Maths · Calculus · chapter 12 of 14
Class 11 Maths Chapter 12: Limits and Derivatives worksheet with answer key
A 20-mark worksheet on Limits and Derivatives for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
Question 1
The value of \limx→4 √x − 2x − 4 is
Show answer
Answer: (b) 14
Write it as (x1/2 − 41/2)/(x − 4). By the remark to Theorem 2 (n = 1/2, a = 4 > 0) the limit is (1/2)·4−1/2 = 1/4.
Question 2
If \limx→2 x2 + ax − 6x − 2 exists (as a real number), then its value is
Show answer
Answer: (a) 5
The denominator is 0 at x = 2, so for the limit to exist the numerator must also vanish there: 4 + 2a − 6 = 0, a = 1. Then x2 + x − 6 = (x + 3)(x − 2), and the limit is 2 + 3 = 5.
Question 3
If \limx→2 (x2 + kx + 1) = 11, then the value of k is
Show answer
Answer: (a) 3
The limit of a polynomial at x = 2 is its value there: 4 + 2k + 1 = 11, so 2k = 6 and k = 3.
Question 4
The value of \limx→2 x5 − 32x − 2 is
Show answer
Answer: (d) 80
32 = 25, so by Theorem 2 the limit is 5(2)4 = 80.
Question 5
Let f(x) = x2 − 1 for x ≠ 2 and f(2) = 0. Then \limx→2 f(x) − f(2) equals
Show answer
Answer: (b) 3
Near x = 2 (but not at 2), f(x) = x2 − 1 approaches 3; the value at 2 is 0. The limit and the value differ: 3 − 0 = 3.
Question 6 · 2 marks
Find the derivative of x√x + 1√x and its value at x = 1.
Show answer
Write it as x3/2 + x−1/2. By the power rule (valid for real powers), the derivative is (3/2)x1/2 − (1/2)x−3/2. At x = 1: 3/2 − 1/2 = 1.
- derivative (3/2)x1/2 − (1/2)x−3/2 (1)
- value 1 at x = 1 (1)
Question 7 · 2 marks
If y = x2 cos x, show that x dydx = 2y − x3 sin x.
Show answer
By the product rule, dy/dx = 2x cos x + x2(−sin x) = 2x cos x − x2 sin x. Multiplying by x: x dy/dx = 2x2 cos x − x3 sin x = 2y − x3 sin x.
- dy/dx = 2x cos x − x2 sin x by the product rule (1)
- multiplies by x and identifies 2y (1)
Question 8 · 2 marks
Evaluate \limx→0 sin 2x + sin 4xx.
Show answer
Split into (sin 2x)/x + (sin 4x)/x = 2·(sin 2x)/(2x) + 4·(sin 4x)/(4x). As x → 0, 2x → 0 and 4x → 0, so each ratio tends to 1 and the limit is 2 + 4 = 6.
- splits and writes each term as k·(sin kx)/(kx) (1)
- limit 6 (1)
Question 9 · 3 marks
Using Theorem 2 on standard limits, evaluate \limx→1 x4 + x2 − 2x − 1.
Show answer
Write x4 + x2 − 2 = (x4 − 1) + (x2 − 1). Then the limit is \lim (x4 − 14)/(x − 1) + \lim (x2 − 12)/(x − 1) = 4(1)3 + 2(1)1 = 4 + 2 = 6.
- splits the numerator as (x4 − 1) + (x2 − 1) (1)
- applies Theorem 2 to each: 4 and 2 (1)
- limit 6 (1)
Question 10 · 3 marks
Differentiate x2 + 1x2 − 1, x ≠ ±1.
Show answer
u = x2 + 1, v = x2 − 1, u′ = v′ = 2x. By the quotient rule, (u′v − uv′)/v2 = [2x(x2 − 1) − 2x(x2 + 1)]/(x2 − 1)2 = −4x/(x2 − 1)2.
- u′ = v′ = 2x and the quotient rule set up (1)
- numerator 2x(x2 − 1) − 2x(x2 + 1) (1)
- −4x/(x2 − 1)2 (1)
Question 11 · 3 marks
Find the derivative of f(x) = √x, x > 0, from the first principle.
Show answer
f′(x) = \limh→0 (√(x + h) − √x)/h. Multiply numerator and denominator by √(x + h) + √x: the numerator becomes (x + h) − x = h, so the quotient is 1/(√(x + h) + √x) for h ≠ 0. As h → 0 this tends to 1/(2√x). So f′(x) = 1/(2√x).
- writes the first-principle limit (1)
- rationalises to 1/(√(x + h) + √x) (1)
- f′(x) = 1/(2√x) (1)
← Chapter 11: Introduction to Three Dimensional GeometryChapter 13: Statistics →