MATHEMATICS-19-11- 11th (PQRS) SOLUTION

REVIEW TEST-6 / 5 Q.1QE Consider the quadratic polynomial f (x) = x2 – 4ax + 5a2 – 6a. (a) Find the smallest positive integral value of 'a' for which f (x) is positive for every real x. (b) Find the largest distance between the roots of the equation f (x) = 0. [2.5 + 2.5] [Ans. (a) 7, (b) a = 6] [Sol. (a) D < 0 16a2 – 4(5a2 – 6a) < 0 4a2 – 5a2 + 6a < 0 a2 – 6a > 0 a(a – 6) > 0  a > 6 or a < 0  smallest +ve integer = 7 Ans. (b) d = |  –  | ( – )2 = ( + )2 – 4 = 16a2 – 4(5a2 – 6a) = – 4a2 + 24a = – 4(a2 – 6a) = – 4[(a – 3)2 – 9] = 36 – 4(a – 3)2  |    |2 = 36 when a = 3 |  –  |max. = 6 Ans. ] Q.2(a) Find the greatest value of c such that system of equations x2 + y2 = 25 x + y = c has a real solution. (b) The equations to a pair of opposite sides of a parallelogram are x2 – 7x + 6 = 0 and y2 – 14y + 40 = 0 find the equations to its diagonals. [2.5+2.5] [Sol.(a)log put x = 5 cos  y = 5 sin  [Ans. (a) cmax = 5 2 ; (b) 5y + 6x = 56; 5y – 6x = 14]  5(cos  + sin ) = c; but (cos  + sin )max = and (cos  + sin )min = – hence, (b) x2 – 7x + 6 = 0 (x – 6)(x – 1) = 0 x = 1, x = 6 Ans. and y2 – 14y + 40 = 0 (y – 4)(y – 10) = 0 y = 4, y = 10 6 AC  y – 10 = – 5 (x – 1) 5y – 50 = – 6x + 6 6x + 5y = 56 (1) 6 BD  y – 4 = 5 (x – 1) 5y – 20 = 6x – 6 6x – 5y + 14 = 0 ....(2) ] Q.6s&p If the third and fourth terms of an arithmetic sequence are increased by 3 and 8 respectively, then the first four terms form a geometric sequence. Find (i) the sum of the first four terms of A.P. (ii) second term of the G.P. [2.5+2.5] [Ans. (i) 54; (ii) 18] [Sol. a, (a + d), (a + 2d), (a + 3d) in A.P. a, a + d, (a + 2d + 3), (a + 3d + 8) are in G.P. hence a + d = ar a  d also r = a a  2d  3 = a  d a  3d  8 = a  2d  3 using a  c  a  c  d  3 d  a  9 a b d d  5 = d  3 a 15 = a  9 b  d  d2 + 6d + 9 = d2 + 5d  d = – 9  a2 – 18a + 81 = a2 – 15a  3a = 81  a = 27 hence A.P. is 27, 18, 9, 0 G.P. is 27, 18, 12, 8 (i) sum of the first four terms of A.P. = Ans. (ii) 2nd term of G.P. = Ans. ] Q.7(a) Let x = 1 or x = – 15 satisfies the equation, log (kx2 + wx + f ) = 2. If k, w and f are relatively log 3 8 prime positive integers then find the value of k + w + f. (b)QE The quadratic equation x2 + mx + n = 0 has roots which are twice those of x2 + px + m = 0 and n m, n and p  0. Find the value of p . [Ans. (a) 96; (b) 8] [2.5+2.5] [Sol.(a) log8(kx2 + wx + f ) = 2  kx2 + wx + f = 64  kx2 + wx + f – 64 = 0 (1) also (1) is identical to (3x – 1)(x + 15)  kx2 + wx + f – 64 = 3x2 + 44x – 15 On comparing Take p = 1 k  w  3 44 f  64 15 = p since they are relatively prime numbers k = 3p  k = 3 w = 44p  w = 44 f = 49  k + w + f = 96 Ans. ] 2  (b) x2 + mx + n = 0 2 and x2 + px + n = 0  2( + ) = – m ....(1) 4  = n ....(2) and  +  = – p ....(3)  = m ....(4)  (1) and (3)  2p = m Q.10log Find the solution set of inequality, [Sol. x(x – 1) > 0 logx3 (x2  x) < 1. [5] Case-I: x + 3 > 1  x > – 2  x  (– 2, 0)  (1, ) Case-II: 0 < x + 3 < 1 – 3 < x < – 2 x2 – x > x + 3 x2 – x < x + 3 x2 – 2x – 3 > 0 x2 – 2x – 3 < 0 (x – 3)(x + 1) > 0 (x – 3)(x + 1) < 0  x  (–1, 0)  (1, 3)  x  (–3, – 2)  (–1, 0)  (1, 3) ]  x  (–3, – 2) Q.11s&p If the first 3 consecutive terms of a geometrical progression are the roots of the equation 2x3 – 19x2 + 57x – 54 = 0 find the sum to infinite number of terms of G.P. [Ans. 27 2 ] [5] [Sol. 2x3 – 19x2 + 57x – 54 = 0 [T/S, Q.4, Ex.2, seq and prog, to be put] a let the roots be r , a, ar 1 r  1  19  a  =  r 2 ....(1)  1  57 a2 1 r  r  = 2 ....(2)   a3 = 27 (3)  from (1) 3(r2 + r + 1) = 19r 2  6r2 + 6r + 6 = 19r  6r2 – 13r + 6 = 0  (2r – 3)(3r – 2) = 0  r = 2 3 or 3 2 (rejected)  Numbers are 9  3 a r , a, ar  27 9 2 , 3, 2 S = = 2 = 2 Ans. ] Q.12 Find the equation to the straight lines joining the origin to the points of intersection of the straight line x  y  1 and the circle 5(x2 + y2 + bx + ay) = 9ab. Also find the linear relation between a and b so that a b these straight lines may be at right angle. [3+2] [Ans. a = 2b or b = 2a] 2n1  S2 r1 2n1 = (r 1)2 r1 = 22 + 32 + 42 + +(2n – 1)2 + (2n)2 = 12 + 22 + 32 + + (2n)2 – 1 = sum of the square of the first (2n) natural numbers (2n)(2n 1)(4n 1) = 6 – 1 = Ans. ] A 5 Q.16 In any triangle if tan 2 = 6 B and tan 2 20 = 37 20 then find the value of tan C. [Ans. 21 ] [5]  A B  [Sol. tan  5  20 = 6 37 ; cot C 185  = 120 305 C 122 = ; tan =  2 2  1 5 20 6  37 2 222 100 122 2 305 244  tan C = 305  122 2 1  305  244 305 = (305 122)(305 122) 244  305 = 183 427 4  5 = 3 7 20 = 21 Ans. ]   Q.17 The radii r1, r2, r3 of escribed circles of a triangle ABC are in harmonic progression. If its area is 24 sq. cm and its perimeter is 24 cm, find the lengths of its sides. [5] [Ans. 6, 8, 10 cms] [REE '99, 6] [T/S, Q.5, Ex-3, ph-3] s  a [Sol.  s  b ,  s  c ,  are in A.P. a, b, c are in A.P. 2b = a + c adding b on both side 3b = a + b + c = 24  b = 8 2s = 24  s = 12 12(12  a)4(12 16  a) = 24  12 × 4(12 – a)(a – 4) = 24 × 24 – a2 + 16a – 48 = 12  a2 – 16a + 60 = 0  (a – 10)(a – 6) = 0 a = 10, a = 6 6, 8, 10 cms Ans.] Q.18cir Find the equation of a circle passing through the origin if the line pair, xy – 3x + 2y – 6 = 0 is orthogonal to it. If this circle is orthogonal to the circle x2 + y2 – kx + 2ky – 8 = 0 then find the value of k. [5] [Ans. x2 + y2 + 4x – 6y = 0; k = 1] [Sol. Line pair (x + 2)(y – 3) = 0 is orthogonal to required circle  centre of circle is (–2, 3) and it passes through (0, 0)  equation of circle is (x + 2)2 + (y – 3)2 = 13  x2 + y2 + 4x – 6y = 0