Q72. x→0((lim 1−cos2(3x)cos3(4x) )( (loge(2x+1))5sin3(4x) )) is equal to (1) 15 (2) 9 (3) 18 (4) 24
What This Question Tests
This question tests the ability to evaluate complex limits by applying standard limit formulas and algebraic manipulation, possibly using L'Hopital's rule or series expansions.
Concepts Tested
Formulas Used
lim (1-cos(ax))/x² = a²/2
lim log(1+ax)/x = a
lim sin(ax)/x = a
📚 NCERT Sections This Tests
1.3 — Define The Following Terms:
Chemistry Class 11 · Chapter 1
1.3 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
12.5 — A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
Physics Class 12 · Chapter 12
12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
5.15 — Discuss The Nature Of Bonding In The Following Coordination Entities On The
Chemistry Class 11 · Chapter 5
5.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i) [Fe(CN)6] 4– (ii) [FeF6] 3– (iii) [Co(C2O4)3]3– (iv) [CoF6] 3–
📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Evaluation of Limits using L'Hopital's Rule or Series Expansion
- Year
- 2023
- Shift
- 08 Apr Shift 1
- Q Number
- Q72
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 13: Limits and Derivatives
More from this Chapter
Q97.The function f : R ∼{0} →R given by f(x) = x1 − e2x−12 can be made continuous at x = 0 by defining f(0) as (1) 2 (2) −1 (3) 0 (4) 1
Q92.Let f(x) = −1) sin ( {(x0, if x = 1 JEE Main 2008 JEE Main Previous Year Paper (1) f is neither differentiable at x = 0 nor at x = 1 (2) f is differentiable at x = 0 and at x = 1 (3) f is differentiable at x = 0 but not at x = 1 (4) f is differentiable at x = 1 but not at x = 0
Q70.Let f : R →R be a positive increasing function with limx→∞ f(3x)f(x) = 1. Then limx→∞ f(2x)f(x) (1) 2 (2) 3 3 2 (3) 3 (4) 1
Q71.Consider the following statements P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement "Suman is brilliant and dishonest if and only if Suman is rich" can be expressed as (1) ∼(Q ↔(P∧∼R)) (2) ∼Q ↔∼P ∧R (3) ∼(P∧∼R) ↔Q (4) ∼P ∧(Q ↔∼R)