Q74.The value of lim n1 ∑nj=1 (2j−1)+4n(2j−1)+8n is equal to: n→∞ (1) 5 + loge( 32 ) (2) 2 −loge( 23 ) (3) 3 + 2 loge( 23 ) (4) 1 + 2 loge( 32 ) π dx is equal to : cos x)(sin4 x+cos4 x)
What This Question Tests
The question requires ensuring continuity of a piecewise function at a specific point by calculating left-hand limit, right-hand limit, and function value, which involves using standard limits and algebraic manipulation.
Concepts Tested
Formulas Used
lim (x→a⁻) f(x) = lim (x→a⁺) f(x) = f(a)
lim x→0 (e^(ax) - 1)/x = a
lim x→0 (1 + x)^(1/x) = e
📚 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.
5.2 — Lists The Kinetic Energies For Various X I
Physics Class 11 · Chapter 5
5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳
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.
📋 Question Details
- Chapter
- Limits & Continuity
- Topic
- Continuity of a piecewise function
- Year
- 2021
- Shift
- 27 Jul Shift 1
- Q Number
- Q74
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 5: Continuity and Differentiability
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