simultaneous equations (e.g., multiply second eq by 2 and add to first) gives 11a=33 => a=3. Substituting a=3 into 3a+b=11 gives 9+b=11 => b=2. Thus, a=3, b=2.
MCQ 17. If y = sin⁻¹(2x / (1+x²)), then dy/dx is:
- A. 2 / (1+x²)
- B. -2 / (1+x²)
- C. 1 / (1+x²)
- D. 2x / (1+x²)
Answer: A. 2 / (1+x²) — Let x = tan θ, so θ = tan⁻¹ x. Then y = sin⁻¹(2tanθ / (1+tan²θ)) = sin⁻¹(sin 2θ) = 2θ = 2tan⁻¹x. Differentiating with respect to x, dy/dx = 2 * (1/(1+x²)) = 2/(1+x²). (Valid for -1 < x < 1).
MCQ 18. The derivative of 5^(3x) with respect to x is:
- A. 5^(3x) log 5
- B. 3 * 5^(3x) log 5
- C. 5^(3x) * 3
- D. 3 * 5^(3x)
Answer: B. 3 * 5^(3x) log 5 — Using the formula d/dx(a^u) = a^u * log a * du/dx. Here a=5 and u=3x. So, d/dx(5^(3x)) = 5^(3x) * log 5 * d/dx(3x) = 5^(3x) * log 5 * 3 = 3 * 5^(3x) log 5.
MCQ 19. If y = e^(sin x), then dy/dx is:
- A. e^(sin x)
- B. cos x * e^(sin x)
- C. -cos x * e^(sin x)
- D. sin x * e^(sin x)
Answer: B. cos x * e^(sin x) — Using the chain rule, if y = e^u where u = sin x, then dy/dx = (dy/du) * (du/dx) = e^u * cos x = e^(sin x) * cos x.
MCQ 20. A function f(x) is said to be continuous at x = a if:
- A. lim (x→a) f(x) exists
- B. f(a) is defined
- C. lim (x→a) f(x) = f(a)
- D. lim (x→a+) f(x) = lim (x→a-) f(x)
Answer: C. lim (x→a) f(x) = f(a) — For a function to be continuous at a point x=a, three conditions must be met: the limit of f(x) as x approaches a must exist, f(a) must be defined, and these two values must be equal.
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.