Solutions for JMET Quant paper-1

Solutions for JMET Quant paper-1 - Solutions for JMET Quant...

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1 Solutions for JMET Quant paper-1 1. Since in A B = {3, 5}, the set B should contain the elements 3 and 5. The element 6, should not be present in B as it does not figure in A B. Now, the remaining elements of µ i.e., 1, 2, 4 and 7 may or may not belong to B 2 4 i.e., 16 such subsets are possible C h o i c e ( 1 ) 2. On solving the determinant, we get a linear equation, Hence the given determinant represents a straight line C h o i c e ( 4 ) 3. x 2 1 x 5 2 + = 0 2 4 20 5 2 x ± = x = 5 + 2, 5 2 Given, angle between these sides is 60 o . From cosine rule, cos A = bc 2 ² a ² c ² b + 1/2 = ) 1 ( 2 ² a 2 5 ( 2 5 ( + + 1 = 18 a 2 or a 2 = 17 i.e. a = 17 Choice (4) 4. Consider the given figure: Let CD be the cliff. tan α = CD/AC tan β = DE/EB = (CD – x) / AC CD – x = AC tan β CD – x = α × β tan CD tan CD α β α tan tan tan = x h = x α β β = β α α cot cot cot x tan tan tan Choice (2) 5. ( ) [ ] c b a c b . a = × = k j i l n m 1 1 1 n m l = l (l n) m (l m) + n (n m) = l 2 + m 2 + n 2 lm mn nl Choice (4) 6. We have f '(0) = 9 9 0 x ) 0 ( f ) x ( f lim 0 x = 9 h ) 0 ( f ) h ( f ) 0 ( f lim or 9 h ) 0 ( f ) h 0 ( f lim 0 x 0 h = = + ( Q f(x + y) = f(x) f(y)) f(0) 9 h 1 ) h ( f lim 0 h = (1) Now, f(x + y) = f(x) × f(y) for all x, y R f(0 + 0) = f(0) × f(0) = f(0) f(0) = 1 (2) from (1) and (2) 9 h 1 ) h ( f lim 0 h = Now, f '(5) = f(5) h 1 ) h ( f lim 0 h = 4 × 9 = 36 Alternate method : The property f(x + y) = f(x) . f(y) is satisfied by exponential functions. Let f(x) = a x , such that a x . a y = a x+y . Given a 5 = 4 and f '(0) = 9 f '(x) = loga . a x , so f '(0) = loga . a 0 = 9 or loga = 9 f '(5) = loga × a 5 = 9 × 4 = 36 Choice (2) 7. The powers of 7 end in 1, 7, 9 or 3, the number x and y should be selected in such a way that 7 x has 1 and 7 y has 9 in units place (or vice versa) or 7 x has 3 and 7 y has 7 in units place (or vice versa) Now the powers 7 1 , 7 5 , 7 6 ………(25 in number) have 7 in units place, similar is a case with the other numbers in units place. Required probability = 100 100 25 25 4 × × × = 4 1 C h o i c e ( 3 ) 8. dx e e e y 0 x cos x cos x cos + = π (1) ( ) () + = π π π π 0 x cos x cos x cos dx e e e y + = π 0 x cos x cos x cos e e dx e y (2) adding (1) and (2) we have π = 0 1 y 2 dx or y = π /2 Choice (3) 9. For the roots to be equal b 2 = 4ac If b = 1, b 2 = 1, ac = ¼ (no integral values of a, c) b = 2, b 2 = 4, ac = 1, a = 1, c = 1 b = 3, b 2 = 9, ac = 9/4 (no integral values of a, c) b = 4, b 2 = 16, ac = 4, (a = 1, c = 4), (a = 2, c = 2), (a = 4, c = 1) b = 5, b 2 = 25, ac = 25/24 (no integral values of a, c) b = 6, b 2 = 36 ac = 9, (a = 3, b = 3) Out of 216 possible equations only 5 satisfy the given condition.
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Solutions for JMET Quant paper-1 - Solutions for JMET Quant...

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