Page 1 :
Subject Code - 041, , Sample Question Paper, CLASS: XII, Session: 2021-22, Mathematics (Code-041), Term - 2, Time Allowed: 2 hours, Maximum Marks: 40, General Instructions:, 1. This question paper contains three sections – A, B and C. Each part is compulsory., 2. Section - A has 6 short answer type (SA1) questions of 2 marks each., 3. Section – B has 4 short answer type (SA2) questions of 3 marks each., 4. Section - C has 4 long answer type questions (LA) of 4 marks each., 5. There is an internal choice in some of the questions., 6. Q14 is a case-based problem having 2 sub parts of 2 marks each., , SECTION – A, 1., , 2, , 𝑙𝑜𝑔𝑥, , Find ∫ (1+𝑙𝑜𝑔𝑥)2 𝑑𝑥, OR, 𝑠𝑖𝑛2𝑥, , Find ∫ √9−𝑐𝑜𝑠4 𝑑𝑥, 𝑥, , 2., , Write the sum of the order and the degree of the following differential, equation:, 𝑑, , 2, , 𝑑𝑦, , ( )=5, 𝑑𝑥 𝑑𝑥, 3., , If 𝑎̂ and 𝑏̂ are unit vectors, then prove that, 𝜃, |𝑎̂ + 𝑏̂| = 2𝑐𝑜𝑠 , where 𝜃 is the angle between them., , 2, , Find the direction cosines of the following line:, 3 − 𝑥 2𝑦 − 1 𝑧, =, =, −1, 2, 4, A bag contains 1 red and 3 white balls. Find the probability distribution of, the number of red balls if 2 balls are drawn at random from the bag one-byone without replacement., Two cards are drawn at random from a pack of 52 cards one-by-one without, replacement. What is the probability of getting first card red and second, card Jack?, , 2, , 2, , 4., 5., , 6., , 2, , 2, , SECTION – B, 𝑥+1, , 7., , Find: ∫ (𝑥 2+1)𝑥 𝑑𝑥, , 3, , 8., , Find the general solution of the following differential equation:, 𝑑𝑦, 𝑦, 𝑥, = 𝑦 − 𝑥𝑠𝑖𝑛( ), 𝑑𝑥, 𝑥, OR, Find the particular solution of the following differential equation, given that, 𝜋, y = 0 when x = 4 :, 𝑑𝑦, 2, + 𝑦𝑐𝑜𝑡𝑥 =, 𝑑𝑥, 1 + 𝑠𝑖𝑛𝑥, ⃗⃗⃗ 𝑎⃗. 𝑏⃗⃗ = 𝑎⃗. 𝑐⃗, 𝑎⃗ × 𝑏⃗⃗ = 𝑎⃗ × 𝑐⃗, then show that 𝑏⃗⃗ = 𝑐⃗., If 𝑎⃗ ≠ 0,, , 3, , 9., , 3

Page 2 :
10., , Find the shortest distance between the following lines:, 𝑟⃗ = (𝑖̂ + 𝑗̂ − 𝑘̂) + 𝑠(2𝑖̂ + 𝑗̂ + 𝑘̂ ), 𝑟⃗ = (𝑖̂ + 𝑗̂ + 2𝑘̂) + 𝑡(4𝑖̂ + 2𝑗̂ + 2𝑘̂), OR, Find the vector and the cartesian equations of the plane containing the point, 𝑖̂ + 2𝑗̂ − 𝑘̂ and parallel to the lines 𝑟⃗ = (𝑖̂ + 2𝑗̂ + 2𝑘̂) + 𝑠(2𝑖̂ − 3𝑗̂ + 2𝑘̂) = 0, and 𝑟⃗ = (3𝑖̂ + 𝑗̂ − 2𝑘̂) + 𝑡(𝑖̂ − 3𝑗̂ + 𝑘̂) = 0, , 3, , SECTION – C, 2, ∫−1|𝑥 3, , 11., , Evaluate:, , 12., , Using integration, find the area of the region in the first quadrant enclosed, by the line x + y = 2, the parabola y 2 = x and the x-axis., OR, Using integration, find the area of the region, {(𝑥, 𝑦): 0 ≤ 𝑦 ≤ √3𝑥, 𝑥 2 + 𝑦 2 ≤ 4}, , 4, , 13., , Find the foot of the perpendicular from the point (1, 2, 0) upon the plane, x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given, plane., , 4, , 2, , − 3𝑥 + 2𝑥|𝑑𝑥, , 4, , CASE-BASED/DATA-BASED, , 14., , Fig 3, An insurance company believes that people can be divided into two classes: those who, are accident prone and those who are not. The company’s statistics show that an, accident-prone person will have an accident at sometime within a fixed one-year period, with probability 0.6, whereas this probability is 0.2 for a person who is not accident, prone. The company knows that 20 percent of the population is accident prone., Based on the given information, answer the following questions., (i)what is the probability that a new policyholder will have an accident, within a year of purchasing a policy?, (ii) Suppose that a new policyholder has an accident within a year of, purchasing a policy. What is the probability that he or she is accident prone?, ---------------------------, , 2, 2

Page 7 :
12., , Using integration, find the area of the region in the first quadrant enclosed by, the line x + y = 2, the parabola y 2 = x and the x-axis., Solution: Solving x + y = 2 and y 2 = x simultaneously, we get the points of, intersection as (1, 1) and (4, -2)., , Fig 1, , 1, , 2, , The required area = the shaded area = ∫0 √𝑥 𝑑𝑥 + ∫1 (2 − 𝑥)𝑑𝑥, 2, , 3, 2, , 2, , 1, , = 3 [𝑥 ]10 + [2𝑥 −, 7, , 1, , 1, , 1, , 𝑥2 2, ], 2 1, , = 3 + 2 = 6 square units, , 1, , OR, Using integration, find the area of the region: {(𝑥, 𝑦): 0 ≤ 𝑦 ≤ √3𝑥, 𝑥 2 + 𝑦 2 ≤, 4}, Solution: Solving 𝑦 = √3𝑥 𝑎𝑛𝑑 𝑥 2 + 𝑦 2 = 4 , we get the points of intersection, as (1, √3) and (-1, −√3), , 1, , Fig 2, , 1

Page 8 :
1, , 2, , The required area = the shaded area = ∫0 √3𝑥 𝑑𝑥 + ∫1 √4 − 𝑥 2 𝑑𝑥, 𝑥2, √3 2 1 1, =, [𝑥 ]0 + [𝑥 √4 − 𝑥 2 + 4 sin−1 ], 2, 2, 21, 𝜋, √3 1, =, + [2𝜋 − √3 − 2 ], 2, 2, 3, 2𝜋, = 3 square units, , 13., , Find the foot of the perpendicular from the point (1, 2, 0) upon the plane, x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given, plane., Solution: The equation of the line perpendicular to the plane and passing, through the point (1, 2, 0) is, 𝑥−1 𝑦−2 𝑧, =, =, 1, −3, 2, The coordinates of the foot of the perpendicular are (𝜇 + 1, −3𝜇 + 2,2𝜇) for, some 𝜇, These coordinates will satisfy the equation of the plane. Hence, we have, 𝜇 + 1 − 3(−3𝜇 + 2) + 2(2𝜇) = 9, ⇒𝜇=1, The foot of the perpendicular is (2, -1, 2)., Hence, the required distance = √(1 − 2)2 + (2 + 1)2 + (0 − 2)2 = √14 𝑢𝑛𝑖𝑡𝑠, , 1, , 1, , 1, ½, , 1, ½, 1

Page 9 :
14., , CASE-BASED/DATA-BASED, , Fig 3, An insurance company believes that people can be divided into two classes: those who are, accident prone and those who are not. The company’s statistics show that an accident-prone, person will have an accident at sometime within a fixed one-year period with probability 0.6,, whereas this probability is 0.2 for a person who is not accident prone. The company knows that, 20 percent of the population is accident prone., Based on the given information, answer the following questions., (i)what is the probability that a new policyholder will have an accident within a, year of purchasing a policy?, (ii) Suppose that a new policyholder has an accident within a year of purchasing a, policy. What is the probability that he or she is accident prone?, Solution: Let E1 = The policy holder is accident prone., E2 = The policy holder is not accident prone., E = The new policy holder has an accident within a year of purchasing a, policy., (i), P(E)= P(E1)× P(E∕E1) + P(E2)× P(E∕E2), 20, 6, 80, 2, 7, =100 × 10 + 100 × 10 = 25, (ii), , By Bayes’ Theorem, 𝑃(𝐸1⁄𝐸) =, =, , 20 6, ×, 100 10, 280, 1000, , 𝑃(𝐸1 )×𝑃(𝐸/𝐸1 ), 𝑃(𝐸), , 3, , =7, , 1, 1, , 1, 1, , ---------------------------