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CBSE · Class 12 · Mathematics

Continuity and Differentiability

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
f(x) = kx + 9 for x ≤ −1, and f(x) = x2 + 8 for x > −1. f is continuous at x = −1 when k equals:
  1. 0
  2. −18
  3. 2
  4. 1
2.
The derivative of sin(6x − 8) with respect to x is:
  1. −6 cos(6x − 8)
  2. 6 sin(6x − 8)
  3. cos(6x − 8)
  4. 6 cos(6x − 8)
3.
If y = (x + 5)3, then dy/dx at x = 0 is:
  1. 76
  2. 125
  3. 75
  4. 375
4.
If 5x2 + 6y2 = 55, then dy/dx is:
  1. 5x/(6y)
  2. −5x/(6y)
  3. −6y/(5x)
  4. −10x/(6y)
5.
f is continuous at x = c when:
  1. f is defined only at c
  2. f(c) = 0
  3. lim (x → c) f(x) = f(c)
  4. f′(c) = 0
6.
If x = 3t2 and y = −4t, then dy/dx at t = −3 is:
  1. 2/9
  2. 4/9
  3. 9/2
  4. −2/9
7.
If y = −4x4 + 3x3 + 2x2 + x − 6, then d2y/dx2 at x = 0 is:
  1. 4
  2. 8
  3. 1
  4. −4
8.
Assertion (A): f(x) = |x − 7| is differentiable at x = 7.
Reason (R): The left-hand and right-hand derivatives of |x − 7| at x = 7 are different.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Is f continuous at x = 1, where f(x) = 5x + 9 for x ≤ 1 and f(x) = 2x + 13 for x > 1?
10.
Find k so that f(x) = (x5 + 1024)/(x + 4) for x ≠ −4, and f(−4) = k, is continuous at x = −4.
11.
Differentiate tan−1(3x) with respect to x.
12.
Find dy/dx if 4x2 + 6xy + 3y2 = 40.
13.
Use logarithmic differentiation to find dy/dx if y = (x + 4)(x + 1)(x + 7).
14.
Find dy/dx if x = 7 cos θ and y = 7 sin θ.
15.
If y = sin 4x − cos 4x, find λ such that d2y/dx2 = λy.
16.
Differentiate y = 4 × 9x + x4 with respect to x.
Section C — Word problems4 × 4 = 16 marks
17.
Find a and b so that f is continuous: f(x) = 18 for x ≤ −3, f(x) = ax + b for −3 < x < 3, and f(x) = −6 for x ≥ 3.
18.
Find dy/dx if y = (x + 5)4(x − 8)5/(x + 1)2.
19.
If y = 8e4x + 3ex, show that d2y/dx2 − 5 dy/dx + 4y = 0.
20.
If x = 4t2 and y = 8t, find dy/dx and d2y/dx2 in terms of t, and their values at t = 1.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Mathematics

Answer key

Continuity and Differentiability · AW-12M-CTD-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) 0k(−1) + 9 = 9
2(d) 6 cos(6x − 8)chain rule: differentiate the outside, then multiply by the derivative of the inside
3(c) 75dy/dx = 3(x + 5)2 = 3 × 52
4(b) −5x/(6y)10x + 12y (dy/dx) = 0
5(c) lim (x → c) f(x) = f(c)—
6(a) 2/9dy/dx = −4 ÷ 6t
7(a) 4d2y/dx2 = −48x2 + 18x + 4
8(d) A is false, but R is true.A is false: the left-hand derivative at x = 7 is −1 and the right-hand derivative is 1. R is true.
9No: LHL = 14 but RHL = 15left limit and f(1) from 5x + 9, right limit from 2x + 13
10k = 1,280lim (xⁿ − aⁿ)/(x − a) = naⁿ−1 = 5 × (−4)4
113/(1 + 9x2)chain rule
12dy/dx = −(4x + 3y)/(3x + 3y)8x + 6y + (6x + 6y) dy/dx = 0
13dy/dx = (x + 4)(x + 1)(x + 7)[1/(x + 4) + 1/(x + 1) + 1/(x + 7)]log y = log(x + a) + log(x + b) + log(x + c)
14dy/dx = −cot θdy/dθ = 7 cos θ, dx/dθ = −7 sin θ
15λ = −16d2/dx2 of sin 4x is −16 sin 4x
16dy/dx = 4 × 9x log 9 + 4x3d/dx ax = ax log a
17a = −4, b = 6(−3)a + b = 18 and 3a + b = −6
18dy/dx = (x + 5)4(x − 8)5/(x + 1)2 × [4/(x + 5) + 5/(x − 8) − 2/(x + 1)]log y = 4 log(x + 5) + 5 log(x − 8) − 2 log(x + 1)
19d2y/dx2 − 5 dy/dx + 4y = 0 (shown)dy/dx = 32e4x + 3ex; d2y/dx2 = 128e4x + 3ex
20dy/dx = 1/(t); d2y/dx2 = −1/(8t3); at t = 1: 1 and −1/8d2y/dx2 = d/dt(dy/dx) ÷ dx/dt