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MARKING SCHEME CLASS XII MATHEMATICS (CODE-041) SECTION: A (Solution of MCQs of 1 Mark each) Q no. ANS HINTS/SOLUTION 1. (D) For a square matri...

MARKING SCHEME CLASS XII MATHEMATICS (CODE-041) SECTION: A (Solution of MCQs of 1 Mark each) Q no. ANS HINTS/SOLUTION 1. (D) For a square matrix A of order n  n , we have A.  adj A   A I n , where I n is the identity matrix of order n  n.  2025 0 0   So, A.  adj A    0 2025 0   2025 I 3  A  2025 & adj A  A   2025  3 1 2  0 0 2025   A  adj A  2025   2025 . 2 2. (A) 3. (C) 𝑑𝑦 𝑦 = 𝑒𝑥 = > = 𝑒𝑥 𝑑𝑥 𝑑𝑦 In the domain (R) of the function, > 0 , hence the function is strictly increasing in (−∞, ∞) 𝑑𝑥   2 4. (B) 2 A  5, B 1 AB  B 1 A B  A  52. 2 5. (B) dy A differential equation of the form  f  x , y  is said to be homogeneous, if f  x , y  is a dx homogeneous function of degree 0. dy  y  dy y   y  Now, x n  y  log e  log e e    n  log e e.     f  x , y  ;  Let . f  x , y  will be a dx  x  dx x   x  homogeneous function of degree 0, if n  1. 6. (A) Method 1: ( Short cut) When the points  x1 , y1  ,  x2 , y2  and  x1  x2 , y1  y2  are collinear in the Cartesian plane then x1  x2 y1  y2 x1  x2 y1  y2  0    x1 y2  x2 y2  x2 y1  x2 y2   0 x1   x1  x2  y1   y1  y2   x2  y2  x2 y1  x1 y2. Page 1 of 15  Method 2: When the points  x1 , y1  ,  x2 , y2  and  x1  x2 , y1  y2  are collinear in the Cartesian plane then x1 y1 1 x2 y2 1 0 x1  x2 y1  y2 1  1.  x2 y1  x2 y2  x1 y2  x2 y2   1 x1 y1  x1 y2  x1 y1  x2 y1    x1 y 2  x2 y 1   0  x2 y1  x1 y2. 7. (A) 0 1 c  A   1 a  b   2 3 0  When the matrix A is skew symmetric then AT   A  aij  a ji ;  c  2; a  0 and b  3 So , a  b  c  0  3  2  1.     8. (C) 1 2 1 P A  ;P B  ;P  A  B  2 3 4 1 1  P  A  ; P  B   2 3 1 1 1 7 Wehave, P  A  B   P  A   P  B   P  A  B      2 3 4 12   7  A P A B P  A  B  1  P  A  B  1  12 5 P      .  B P B P B  P B   2 8   3 9. (B) For obtuse angle, cos 𝜃 < 0 => 𝑝⃗. 𝑞⃗ < 0 𝟐𝜶𝟐 − 𝟑𝜶 + 𝜶 < 𝟎 => 𝟐𝜶𝟐 − 𝟐𝜶 < 𝟎 => 𝜶 ∈ (𝟎, 𝟏) 10. (C) a  3, b  4, a  b  5 2 We have , a  b  a  b  2 a  b 2  2 2   2 9  16  50  a  b  5. 11. (B) Corner point Value of the objective function Z  4 x  3 y 1. O  0,0  z0 2. R  40,0  z  160 3. Q  30, 20  z  120  60  180 4. P  0,40  z  120 Since , the feasible region is bounded so the maximum value of the objective function z  180 is at Q  30,20. Page 2 of 15 12. (A) 𝑑𝑥 𝑑𝑥 ∫ 1 =∫ 1 𝑥 3 (1 + 𝑥 4 ) 2 𝑥 5 (1 + 1 2 ) 𝑥4 1 4 𝑑𝑥 1 ( Let 1 + 𝑥 −4 = 1 + = 𝑡, 𝑑𝑡 = −4𝑥 −5 𝑑𝑥 = − 𝑑𝑥 ⇒ = − 𝑑𝑡 ) 𝑥4 𝑥5 𝑥5 4 1 𝑑𝑡 1 = −4∫ 1 = − 4 × 2 × √𝑡 + 𝑐, where ' c ' denotes any arbitrary constant of integration. 𝑡2 1 1 1 = − 2 √1 + 𝑥 4 + 𝑐 = − 2𝑥 2 √1 + 𝑥 4 + 𝑐 13. (A) We know, 0 f  x  dx  0, if f  2a  x    f  x  2a Let f  x   cos ec7 x. Now, f  2  x   cos ec7  2  x    cos ec 7 x   f  x  2   cos ec 7 x dx  0; Using the property 0 f  x  dx  0, if f  2a  x    f  x . 2a 0 14. (B) ′ 𝑑𝑦 The given differential equation 𝑒 𝑦 = 𝑥 => 𝑑𝑥 = log 𝑥 𝑑𝑦 = log 𝑥 𝑑𝑥 => ∫ 𝑑𝑦 = ∫ log 𝑥 𝑑𝑥 𝑦 = 𝑥 log 𝑥 − 𝑥 + 𝑐 hence the correct option is (B). 15. (B) The graph represents y  cos 1 x whose domain is   1,1 and range is  0,  . 16. (D) Since the inequality Z  18 x  10 y  134 has no point in common with the feasible region hence the minimum value of the objective function Z  18 x  10 y is 134 at P  3,8 . 17. (D) The graph of the function 𝑓: 𝑅 → 𝑅 defined by f  x    x  ;  where . denotes G. I.F  is a straight line  x   2.5  h,2.5  h , ' h ' is an infinitesimally small positive quantity. Hence, the function is continuous and differentiable at x  2.5. 18. (B) The required region is symmetric about the y  axis. 4  3  y2 4 64 So, required area (in sq units ) is  2 2 ydy  4   . 3 3 0    2 0 19. (A) Both (A) and (R) are true and (R) is the correct explanation of (A). 20. (A) Both (A) and (R) are true and (R) is the correct explanation of (A). Section –B [This section comprises of solution of very short answer type questions (VSA) of 2 marks each] Page 3 of 15 21 𝜋 1 cot −1 (3𝑥 + 5) > = cot −1 1 4 2 1 =>3x + 5 < 1 ( as cot −1 𝑥 is strictly decreasing function in its domain) 2 => 3x < – 4 4 => 𝑥 < − 3 4 ⸫ 𝑥 ∈ (−∞, − ) 1 3 22. The marginal cost function is C '  x   0.00039 x 2  0.004 x  5. 1 C '  150   ₹ 14.375. 1 23.(a) y  tan1 x and z  loge x. dy 1 1 Then  dx 1  x 2 2 dz 1 1 and  dx x 2 dy dy dx  1 dz dz 2 dx So, 1  1 x  2 x 1. 1 1  x2 2 x OR Let y  (cos x ) x. Then, y  e x logecosx 23.(b) dy d  e x loge cos x ( x log e cos x ) 1 On differentiating both sides with respect to x , we get dx dx 2 dy  d d    (cos x ) x  log e cos x ( x )  x (log e cos x )  1 dx  dx dx  2 dy  1  dy   (cos x ) x  log e cos x  x. (  sin x )    (cos x ) x (log e cos x  x tan x ). 1 dx  cos x  dx 24.(a) ⃗⃗ + λc⃗ = (−1 + 3λ)î + (2 + λ )ĵ + k̂ We have b 1 2 ⃗⃗ + λc⃗). a⃗⃗ = 0 => 2(−1 + 3λ ) + 2 (2 + λ ) + 3 = 0 (b 1 1 5 2 λ = −8 OR 1 24.(b) ⃗⃗⃗⃗⃗⃗ = 𝑂𝐴 𝐵𝐴 ⃗⃗⃗⃗⃗⃗ = (4𝑖̂ + 3𝑘̂ ) − 𝑘̂ = 4𝑖̂ + 2𝑘̂ ⃗⃗⃗⃗⃗⃗ − 𝑂𝐵 2 Page 4 of 15  4 2 2 1 1 ̂ = 𝐵𝐴 𝑖̂ + 𝑘̂ = 𝑖̂ + 𝑘̂ 2√5 2√5 √5 √5 2 ⃗⃗⃗⃗⃗⃗ with the x , y and the z axes are respectively So, the angles made by the vector 𝐵𝐴 1 −1 2 𝜋 −1 1 𝑐𝑜𝑠 ( ) , 2 , 𝑐𝑜𝑠 ( ). √5 √5 25. ⃗⃗⃗⃗⃗ 𝑑1 = 𝑎⃗ + 𝑏⃗⃗ = 4𝑖̂ − 2𝑗̂ − 2𝑘̂ , ⃗⃗⃗⃗⃗ 𝑑2 = 𝑎⃗ − 𝑏⃗⃗ = −6𝑗̂ − 8𝑘̂ 1 2 1 1 𝑖̂ 𝑗̂ 𝑘̂ Area of the parallelogram = |𝑑1 × 𝑑2 | = ||4 −2 −2|| = 2|𝑖̂ + 8𝑗̂ − 6𝑘̂ | ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 1 2 2 0 −6 −8 1 Area of the parallelogram = 2√101 sq. units. 2 Section –C [This section comprises of solution short answer type questions (SA) of 3 marks each] 26. y 1 3 2 3 1 x 𝑥 2 + 32 = 𝑦 2 2 𝑑𝑥 𝑑𝑦 𝑊ℎ𝑒𝑛 𝑦 = 5 𝑡ℎ𝑒𝑛 𝑥 = 4, 𝑛𝑜𝑤 2𝑥 = 2𝑦 1 𝑑𝑡 𝑑𝑡 𝑑𝑦 𝑑𝑦 4 (200) = 5 => = 160 cm/s 1 𝑑𝑡 𝑑𝑡 1 27. 1 𝑑𝐴 1 1 1 𝐴 = 3 √𝑡 ∴ 𝑑𝑡 = 6 𝑡 −2 = 6 𝑡 ; ∀𝑡 ∈ (5,18) √ 𝑑𝐴 1 𝑑2 𝐴 1 1 𝑑𝑡 =6 𝑡 ∴ 𝑑𝑡 2 = − 12𝑡 √ √𝑡 1 𝑑2 𝐴 2 So, 𝑑𝑡 2 < 0, ∀𝑡 ∈ (5,18) This means that the rate of change of the ability to understand spatial concepts decreases 1 2 (slows down) with age. 28(a) ⃗⃗⃗⃗ 𝒍.𝒍⃗⃗⃗⃗ ̂).(3ı̂−2ȷ̂ + 𝑘 (ı̂−2ȷ̂+3k ̂) 1 (i) 𝜽 = 𝐜𝐨𝐬 −𝟏 ( ⃗⃗⃗⃗𝟏 𝟐 ) = 𝐜𝐨𝐬 −𝟏 (|(ı̂−2ȷ̂+3k̂)|| (3ı̂−2ȷ̂ + 𝑘̂)|) |𝒍𝟏 |.|𝒍⃗⃗⃗⃗ 𝟐| 1 𝟑+𝟒+𝟑 𝟏𝟎 𝟓 2 = 𝒄𝒐𝒔−𝟏 ( ) = 𝒄𝒐𝒔−𝟏 (𝟏𝟒) = 𝒄𝒐𝒔−𝟏 (𝟕). √𝟏+𝟒+𝟗√𝟗+𝟒+𝟏 ⃗⃗⃗⃗ 𝒍𝟏.𝒍⃗⃗⃗⃗ 𝟐 ̂).(3ı̂−2ȷ̂ + 𝑘 (ı̂−2ȷ̂+3k ̂) 1 (ii) Scalar projection of ⃗⃗⃗⃗ 𝒍𝟏 on ⃗⃗⃗⃗ 𝒍𝟐 = = |𝒍⃗⃗⃗⃗ 𝟐 | ̂ | (3ı̂−2ȷ̂ + 𝑘 )| 1 = 3+4+3 = 10. 2 √9+4+1 √14 Page 5 of 15 28(b) Line perpendicular to the lines 𝑟⃗ = 2ı̂ + ȷ̂ − 3k̂ + λ(ı̂ + 2ȷ̂ + 5k̂) and 𝑟⃗ = 3ı̂ + 3ȷ̂ − 7k̂ + μ(3ı̂ − 2ȷ̂ + 5k̂). 𝑖̂ 𝑗̂ 𝑘̂ has a vector parallel it is given by 𝑏⃗⃗ = ⃗⃗⃗⃗ 𝑏1 × ⃗⃗⃗⃗⃗ 𝑏2 = |1 2 5| = 20î + 10ĵ − 8k̂ 1 3 −2 5 ⸫ equation of line in vector form is 𝑟⃗ = − ı̂ + 2 ȷ̂ + 7k̂ + a(10ı̂ + 5ȷ̂ − 4k̂) 1 𝑥+1 𝑦−2 𝑧−7 And equation of line in cartesian form is = = 10 5 −4 1 29.(a) 1 1 ∫{ − } 𝑑𝑥 𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2 𝑑𝑥 1 1 𝑑 1 1 =∫ −∫ 2 𝑑𝑥 = ∫ 𝑑𝑥 − ∫ { ( ) ∫ 𝑑𝑥} 𝑑𝑥 − ∫ 𝑑𝑥 𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥) 𝑙𝑜𝑔𝑒 𝑥 𝑑𝑥 𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2 1 𝑥 1 1 1 = +∫. 𝑥. 𝑑𝑥 − ∫ 𝑑𝑥 1 𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2 𝑥 1 𝑑𝑥 𝑥 = +∫ 2 𝑑𝑥 − ∫ 2 = + 𝑐; 1 𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥) (𝑙𝑜𝑔𝑒 𝑥) 𝑙𝑜𝑔𝑒 𝑥 where′𝑐′is any arbitary constant of integration. OR 1  x  1  x  dx n 29.(b) 0 1 𝑎 𝑎 = ∫ (1 − 𝑥){1 − (1 − 𝑥)}𝑛 𝑑𝑥, (𝑎𝑠, ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥 ) 0 0 0 1 1 = ∫ 𝑥 𝑛 (1 − 𝑥)𝑑𝑥 0 1 1 1 = ∫ 𝑥 𝑛 𝑑𝑥 − ∫ 𝑥 𝑛+1 𝑑𝑥 2 0 0 1 1 1 = [𝑥 𝑛+1 ]1 0 − [𝑥 𝑛+2 ]0 1 𝑛+1 𝑛+2 2 1 1 1 = 𝑛+1 − 𝑛+2 = (𝑛+1)(𝑛+2). 1 30. The feasible region determined by the constraints, 2 x  y  3, x  2 y  6, x  0, y  0 is as shown. Page 6 of 15  1 The corner points of the unbounded feasible region are A( 6, 0) and B ( 0, 3). The values of Z at these corner points are as follows: Value of the objective function Corner point Z  x  2y 1 A( 6, 0) 6 B ( 0, 3) 6 We observe the region x  2 y  6 have no points in common with the unbounded feasible region. Hence 1 the minimum value of z  6. 2 It can be seen that the value of Z at points A and B is same. If we take any other point on the line x  2 y  6 such as (2,2) on line x  2 y  6, then Z  6. 1 Thus, the minimum value of Z occurs for more than 2 points, and is equal to 6. 2 31.(a) Since the event of raining today and not raining today are complementary events so if the probability that it rains today is 0.4 then the probability that it does not rain today is 1  0.4  0.6  P1  0.6 Page 7 of 15  If it rains today, the probability that it will rain tomorrow is 0.8 then the probability that it will not rain tomorrow is 1  0.8  0.2. If it does not rain today, the probability that it will rain tomorrow is 0.7 then the probability that it will not rain tomorrow is 1  0.7  0.3 (i) P1  P4  P2  P3  0.6  0.3  0.2  0.7  0.04. 1 (ii) Let E1 and E2 be the events that it will rain today and it will not rain today respectively. 1 P  E1   0.4 & P  E2   0.6 𝐴 𝐴 1 A be the event that it will rain tomorrow. 𝑃 (𝐸 ) = 0.8 & 𝑃 (𝐸 ) = 0.7 2 1 2 𝐴 𝐴 We have, 𝑃(𝐴) = 𝑃(𝐸1 )𝑃 (𝐸 ) + 𝑃(𝐸2 )𝑃 (𝐸 ) = 0.4 × 0.8 + 0.6 × 0.7 = 0.74. 1 2 1 The probability of rain tomorrow is 0.74. 2 1 OR Given 𝑃(𝑋 = 𝑟)𝛼 5𝑟 1 31.(b) 1 2 𝑃(𝑋 = 𝑟) = 𝑘 5𝑟 ( where k is a non-zero constant ) , 1 𝑃(𝑟 = 0) = 𝑘. 0 5 1 𝑃(𝑟 = 1) = 𝑘. 1 5 1 1 𝑃(𝑟 = 2) = 𝑘. 2 5 2 1 𝑃(𝑟 = 3) = 𝑘. 3 5 …………………………. …………………………. We have, 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)+............... = 1 1 2 Page 8 of 15  1 1 1 ⇒ 𝑘 (1 + + 2 + 3 +............. ) = 1 5 5 5 1 1 4 ⇒ 𝑘( 1) = 1 ⇒ 𝑘 = 2 1− 5 5 So, 𝑃(𝑋 < 3) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) 4 1 1 4 25 + 5 + 1 124 = (1 + + 2 ) = ( )=. 1 5 5 5 5 25 125 Section –D [This section comprises of solution of long answer type questions (LA) of 5 marks each] 32. 1 𝜋 𝜋 Required area = 20 ∫𝜋4 𝑐𝑜𝑠2𝑥 𝑑𝑥 + |20 ∫𝜋3 𝑐𝑜𝑠2𝑥 𝑑𝑥 | 1+1 6 4 𝜋 𝜋 sin 2𝑥 4 sin 2𝑥 3 1 = 20 [ ]𝜋 + |20 [ ] | 2 2 𝜋 6 4 1 √3 √3 √3 = 10 (1 − 2 )+ 10 (1 − 2 ) = 20 (1 − 2 ) sq. units. 33. y  ax 2  bx  c 15  4a  2b  c 1 25  16a  4b  c 15  196a  14b  c The set of equations can be represented in the matrix form as AX  B , 1 4 2 1 𝑎 15 4 2 1 𝑎 15 2 where 𝐴 = [ 16 4 1]’ 𝑋 = [𝑏] and 𝐵 = ⇒ [ 16 4 1] [𝑏] =. 196 14 1 𝑐 15 196 14 1 𝑐 15 1 1 |𝐴| = 4(4 − 14) − 2(16 − 196) + (224 − 784) = −40 + 360 − 560 = −240 ≠ 0. Hence A 2 exists. Page 9 of 15  −10 180 −560 𝑇 −10 12 −2 1 Now,𝑎𝑑𝑗(𝐴) = [ 12 −192 336 ] = [ 180 −192 12 ] −2 12 −16 −560 336 −16 𝑎 1 −10 12 −2 15 5 −10 12 −2 3 5 24 1 [𝑏 ] = − [ 180 −192 12 ] = − [ 180 −192 12 ] = − [−384] 𝑐 240 240 240 −560 336 −16 15 −560 336 −16 3 −48 1 1  a   , b  8, c  1 2 2 1 1 So, the equation becomes y   x 2  8 x  1 2 2 34.(a) 𝑥 3 ,if 𝑥 ≥ 0 We have, 𝑓(𝑥) = |𝑥|3 , { (−𝑥)3 = −𝑥 3 ,if𝑥 < 0 1 𝑓(𝑥)−𝑓(0) −𝑥 3 −0 2 Now, (𝐿𝐻𝐷 𝑎𝑡 𝑥 = 0) = 𝑙𝑖𝑚− 𝑥−0 = 𝑙𝑖𝑚− ( 𝑥 ) = 𝑙𝑖𝑚−(−𝑥 2 ) = 0 𝑥→0 𝑥→0 𝑥→0 1 𝑓(𝑥)−𝑓(0) 𝑥 3 −0 (𝑅𝐻𝐷 𝑎𝑡𝑥 = 0) 𝑙𝑖𝑚+ 𝑥−0 = 𝑙𝑖𝑚+ ( 𝑥 ) = 𝑙𝑖𝑚(−𝑥 2 ) = 0 2 𝑥→0 𝑥→0 𝑥→0 1 ∴ (𝐿𝐻𝐷 𝑜𝑓 𝑓(𝑥) 𝑎𝑡 𝑥 = 0) = (𝑅𝐻𝐷 𝑜𝑓 𝑓(𝑥) 𝑎𝑡 𝑥 = 0) 2 So, f  x  is differentiable at x  0 and the derivative of f  x  is given by 3𝑥 2 ,if𝑥 ≥ 0 1 𝑓′(𝑥) = { −3𝑥 2 ,if𝑥 < 0 1 𝑓′(𝑥)−𝑓′(0) −3𝑥 2 −0 Now, (𝐿𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0) = 𝑙𝑖𝑚− 𝑥−0 = 𝑙𝑖𝑚− ( 𝑥 ) = 𝑙𝑖𝑚−(−3𝑥) = 0 2 𝑥→0 𝑥→0 𝑥→0 𝑓′(𝑥)−𝑓′(0) 3𝑥 2 −0 1 (𝑅𝐻𝐷 𝑜𝑓𝑓′(𝑥) 𝑎𝑡 𝑥 = 0) = 𝑙𝑖𝑚+ = 𝑙𝑖𝑚+ ( ) = 𝑙𝑖𝑚+(3𝑥) = 0 𝑥→0 𝑥−0 𝑥→0 𝑥−0 𝑥→0 2 ∴ (𝐿𝐻𝐷 𝑜𝑓𝑓′(𝑥)𝑎𝑡 𝑥 = 0) = (𝑅𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0) 1 2 So, 𝑓′(𝑥)is differentiable at x  0. 1 2 6𝑥,if𝑥 ≥ 0 Hence, 𝑓′′(𝑥) = { −6𝑥,if𝑥 < 0. 1 2 OR Given relation is (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑐 2 , 𝑐 > 0. 34.(b) 1 Let x  a  c cos  and 𝑦 − 𝑏 = 𝑐 𝑠𝑖𝑛 𝜃. 2 𝑑𝑥 𝑑𝑦 Therefore, 𝑑𝜃 = −𝑐 𝑠𝑖𝑛 𝜃 And = 𝑐 𝑐𝑜𝑠𝜃 1 𝑑𝜃 2 𝑑𝑦 ∴ 𝑑𝑥 = − 𝑐𝑜𝑡 𝜃 1 𝑑 𝑑𝑦 𝑑 Differentiate both sides with respect to , we get 𝑑𝜃 (𝑑𝑥 ) = 𝑑𝜃 (− 𝑐𝑜𝑡 𝜃) 1 2 Page 10 of 15  𝑑 𝑑𝑦 𝑑𝑥 Or, 𝑑𝑥 (𝑑𝑥 ) 𝑑𝜃 = 𝑐𝑜𝑠 𝑒 𝑐 2 𝜃 1 𝑑2 𝑦 2 Or, (−𝑐 𝑠𝑖𝑛 𝜃) = cosec 𝜃 2 𝑑𝑥 2 1 𝑑2𝑦 𝑐𝑜𝑠𝑒𝑐 3 𝜃 = − 2 𝑑𝑥 2 𝑐 3 𝑑𝑦 2 2 3 3 1 [1+( ) ] 𝑐[1+𝑐𝑜𝑡 2 𝜃]2 − 𝑐(𝑐𝑜𝑠 𝑒𝑐 2 𝜃)2 𝑑𝑥 ∴ 𝑑2 𝑦 = = = −𝑐, − 𝑐𝑜𝑠 𝑒𝑐 3 𝜃 cosec 3 𝜃 𝑑𝑥2 1 Which is constant and is independent of a and b. 2 35.(a) Given that equation of lines are 𝑟⃗ = (−𝑖̂ − 𝑗̂ − 𝑘̂ ) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘̂ )................ (𝑖) and 𝑟⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘̂ ) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘̂ ).................. (𝑖𝑖) The given lines are non-parallel lines as vectors 7𝑖̂ − 6𝑗̂ + 𝑘̂and 𝑖̂ − 2𝑗̂ + 𝑘̂ are not parallel. There is a unique line segment PQ ( P lying on line  i  and Q on the other line  ii  ), which is at right angles to both the lines PQ is the shortest distance between the lines. Hence, the shortest possible distance between the lines  PQ. Let the position vector of the point P lying on the line𝑟⃗ = (−𝑖̂ − 𝑗̂ − 𝑘̂ ) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘̂ ) where '  ' 1 is a scalar, is (7𝜆 − 1)𝑖̂ − (6𝜆 + 1)𝑗̂ + (𝜆 − 1)𝑘̂ , for some  and the position vector of the point Q 2 1 lying on the line 𝑟⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘̂ ) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘̂ )where '  ' is a scalar, is 2 (𝜇 + 3)𝑖̂ + (−2𝜇 + 5)𝑗̂ + (𝜇 + 7)𝑘̂ , for some . Now, the vector ⃗⃗⃗⃗⃗⃗ 𝑃𝑄 = 𝑂𝑄 𝑂𝑃 = (𝜇 + 3 − 7𝜆 + 1)𝑖̂ + (−2𝜇 + 5 + 6𝜆 + 1)𝑗̂ + (𝜇 + 7 − 𝜆 + 1)𝑘̂ ⃗⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ = (𝜇 − 7𝜆 + 4)𝑖̂ + (−2𝜇 + 6𝜆 + 6)𝑗̂ + (𝜇 − 𝜆 + 8)𝑘̂ ; (where ' O ' is the origin), is 1 𝑖. 𝑒. , 𝑃𝑄 ⃗⃗⃗⃗⃗⃗ is perpendicular to both the vectors 7𝑖̂ − 6𝑗̂ + 𝑘̂ and perpendicular to both the lines, so the vector 𝑃𝑄 𝑖̂ − 2𝑗̂ + 𝑘̂.  (𝜇 − 7𝜆 + 4). 7 + (−2𝜇 + 6𝜆 + 6). (−6) + (𝜇 − 𝜆 + 8). 1 = 0 Page 11 of 15  &(𝜇 − 7𝜆 + 4). 1 + (−2𝜇 + 6𝜆 + 6). (−2) + (𝜇 − 𝜆 + 8). 1 = 0  𝟐𝟎𝝁 − 𝟖𝟔𝝀 = 𝟎 => 𝟏𝟎𝝁 − 𝟒𝟑𝝀 = 𝟎&6𝜇 − 20𝜆 = 0 ⇒ 3𝜇 − 10𝜆 = 0 1 On solving the above equations, we get     0 1 So, the position vector of the points P and Q are −𝑖̂ − 𝑗̂ − 𝑘̂ and 3𝑖̂ + 5𝑗̂ + 7𝑘̂ respectively. 2 1 𝑃𝑄 = 4𝑖̂ + 6𝑗̂ + 8𝑘̂ and ⃗⃗⃗⃗⃗⃗ 2 ⃗⃗⃗⃗⃗⃗ | = √42 + 62 + 82 = √116 = 2√29 𝑢𝑛𝑖𝑡𝑠. |𝑃𝑄 1 OR 35.(b) Let P 1, 2 , 1 be the given point and L be the foot of the perpendicular from P to the given line AB (𝑎𝑠 𝑠ℎ𝑜𝑤𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒 𝑎𝑏𝑜𝑣𝑒). x  3 y 1 z 1 Let’s put    . Then, x    3, y  2  1, z  3  1 1 2 3 1 2 Let the coordinates of the point L be    3,2  1,3  1. So, direction ratios of PL are(𝜆 + 3 − 1,2𝜆 − 1 − 2,3𝜆 + 1 − 1)𝑖. 𝑒. , (𝜆 + 2,2𝜆 − 3,3𝜆) 1 Direction ratios of the given line are 1, 2 and 3, which is perpendicular to PL. Therefore, we have, 2 2 (𝜆 + 2). 1 + (2𝜆 − 3). 2 + 3𝜆. 3 = 0 ⇒ 14𝜆 = 4 ⇒ 𝜆 = 1 7 2 2 23 2 3 2 13 Then, 𝜆 + 3 = 7 + 3 = 7 ; 2𝜆 − 1 = 2 (7) − 1 = − 7 ; 3𝜆 + 1 = 3 (7) + 1 = 7 1 23 3 13 2 Therefore, coordinates of the point L are ( 7 , − 7 , 7 ). Let 𝑄(𝑥1 , 𝑦1 , 𝑧1 )be the image of P 1, 2 , 1 with respect to the given line. Then, L is the mid-point 1 of PQ. 1+𝑥1 23 2+𝑦1 3 1+𝑧1 13 39 20 19 Therefore, 2 = 7 , 2 = −7, 2 = 7 ⇒ 𝑥1 = 7 , 𝑦1 = − 7 , 𝑧1 = 7 Hence, the image of the point P  1, 2,1 with respect to the given line 𝑄 ( , − 39 20 19 , ). 1 7 7 7 The equation of the line joining P  1, 2,1 and 𝑄 ( 7 , − 39 20 19 7 , 7 )is Page 12 of 15  𝑥−1 𝑦−2 𝑧−1 𝑥−1 𝑦−2 𝑧−1 1 = = ⇒ = =. 32/7 −34/7 12/7 16 −17 6 Section –E [This section comprises solution of 3 case- study/passage based questions of 4 marks each with two sub parts. Solution of the first two case study questions have three sub parts (i),(ii),(iii) of marks 1,1,2 respectively. Solution of the third case study question has two sub parts of 2 marks each.) 36. (i) 𝑉 = (40 − 2𝑥)(25 − 2𝑥)𝑥𝑐𝑚3 1 𝑑𝑉 (ii) = 4(3𝑥 − 50)(𝑥 − 5) 1 𝑑𝑥 𝑑𝑉 𝟏⁄ (iii) (a) For extreme values = 4(3𝑥 − 50)(𝑥 − 5) = 0 𝟐 𝑑𝑥 50 𝟏⁄ ⇒𝑥= 3 or 𝑥 = 5 𝟐 𝑑2 𝑉 = 24𝑥 − 260 𝟏⁄ 𝑑𝑥 2 𝟐 𝑑2 𝑉 ∴ 𝑑𝑥 2 at 𝑥 = 5 is − 140 < 0 𝟏⁄ 𝟐 ∴ 𝑉 is max 𝑤ℎ𝑒𝑛 𝑥 = 5 (iii) OR 𝟏⁄ 𝑑𝑉 (b) For extreme values 𝑑𝑥 = 4(3𝑥 2 − 65𝑥 + 250) 𝟐 𝑑2 𝑉 𝟏⁄ = 4(6𝑥 − 65) 𝟐 𝑑𝑥 2 𝑑𝑉 65 𝑑2 𝑉 65 𝑎𝑡 𝑥 = exists and 𝑑𝑥 2 𝑎𝑡 𝑥 = 𝑖𝑠 0. 𝑑𝑥 6 6 𝟏⁄ 𝑑2𝑉 65 − 𝑑2𝑉 65 + 𝟐 𝑎𝑡 𝑥 = ( ) is negative and 𝑎𝑡 𝑥 = ( ) is positive 𝑑𝑥 2 6 𝑑𝑥 2 6 65 ⸫𝑥= is a point of inflection. 𝟏⁄ 6 𝟐 Number of relations is equal to the number of subsets of the set B  G  2   37. n BG (i) n B   n G  1 2  23 2  26 ( 𝑾𝒉𝒆𝒓𝒆𝒏(𝑨) 𝒅𝒆𝒏𝒐𝒕𝒆𝒔 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒇𝒊𝒏𝒊𝒕𝒆 𝒔𝒆𝒕 A ) (ii) Smallest Equivalence relation on G is {(𝒈𝟏 , 𝒈𝟏 ), (𝒈𝟐 , 𝒈𝟐 )} 1 (iii) (a) (A) reflexive but not symmetric = {(𝒃𝟏 , 𝒃𝟐 ), (𝒃𝟐 , 𝒃𝟏 ), (𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 )}. Page 13 of 15  So the minimum number of elements to be added are (𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 ) 1 {Note : it can be any one of the pair from, (𝒃𝟑 , 𝒃𝟐 ), (𝒃𝟏 , 𝒃𝟑 ), (𝒃𝟑 , 𝒃𝟏 ) in place of (𝒃𝟐 , 𝒃𝟑 ) 𝒂𝒍𝒔𝒐} (B) reflexive and symmetric but not transitive = {(𝒃𝟏 , 𝒃𝟐 ), (𝒃𝟐 , 𝒃𝟏 ), (𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 ), (𝒃𝟑 , 𝒃𝟐 ) }. 1 So the minimum number of elements to be added are (𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 ), (𝒃𝟑 , 𝒃𝟐 ) OR (iii) (b) One-one and onto function 𝟐 𝒙𝟐 𝒙 = 𝟒𝒚. let𝒚 = 𝒇(𝒙) = 𝟒 𝒙𝟏 𝟐 𝒙𝟏 𝟐 Let 𝒙𝟏 , 𝒙𝟐 ∈ [𝟎, 𝟐𝟎√𝟐] such that 𝒇(𝒙𝟏 ) = 𝒇(𝒙𝟐 ) ⇒ = 𝟒 𝟒 1 ⇒ 𝒙𝟏 𝟐 = 𝒙𝟐 𝟐 ⇒ (𝒙𝟏 − 𝒙𝟐 )(𝒙𝟏 + 𝒙𝟐 ) = 𝟎 ⇒ 𝒙𝟏 = 𝒙𝟐 as 𝒙𝟏 , 𝒙𝟐 ∈ [𝟎, 𝟐𝟎√𝟐] ∴ 𝒇 is one-one function Now, 𝟎 ≤ 𝒚 ≤ 𝟐𝟎𝟎 hence the value of 𝒚 is non-negative and 𝒇(𝟐√𝒚) = 𝒚 ∴ for any arbitrary 𝒚 ∈ [𝟎, 𝟐𝟎𝟎], the pre-image of 𝒚 exists in [𝟎, 𝟐𝟎√𝟐] 1 hence 𝒇 is onto function. 38. Let E1 be the event that one parrot and one owl flew from cage –I 𝐸2 be the event that two parrots flew from Cage-I A be the event that the owl is still in cage-I (i) Total ways for A to happen From cage I 1 parrot and 1 owl flew and then from Cage-II 1 parrot and 1 owl flew back + From cage I 1 parrot and 1 owl flew and then from Cage-II 2 parrots flew back + From cage I 2 parrots flew and then from Cage-II 2 parrots came 1 back. 2 =(5𝐶1 × 1𝐶1 )(7𝐶1 × 1𝐶1 ) + (5𝐶1 × 1𝐶1 )(7𝐶2 ) + (5𝐶2 )(8𝐶2 ) Probability that the owl is still in cage –I = P(𝐸1 ∩ 𝐴) + P(𝐸2 ∩ 𝐴) (5𝐶1 × 1𝐶1 )(7𝐶1 × 1𝐶1 ) + (5𝐶2 )(8𝐶2 ) 1 (5𝐶1 × 1𝐶1 )(7𝐶1 × 1𝐶1 ) + (5𝐶1 × 1𝐶1 )(7𝐶2 ) + (5𝐶2 )(8𝐶2 ) 35 + 280 315 3 1 = = = 35 + 105 + 280 420 4 2 Page 14 of 15 (i) The probability that one parrot and the owl flew from Cage-I to Cage-II given 1 2 that the owl is still in cage-I is 𝑃 (𝐸1⁄𝐴) 1 𝐸 P(𝐸1 ∩ 𝐴) 𝑃 ( 1⁄𝐴) = P(𝐸1 ∩ 𝐴)+P(𝐸2 ∩ 𝐴) (by Baye’s Theorem) 2 35 420 1 = 315 = 1 9 420 Page 15 of 15