EAD Mathematics -12 Sample Paper 1

sample Paper 1 [Easy Concept] Time Allowed : 3 hrs.] [Max. Marks : 80 General Instructions: 1. This question paper contains two parts A and B. Each part is compulsory. Part A carries 24 marks and Part B carries 56 marks 2. Part-A has Objective Type Questions and Part-B has Descriptive Type Questions 3. Both Part A and Part B have choices. Part – A: 1. It consists of two sections-I and II. 2. Section I comprises of 16 very short answer type questions. 3. Section II contains 2 case studies. Each case study comprises of 5 case-based MCQs. An examinee is to attempt any 4 out of 5 MCQs. Part – B: 1. It consists of three sections-III, IV and V. 2. Section III comprises of 10 questions of 2 marks each. 3. Section IV comprises of 7 questions of 3 marks each. 4. Section V comprises of 3 questions of 5 marks each. 5. Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions. Part – A Section – I All questions are compulsory. In case of internal choices attempt any one. 1. Check whether the function f : R → R defined as f(x) = x2 is one-one or not. OR How many one-one relations are possible in set A = {1, 2, 3}? 2. Given a relation R in set A = {1, 2, 3}. Is relation R = {(1, 1), (2, 2), (2, 3)} reflexive? 3. A relation R in set of natural numbers is defined as R = {(x, y) : y = x }. Is R a function in N? OR Under what conditions a relation is said to be an equivalence relation? omfaotrrdixerTSRSSS−3241×WWWWVX RTSSSS1053VXWWWW.lI.f 4. Find the order of |A| = 3, then find |A.Adj A|. 5. Given a matrix A OR If matrix A = cos a sin aaG, then find the value of a for which A is an identity matrix. =– sin a cos 6. Find the value of x, for which the matrix 5 – x x + 1 is singular. = 2 4 G y 7. If x2 dx = x + f(x) + C, then find f(x). OR x2 + 1 Evaluate y 1 dx. 1 – 4x2 52 Together with®  EAD Mathematics—12

8. Find the area bounded by the curve y = 1 , the x-axis and between x = 1 and x = 3. x 9. If differential equation corresponding to a function y = Aex + Be–x, A and B being arbitrary constants is formed then what is order of differential equation? OR Find the value of 2m – n, if m and n represent degree and order of the differential equation d2y + x = 0. dx2 10. Find a vector of magnitude 3 units along the direction of vector 3 a . 11. Evaluate |2 b × a |, if it is given that a = 2it – tj + 3kt and b = 3tj + kt . 12. Find a vector in direction of vector 2it – tj + 2kt and of magnitude 3. 13. Write the equation of a line passing through point (0, 1, 3) and parallel to the line r = it − tj + kt + λ (2it − 3kt). 14. Find the direction ratios of line 2x = 3y = 5 – 4z. 15. If odds in favour of an event A are 10 : 11, then find the probability of event A. 16. A four digit number is formed by using the digits 1, 2, 3, 5 with no repetition. Find the probability that number is divisible by 5. Section – II Both the Case study based questions are compulsory. Attempt any 4 sub parts from each question. Each sub- part carries 1 mark. 17. To conserve water for emergency needs in a village, the gram panchyat decides to construct an open tank with a square base from a sheet of metal which has been donated by a company, from the account of social welfare account. The area of metal sheet donated is to be least and water to be stored is 4000 m3. h y Metal Sheet x Based on the above information answer the following: (i) If x, y and h represent the sides of the base and hight of the tank respectively then area A of metal sheet used is (a) 2(x + y) × h = A (b) (2x + y) × h = A (c) x + y + h = A (d) x2 + 4xh = A (ii) The volume of water is (a) x2yh = 4000 (b) xy2h = 4000 (c) xyh2 = 4000 (d) x2h = 4000 (iii) Area of metal sheet in terms of x only is (a) A = 500 (b) A = 6000 (c) A = 2x + 2000 (d) A = x2 + 16000 x x x2 x (iv) Area of metal sheet used is least when side of base is (a) 10 m (b) 20 m (c) 15 m (d) 200 m (v) Least area of metal sheet used is (a) 1200 m2 (b) 1000 m2 (c) 900 m2 (d) 1400 m2 18. In the times of COVID-19, testing is very important. On testing the probability that a person is diagnosed correctly, when a person is actually suffering from COVID-19 is 0.99. The probability that doctor diagnoses incorrectly that a person is suffering from covid-19 is 0.001. In a certain city it was deducted that there is 0.001 chance that a person suffer from covid-19. Sample Papers 53

Based on the above information answer the following: (i) If the population of city is 200,000. How many persons are expected to suffer from covid-19? (a) 2000 (b) 200 (c) 20 (d) 2 (ii) What is the probability that a person is diagnosed correctly for covid-19? (a) 0.99 (b) 0.001 (c) 0.01 (d) 0.999 (iii) The probability of a person diagnosed having covid-19 is (a) 0.1989 (b) 0.00099 (c) 0.000999 (d) 0.001989 (iv) The probability that a person actually has a COVID-19, when he is diagnosed to have COVID-19 (a) 110 (b) 10 (c) 11 (d) 100 221 221 221 221 (v) The probability that error occurred in diagnosing the COVID-19 is (a) 0.99 (b) 0.01 (c) 0.999 (d) 0.001 Part – B Section – III All questions are compulsory. In case of internal choices attempt any one. 19. Express sin–1f 2 sin x +3 cos x p, in the simplest form. 13 SRx + y + zWV SR9VW 20. Find the values of x, y and z, if S x+ z W = SSTS75WWWX. SS y+ z WW T X OR If A is a square matrix such that A2 = A, then write the value of (I + A)2 – 3A. Z]] 21. Find the value of k, such that the function f(x) = [ x2 – 255, x ! 5 is continuous at x = 5. ] x – \\ 2k, x=5 22. If tangent to the curve y + 3x2 + 7 = 0 at the point (h, k) is parallel to the line y + 6x – 1 = 0, then find the value of h. em tan−1x dx . y 23. Evaluate 1 + x2 OR Evaluate y sin x sin 3x dx 24. Find the area of the region bounded by the curve y = x2 and the line y = 2 in first quadrant. 25. Find the particular solution of the differential equation dy = y tan x, given that y = 1 when x = 0. dx 26. Find the area of a parallelogram whose one side and one diagonal are along the vectors it + 2 tj – kt and 2 tj + 5 kt respectively. 27. Find the direction cosines of line x − 1 = y + 3 ; z = 1. 2 3 28. A pair of coins is tossed once. Event E is “getting at least one head” and event F is “getting at most one head”. Are the events E and F independent? OR A die is thrown twice and the sum of the numbers appearing is observed to be 7. What is the conditional probability that the number 2 has appeared at least once? 54 Together with®  EAD Mathematics—12

Section – IV All questions are compulsory. In case of internal choices attempt any one. 29. Show that the relation S in the set R of real numbers, defined as S = {(a, b) : a, b ∈ R and a ≤ b3} is neither reflexive, nor symmetric, nor transitive. 30. If x16 y9 = (x2 + y)17, prove that dy = 2y . dx x 31. If x = a(2q – sin 2q) and y = a(1 – cos 2q), find dy when q = π . dx 3 OR Show that the function f (x) = |x – 3|, x ∈ R is continuous but not differentiable at x = 3. 32. Find the intervals in which the function f given by f(x) = 2x3 – 9x2 + 12x + 15 is strictly increasing or strictly decreasing. y 33. Evaluate 2| x3 − x| dx . –1 34. Find the area of the region enclosed between the parabola y2 = 4ax and the line y = mx. OR Using method of integration find the area bounded by the curve | x | + | y | = 1. 35. Solve the differential equation, dy = x5tan–1(x3). dx Section – V All questions are compulsory. In case of internal choices attempt any one. SR1 – 1 0WV SR 2 2 – 4WV 36. Given A f=roSSSTm20 3 eq24uWWXWaatinodnBs. = SSS– 4 2 – 45WWXW verify that BA = 6I, how we can use the result to find the values of x, y, z 1 T 2 –1 given x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7. SR2 5VW OR If A = TSSS31 24WXWW, –3 – find A–1. How we can use A–1 to find x, y, z for the following system of equations: 2 – 1 2x – 3y + 5z = 16; 3x + 2y – 4z = – 4; x + y – 2z = –3 37. Find the equation of the plane determined by the points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6). Also find the distance of the point P(6, 5, 9) from the plane. OR Find the equation of a line passing through the point (1, 2, –4) and perpendicular to two lines r = (8it – 19tj + 10kt) + λ (3it – 16tj + 7kt) and r = (15it + 29tj + 5kt) + µ (3it + 8tj – 5kt). 38. Solve the following problem graphically: Minimise and Maximise Z = 3x + 9y subject to the constraints: x + 3y ≤ 60 x + y ≥ 10 x ≤ y x ≥ 0,  y ≥ 0 Sample Papers 55

OR The feasible region of the system of linear constraints is given as Answer each of the following (i) Find the constraints for the LPP taking x-along x-axis and y-along y-axis. (ii) If Z = 400x + 300y be objective function, find value of maximum Z and the point of maximum. Solutions 1. Not one-one as let x1 = 2 and x2 = –2 Then f(x1) = 4, f(x2) = 4 \\  x1 ≠ x2 but f(x1) = f(x2). Hence, not one-one. OR R : {1, 2, 3} → {1, 2, 3} For one-one relation x1 ≠ x2 ⇒ f(x1) ≠ f(x2) or f(x1) = f(x2)  ⇒  x1 = x2 \\  Total one-one relations are 3 × 2 × 1 = 6. 2. Relation R is set A is not reflexive as for a ∈ A, (a, a) ∉ R, e.g. (3, 3) ∉ R but 3 ∈ A. 3. Given R : N → N defined as R = {(x, y) : y = x } This is not a function as for x = 2, there is no y ∈ N (co-domain) such that y = f(x). OR A relation is an equivalence relation if it is reflexive, symmetric and transitive. SR 2VW RS0WVl SR 2VW 4. Order is 3 × 3 as SSST−41XWWW TSSS15WXWW = SSST−41WXWW[0 1 5] 5. As |A . Adj A| = |A|3 = (3)3 = 27. OR As =–csoisnaa csoinsaaG = 1 0 =0 1G ⇒ cos a = 1, sin a = 0 ⇒ a = 0° 56 Together with®  EAD Mathematics—12

6. As matrix is singular, then 5– x x+1 =0 2 4 ⇒  20 – 4x – 2x – 2 = 0  ⇒ 6x = 18  ⇒  x = 3 7. As y y y yx2 = (x2 + 1) – 1 dx = 1 dx = x – tan–1x + C ⇒ x2 +1 x2 + 1 x2 + 1 dx 1 $ dx – f(x) = – tan–1x OR y 1 dx = y 1 dx Let 2x = t 1 – 4x2 1 – (2x)2 ⇒ 2dx = dt y = 21 1 t2 dt = 21 sin–1t + C = 1 sin–1(2x) + C 1– 2 8. Area = y 3 y dx y 1 y= 1 y = 13 1x 3 x dx = >log | x |H 1 = log 3 – log 1 = log 3 sq units 01 3x 9. As there are two arbitrary constants. Hence, differential equation is of order 2. OR Given equation is d2y +x=0 dx2 Order (n) = 2,  degree (m) = 1 \\ 2m – n = 2 × 1 – 2 = 2 – 2 = 0. 10. Direction along vector 3 a is given by at. \\ Required vector is 3 at. 11. As given vectors are a = 2it – tj + 3kt and b = 3tj + kt it tj kt 2 b × a = 0 6 2 = 20it + 4 tj − 12 kt = 4(5it + tj − 3kt) 2 −1 3 |2 b × a | = 4 25 + 1 + 9 = 4 35 12. As vector is 3 at, where a = 2it − tj + 2kt = 3e 2 it − 1 tj + 2 kt o = 2it − tj + 2kt . 3 3 3 13. As required line is parallel to the given line so direction is along vector (2 it – 3 kt ) and passes through point (0, 1, 3) \\ line is r = ( tj + 3 kt ) + m(2 it – 3 kt ). 14. As line is 2x = 3y = 5 – 4z LCM (2, 3, 4) = 12, Now 2x = 3y = − 4e z – 5 12 12 12 4o x y z – 5 6 4 4 ⇒ = = is the line. –3 \\  Direction ratios are 6, 4, –3. Sample Papers 57

15. As for odds in favour of A we have ratio P(A) : P(A), which is given as 10 : 11 \\ P(A) = 10 = 10 . 10 + 11 21 16. As total possibilities for four digit number = n(s) = 4! Favourable possibilities for number to be divisible by 5, n(A) = 3! × 1 = 3! \\ Required probability = n (A) = 3! = 1 . n (S) 4! 4 17. As base is a square and tank is open at the top. Hence x = y. (i) (d), as area of metal sheet A = area of base + area of four sides = x × x + 4x × h A = x2 + 4xh (ii) (d), as volume of cuboid = l × b × h = x × x × h = x2h = 4000 (iii) (d), as we eliminate h, h = 4000 x2 4000 16000 \\ A = x2 + 4x . x2 = x2 + x (iv) (b), For least area dA =0 dx ⇒ 2x – 16000 = 0 ⇒  x3 = 8000  ⇒  x = 20 m x2 d2A > 0 dx2 d2A = 2 + 32000 > 0  for x = 20 dx2 x3 (v) (a), We substitute x = 20 Least area is A = x2 + 16000 x A = 400 + 16000 = 400 + 800 = 1200 m2 20 18. (i) (b), as 0.001 are expected to suffers from COVID-19 \\ Number of persons out of 200,000 = 200000 × 0.001 = 200 (ii) (a), according to the statement. (iii) (d), Probability that a person has COVID-19 = 1 × 99 + 999 × 1 = 0.001989 1000 100 1000 1000 (iv) (a), Using Baye’s theorem 1 × 99 990 110 1000 100 1989 221 P robability = = = 0.001989 (v) (d), as a statement 19. Consider sin–1 2 sin x + 3 cos x = sin−1 > 2 sin x + 3 cos xH f 13 p 13 13 Let 2 = cos a, 3 = sin a [from (i)] 13 13 58 Together with®  EAD Mathematics—12

We notice cos2a + sin2 a = 1, True sin−1 2 sin x +3 cos x p = sin–1 [sin x cos a + cos x sin a] = sin–1 [sin (x + a)] f 13 = x + a = x + cos–1 2  [from (i)] 13 20. x + y + z = 9, x + z = 5, y + z = 7, on solving we get x = 2, y = 4, z = 3 OR (I + A)2 – 3A = I2 + IA + AI + A2 – 3A = I + A + A + A – 3A = I 21. If function is continuous at x = 5, then lim f(x) = f(5) x\"5 x2 25 ⇒ lim x − 5 = 2k  ⇒  10 = 2k  ⇒  k = 5. − x\"5 22. As curve is y + 3x2 + 7 = 0 dy dy dx dx ⇒ + 6x = 0  ⇒  = –6x ⇒ ddxyG(h, k) = –6h (slope of the tangent) If tangent is parallel to the line y + 6x – 1 = 0 then –6h = –6  ⇒  h = 1 Let m tan–1 x = t em tan−1x 1 1 1 1 + x2 m m m y y 23. Consider dx = et dt = e t + C = e m tan –1x + C m dx dt ⇒ x2 = 1 OR + Consider y sin x sin 3x dx = 1 y 2 sin 3x sin x dx = 1 y (cos 2x − cos 4x) dx = 1 sin 2x − sin 4x + C. 2 2 2= 2 4G 24. Required area is shaded area = y 2 x dy = y 2 y dy y y = x2 00 = >23 y3/2G02 = 2 .2 2 2 3 y=2 = 4 32 sq units Ox 25. y dy = y tan x dx ⇒ log |y| = log|sec x| + log C y ⇒ y = C sec x...(i) Given y = 1, x = 0 ⇒ 1 = C sec 0 ⇒ C = 1 ∴  solution is y = sec x [from (i)] 26. Area of parallelogram = |( it + 2 tj – kt ) × (2 tj + 5 kt )| D 2tj + 5tk C A it + 2 tj – kt B it tj kt Consider ( it + 2 tj – kt ) × (2 tj + 5 kt ) = 1 2 − 1 = |12 it – 5 tj + 2 kt | 173 sq units. 02 5 \\ Area of parallelogram = |12 it – 5 tj + 2 kt | = 144 + 25 + 4 = 27. As line is x − 1 = y + 3 ; z = 1, i.e. x−1 = y+3 z−1 2 3 2 3= 0 \\ Direction ratios are 2, 3, 0 Dividing by 4 + 9 + 0 = 13 \\ Direction cosines are 2 , 3 , 0. 3 13 Sample Papers 59

28. Pair of coins is tossed \\ n(S) = 4 E : Getting at least one head = {HT, TH, HH} F : Getting at most one head = {TT, HT, TH}  E ∩ F : {HT, TH} 3 3 2 4 4 4 P(E) = , P(F) = , P(E ∩ F) =  As 3 × 3 ≠ 2 4 4 4  i.e. P(E) . P(F) ≠ P(E ∩ F) Hence, events E and F are not independent. OR n(S) = 36 A :  sum of numbers appeared is 7, i.e. (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1); n(A) = 6 B : number 2 has appeared at least once, i.e. (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6); n(B) = 11 A ∩ B : (2, 5), (5, 2); n(A ∩ B) = 2 2 \\ P(B/A) = P] A ∩ Bg = 36 = 1 . 6 3 P (A) 29. Given S = {(a, b) ∈ R × R | a ≤ b3} 36 We can consider counter example. For reflexive: Let (–2, –2) ∈ S ⇒ –2 ≤ (–2)3 ⇒ –2 ≤ –8, false, Hence, not reflexive. For symmetric: Let (–1, 2) ∈ S ⇒ –1 ≤ (2)3 ⇒ –1 ≤ 8 true, If symmetric then (2, –1) ∈ S ⇒ 2 ≤ (–1)3 ⇒ 2 ≤ –1, false, Hence, not symmetric. For transitive: Let (25, 3) ∈ S and (3, 2) ∈ S ⇒ 25 ≤ (3)3 and 3 ≤ (2)3 ⇒ 25 ≤ 27 and 3 ≤ 8, true in both cases. If transitive then (25, 2) ∈ S ⇒ 25 ≤ (2)3 ⇒ 25 ≤ 8, false Hence, not transitive. 30. Consider x16  y9 = (x2 + y)17 ⇒ 16 log x + 9 log y = 17 log (x2 + y) (on taking log on both sides) Differentiating both sides w.r.t. x, we get 16 + 9 $ dy = 17⋅ 1 y f2x + dy p ⇒ x y dx x2 + dx ⇒ 16 + 9 $ dy = 34x + 17 y $ dy x y dx x2 + y x2 + dx 9 − 17 y p dy = 34 x − 16 fy x2 + dx x2 + y x 60 Together with®  EAD Mathematics—12

⇒ * 9x2 + 9y − 17y 4 $ dy = 34x2 − 16x2 − 16y y(x2 + y) dx x(x2 + y) ⇒ * 9x2 – 8y $ dy = 18x2 − 16y dx x(x2 + y) y(x2 + y)4 ⇒ dy = 2(9x2 − 8y) y = 2y . dx x(9x2 − 8y) x 31. Consider x = a(2q – sin 2q) and y = a(1 – cos 2q) dx = a(2 – 2 cos 2q) and dy = a(2sin 2q) dθ dθ = 2a(1 – cos 2q)     = 2a sin 2q = 2a(2sin2 q) = 4a sin2q \\ dy = dy ÷ dx = 2a sin 2q = 2 sin q cos q = cot q dx dθ dθ 4a sin2q 2 sin2q \\ dy G π = cot π = 1. dx 3 3 3 θ = OR Given function f (x) = |x – 3| = *− x − 3, x$3 x + 3, x<3 For continuity at x = 3, LHL = lim f (3 – h) = lim {–(3 – h) + 3} x=3 h\"0 h\"0 = lim  h = 0 h\"0 RHL = lim f (3 + h) = lim {(3 + h) – 3} x=3 h\"0 h\"0 = lim  h = 0 h\"0 f (3) = 3 – 3 = 0 As LHL = RHL = f (3), x=3 x=3 Hence, function is continuous at x = 3. For differentiability at x = 3, LHD = lim f (3 − h) − f (3) lim (− 3 + h + 3) − (0) −h = −h x=3 h\"0 h\"0 = lim h = lim (− 1) = − 1 −h h\"0 h\"0 RHD = lim f (3 + h) − f (3) lim (3 + h − 3) − (0) h = h x=3 h\"0 h\"0 = lim h = lim (1) = 1 h h\"0 h\"0 As LHD ≠ RHD. Hence, function is not derivable (differentiable) at x = 3. x=3 x=3 32. Consider f(x) = 2x3 – 9x2 + 12x + 15 f ′(x) = 6x2 – 18x + 12 = 6(x2 – 3x + 2) = 6(x – 1) (x – 2) ...(i) For critical points f ′(x) = 0 ⇒ x = 1, 2 Sample Papers 61

x<1 1<x<2 x>2 6+ + + x–1 – ++ x–2 – – + f′(x) + – + ↑↓↑ Strictly increasing for (–∞, 1) ∪ (2, ∞), strictly decreasing in (1, 2) y 33. Consider  2 x3 – x dx x3 – x = 0 x(x2 – 1) = 0 Now, –1 ⇒ ⇒  x(x – 1) (x + 1) = 0 ⇒  x = 0, –1, 1 –1 0 1 2 x3 – x is positive for –1 < x < 0, x3 – x is negative for 0 < x < 1, x3 – x is positive for 1 < x < 2, y y y y \\ 2 x3 – x dx = 0 (x3 – x) dx – 1(x3 – x) dx + 2 (x3 – x) dx –1 –1 0 1 = > x44 – x2 0 – x4 – x2 1 + x4 – x2 2 2 >4 H >4 H H 2 2 0 1 –1 1 1 1 1 16 4 1 1 1 1 1 3 11 = (0 – 0) – e4 – 2o – e4 – 2 o + (0 – 0) + e4 – 2o – e4 – 2o = 4 + 4 + 2 + 4 = 2 + 4 = 4 34. Given curves are y2 = 4ax and y = mx, plotting the graph of curves we notice that we have to find the shaded area. Eliminating y from the equations, we get (mx)2 = 4ax ⇒ m2x2 = 4ax ⇒ m2 x2 – 4ax = 0 ⇒ x(m2x­ – 4a) = 0 y1 : y2 = 4ax ⇒ x = 0, x = 4a y2 : y = mx m2 ∴ 4a Y yArea = m2 (y1 – y2)dx y = mx 0 4a y = m2 _ 4ax – mxidx 0 y2 = 4ax = >2 a  ·  2 x3/2 – mx2 4a 3 2 Hm2 3 2 0 = 4 3 a 4a p m 4a 2 4a X m2 2 f m2 p O m2 f – = 32a2 – 8a2 = 8a2 sq units 3m3 m3 3m3 OR Curves are x + y = 1, –x – y = 1, – x + y = 1, x – y = 1 Area = 4 × area in 1st quadrant y = 4 1(1 – x)dx = 4>x − x2 1 H 0 2 0 1 = 4d1 − 2 n – 0 = 2 sq units 62 Together with®  EAD Mathematics—12

35. Consider equation, dy = x5tan–1(x3) ⇒ dx dy = x5tan–1(x3)dx Integrating both sides, we get y y ydy = x5 tan–1(x3) dx = x3 $ x2 tan–1(x3) dx Let x3 = t ⇒ 3x2dx = dt 1 yy = 3 t · tan–1t dt 21 y y = 13>tan–1t t2 1 t2 (t2 + 1) – 1 dt · 2 – 1 + t2 · 2 dtH = 61 t2 tan–1t – 1 1 + t2 6 y y = 16 t2tan–1t – 1 1 · dt + 1 1 dt = 61 t2tan–1t – 1 t + 1 tan –1 t + C 6 6 + 6 6 1 t2 = 16 x6 · tan–1(x3) – 1 x3 + 1 tan –1(x3) + C. 6 6 RS 2 2 − 4VW RS1 − 1 0WV 36. Consider BA = TRSSSS− 242 2 − 45WXWW STSS20 3 24XWWW −1 1 0 8WV + 4− 0 −2 + 6 − 4 + 8 − = SSS− 0WV00−+48+−108XWWW TSR6 4 +4 −0 4+6−4 2 −2 +0 −2 − 3 + 5 0VW 0 RS1 0 BA = SSST00 60XWWW = 6SSST00 10WWXW = 6I...(i) 6 1 0 0 Given equations are x – y = 3 2x + 3y + 4z = 17 y + 2z = 7 SR1 0VW RSxVW SR 3VW −1 24WWW S zyWWW = SSS177WWW Matrix equation is SSST20 3 SS ⇒ 1 XT X T X From (i), we have AX = C  ⇒  X = A–1 C. ...(ii) BA = 6I  ⇒  B = 6IA– 1 ⇒ A– 1 = 1 B 6 Now we can substitution A–1 in (ii) and solve for x, y, z OR Given 2 –3 5 | A| = 3 2 –4 1 1 –2 2 –3 5 | A| = 3 2 –4 = 2(0) + 3(–2) + 5(1) = –1 ≠ 0 1 1 –2 SR 0 2 1WVT RS0 –1 2VW Adj A = TSSS–21 –9 –153XWWW = TSSS21 –9 1233WWWX 23 –5 Sample Papers 63

A–1 = | 1 1 SR0 –1 2VW SR 0 1 − 2VW A| adj A = − 1 SSST21 –9 1233XWWW = SSTS−−21 9 − 2133XWWW...(i) Consider equations, –5 5 − 2x – 3y + 5z = 16 3x + 2y – 4z = –4 x + y – 2z = –3 Corresponding matrix equation is SRSSS zxyWXVWWW = SSSSTR––1643XWWWVW SR2 5WV T SSTS31 –3 – 24WXWW 2 – 1 i.e., AX = B, we can write to find x, y, z is X = A–1B ...(ii) Now as we have A–1 we can substitute from (i) in (ii) and solve for x, y, z. 37. Plane through the points A(3, – 1, 2), B(5, 2, 4) and C(–1, –1, 6) is x−3 y+1 z−2 5 − 3 2 + 1 4 − 2 = 0 −1 − 3 −1 + 1 6 − 2 x−3 y+1 z−2 ⇒ 2 3 2 = 0 −4 0 4 ⇒ 12(x – 3) – 16(y + 1) + 12(z – 2) = 0 ⇒ 12x – 36 – 16y – 16 + 12z – 24 = 0 ⇒ 12x – 16y + 12z – 76 = 0 ⇒ 3x – 4y + 3z – 19 = 0 is the required equation of the plane. Distance of the point P(6, 5, 9) from the plane 3x – 4y + 3z – 19 = 0 is 18 − 20 + 27 − 19 = 6 6 units 9 + 16 + 9 34 = 34 OR Let line through the point (1, 2, –4) be r = (it + 2tj − 4kt) + λlm , l′ is a scalar ...(i) Line (i) is perpendicular to lines r = (8it − 19tj + 10kt) + λ (3it − 16tj + 7kt) and r = (15it + 29tj + 5kt) + µ(3it + 8tj − 5kt) \\ (3it − 16tj + 7kt) $ m = 0 and (3it + 8tj − 5kt) $ m = 0 tj kt ⇒ − 16 7 = 24it + 36tj + 72kt it m = 3 8 −5 3 \\ From (i) line is r = (it + 2tj − 4kt) + λl (24it + 36tj + 72kt) or r = (it + 2tj − 4kt) + λll (2it + 3tj + 6kt) where l″ = 12l′, is a scalar 64 Together with®  EAD Mathematics—12

38. Plotting the inequations x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0. We notice common shaded portion is the feasible solution. Possible points for maximum and minimum z are A(5, 5), B(15, 15), C(0, 20), D(0, 10) Points Z = 3x + 9y Values ← Minimum A(5, 5) 15 + 45 60 ← Maximum B(15, 15) 45 + 135 180 ← Maximum C(0, 20) 0 + 180 180 D(0, 10) 0 + 90 90 Minimum Z is at A(5, 5), i.e. x = 5, y = 5, Minimum Z = 60. Maximum Z is at B(15, 15), i.e. x = 15, y = 15 and C(0, 20), i.e. x = 0, y = 20, Maximum Z = 180. OR (i) BC passes through (0, 200) and (200, 0). \\ Equation of BC is x + y = 1 ⇒ x + y = 200 200 200 Equation of AC is x = 20, as AC | | the y-axis and x-coordinate of A is 20. Equation of AB is y = 4x as line passes through O(0, 0) and A(20, 80). Maximum Z = 64000 at B(40, 160). Taking feasible solution into account the constraints are x ≥ 20,  y ≥ 0 x + y ≤ 200 y ≥ 4x (ii) On solving we get coordinates of B as (40, 160) and those of C as (20, 180). Points z = 400x + 300y Values A(20, 80) 8,000 + 24,000 32,000 B(40, 160) 16,000 + 48,000 64,000 ← Maximum C(20, 180) 8,000 + 54,000 62,000 Sample Papers 65