Class 12 Mathematics Sample Paper 01 (2022-2023 PDF)

myCBSEguide Class 12 - Mathematics Sample Paper - 01 (2022-23) Maximum Marks: 80 Time Allowed: : 3 hours General Instructions: 1. This Question paper contains - five sections A, B, C...

myCBSEguide Class 12 - Mathematics Sample Paper - 01 (2022-23) Maximum Marks: 80 Time Allowed: : 3 hours General Instructions: 1. This Question paper contains - five sections A, B, C, D and E. Each section is compulsory. However, there are internal choices in some questions. 2. Section A has 18 MCQ’s and 02 Assertion-Reason based questions of 1 mark each. 3. Section B has 5 Very Short Answer (VSA)-type questions of 2 marks each. 4. Section C has 6 Short Answer (SA)-type questions of 3 marks each. 5. Section D has 4 Long Answer (LA)-type questions of 5 marks each. 6. Section E has 3 source based/case based/passage based/integrated units of assessment (4 marks each) with sub parts. Section A 1. ∫ cot xdx =? 2 a) -cot x + x + C b) cot x - x + C c) cot x + x + C d) -cot x - x + C 2. The direction cosines of X -axis are a) < 0 , 1 , 1 > b) < 0 , 0 , 1 > c) < 1 , 0 , 0 > d) < 0 , 1 , 0 > 3. Find |a⃗| and ∣∣b∣∣⃗ , if ⃗ ⃗ (a⃗ + b). (a⃗ − b) = 8 and |a⃗| = 8 ∣ ∣⃗ b ∣ ∣. 16√2 2√2 a) , 3√7 3√7 19√2 2√5 b) , 3√7 3√7 17√2 2√3 c) , 3√7 3√7 21√2 2√6 d) , 3√7 3√7 4. A machine operates only when all of its three components function. The probabilities of the failures of the first, second and third component are 0.2, 0.3 and 0.5 respectively. What is the probability that the machine will fail? a) None of these b) 0.07 c) 0.72 d) 0.70 sin(x−α) 5. ∫ sin(x+α) dx =? a) None of these b) x cos 2α + sin 2α. log|sin (x + α)| + C c) x cos 2α + sin α. log |sin (x + α)| + C d) x cos 2α - sin 2α. log |sin (x + α)| + C Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 1 / 21 myCBSEguide 6. Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P (A|B) a) 0.27 b) 0.3 c) 0.2 d) 0.33 To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to create similar papers with their own name and logo. 7. The area bounded by the curves y = √− − x , 2y + 3 = x and the x-axis in the first quadrant is a) 36 b) 18 c) 9 d) none of these 8. The angle between a line with direction ratios 2 : 2 : 1 and a line joining (3, 1, 4) to (7, 2, 12) a) cos ( ) −1 2 3 b) tan −1 (− 2 3 ) c) none of these d) cos ( ) −1 3 2 −→ − − −→ 9. In a hexagon ABCDEF AB⃗ = ⃗ a, BC = b ⃗ and C D = c.⃗ Then AE = a) a⃗ + 2b + 2c ⃗ ⃗ b) 2a⃗ + b ⃗ + c ⃗ c) a⃗ + b ⃗ + c ⃗ d) b ⃗ + c ⃗ 10. General solution of y log y dx – x dy = 0 a) y = e −cx b) y = e cx c) y = e 2 cx d) y = e + e cx −cx 11. The area bounded by y = 2cosx , x = 0 to x = 2π and the axis of x in square units is - a) 4 b) 6 c) 8 d) 7 3 12. ∫ |x| dx is equal to a) sin √− − x + C 4 b) −x 4 + C c) none of these 4 |x| d) 4 + c 13. The function f(x) = |x| has a) only one maxima b) only one minima c) no maxima or minima d) none of these 14. If A = [aij] is a square matrix of even order such that aij = i2 - j2, then a) A is symmetric matrix and |A| is a square Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 2 / 21 myCBSEguide b) none of these. c) A is a skew-symmetric matrix and |A| = 0 d) A is symmetric matrix and | A | = 0 1 1 2 ⎡ ⎤ 15. If A = satisfies ATA = I, then x + y = 1 ⎢2 1 −2 ⎥ 3 ⎣ ⎦ x 2 y a) -3 b) none of these c) 0 d) 3 ∣ a + ib c + id ∣ 16. ∣ ∣ =? ∣ −c + id a − ib ∣ a) None of these b) (a2 + b2 + c2 - d2) c) (a2 - b2 + c2 - d2) d) (a2 + b2 + c2 + d2) 17. One branch of cos–1 other than the principal value branch corresponds to a) [2π, 3π] b) [π, 2π] − { 3π 2 } c) [ π 2 , 3π 2 ] d) (0, π) dy 18. General solution of dx = ( 1 + x ) (1 + y ) 2 2 is 3 a) tan −1 y =x+ x 3 + C 3 b) cos −1 y =x+ x 3 + C 3 c) cot −1 y =x+ x 3 + C 3 d) sin −1 y =x+ x 3 + C 19. Assertion (A): The function f(x) = x2 - 4x + 6 is strictly increasing in the interval (2, ∞). Reason (R): The function f(x) = x2 - 4x + 6 is strictly decreasing in the interval (−∞ , 2). a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true. 2 −2 3 −2 20. Assertion (A): If A = [ ] , then A-1 = [ ] 4 3 4 3 2 5 −1 5 − Reason (R): A = [ , then A-1 = [ 13 13 ] ] 3 1 −3 2 − 13 13 a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true. Section B – 21. Evaluate:- tan −1 (− 1 ) + tan −1 (− √3) + tan −1 (sin(− π 2 )) √3 −−−−−−−− 2 dy dy 22. Write the order and degree of the differential equation y = x dx + a√1 + ( dx ) Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 3 / 21 myCBSEguide ∣2 −3 5 ∣ ∣ ∣ 23. In the determinant 6 0 4. Verify that a 11 A31 + a12 A32 + a12 A33 = 0 ∣ ∣ ∣1 5 −7 ∣ OR 2 1 5 3 1 0 Find the matrix X satisfying the equation: [ ]X[ ] = [ ] 5 3 3 2 0 1 24. Show that the four points whose position vectors are 6^i − 7^j , 16^i − 19^j − 4k ^ , 3 i − 6k, 2 i − 5j + 10k coplanar. ^ ^ ^ ^ ^ 25. An urn contains 5 white and 8 black balls. Two successive drawings of three balls at a time are made such that the balls are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and the second draw gives 3 black balls. Section C 3 26. Evaluate the integral: ∫ (sin −1 x) dx dy 27. If y(x) is a solution of the differential equation ( and y(0) = 1, then find the value of y (. 2+sin x π ) = − cos x ) 1+y dx 2 OR dy Find the particular solution of the following differential equation, given that x = 2, y = 1 x dx 2 + 2y = x , (x ≠ 0). 28. If a⃗ + b + c ⃗ = ⃗ 0 , show that a × b = ⃗ ⃗ b × c ⃗ = c ⃗ × a⃗. Interpret the result geometrically? OR Verify that a⃗ × (b ⃗ + c )⃗ = (a⃗ × b)⃗ + (a⃗ × c )⃗ , when a⃗ = ^ ^ ^ ⃗ ^ ^ ^ 4 i − j + k, b = i + j + k and c ⃗ = ^ ^ ^ i − j + k. −−−−−−−−−− 29. Evaluate the integral: ∫ (3x + 1)√4 − 3x − 2x dx 2 OR Evaluate: ∫ 1 dx sin x+sec x 2 λ (x + 2) , if x ≤ 0 30. For what value of λ , the function defined by f(x) = { is continuous at x = 0? Hence, check the 4x + 6, if x > 0 differentiability of f(x) at x = 0. − − 31. Calculate the area under the curve y = 2√x included between the lines x = 0 and x = 1. Section D 32. Solve the following LPP graphically: Maximize Z = 5x + 7y Subject to x+y≤4 3x + 8y ≤ 24 10x + 7y ≤ 35 x, y ≥ 0 33. Let R be relation defined on the set of natural number N as follows: R = {(x, y): x ∈ N, y ∈ N, 2x + y = 41}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive. OR Prove that the relation R on the set N × N defined by (a, b) R (c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N × N is an equivalence relation. Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 4 / 21 myCBSEguide Also, find the equivalence classes [(2, 3)] and [(1, 3)]. → 34. Find the shortest distance between the lines given by ^ ^ ^ r = (8 + 3λ) i − (9 + 16λ) j + (10 + 7λ) k and → ^ ^ ^ ^ ^ ^ r = 15 i + 29j + 5k + μ(3 i + 8j − 5k). OR By computing the shortest distance determine whether the pairs of lines intersect or not: ^ ^ ^ ^ r ⃗ = ( i − j ) + λ(2 i + k) and r ⃗ = ^ ^ ^ ^ ^ (2 i − j ) + μ( i + j − k) 2 35. Differentiate sin with respect to cos , if 0 < x < 1. −1 2x −1 1−x ( ) ( ) 2 2 1+x 1+x Section E 36. Read the text carefully and answer the questions: Mrs. Maya is the owner of a high-rise residential society having 50 apartments. When he set rent at ₹10000/month, all apartments are rented. If he increases rent by ₹250/ month, one fewer apartment is rented. The maintenance cost for each occupied unit is ₹500/month. i. If P is the rent price per apartment and N is the number of rented apartments, then find the profit. ii. If x represents the number of apartments which are not rented, then express profit as a function of x. iii. Find the number of apartments which are not rented so that profit is maximum. OR Verify that profit is maximum at critical value of x by second derivative test. To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to create similar papers with their own name and logo. 37. Read the text carefully and answer the questions: Three schools A, B and C organized a mela for collecting funds for helping the rehabilitation of flood victims. They sold handmade fans, mats, and plates from recycled material at a cost of ₹ 25, ₹ 100 and ₹ 50 each. The number of articles sold by school A, B, C are given below. Article School A B C Fans 40 25 35 Mats 50 40 50 Plates 20 30 40 i. Represent the sale of handmade fans, mats and plates by three schools A, B and C and the sale prices (in ₹) of given products per unit, in matrix form. ii. Find the funds collected by school A, B and C by selling the given articles. iii. If they increase the cost price of each unit by 20%, then write the matrix representing new price. Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 5 / 21 myCBSEguide OR Find the total funds collected for the required purpose after 20% hike in price. 38. Read the text carefully and answer the questions: In pre-board examination of class XII, commerce stream with Economics and Mathematics of a particular school, 50% of the students failed in Economics, 35% failed in Mathematics and 25% failed in both Economics and Mathematics. A student is selected at random from the class. i. Find the probability that the selected student has failed in Economics, if it is known that he has failed in Mathematics? ii. Find the probability that the selected student has failed in Mathematics, if it is known that he has failed in Economics? Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 6 / 21 myCBSEguide Class 12 - Mathematics Sample Paper - 01 (2022-23) Solution Section A 1. (d) -cot x - x + C Explanation: GIven : ∫ cot 2 xdx 2 2 ⇒ ∫ cot xdx = ∫ ((csc x) − 1) dx = -cot x - x + c 2. (c) < 1 , 0 , 0 > Explanation: As we know that if a lines makes angles a , b and c with X-axis , Y-axis and Z-axis respectively then direction cosines are given by < cos a , cos b ,cos c > In our case line is X-axis itself which we know makes angle of 0 , 90 , 90 with X-axis , Y-axis and Z-axis ∘ ∘ ∘ respectively then direction cosine will be ∘ ∘ ∘ = < 1, 0, 0 > 16√2 2√2 3. (a) , 3√7 3√7 → → → → Explanation: ( a + b ). ( a − b ) = 8 → → 2 2 ⇒ | a | − | b | = 8 → → → → −− 2 2 2 8 2√2 ⇒ 64| b | − | b | = 8 ⇒ 63| b | = 8 ⇒ | b | = √ = 63 3√7 → → → 16√2 ⇒ | a | = 8| b | ⇒ | a | = 3√7 4. (c) 0.72 Explanation: The probability of failure of the first component = 0.2 = P(A) The probability of failure of second component = 0.3 = P(B) The probability of failure of third component = 0.5 = P(C) As the events are independent, The machine will operate only when all the components work, i.e., (1 - 0.2)(1 - 0.3)(1 - 0.5) = P(A’)P(B’)P(C’) In rest of the cases, it won’t work, So P(A ∪ B ∪ C) = 1 - P(A’ ∩ B’ ∩ C’) = 1 - (0.8).(0.7).(0.5) ⇒ 1 - 0.28 = 0.72 To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to create similar papers with their own name and logo. 5. (d) x cos 2α - sin 2α. log |sin (x + α)| + C sin(x−α) Explanation: The given integral is ∫ sin(x+α) dx n+1 since we know that ∫ x n x dx = + c n+1 sin (a + b) = sin a cos b + cos a sin b ∫ cot x = log (sin x) + c Therefore , sin(x+α−2α) ⇒∫ dx sin(x+α) Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 7 / 21 myCBSEguide sin(x+α) cos(−2α)+cos(x+α) sin(−2α) ⇒ ∫ dx sin(x+α) ⇒ ∫ cos(2α)dx − sin 2α ∫ cot(x + α)dx ⇒ cos (2 α) x - sin 2 α log |sin (x+α)|+c 6. (b) 0.3 Explanation: Let A and B be independent events with P (A) = 0.3 and P(B) = 0.4 P(A/B) = P(A) = 0.3. 7. (c) 9 Explanation: Required area: 9 3 9 9 9 2 − − sq.units x−3 x 2 1 x ∫ √x dx − ∫ ( ) dx = [ ] − [ − 3x] = 9 2 3/2 2 2 3 0 3 0 8. (a) cos −1 ( 2 3 ) Explanation: The angle between a line with direction ratios 2 : 2 : 1 and a line joining (3, 1, 4) to (7, 2, 12) Direction ratios of the line joining the points A(3, 1, 4), B(7, 2, 12) is = < 7-3 , 2-1 , 12-4> = Now as the angle between two lines having direction ratios and is given by −1 a1a2+b1b2+c1c2 Cos √ 2 2 2√ 2 2 2 a 1 +b 1 +c 1 a 2 +b 2 +c 2 Using the vuales we have −1 2×4+2×1+1×8 −1 18 −1 2 cos = cos = cos √ 2 2 2√ 2 2 2 27 3 2 +2 +1 4 +1 +8 9. (d) b ⃗ + c ⃗ Explanation: In △BCD, −→ − −→ − −→ BC + CD = BD −→ → − −→ → Given that BC = b , CD = c − −→ −→ And BD is parallel to AE −→ → ¯¯ ⇒ AE = b + c 10. (b) y = e cx Explanation: y log y dx = x dy 1 1 ∫ dx = ∫ dy x y log y ′ f (x)dx log|x| = log|log y| + log C Since ∫ = log|f(x)| + c and 1 = c a new constant f(x) C log x = log(C log y) x = C log y 1 log y = x C log y = cx cx y = e 11. (c) 8 Explanation: Required area: Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 8 / 21 myCBSEguide π π π 2 2 =∫ ydx =∫ 2 cos xdx = [2 sin x] 0 2 =2 0 0 Therefore, total area from x = 0 to x = 2π is 4 × 2 = 8 sq. units. 12. (c) none of these Explanation: ∫ |x| , 3 3 dx = ∫ x dx If x > 0 4 x = + c 4 And, ∫ |x| , If x < 0 3 3 dx = ∫ − x dx 4 x = − + c 4 13. (b) only one minima −x, x < 0 Explanation: Given,f(x) = |x| = x, x > 0 ⇒ f (x) = −1 ′ when x< 0 and 1 when x > 0 But, we have f'(x) does not exist at x = 0, hence we have x = 0 is a critical point At x = 0, we get f(0) = 0 For any other value of x, we have f(x) > 0 , hence f(x) has a minimum at x= 0. 14. (b) none of these. Explanation: aij = i2 – j2 a11 = 12 – 12 = 0 a12 = 12 – 22 = -3 a21 = 22 – 12 = 3 a22 = 22 – 22 = 0 0 −3 ∴ A=[ ] 3 0 0 3 AT = [ ] −3 0 0 3 -A = [ ] −3 0 So, AT = -A |A| = 0(0) – (-3)(3) = 0 + 9 = 9 ≠ 0 So, none of these. 15. (b) none of these 1 1 2 ⎡ ⎤ Explanation: We have, A = 1 3 ⎢2 1 −2 ⎥ ⎣ ⎦ x 2 y 1 2 x ⎡ ⎤ T 1 ⇒ A = ⎢1 1 2⎥ 3 ⎣ ⎦ 2 −2 y Now, ATA = I 2 x + 5 2x + 3 xy − 2 9 0 0 ⎡ ⎤ ⎡ ⎤ ⇒ ⎢ 3 + 2x 6 2y ⎥ = ⎢0 9 0⎥ ⎣ 2 ⎦ ⎣ ⎦ xy − 6 2y y + 8 0 0 9 The corresponding elements of two equal matrices are not equal. Thus, the matrix A is not orthogonal. 16. (d) (a2 + b2 + c2 + d2) Explanation: Δ = (a + ib)(a - ib) + (c - id)(c + id) = (a2 + b2 + c2 + d2) Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 9 / 21 myCBSEguide 17. (a) [2π, 3π] Explanation: [2π, 3π] 3 18. (a) tan =x+ −1 x y + C 3 dy Explanation: dx = ( 1 + x ) (1 + y ) 2 2 ∫ 1 2 dy = ∫ (1 + x )dx 2 Since∫ dx 2 −1 = tan x+ c 1+y 1+x 3 −1 x tan y = x+ + C 3 19. (b) Both A and R are true but R is not the correct explanation of A. Explanation: We have, f(x) = x2 - 4x + 6 or f'(x) = 2x - 4 = 2 (x - 2) Therefore, f'(x) = 0 gives x = 2. Now, the point x = 2 divides the real line into two disjoint intervals namely, (−∞ , 2) and (2, ∞). In the interval (−∞ , 2), f'(x) = 2x - 4 < 0. Therefore, f is strictly decreasing in this interval. Also, in the interval (2, ∞), f'(x) > 0 and so the function f is strictly increasing in this interval. Hence, both the statements are true but Reason is not the correct explanation of Assertion. 20. (d) A is false but R is true. 2 −2 Explanation: Assertion: Let A = [ ] 4 3 ∣2 −2 ∣ We have, |A| = ∣ ∣ = 6 - (-8) = 14 ∣4 3 ∣ Cofactors of |A| are A11 = 3, A12 = -4, A21 = 2 and A22 = 2 ′ 3 −4 3 2 ∴ adj (A) = [ ] = [ ] 2 2 −4 2 3 2 Now, A −1 = 1 (adj A) = 1 14 [ ] |A| −4 2 3 2 3 1 −1 14 14 14 7 ⇒ A = [ ] = [ ] 4 2 2 1 − − 14 14 7 7 −1 5 Reason: Let A = [ ] −3 2 ∣ −1 5∣ We have, |A| = ∣ ∣ = -2 - (-15) = 13 ∣ −3 2∣ Now, cofactors of |A| are A11 = 2, A12 = 3, A21 = -5 and A22 = -1 ′ 2 3 2 −5 ∴ adj (A) = [ ] = [ ] −5 −1 3 −1 2 −5 Now, A −1 = 1 |A| (adj A) = 1 13 [ ] 3 −1 2 5 − 13 13 = [ ] 3 1 − 13 13 Section B – 21. tan −1 (− 1 ) + tan −1 (− √3) + tan −1 (sin(− π 2 )) √3 −1 1 −1 – −1 = tan (− ) + tan (− √3) + tan (−1) √3 Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 10 / 21 myCBSEguide We know that, for any x ∈ R, tan-1x represents an angle in ( whose tangent is x. −π π , ) 2 2 −1 1 π ∴ tan (− ) = − √3 6 −1 – π tan (− √3) = − 3 −1 π tan (−1) = − 4 −1 1 −1 – −1 π π π 3π ⇒ tan (− ) + tan (− √3) + tan (−1) = − − − = − √3 6 3 4 4 – Principle value of tan is −. −1 1 −1 −1 π 3π ∴ (− ) + tan (− √3) + tan (sin(− )) √3 2 4 −−−−−−−− 2 dy dy 22. y = x dx + a√1 + ( dx ) −−−−−−−− 2 2 2 dy dy (y − x ) = (a√1 − ( ) ) dx dx 2 2 dy dy 2 (y − x ) = a [1 + ( ) ] dx dx 2 2 dy dy dy 2 2 2 2 y + x ( ) − 2.xy = a + a ( ) dx dx dx 2 dy dy 2 2 2 2 (x − a )( ) − 2xy + y − a = 0 dx dx order = 1 degree = 2 23. a 11 = 2, a12 = −3, a13 = 5 A31 = −12, A32 = 22, A33 = 18 L.H.S = a11 A31 , a12 A32 + a13 A33 = 2 (-12) + (-3) (22) +5 (18) = -24 - 66 + 90 = -90 + 90 =0 OR 2 1 5 3 1 0 Let A = [ ]B =[ ]C =[ ] 5 3 3 2 0 1 Then the given equation becomes as AXB = I ⇒ X = A – 1B – 1 now |A| = 6 – 5 = 1 and |B| = 10 – 9 = 1 3 −1 A–1= adj(A) 1 ∴ = [ ] |A| 1 −5 2 2 −3 and B – 1 = adj(B) 1 = [ ] |B| 1 −3 5 3 −1 2 −3 ⇒ X = A – 1B – 1 = [ ][ ] −5 2 −3 5 6+ 3 −9 − 5 9 −14 = [ ] = [ ] −10 − 6 15 + 10 −16 25 9 −14 Hence, x = [ ] −16 25 24. ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ Let OA = 6 i − 7j , OB = 16 i − 19j − 4k, OC = 3 i − 6k, OD = 2 i − 5j + 10k Therefore, AB = OB - OA = 10^i − 12^j − 4k ^ Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 11 / 21 myCBSEguide AC = OC - OA =−3ˆi + 7ˆj − 6k ˆ ^ ^ ^ AD = OD − OA = −4 i + 2j + 10k The four points are coplanar if vectors AB, AC and AD are coplanar. Therefore,we have ∣ 10 −12 −4 ∣ ∣ ∣ −3 7 −6 = 10(70 + 12) + 12(-30 - 24) -4(-6 + 28) ∣ ∣ ∣ −4 2 10 ∣ = 820 - 648 - 88 =84 25. Consider the following events: A = Drawing 3 white balls in first draw, B = Drawing 3 black balls in the second draw. Required probability = P (A ∩ B) = P (A) P( )...(i) B A Since there are 13 balls out of which 3 balls can be drawn in 13 C3 ways and 3 white balls out of 5 white balls can be drawn in 5 C3 ways. Therefore, we have, 5 C3 P(A) = 10 5 = = 13 C3 286 143 After drawing 3 white balls in first draw 10 balls are left in the bag, out of which 8 are black balls. 8 C3 P(B/A) = 56 7 ∴ = = 10 C3 120 15 Substituting these values in (i), we obtain Required probability = P (A nB) = P(A) P( ) = B 5 7 7 × = A 143 15 429 Section C 26. Let the given integral be, 3 −1 I = ∫ (sin x) dx Let sin-1x = t −−−−− ⇒ sin t = x ⇒ cos t = √1 − x 2 Differentiating both sides we get cos t dt = dx Now, the integral becomes 3 −1 I = ∫ (sin x) dx 3 = ∫ t cos tdt = t 3 sin t − ∫ 3t 2 sin tdt (Using by parts) 3 2 = t sin t − 3 ∫ t sin tdt 3 2 = t sin t − 3 [− t cos t − ∫ −2t cos tdt] 3 2 = t sin t + 3t cos t − 6 ∫ t cos tdt 3 2 = t sin t + 3t cos t − 6 [t sin t − ∫ sin tdt] 3 2 = t sin t + 3t cos t − 6[t sin t + cos t] + C 3 2 −−−−− −−−−− −1 −1 2 −1 2 = (sin x) x + 3(sin x) √1 − x − 6 (sin x) x − 6√1 − x + C 2 2 −−−−− −1 −1 −1 2 = x (sin x) [(sin x) − 6] + 3 [(sin x) − 2] √1 − x + C 27. We have, 2+sin x dy ( ) = − cos x 1+y dx 1 cos x ⇒ dy = − dx 1+y 2+sin x On integrating both sides, we get, 1 cos x ∫ dy = − ∫ dx 1+y 2+sin x put 2 + sin x = t ⇒ cos xdx = dt ⎡ ⎤ cos x dt ⇒ log |1 + y| = − log |2 + sin x| + log C ⎢ then ∫ dx = ∫ = log |t| + C ⎥ ⎢ 2+sin x t ⎥ ⎣ ⎦ = log |2 + sin x| + C Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 12 / 21 myCBSEguide ⇒ log |1 + y| + log |2 + sin x| = log C ⇒ log(|1 + y||2 + sin x|) = log C [∵ log m + log n = log mn] ⇒ (1 + y)(2 + sin x) = C...(i) Also, given that at x = 0, y = 1. On putting x = 0 and y = 1 in Eq. (i), we get, ( 1+ 1) (2+ sin 0) = C ⇒ C=4 On putting C = 4 in Eq. (i), we get, (1 + y ) (2 + sin x) = 4 4 ⇒ 1+ y = 2+sin x 4 ⇒ y = − 1 2+sin x 4−2−sin x 2−sin x ⇒ y = ⇒ y = 2+sin x 2+sin x π 2−sin Now, at x = π 2 ,y( π 2 ) = 2 π ∴ y( π 2 ) = 1 3 [∵ sin π 2 = 1] 2+sin 2 OR dy We have, x + 2y = x (x ≠ 0) 2 dx dy...(i) 2 ⇒ + ( ) ⋅y = x dx x This is linear differential equation of the form dy + Py = Q , here P = 2 x and Q = x. dx 2 ∫ Pdx ∫ (2/x)dx 2 log x log x 2 ∴ IF = e = e = e = e = x The general solution is given by y ⋅ I F = ∫ (IF × Q)dx + C 2 2 ⇒ y⋅x = ∫ x × xdx + C 2 3 ⇒ y⋅x = ∫ x dx + C 4...(ii) 2 x ∴ y⋅x = + C 4 On putting x = 2, y = 1 in Eq. (ii), we get 4 2 2 1⋅2 = + C ⇒ 4 = 4+ C ⇒ C = 0 4 4 [from Eq. (ii)] 2 x ∴ y⋅x = 4 2 x ⇒ y = 4 which is the required particular solution. 28. Since, a⃗ + b⃗ + c⃗ = 0 ⃗ ⇒ b = − c ⃗ − a⃗ Now, a⃗ × b ⃗ = a⃗ × (− c ⃗ − a⃗) = a⃗ × − c ⃗ − a⃗ × a⃗ = a⃗ × − c ⃗ ⃗ ⇒ a⃗ × b = c ⃗ × a⃗...(i) Also, ¯ b × c ⃗ = (− c ⃗ − a⃗) × c ⃗ = − c ⃗ × c ⃗ − a⃗ × c ⃗ = − a⃗ × c ⃗ = − c ⃗ × c ⃗ − a⃗ × c ⃗ = − a⃗ × c ⃗ = c ⃗ × a⃗...(ii) From (i) and (ii) ⃗ ⃗ a⃗ × b = b × c ⃗ = c ⃗ × a⃗ Geometrical interpretation of the result → → If ABCD is a parallelogram such that AB = a⃗ and AD = b ⃗ and these adjacent sides are making angle θ between each other, then Area of the parallelogram ABC D = ⃗ ⃗ |a⃗||b|| sin θ = |a⃗ × b| Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 13 / 21 myCBSEguide since parallelogram on the same base and between the same parallels are equal we have |a⃗ × b|⃗ = ⃗ |a⃗ × c |⃗ = |b × c |⃗ This also implies that, a⃗ × b ⃗ = ⃗ a⃗ × c ⃗ = b × c ⃗ So, the area of the parallelograms formed by taking any two sides represented by a⃗ , b ⃗ and c ⃗ as adjacent are equal. OR Given vectors are, ^ ^ ^ ⃗ ^ ^ ^ a⃗ = 4 i − j + k, b = i + j + k and c ⃗ = ^ ^ ^ i − j + k ⃗ ⃗ = (^ ^ ^ ^ ^ ^ ^ ^ ⇒ (b + c ) i + j + k) + ( i − j + k) = (2 i + 2k) ∣^ ^ ^∣ i j k ∣ ∣ ∣ −1 1∣ ∣4 1∣ ∣4 −1 ∣ ⃗ ^ ^ ^ ⇒ ( a⃗ × b) = ∣ 4 −1 1∣ = i ∣ ∣ − j∣ ∣ + k∣ ∣ ∣ ∣ ∣ 1 1∣ ∣1 1∣ ∣1 1 ∣ ∣1 1 1∣ = ^ i (- 1 - 1) - (4 - 1) + ^ j ^ k (4 + 1) = (-2^i - 3^j + 5k ^ ) ∣ ^∣ i j k ∣ ∣ ∣ −1 1∣ ∣4 1∣ ∣4 −1 ∣ ^ ^ ^ ⇒ ( a⃗ × c̄ ) = ∣ 4 −1 1∣ = i ∣ ∣ − j∣ ∣ + k∣ ∣ ∣ ∣ ∣ −1 1∣ ∣1 1∣ ∣1 −1 ∣ ∣1 −1 1∣ = ^ i (- 1 + 1) - (4 - 1) + (-4 + 1) = (-3^j - 3k ^ j ^ k ^ ) ∣^ ^ ^∣ i j k ∣ ∣ ∣ −1 1∣ ∣4 1∣ ∣4 −1 ∣ L.H.S. = ⃗ a⃗ × ( b + c ) ⃗ = ∣ 4 −1 ^ 1∣ = i ∣ ^ ∣ − j∣ ^ ∣ + k∣ ∣ ∣ ∣ ∣ 0 2∣ ∣2 2∣ ∣2 0 ∣ ∣2 0 2∣ = ^ i (- 2 - 0) - (8 - 2) + (0 + 2) = (-2^i - 6^j + 2k ^ j ^ k ^ ) R.H.S. = ⃗ ⃗ = (−2^ ( a⃗ × b) + ( a⃗ × c ) ^ ^ ^ ^ ^ ^ ^ i − 3j + 5k) + (−3j − 3k) = (−2 i − 6j + 2k) Hence, a⃗ × (b ⃗ + c )⃗ ⃗ = ( a⃗ × b) + ( a⃗ × c ) ⃗ 29. Let the given integral be, − −−−−−−−−− I = ∫ (3x + 1)√4 − 3x − 2x dx 2 Let (3x + 1) = A d (4 − 3x − 2x ) 2 +B dx ⇒ (3x + 1) = A(-3 - 4x) + B ⇒ (3x + 1) = -4Ax + (B - 3A) ⇒ 3 = -4A and (B - 3A) = 1 ⇒ A = - and B = - 3 5 4 4 −−−−−−−−−− −−−−−−−−−− I=- - 3 5 2 2 ⇒ ∫ (−3 − 4x) √4 − 3x − 2x dx ∫ √4 − 3x − 2x dx 4 4 Let I = - I1 - I2.....(i) 3 5 4 4 Now, −−−−−−−−−− I1 = ∫ (−3 − 4x)√4 − 3x − 2x 2 dx Let (4 - 3x - 2x2) = t or (-3 - 4x)dx = dt ⇒ I1 = ∫ √tdt 3 = 2 3 t 2 + c1 3 I1 = + c1 2 2 ⇒ (4 − 3x − 2x ) 2 3 −−−−−−−−−− Now, I2 = 2 ∫ √4 − 3x − 2x dx Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 14 / 21 myCBSEguide −−−−−−−−−−−−− = ∫ √2 (2 − 3 2 x − x ) dx 2 −−−−−−−−−−−−−−−− – = √2 ∫ √( 17 9 3 2 − − x − x ) dx 4 4 2 −−−−−−−−−−−−−−−−−−−−−− 2 – √17 = 9 3 2 √2 ∫ √[( ) − ( + x + x )] dx 2 4 2 −−−−−−−−−−−−−−−−−− 2 2 – √17 = 3 √2 ∫ √[( ) − (x + ) ] dx 2 2 3 x+ – = √2 sin( √17 2 ) + c2 2 – = √2 sin( + c2 2x+3 ) √17 Using (i), we get 3 – I=− - +C 3 2 2 5 2x+3 × (4 − 3x − 2x ) 2 × √2 sin( ) 4 3 4 √17 3 5√2 I=− - +C 1 2 2x+3 ∴ (4 − 3x − 2x ) 2 sin( ) 2 4 √17 OR The given integral is I=∫ 1 dx sin x+sec x cos x 2 cos x I = ∫ dx = ∫ dx 1+sin x cos x 2+2 sin x cos x (cos x+sin x)+(cos x−sin x) ⇒ I = ∫ dx 2+2 sin x cos x cos x+sin x cos x−sin x ⇒ I = ∫ dx + ∫ dx 2+2 sin x cos x 2+2 sin x cos x cos x+sin x cos x−sin x ⇒ I = ∫ dx + ∫ dx 3−(1−2 sin x cos x) 1+(1+2 sin x cos x) (cos x+sin x) cos x−sin x ⇒ I = ∫ dx + ∫ dx 2 2 3−(sin x−cos x) 1+(sin x+cos x) ⇒ I = ∫ 1 2 2 du + ∫ 1 2 dv, where u = sin x - cos x and v = sin x + cos x (√3) −u 1+v 1 ∣ √3+u ∣ −1 ⇒ I = log + tan v+ C 2√3 ∣ √3−u ∣ + tan-1 (sin x + cos x) + C 1 ∣ √3+(sin x−cos x) ∣ ⇒ I = log 2√3 ∣ √3−(sin x−cos x) ∣ 2 λ (x + 2) , if x ≤ 0 30. Given f(x) = { is continuous at x = 0 4x + 6, if x > 0 Since f(x) is continuous at x=0.Therefore, we have, LHL = RHL = f(0).......(i) Here, RHL = lim f(x) = lim (4x + 6) + + x→0 x→0 = lim [4(0 + h) + 6] h→0 = lim (4h + 6) h→0 = 4× 0 +6=6 From Eq(i), RHL = f(0) ⇒ 2λ = 6 ⇒ λ = 3 Now, given function becomes 2 3 (x + 2) , if x ≤ 0 f(x) = { 4x + 6, if x > 0 Now, let us check the differentiability at x = 0. f(0−h)−f(0) LHD = lim −h h→0 Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 15 / 21 myCBSEguide 2 3[(0−h) +2]−3(0+2) = lim −h h→0 2 3[h +2]−6 = lim = lim (−3h) =0 −h h→0 h→0 f(0+h)−f(0) and RHD = lim h h→0 [4(0+h)+6]−3(0+2) = lim = lim 4h =4 h h h→0 h→0 ∵ LHD ≠ RHD ∴ f(x) is not differentiable at x = 0. −− 31. We have, y = 2√x. x = 0 and x = 1. 1 − − ∴ Area of shaded region, A = ∫ 0 (2√x ) dx 1 3/2 x = 2.[.2] 3 0 = 2( 2 3.1 − 0) = 4 3 sq units Section D 32. Converting the inequations into equations, we obtain the following equations: x + y = 4, 3x + 8y = 24, 10x + 7y = 35, x = 0 and y = 0. These equations represent straight lines in XOY-plane. The line x + y = 4 meets the coordinate axes at A1 (4, 0) and B1 (0, 4). Join these points to obtain the line x + y = 4. The line 3x + 8y = 24 meets the coordinate axes at A2 (8, 0) and B2 (0, 3). Join these points to obtain the line 3x + 8y = 24. The line 10x + 7y = 35 cuts the coordinates axes at A3 (3.5, 0) and B3 (0,5). These points are joined to obtain the line 10x + 7y = 35. Also, x = 0 is the y-axis and y = 0 is the x-axis. The feasible region of the LPP is shaded in Figure. The coordinates of the corner points of the feasible region OA3PQB2 are O (0, 0), A3(3.5, 0), P( , ), Q( , ) and 7 5 8 12 5 3 5 5 B2(0, 3). Now, we take a constant value, say 10 for Z. Putting Z = 10 in Z = 5x + 7y, we obtain the line 5x + 7y = 10. This line meets the coordinate axes at P1 (2, 0) and Q(0, ). Join these points by a dotted line. Now, move this line parallel to 10 7 itself in the increasing direction away from the origin. P2Q2 and P3Q3 are such lines. Out of these lines locate a line Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 16 / 21 myCBSEguide farthest from the origin and has at least one common point to the feasible region OA3 PQB2. Clearly, P3 Q3 is such line and it passes through the vertex Q ( , ) of the feasible region. 8 12 5 5 Hence x = 8 5 and y = 12 5 gives the maximum value of Z. The maximum value of Z is given by Z = 5× + 7× = 24.8 8 12 5 5 33. Given that, R = {(1, 39), (2, 37), (3, 35).... (19, 3), (20, 1)} Domain = {1,2,3,.......,20} Range = {1,3,5,7......,39} R is not reflexive as (2, 2) ∉ R as 2 × 2 + 2 ≠ 41 R is not symmetric as (1, 39) ∈ R but (39, 1) ∉ R R is not transitive as (11, 19) ∈ R, (19, 3) ∈ R But (11, 3) ∉ R Hence, R is neither reflexive, nor symmetric and nor transitive. OR We observe the following properties of relation R. Reflexivity: Let (a, b) be an arbitrary element of N × N. Then, (a, b) ∈ N × N ⇒ a, b ∈ N ⇒ a + b = b + a [by commutativity of addition on N] ⇒ (a, b) R (a, b) Thus, (a, b) R (a, b) for all (a, b) ∈ N × N. So, R is reflexive on N × N. Symmetry: Let (a, b), (c, d) ∈ N × N. be such that (a, b) R (c, d) Then, (a, b) R (c, d) ⇒ a+d=b+c ⇒ c + b = d + a [By commutativity of addition on N] ⇒ (c, d) R (a, b) [By definition of R] Thus (a, b) R (c, d) ⇒(c, d) R (a, b) for all (a, b), (c, d) ∈ N × N. So, R is symmetric on N × N. Transitivity: Let (a, b), (c, d), (e, f) ∈ N × N such that (a, b) R (c, d) and (c, d) R (e, f). Then, (a, b)R(c, d) ⇒ a + d = b + c }⇒ (a + d) + (c + f) = (b + c) + (d + e) (c, d)R(e, f) ⇒ c + f = d + e ⇒ a + f = b + e ⇒ (a, b) R (e, f) Thus, (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N. So, R is transitive on N × N. Hence, R being reflexive, symmetric and transitive, is an equivalence relation on N × N. [(2, 3)] = {(x, y) ∈ N × N : (x, y) R (2, 3)} ⇒ [(2, 3)] = {(x, y) ∈ N × N : x + 3 = y + 2} = {(x, y) ∈ N × N: x - y = 1} = {(x, y) ∈ N × N: y = x + 1} = {(x, x + 1) : x ∈ N] = {(1, 2),(2, 3),(3, 4),(4, 5),..) {(7, 3)] = {(x, y) ∈ N × N : ( x , y ) R(7, 3)} Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 17 / 21 myCBSEguide = {{x, y) ∈ N× N : x + 3 = y + 7] = {(x,y) ∈ N × N : y = x - 4} = {(x, x - 4 ) ∈ N × N :x ∈ N] = {(5, 1), (6, 2), (7, 3), (8, 4), (9, 5),..} → 34. We have, ^ ^ ^ r = (8 + 3λ) i + (−9 − 16λ) j + (10 + 7λ) k ^ ^ ^ ^ ^ ^ = 8 i − 9j + 10k + 3λ i − 16λ j + 7λ k ^ ^ ^ ^ ^ ^ = 8 i − 9j + 10k + λ(3 i − 16j + 7k) → → ⇒ ^ ^ ^ a 1 = 8 i − 9j + 10k and b 1 ^ ^ ^ = 3 i − 16j + 7k.... (i) → Also, ^ ^ ^ ^ ^ ^ r = 15 i + 29j + 5k + μ(3 i + 8j − 5k) → → ⇒ ^ ^ ^ a 2 = 15 i + 29j + 5k and ^ ^ ^ b2 = 3 i + 8j − 5k....... (ii) ∣ → → ∣ → → ∣ ( b 1× b 2 )⋅( a 1− a 2) ∣ Now, shortest distance between two lines is given by → → ∣ ∣ ∣ ∣ ∣ b 1× b 2∣ ∣ ∣ ∣ ∣ ∣^ ^ ^ ∣ i j k → → ∣ ∣ ∴ b 1 × b 2 = ∣3 −16 7∣ ∣ ∣ ∣3 8 −5 ∣ ^ ^ ^ = i (80 − 56) − j (−15 − 21) + k(24 + 48) ^ ^ ^ = 24 i + 36j + 72k → → −−−−−−−−−−−−−−−−− ∣ ∣ Now, ∣ 2 2 2 b 1 × b 2 ∣ = √(24) + (36) + (72) ∣ ∣ − − −−−− −− −− 2 2 2 = 12√2 + 3 + 6 = 84 → → And ( a 2 ^ ^ ^ − a 1 ) = (15 − 8) i + (29 + 9) j + (5 − 10) k ^ ^ = 7 i + 38j − 5k ^ ^ ^ ^ ^ ∣ (24i+36 j +72k)⋅(7 i +38 j −5k) ∣ ∴ Shortest distance = ∣ 84 ∣ ∣ ∣ units 168+1368−360 1176 = ∣ ∣ = ∣ ∣ ∣ = 14 ∣ ∣ 84 ∣ 84 OR Equation of line in vector form Line I: r⃗ = (^ ı − ^ ^ ȷ + 0k) + λ(2 ^ ı + 0^ ^ ȷ + k) Line II: r ⃗ = ^ ^ ^ ^ ^ (2 f − j ) + μ( i + j − k) Here, → ^ a1 = ^ ı − ^ ȷ + 0k → a2 = 2 ^ ı − ^ ȷ → ^ ^ ^ b1 = 2 i + 0 j + k → ^ ^ ^ b2 = i + j − k We know that the shortest distance between lines is → → → → |(a2 −a1 )(b1 ×b2 )| d= → → |b1 ×b2 | → → ( a2 − a1 ) = (2^ı − ^ȷ ) − ( ^ı − ^ȷ + 0k ^ ) → → ^ ( a2 − a1 ) = ^ ı + 0^ ȷ + 0k ∣^ ^ ^ ∣ 1 ȷ k ∣ ∣ ⃗ b1 × b2 ⃗ = ∣2 0 1 ∣ ∣ ∣ ∣1 1 −1 ∣ Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 18 / 21 myCBSEguide ⃗ ⃗ ^ ^ ^ b1 × b2 = (0 − 1) i − (−2 − 1) j + (2 − 0) k ⃗ ⃗ ^ ^ ^ ⇒ b1 × b2 = − 1 + 3 j + 2k → → −−−−−−−−−−−−− 2 2 2 | b1 × b2 | = √(−1) + 3 + 2 ⃗ ⃗ −− ⇒ |b1 × b2 | = √14 → → → → |( a2 − a1 )( b1 × b2 )| = |( ^ı + 0^ȷ + 0k ^ )(− ^ ı + 3^ ^ ȷ + 2k)| → → → → ⇒ |( a2 − a1 )( b1 × b2 )| = 1 Substituting these values in the expression, ⃗ ⃗ |(a⃗2 −a⃗1 )(b1 ×b2 )| d= ⃗ ⃗ |b1 ×b2 | d= 1 √14 d= units 1 √14 Shortest distance d between the lines is not 0. Hence the given lines are not intersecting. 35. Let u = sin −1 ( 2x 2 ) 1+x Put x = tan θ −1 2 tan θ ⇒ u = sin ( ) 2 1+tan θ ⇒ u = sin-1(sin 2θ )...(i) 2 Let v = −1 1−x cos ( ) 2 1+x 2 −1 1−tan θ ⇒ v = cos ( ) 2 1+tan θ ⇒ v = cos-1(cos 2θ )...(ii) Here, 0 < x < 1 ⇒ 0 < tan θ < 1 π ⇒ 0 < θ < 4 So, from equation (i), −1 π π u = 2θ [ since , sin (sin θ) = θ, if θ ∈ (− , )] 2 2 ⇒ u = 2tan-1x...[Since, x = tan θ ] Differentiating it with respect to x, du = 2 2...(iii) dx 1+x from equation (ii), −1 v = 2θ [ since , cos (cos θ) = θ, ifθ ∈ [0, π]] ⇒ v = 2tan-1x...[Since, x = tan θ ] Differentiating it with respect to x, dv dx =...(iv) 2 2 1+x Dividing equation (iii) by (iv), du 2 =1 dx 2 1+x = × dv 1+x 2 2 dx To practice more questions & prepare well for exams, download myCBSEguide App. It provides complete study material for CBSE, NCERT, JEE (main), NEET-UG and NDA exams. Teachers can use Examin8 App to create similar papers with their own name and logo. Section E 36. i. If P is the rent price per apartment and N is the number of rented apartments, the profit is given by NP - 500 N = N(P - 500) [∵ ₹500/month is the maintenance charge for each occupied unit] Copyright © myCBSEguide.com. Mass distribution in any mode is strictly prohibited. 19 / 21 myCBSEguide ii. Let R be the rent price per apartment and N is the number of rented apartments. Now, if x be the number of non-rented apartments, then N(x) = 50 - x and R(x) = 10000 + 250x Thus, profit = P(x) = NR = (50 - x) (10000 + 250 x - 500) = (50 - x) (9500 + 250 x) = 250(50 - x) (38 + x) iii. We have, P(x) = 250(50 - x) (38 + x) Now, P'(x) = 250[50 - x - (38 + x)] = 250[12 - 2x] For maxima/minima, put P'(x) = 0 ⇒ 12 - 2x = 0 ⇒ x = 6 Number of apartments are 6. OR P′(x) = 250(12 - 2x) P′′(x) = - 500 < 0 ⇒ P(x) is maximum at x = 6 F ans Mats P lates A 40 50 20 ⎡ ⎤ 37. i. P = B ⎢ 25 40 30 ⎥ ⎣ ⎦ C 35 50 40 25 F ans ⎡ ⎤ Q = ⎢ 100 ⎥ M ats ⎣ ⎦ 50 P lates ii. Clearly, total funds collected by each school is given by the matrix 40 50 20 25 ⎡ ⎤⎡ ⎤ PQ = ⎢ 25 40 30 ⎥ ⎢ 100 ⎥ ⎣ ⎦⎣ ⎦ 35 50 40 50 1000 + 5000 + 1000 7000 ⎡ ⎤ ⎡ ⎤ =⎢ 625 + 4000 + 1500 ⎥ = ⎢ 6125 ⎥ ⎣ ⎦ ⎣ ⎦ 875 + 5000 + 2000 7875 ∴ Funds collected by school A is ₹7000. Funds collected by school B is ₹6125. Funds collected by school C is ₹7875. 25 F ans ⎡ ⎤ N ew price matrix Q = 20% × ⎢ 100 ⎥ M ats ⎣ ⎦ 50 P lates 25 + 25 × 0.20 F ans ⎡ ⎤ iii. ⇒ Q = ⎢ 100 + 100 × 0.20 ⎥ M ats ​ ⎣ ⎦ 50 + 50 × 0.20 P lates 30 F ans ⎡ ⎤ Q = ⎢ 120 ⎥ M ats ⎣ ⎦