Q84.If โซ ๐๐ฅ 2 = ๐ฅ๐๐ฅ1 + ๐ฅ6 3 + ๐ถ, where ๐ถ is a constant of integration, then the function ๐๐ฅ is equal to ๐ฅ31 + ๐ฅ6 3 (1) 3 (2) - 1 ๐ฅ2 2๐ฅ3 1 1 (3) - (4) - 6๐ฅ3 2๐ฅ2 ๐ฅ ๐ฅ
What This Question Tests
This question requires recognizing a suitable substitution for a complex integrand and then carefully comparing the resulting antiderivative with the given format to determine the function f(x).
Concepts Tested
Formulas Used
โซ u^n du = u^(n+1)/(n+1) + C
๐ NCERT Sections This Tests
2.1 โ Two Charges 5 ร 10โ8 C And โ3 ร 10โ8 C Are Located 16 Cm Apart. At
Physics Class 11 ยท Chapter 2
2.1 Two charges 5 ร 10โ8 C and โ3 ร 10โ8 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.
5.2 โ Lists The Kinetic Energies For Various X I
Physics Class 11 ยท Chapter 5
5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position โณ 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 )โxAnswer The initial kinetic energy of the bullet โ x โ 0 โ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1ร1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = โซF ( i 1 2 x mv f = 100 J where โlimโ stands for the limit of the sum when 2 โx tends to zero. Thus, for a varying force 2 ร 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 sโ1 The speed is reduced by approximately 68% (not 90%). โณ
3.10 โ In A Reaction Between A And B, The Initial Rate Of Reaction (R0) Was Measured
Chemistry Class 11 ยท Chapter 3
3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol Lโ1 0.20 0.20 0.40 B/ mol Lโ1 0.30 0.10 0.05 r0/mol Lโ1sโ1 5.07 ร 10โ5 5.07 ร 10โ5 1.43 ร 10โ4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ยฎ C + D Experiment [A]/mol Lโ1 [B]/mol Lโ1 Initial rate of formation of D/mol Lโ1 minโ1 I 0.1 0.1 6.0 ร 10โ3 II 0.3 0.2 7.2 ร 10โ2 III 0.3 0.4 2.88 ร 10โ1 IV 0.4 0.1 2.40 ร 10โ2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol Lโ1 [B]/ mol Lโ1 Initial rate/ mol Lโ1 minโ1 I 0.1 0.1 2.0 ร 10โ2 II โ 0.2 4.0 ร 10โ2 III 0.4 0.4 โ IV โ 0.2 2.0 ร 10โ2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 sโ1 (ii) 2 minโ1 (iii) 4 yearsโ1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ยฎ 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 ร [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol Lโ1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
๐ Question Details
- Chapter
- Indefinite Integration
- Topic
- Integration by Substitution
- Year
- 2019
- Shift
- 08 Apr Shift 2
- Q Number
- Q84
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 7: Integrals
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