KEY FOR MATHS EXAMINATION PART - I 25. b d inverse axiom 27. d [3] 1. a 1 2. c k a 1 4. c

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1 MODEL HIGHER SECONDARY EXAMINATION KEY FOR MATHS PART - I Q. No. Key. a 2. c k - 3. a. c u = 0 5. a tan - /3 6. d abc 7. a (0, 0, -) 8. a (-, - 8) 2 9. a purely imaginary 0. a the straight line x = /. c c 2 tan c x = - x. c b 5. c 3 θ = 27t a - π 7. b an asymptote parallel to y-axis 8. b 9. b Answer b 2-2. a 20π 22. d -tan x Q. No. Key 25. b Answer 26. d inverse axiom 27. d [3] 28. c d a 3. c - P(x < a) 32. d Q - {0} 33. d 2, 2 d 3. b - x dy c 35. a 36. b 8/3 37. b q 38. c r = a 39. a π/ 0. c consistant 23. b A cos x + B sin x 2. b d 2 y dx 2 = 0

2 - -2 [ - -] 2 0 [ 0 2] 2 0 SECTION - B. A = 2 adj A = A (adj A) = = A I (adj A) A = = A I 0 2 A (adj A) = (adj A) A = A I A ~ 0 2 R R ~ R R R 3 R 3 + 2R ~ R 0 0 R 2 R 3 R 3-7R 2 ρ (A) = 3 i j k 3. (a) a x b = 2 3 = 8i - 0j + k -3-2 Area = a x b = (8) 2 + (-0) 2 + () 2 = 80 Area = 6 5 sq units (b) V = [ a b c ] -2 0 λ = 56 λ = r = x 2 + y 2 + z 2 r x 2 + y 2 + z 2-8x + 6y - 0z - 50 = 0 Centre = (, -3, 5) Radius = 0 units 2

3 5. Z + Z 2 Z + Z 2 Z + Z 2 2 = (Z + Z 2 ) (Z + Z 2 ) = Z 2 + Z Re (Z Z 2 ) Z 2 + Z Z Z 2 Z + Z 2 Z + Z 2 6. r = 2 θ = π/6 ( 3 + i) n = 2 n ( cosn π/6 + i sin n π/6) ( 3 - i) n = 2 n (cos n π/6 - i sin n π/6) ( 3 + i) n + ( 3 - i) n = 2 n + cos n π/6 7. a = 2 c 2 = 6 The centre is the midpoint of AA' = (, 3) Equation of the R.H with centre (, 3) is (x - ) (y - 3) = 6 The combined equation of the asymptotes is (x - ) (y - 3) = 0 Separate equations are x - = 0 and y - 3 = 0 8. Let f(x) = log e ( + x) ; f(o) = 0 f '(x) = + x ; f'(o) = f ''(x) = - ( + x) 2 ; f''(o) = - f '''(x) = 2 ( + x) 3 ; f '''(0) = 2 f(x) = f(0) + x x f '(0) + x f ''(0) + f ''' (0) +...! 2! 3! x 2 x 3 x log ( + x) = x Diagram b A = y dx a a b A = a 2 - x 2 dx 0 a A = πab sq. unit 50. (x - a) 2 + y 2 = (x - a) + yy' = 0 = (x - a) = -yy' y 2 [(y') 2 + ] = 3

4 5. p q r p q ~ r (p q) (~r) T T T T F T T T F T T T T F T F F F T F F F T T F T T F F F F T F F T T F F T F F F F F F F T T th Column 5 th Column 6 th Column 52. Truth table p q p q p q T T T T F F F T F F F T Truth table for ((~p) q) ((~q) p) p q ~ p ~ q ~ p q (~q) p ((~ p) q) (~q) p) T T F F T T T T F F T F T F F T T F T F F F F T T T T T 3 rd Column th Column 5 th Column 6 th Column 7 th Column p q ((~p) q) ((~q) p) e -λ λ x 53. P (X = x) = x! e -λ λ 2 e -λ λ 3 2! = 3! λ = 3 e -3 (3) 5 P (x = 5) = 5! = 0.0

5 5. P = 5 q = 5 P[Atleast 5 pass] = P(x 5) = P(x = 5) + P (x = 6) 208 P (X 5) = (a) x + y = 00 P = xy = x (00 - X) = 00x - x 2 P ' (x) = 00-2 x P '' = -2 < 0 P is maximum, when x = 50 y = 50 SECTION - C 56. The system of equations can be written as Ax = B [A, B] = 55. (b) dw δw dx δw dy δw dz = δx + δy + δz δw dx δx = ; = - sin t δw δy = 2; = cos t δw = 2z ; dz δy = dw = (-sint) + 2 cost + 2z = - sint + 2 cos t + 2t µ ~ 3 0 R 2 R 2 - R -µ 2 0 R 3 R 3-2R ~ 3 0 R 2 R 3 + R 2 0 -µ µ 0 Case (i) If µ 8, ρ [A] = ρ[a, B] = 3 The system has the trivial solution 5

6 Case (ii) If µ = 8, ρ(a) = ρ [A, B] = 2 < number of unknowns The given system is equivalant to x + y + 3z = 0; y + z = 0 Taking z = k, we get x = k, y = -k, z = k [ k R - {0}] which are non-trivial solutions 57. Diagram 58. OP = cos A i + sin A j OQ = cos B i + sin B j OQ x OP = Sin (A - B) k OQ x OP = k [sin A cos B - cos A sin B] sin (A - B) = sin A cos B - cos A sin B a = 2i + 2j - k, b = 3i + j + 2k, c = 7i + 6k r = ( - s - t) (2i + 2j - k) + s (3i + j + 2k) + t(7i + 6k) x - 2 y - 2 z + Cartesian equation of the plane is 2 3 = x + 2y - 3z = arg z - = π/2 r arg (z - ) - arg (z + 3) = π/2 z + 3 y y tan ( - - tan - = π/2 x - ) ( x + 3) y - y tan - x - x + 3 = π/2 y y + x - x + 3 y y r + = 0 x - x + 3 locus of P is x 2 + y 2 + 2x - 3 = Diagram 6

7 x 2 = -ay a = 9/0 x 2 = - x 9/0 y The point (x, -7.5) lies on the parabola -2 marks x = Diagram Closest distance F A = CF - CF = a - ae = 36 ( ) = 36 x 0.79 = million miles The forthest position F A = F C + CA = ae + a = a (e + ) = 36 ( ) = 36 x.206 = 3.6 million miles 62. lim x sinx is of the form 0 0 x 0 y = x sinx r logy = sin x log x log x log y = cosec x lim log y = lim log x which is of the type - x 0 + x 0 + cosec x = lim -sin 2 x (of the type 0 ) x 0 x cos x 0 lim x 0 + log y = 0 By composite function theorem, we have log lim y = 0 r lim y = e 0 = x 0 + x (i) Domain, extent, Intercept and origin : When x > 0, y is well defined, As x, y ± The curve exists in first and fourth quadrant only The intercepts with the axes are given by x = 0, y = 0 and when y = 0, x = 0 clearly the curve passes through origin (ii) Symmetry : The curve is symmetric about x -axis only 7

8 (iii) Asymptotes : The curve does not admit asymptotes (iv) Monotonicity : For the branch y = 2x /2 of the curve is increasing For the branch y = - 2 x 3/2 of the curve is decreasing (v) Special points : (0, 0) is not a point of inflexion Diagram : 6. Diagram dx dy = - 3a cos 2 t sint = 3a sin 2 t cos t dx 2 dy = 3a sint cost π/2 Length of the entire curve = 3a sint cost 0 = 6a 65. Diagram Equation of OB is y = 3/3 x r y = x V = π y 2 dx 0 = π x 2 dx 0 = 9π 66. Put x + y = z r + dy/dx = dz/dx z 2 r dz = dx + z 2 r ( - ) dz = dx + c + z 2 r z - tan - z = x + c r y - tan - (x + y) = c 8

9 67. da = KA A = ce kt - mark when t = 0, A = r C = A = e kt when t = 30, A = 60000, e 30k = 6/3 when t = 60, A = x e 60k The approximate population is 2020 is (i) closure axion : ( x x ) x x y y A = G, B = y y G AB = 2xy 2xy G ( therefore x 0, y 0 r 2xy 0) 2xy 2xy (ii) Matrix multiplication is always associative /2 /2 (iii) E = /2 /2 G (iv) A - = /x /x G /x /x G is a group under matrix multiplication 69. Given µ = 3 and σ = 0.2 (i) (a) when x = 30.5, Z = 2.5 when x = 3.5, Z = 2.5 Required probability p (30.5 < x < 3.5) = (b) when x = 30, z = 5 when x = 32, z = 5 p (30 < x < 32) = p ( 5 < z < 5 ) = (app) (ii) when x = 30.5, z = 2.5 p ( x > 30.5) = p (z > 2.5) = p ( 0 < z < 2.5) = =

10 70. (a) Point of intersection as (K 2/3, K /3 ) M = 2K /3 M 2 = K /3 But they are orthogonal and therefore M x M 2 = ) r ( = 2 K /3 K /3 8 k 2 = (b) The condition for y = mx + c to be a tangent to a x 2 y hyperbola - 2 = is c 2 = a 2 m 2 b 2 a 2 b 2 m = c = ¾ 5 2 a 2 = 9, b 2 = 9 c 2 =, a 2 m 2 b 2 = 6 c 2 = a 2 m 2 b 2 it touches the hyperbola 9 6 -a The point of contact is 2 m - b 2, c c - The point of contact is 5, 3 Entrance Coaching Programme for AIEEE, JIPMER, CENTAC & AIPMT Separate Batch for AIEEE (with material) 0

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola)

QUESTION BANK ON. CONIC SECTION (Parabola, Ellipse & Hyperbola) QUESTION BANK ON CONIC SECTION (Parabola, Ellipse & Hyperbola) Question bank on Parabola, Ellipse & Hyperbola Select the correct alternative : (Only one is correct) Q. Two mutually perpendicular tangents