Chapter-12 Limits and Derivatives — Online MCQ Test
MATHS · Grade 11 · CBSE(NCERT)
Practice Chapter-12 Limits and Derivatives with a free chapter-wise online MCQ test.
This chapter covers: Left-hand limit - Right-hand limit - Existence of limit - Algebra of limits - Standard limits - Derivative - Rate of change - First principle - Product rule - Quotient rule.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-12 Limits and Derivatives — Important Questions & Answers
What is the left-hand limit of a function f(x) at x = a denoted as?
- A. lim(x→a⁺) f(x)
- B. lim(x→a⁻) f(x)
- C. lim(x→a) f(x)
- D. lim(x→0) f(x)
Answer: B. lim(x→a⁻) f(x)
The left-hand limit is denoted by lim(x→a⁻) f(x), which means x approaches a from values less than a.
The left-hand limit is denoted by lim(x→a⁻) f(x), which means x approaches a from values less than a.
For a limit to exist at x = a, which condition must be satisfied?
- A. Only left-hand limit should exist
- B. Only right-hand limit should exist
- C. Left-hand limit = Right-hand limit
- D. Either left or right-hand limit should exist
Answer: C. Left-hand limit = Right-hand limit
For a limit to exist at x = a, the left-hand limit and right-hand limit must be equal: lim(x→a⁻) f(x) = lim(x→a⁺) f(x).
For a limit to exist at x = a, the left-hand limit and right-hand limit must be equal: lim(x→a⁻) f(x) = lim(x→a⁺) f(x).
Find lim(x→2) (x² + 3x + 2):
- A. 12
- B. 10
- C. 8
- D. 14
Answer: A. 12
By direct substitution: lim(x→2) (x² + 3x + 2) = 2² + 3(2) + 2 = 4 + 6 + 2 = 12.
By direct substitution: lim(x→2) (x² + 3x + 2) = 2² + 3(2) + 2 = 4 + 6 + 2 = 12.
If f(x) = |x| at x = 0, which statement is true?
- A. Left-hand limit = 0 and Right-hand limit = 0, limit exists
- B. Left-hand limit ≠ Right-hand limit, limit does not exist
- C. Only left-hand limit exists
- D. The function is differentiable at x = 0
Answer: A. Left-hand limit = 0 and Right-hand limit = 0, limit exists
For f(x) = |x|: lim(x→0⁻) |x| = 0 and lim(x→0⁺) |x| = 0, so the limit exists and equals 0. However, the function is not differentiable at x = 0.
For f(x) = |x|: lim(x→0⁻) |x| = 0 and lim(x→0⁺) |x| = 0, so the limit exists and equals 0. However, the function is not differentiable at x = 0.
Given f'(x) = 3x² and f(1) = 2, which statement about f is true?
- A. f(x) = x³ + 1
- B. f(x) = x³ + c where c is any constant
- C. f cannot be determined uniquely
- D. f(x) = x³ + 5
Answer: A. f(x) = x³ + 1
If f'(x) = 3x², then f(x) = x³ + c. Using f(1) = 2: 1³ + c = 2, so c = 1, giving f(x) = x³ + 1.
If f'(x) = 3x², then f(x) = x³ + c. Using f(1) = 2: 1³ + c = 2, so c = 1, giving f(x) = x³ + 1.