MATHEMATICS-13-08- 11th (PQRS) SOLUTION

REVIEW TEST-3 Class : XI (PQRS) Time : 100 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain 15 questions and 24 pages. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. 2. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 3. Use of Calculator, Log table and Mobile is not permitted. 4. Legibility and clarity in answering the question will be appreciated. 5. Put a cross ( × ) on the rough work done by you. 6. Last page is an Extra page. You may use it for any unfinished question with a specific remark: "continued on back page". Name Father's Name Class : Batch : B.C. Roll No. Invigilator's Full Name For Office Use ……………………………. Total Marks Obtained………………… Q.No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Marks XI(PQRS) MATHEMATICS REVIEW TEST-3 Q.1 The sum of the first five terms of a geometric series is 189, the sum of the first six terms is 381, and the sum of the first seven terms is 765. What is the common ratio in this series. [4] [Ans. 2] [Sol. S5 = 189; S6 = 381; S7 = 765 t6 = S6 – S5 = 381 – 189 = 192 t7 = S7 – S6 = 765 – 381 = 384 t7 now common ratio = t6 384 = 192 = 2 Ans. ] Q.2 Form a quadratic equation with rational coefficients if one of its root is cot218°. [4] 1+ cos 36 1+ 5 +1 (5 + 5 )(3 + 5 ) [Sol. cot218° = 1 cos 36 = 4 = 1 5 +1 4 (3  5 )(3 + 5 ) 20 + 5 5 + 3 = 9  5 = 20 + 8 4 = 5 + 2 Hence if  = 5 + 2  = 5  2   +  = 10;  = 25 – 20 = 5  quadratic equation is x2 – 10x + 5 = 0 Ans. ] Q.3 Let  and  be the roots of the quadratic equation (x – 2)(x – 3)+(x – 3)(x + 1)+(x + 1)(x – 2)=0. Find the value of 1 ( +1)( +1) 1 + (  2)(  2) 1 + (  3)(  3) . [4] [Sol. Quadratic equation is 3x2 – 8x + 1 = 0  +  = 8 1 3 and  = 3 1 1 1 to find the value of  + ( + ) +1 + 1 3   2( + ) + 4 +   3( + ) + 9 = 4 – 1 + 4 3 3 = – 4 + 4 = 0 Ans. ] Q.4 If a sin2x + b lies in the interval [–2, 8] for every x  R then find the value of (a – b). [4] [Sol. f (x) = a sin2x + b [Ans. 12] f (x) has a maximum value of 8 which occurs when sin2x = 1  a + b = 8 (1) |||ly f (x) has a minimum value of – 2 which occurs where sin x = 0  b = – 2 (2) from (1) and (2) a = 10; b = – 2  a – b = 12] Q.5 For x  0, what is the smallest possible value of the expression log(x3 – 4x2 + x + 26) – log(x + 2)? [4] [Sol. log log (x3  4x2 + x + 26) (x + 2) (x 2  6x2 +13)(x + 2) (x + 2) [Ans. log 4] log (x2 – 6x + 13) [ x  – 2] log(x – 3)2 + 4  Minimum value of log 4 when x = 3 ] Q.6 The coefficients of the equation ax2 + bx + c = 0 where a  0, satisfy the inequality (a + b + c)(4a – 2b + c) < 0. Prove that this equation has 2 distinct real solutions. [4] [Sol. ax2 + bx + c = 0 (a  0) given (a + b + c)(4a – 2b + c) < 0 f (1) = a + b + c f (–2) = 4a – 2b + c f (1) · f (–2) < 0 since f (1) and f (–2) have opposite signs  exactly one root lies between – 2 and 1  this quadratic equation has 2 distinct real roots.] Q.7s&p In an arithmetic progression, the third term is 15 and the eleventh term is 55. An infinite geometric progression can be formed beginning with the eighth term of this A.P. and followed by the fourth and second term. Find the sum of this geometric progression upto n terms. Also compute S if it exists. [5] [Sol. Given a + 2d = 15 (1) a + 10d = 55 (2) solving (1) and (2)  8d = 40 d = 5 & a = 5  t8 = 40 & t4 = 20 t2 = 10 hence the G.P. is 40, 20, 10, .......  ( | r | < 1) hence S exists. 40  S = 1 (1 2) = 80 Ans. 401 1   S =  2n  = 801 1  Ans. ] n 1 1 2  2n  Q.8 Find the solution set of this equation log|sin x|(x2 – 8x + 23) > log|sin x|(8) in x  [0, 2]. [5] [Sol. Note that x2 – 8x + 23 > 0  x  R x  n and x  (2n + 1)  2 | sin x | < 1 hence above inequality holds if  x  0, , 2,   3 ,  2   x2 – 8x + 23 < 8 x2 – 8x + 15 < 0 (x – 3)(x – 5) < 0 , 3   3 ,5  x  (3, )   2    2  ]      Q.9 Find the positive integers p, q, r, s satisfying tan 24 = ( – q )( – s). [5]  [Sol. Solving using the Identity tan 2 1 cos = sin   where  = 12  1 cos15 4 4  ( 6 + 2 ) [4  ( + 2)][ + 2] tan 24 = sin15 = 6  2 = 4 6  2 = 4 = 4( + 2)  (8 + 4 3) =( 4 + ) – ( 2 + ) = (  3 ) (2  2 ) = 3( 1)  2 ( 1) = ( – 2 )( 1) hence p = 3, q = 2; r = 2; s = 1 ] Q.10 Find the sum to n terms of the series. 1 + 2 + 3 + 4 + 5 + ........ 2 4 8 16 32 Also find the sum if it exist if n  . [Ans. Sn = 2n+1  n  2 2n ; S = 2] [5] [Sol. S = 1 + 2 + 3 + 4 + 5 +........ + n ....(1) 2 4 8 16 32 2n S = 1 + 2 + 3 + .............. + n 1 + n ....(2) 2 4 8 16 2n 2n+1 – – – – – – —————————————————— S = 1 + 1 2 2 22 + 1 + 1 23 24 + ........ + 1 2n – 2n+1 1 1 1   2n  = – n n+1 = 1 1  n 1 1 2 2  2n+1  2  n  2n 2n+1 2n+1  2  n S = 2   In n   2n+1   =  2n 1 n  Ans. S = Lim 2  n  2n1  2n  = 2 Ans. ] Q.11s&p If sin x, sin22x and cos x · sin 4x form an increasing geometric sequence, find the numerial value of cos 2x. Also find the common ratio of geometric sequence. [5] [Sol. Given sin x, sin22x and cos x · sin 4x are in G.P. (r > 1 as G.P. is increasing)  sin42x = (sin x) (cos x) (sin 4x)  16 sin4x cos4x = sin x cos x sin 4x  16 sin3x cos3x = sin 4x (sin x  0, cos x  0)  16(sin x cos x)3 = 2 sin 2x · cos 2x  (sin 2x)3 = sin 2x · cos 2x  sin22x = cos 2x (sin 2x  0) 1 – cos22x = cos 2x y2 + y – 1 = 0 cos 2x = cos 2x cannot be  cos 2x = ; 2 hence rejected 2 2 with this value of cos 2x 1 + cos 2x = 1 + 5 1 = 2 5 +1 2 sin x =  cos 2x = sin2 2x = = = 5 1 2 r = sin x = 4 sin x cos2x = 2 sin x(1 + cos 2x) r = · = = Ans. ] Q.1254/1 Find all possible parameters 'a' for which, f (x) = (a2 + a – 2)x2 – (a + 5)x – 2 is non positive for every x  [0, 1]. [5] [Ans. a  [ 3, 3] ] [Sol. f(x) = (a + 2) (a – 1)x2 – (a + 5)x – 2 < 0 for a = –2 f(x) = –3x – 2 which is negative in [0, 1] for a = 1 f(x) = –6x – 2 which is negative in [0, 1] Case I : for a2 + a – 2 > 0 i.e. a > 1 or a < –2 (1) 1. f(0) < 0 2. f(1) < 0  a  [–3, 3] (2) taking (1)  (2) a  [–3, –2) U (1, 3] Case II : for a2 + a – 2 < 0 i.e. –2 < a < 1 (3) 1. D > 0 – b < 0 take 2a f(0) < 0 a  (4) 2. D > 0 f(1) < 0 take n – b 1 2a .........(5) take union (4) U (5) (6) take  of (3) & (6) gives common solution a  (–2, 1) Case III : when a2 + a – 2 < 0 and D < 0 Since D > 0  no solution in this case combining all a  [–3, 3] Ans. ] Q.13 The 1st, 2nd and 3rd terms of an arithmetic series are a, b and a2 where 'a' is negative. The 1st, 2nd and 3rd terms of a geometric series are a, a2 and b find the (a) value of a and b (b) sum of infinite geometric series if it exists. If no then find the sum to n terms of the G.P. (c) sum of the 40 term of the arithmetic series. [5] 1 1 1 545 [Sol. a, b, a2 are in A.P. (a < 0) a, a2, b are in G.P. [Ans. (a) a = – 2 , b = – 8 ; (b) – 3 ; (c) 2 ] (a) 2b = a + a2  b = and a4 = ab a (a + 1) 2 a3 = b since a < 0 [  a  0] a3 = a (a + 1)  2a2 – a – 1 = 0  (2a + 1)(a – 1) = 0 2 1 a = 1 (rejected as a < 0) or a = – 2 Ans. 1 1  1  1  a2 = 4 ;  b = – 4  2 +1 = – 8 Ans.    common difference b – a = – 1 + 1 = 3 a 2 1 2 8 2 8 1 (b) common ratio r = a ; r = 4  (1) = – 2 ; Since | r | < 1  S exists  S = – 1 = – 21+ 1  1 3 Ans.     (c) S = 20 1+ 39  3 = 20 { 8 + 117}= 5 × 109 = 1+ 39  3 = 545 Ans. ] 40     2   2 Q.14 The nth term, an of a sequence of numbers is given by the formula an = an – 1 + 2n for n  2 and a1 = 1. Find an equation expressing an as a polynomial in n. Also find the sum to n terms of the sequence. [8] [Sol. a1 = 1 a2 = a1 + 4 = 1 + 4 = 5 a3 = a2 + 6 = 5 + 6 = 11 a4 = a3 + 8 = 11 + 8 = 19 and so on hence S = 1 + 5 + 11 + 19 + + an S = + 1 + 5 + 11 + + an – 1 + an ( – ) ———————————————— 0 = 1 + 4 + 6 + 8 + + (an – an – 1) – an an = 1 + 2 (12 +4 34+ 44 +4.4...2 4+ (4a n4 4a n431)) (n1) terms a = 1 + 2· n 1 [4 + (n – 2)] = 1 + (n – 1) (n + 2) = n2 + n – 1 Ans. n 2 sum = an = n2 + n – 1 = n(n + 1)(2n + 1) 6 n(n + 1) + 2 – n n [(n + 1)(2n + 1) + 3(n + 1) – 6] = 6 n [2n2 + 3n + 1 + 3n + 3 – 6] 6 = n [2n2 6 n(n 2 + 3n 1) + 6n – 2] = 3 Ans. ]  2x x Q.15 Let f (x) denote the sum of the infinite trigonometric series, f (x) = sin n=1 3n sin 3n . Find f (x) (independent of n) also evaluate the sum of the solutions of the equation f (x) = 0 lying in the interval (0, 629). [8]  2x x 1  2x x 1   x x  [Sol. f (x) = sin n sin n = n=1 3 3  2 sin n sin n n=1 = 2  cos 3n  cos 3n1  now substituting n = 1, 2, 3, 4........ f (x) = 1 cos x  cos x 2  3 + 1 cos x 2  32  – cos x  3  + 1 cos x  cos x  2  33 32  □  + 1 cos x  cos x  2  3n 3n1  ———————————— f (x) = Lim n  1 1 cos x 2  3n – cos x  f (x) = 2 [1 – cos x] now f (x) = 0  cos x = 1 x = 2n, n  I sum of the solutions in (0, 629) S = 2[ + 2 + 3 + + 100] = 2 · 5050 = 10100 Ans. ]