KENDRIYA VIDYALAYA SANGATHAN, CHENNAI REGION CLASS XII-COMMON PRE-BOARD EXAMINATION. Answer key (Mathematics) Section A

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1 KENDRIYA VIDYALAYA SANGATHAN, CHENNAI REGION CLASS XII-COMMON PRE-BOARD EXAMINATION Answer key (Mathematics) Section A. x =. x + y = 6. degree =. π x + y + z = Section B. Proving Reflexive Proving Symmetric Proving Transitive Conclusion. tan + tan 5 + tan 7 + tan 8 tan + 5 Χ 5 tan 7 + tan tan Χ + tan Χ 8 tan π Page of 5

2 OR tan x x + tan x+ x+ = π tan x x + tan x+ x+ = tan x tan x = tan x+ tan x+ By applying formula on the R.H.S. x tan x = tan x+ Applying tan both sides and solving x = ±. x y 7 = x + y mark log x y 7 = log x + y logx + 7 log y = log x + y Differentiating with respect to x y 7x = y 7x x x+y y x+y dy = y x dy mark Let x = cosθ cos x = θ Let y = tan +x x +x + x +cos θ cos θ y = tan +cos θ+ cos θ OR Page of 5

3 y = tan ( tanθ +tanθ ) y = (π θ) y = ( π cos x ) Let z = cos x y = π z dy dz =. For calculating LHL = 8 For calculating RHL = 8 For calculating K = 8 5. ex sin x cosx = ex sin xcos x cos x = e x [tanx sec x] = e x tanx + c 6. y x + = ln x + x Differentiating with respect to x y x + x + x + dy = x + x x + x xy + x + x + dy = x x + x + x + x xy + x + x + dy = x + x + dy = (+xy ) x + x + dy + xy + = Page of 5

4 7. x sinπx = x sinπx = x sinπx + x sinπx x sinπx On integrating both integrals on right-hand side, we get = xcosπx π = π + π + sinπx xcosπx π π + sinπx π 8. sin x sin x+α = sin x sinxcosα +cosxsinα = sin x cosα +cotxsinα = cose c x cosα +cotxsinα marks On substitution of cosα + cotxsinα = t = sinα t dt = t sinα + C On substitution of t t = cosα + cotxsinα = sinα sin x+α sinx + C Page of 5

5 9. a = i + j + k, b = i + j 5k, c = λi + j + k aχ b+c b+c = (i) b + c = + λ i + 6j k b + c = λ + λ (ii) aχ b + c = i j k + λ 6 By equation (i), (ii) & (iii) = 8i + + λ j + λ k (iii) 8i + +λ j + λ k λ +λ+ = On solving we will get λ = OR a + b + c = a + b = c a + b. a + b = c. c a + b + a. b = c By Substitution of values a. b = 9 a. b = 5 Page 5 of 5

6 a b cosθ = 5 By Substitution of values cosθ = θ = 6 mark. x+ = y+ = z+ 7 6 a = i j k a = i + 5j + 7k b = 7i 6j + k b = i j + k and x = y 5 = z 7 a a = i + 6j + 8k b Χ b = i 6j 8k b Χ b = 6 Shortest distance = a a. b Χ b b Χ b = 6 6 = 6 OR Equation of plane passing through,, is a x + b y + c z + = (i) (i) Passes through,, Page 6 of 5

7 a + b + 5c = (ii) (i) is perpendicular to x y + z = a b + c = (iii) On solving (ii) and (iii) a = 8, b = 7, c = From (i) equation of plane 8x + 7y + z 9 =. Let X is the random variable denoting the number of selected scouts, X takes values,,. P X = = C 5 C = 8 5 P X = = C Χ C 5 C = 5 P X = = C 5 C = 87 5 Now mean = P i X i = 9 Relevant Value 5 Page 7 of 5

8 . x x + y x + y x + y x x + y x + y x + y x Applying R R + R + R = (x + y) (x + y) (x + y) x + y x x + y x + y x + y x = x + y x + y x x + y x + y x + y x Applying C C C, C C C = (x + y) y y x + y y y x Expanding along R we get = 9y x + y Y x. Diagram Volume of the tank = 8m = xy----- (i) Let C be the cost of making the tank C = 7xy + 5 Χ (x + y) C = 7xy + 8(x + y) Page 8 of 5

9 From equation (i) C = 7x. x + 8(x + x ) C = x + x (ii) dc = 8 x For maxima and minima, dc = 8 x = x = as x -, x= Now d C Dx = 8 Χ 8 x d C Dx x= = 8 > C is minimum at x = By equation (ii) C x= = Rs. OR Area of ABC = Χ AB Χ DC = Χ bsinθ(a acosθ) = absinθ( cosθ)-----(i) Page 9 of 5

10 da dθ = ab(sin θ + cos θ cos θ) For maxima and minima da dθ = ab(sin θ + cos θ cos θ) = cosθ cosθ = cosθ = cosθ θ = nπ ± θ θ = nπ ± θ ----(ii) As θ (, π) by equation (ii) θ = π θ θ = π d A = ab(sinθ sinθ) dθ d A dθ θ= π < A is maximum. By equation (i) A max = ab sq.unit. Let x, y, z be the amount of prize to be awarded in the field of agriculture, education & social science respectively. The given situation Can be written in the matrix form as: AX = B Page of 5

11 Where A = , X = x y z, B = AX = B X = A B A = 75 adj A = A = adj A A = marks x y z = x y z = x =, y =, z = Value based relevant answer 5. Page of 5

12 Point of intersection of given curves is x = Required Area = x + 9 x =. x + x 9 x sin x = 9π sin OR I = x + x + x ---- (i) I = x + x + x Let I = x = ( x) + x = x + x = 5 I = x = ( x) + x = x + x = I = x = ( x) + x = x + x = 5 By equation (i) I = I + I + I = = Page of 5

13 6. Let x and y be the number of pieces of type A and B manufactured per week respectively. If z is the profit then, z = 8x + y Maximize z subject to the constraint 9x + y 8 x + y 6----(i) x + y ----(ii) x, y ----(iii) Mark mark Page of 5

14 Corner Points (,) A(,) B(,6) C(,) Relevant answer z = 8x + y 6 68 maximum 7. Let the event b defined as E = The examinee guesses the answer E = The examinee copies the answer E = The examinee knows the answer A = The examinee answers correctly P E = 6, P E = 9, P E = = 8 s P A E = P A E = 8 (Out of choices is correct) s s P A E = (If the answer is known it is always correct) s P E A = Required P E A = P E.P A E P E.P A E +P E.P A E +P E.P A E On substitution P E A = Yes the probability of copying is less than other probability. 8. The given planes are x + y 6z = ---- (i) 5x y + z + 9 = ---- (ii) Equation of the plane passing through the intersection of (i) and (ii) x + y 6z + λ 5x y + z + 9 = ---- (iii) mark Given that plane (iii) is parallel to x = y = z 5 + 5λ. + λ. + λ 6. = On solving λ = 8 7 On substitution of λ in (iii) equation of plane 5x 7y z + 5 = Page of 5

15 9. x dy + y x + y = dy xy +y = (i) x Let y = vx dy dv = v + x By equation (i) v + x dv = vx +v x x dv = v +v x dv = v +v x dv = v v+ x log v log v + = log x + C log vx vx v+ = log k = k v+ Putting v = y we get x x y = k y + x For Particular solution k = Therefore Particular Solution is x y = y + x Page 5 of 5

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