Class 11 Maths Chapter 12: Limits and Derivatives MCQs with answers
10 multiple-choice questions on Limits and Derivatives, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern and JEE Main, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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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.
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Answer: (d) 4
For x slightly greater than 1, f(x) = 5 − x, whose values approach 5 − 1 = 4. (The left hand limit is 2(1) + 1 = 3.)
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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.
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Answer: (a) 2
By the algebra of limits, 2 \lim sin x + \lim cos x = 2(1) + 0 = 2.
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Answer: (d) 80
32 = 25, so by Theorem 2 the limit is 5(2)4 = 80.
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Answer: (d) 3
By Theorem 2 the limit is 5a4. So 5a4 = 405, a4 = 81 and, with a > 0, a = 3. (c) 9 forgets the fourth root; (b) 81 stops at a4.
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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.
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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.
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Answer: (a) 2
Split: x/x + (sin x)/x = 1 + (sin x)/x, which tends to 1 + 1 = 2.
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Answer: (c) 8
1 − cos 4x = 2 sin2 2x, so the expression is 2·(sin 2x / x)2 = 2·4·(sin 2x / 2x)2, which tends to 8.
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