Q67. lim sin2(π cos4 x) is equal to : x4 x→0 (1) 2π2 (2) π2 (3) 4π2 (4) 4π
What This Question Tests
This question requires evaluating a trigonometric limit involving multiple functions, often simplified using standard limit formulas or series expansions.
Concepts Tested
Formulas Used
lim (sin(ax))/(ax) = 1 as x->0
Taylor series expansion for cos(x) ~ 1 - x^2/2
📚 NCERT Sections This Tests
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.
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.
1.27 — If The Solubility Product Of Cus Is 6 × 10–16, Calculate The Maximum Molarity Of
Chemistry Class 11 · Chapter 1
1.27 If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Evaluation of limits
- Year
- 2021
- Shift
- 31 Aug Shift 1
- Q Number
- Q67
- 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)