Q65.The integral ∫ 8dx 6 is equal to: (where C is a constant of integration) (x+4) 7 (x−3) 7 (1) x−3 71 (2) x−3 −17 ( x+4 ) + C ( x+4 ) + C (3) 1 x−3 73 (4) x−3 −137 2 ( x+4 ) + C −113 ( x+4 ) + C
What This Question Tests
This question involves a challenging integration by substitution, requiring clever algebraic manipulation to transform the integrand into a form suitable for direct integration after substitution.
Concepts Tested
Formulas Used
∫ t^n dt = t^(n+1)/(n+1) + C
d/dx ((x-a)/(x+b)) = (a+b)/(x+b)^2
📚 NCERT Sections This Tests
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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.
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%). ⊳
📋 Question Details
- Chapter
- Indefinite Integration
- Topic
- Integration by substitution
- Year
- 2020
- Shift
- 09 Jan Shift 1
- Q Number
- Q65
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 7: Integrals
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