MATHEMATICS-17-09-11th (PQRS) SOLUTION

REVIEW TEST-4 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-4 n n  r s if r  s  Q.1 Evaluate   rs·2 ·5 where rs =  1 . if r = s r=1 s=1 Will the sum hold if n  ? [4] r 2 n 10 [Sol. A =  2 [r ·5 + r ·5 + ......... + r ·5 ] [Ans. (10n – 1)] r=1 1 2 n 9 n = (r r=1 1 ·2r ·5) n + (r r=1 2 n ·2r ·52 ) + ........ + ( r=1 ·2r ·5n ) = 2.5 + 22 · 52 + 23 · 53 + + 2n · 5n 10(10n 1) = 10 + 102 + 103 + + 10n = 9 No because common ratio > 1. ] Q.2 Find the general solution of the equation, 2 + tan x · cot 2 tan(x 2)·cot(x 2) x + cot x · tan 2 x = 0. [4] 2 2 [Sol. 2 + 1 tan2(x 2) 2 + 1  tan 2 (x 2) = 0 [Ans. 2n ± 3 , n  N] 2 + 1 tan2 (x 2) + 2 = 0 x   or 2 [Q.6, Ex-1, ph-2] 1 tan  x  + 4 + 1 tan   x 2  = 0  1 – tan2 x = y 2 y2 + 4y + 4 = 0 (y + 2)2 = 0  y = – 2 1 – tan2 x = – 2  tan2 x = 3 = tan2   x = n ±   2n ± 2 , n  N] 2 2 3 2 3 3 Q.3 Given that 3 sin x + 4 cos x = 5 where x  (0,  2). Find the value of 2 sin x + cos x + 4 tan x. [4] [Sol. 3 sin x + 4 cos x = 5 [Ans. 5] 6t 1+ t 2 4(1  t2 ) + 1 + t2 = 5 where t = tan x 2 6t + 4(1 – t2) = 5 + 5t2 9t2 – 6t + 1 = 0  (3t – 1)2 = 0 t = 1/3 now y = 2 sin x + cos x + 4 tan x 2(2t) = 2 1 t2 8t + 2 + 2 = 4 31 + 8 91 + 8·1 3 1 = 4 · 9 + 8 · 9 + 8 ·9 1+ t 6 4 1+ t 1 t 1+ 1+ 1 9 9 9 3 10 9 10 3 8 = 5 + 5 + 3 = 2 + 3 = 5 Ans. ] Q.4 Find the integral solution of the inequality  0. [4] [Sol. 2x – x2 + 8 > 0 [Ans. {2, 3}]  x2 – 2x – 8 < 0 (x – 4)(x + 2) < 0 and x – 1 > 0 x > 1  x  (1, 4) log0.3(x – 1)  0  x – 1  1 x  2  x  [2, 4)  integers are {2, 3} ]  Q.5ph-3 In  ABC, suppose AB = 5 cm, AC = 7 cm,  ABC = 3 (a) Find the length of the side BC. (b) Find the area of ABC. [4] [Ans. (a) a = 8; (b) 10 sq. units] [Sol.(a) Use cosine law cos 60° = 25 + a 2  49 10a 1 25 + a 2  49 2 = 10a  a2 – 5a – 24 = 0  (a – 8)(a + 3) = 0  a = 8 or a  – 3 1 (b) A = 2 · 5 · 8 sin 60° = 10 sq. units ] Q.6ph-3 The sides of a triangle are n – 1, n and n + 1 and the area is n . Determine n. [4] [Sol. Area = n =  [Ans. n = 6]  = 3n s = 2  n =  3n  n + 2  n  n  2   n3 =  2     2 2 2  16n = (n2 – 4)3       3n2 – 16n – 12 = 0  (3n + 2)(n – 6) = 0  n = 6 Ans. ] Q.7 With usual notions, prove that in a triangle ABC, r + r1 + r2 – r3 = 4R cos C. [5] [Sol. LHS = 4R sin C sin B sin A  cos A cos B + 4R cos C sin A cos B + sin B cos A  2  2 2 2 2  2  2 2 2 2  = 4R sin C  sin C + 4R cos C cos C 2  2  2 2  2 C 2 C = 4R cos – sin 2 2  = 4R cos C. Hence proved.  Alternatively: LHS = s  + s a  + s b  – s  c  1  1 1  1   2s  (a + b)   1 1  =  s +  +    s  a s  b s  c =  (s  a)(s  b)    s  –    ~~    ~~~~   c s  =  c  c  = cs(s  c)  (s  a)(s  b)  (s  a)(s  b) s(s  c)   s(s  a)(s  b)(s  c)  = c [s2 – cs – s2 + as + bs – ab] =  c  [s(a + b – c) – ab] c (a + b + c)(a + b  c)  2ab c =   2  = 2 [(a + b)2 – c2 – 2ab] c(a 2 + b2  c2 )ab = 2ab abc cos C =  = 4R cos C hence proved ] Q.8 Find the general solution of the equation, sin x + cos x = 0. Also find the sum of all solutions in [0, 100]. [5]   + x  1 [Sol. cos x = – sin x = cos      + x  [Q.18, Ex-1, ph-2] [Ans. x = n – 4 , n  I; sum = 5025]  x = 2n ±     + ve sign gives no solution  – ve sign gives 2x = 2n – 2 1 , n  I 1 100  2x = 2n – 2 1   x = n – 4 , n  I Ans. now sum =  n  4  = 5050 – 25 = 5025 Ans. ] n=1   Q.9 Find all negative values of 'a' which makes the quadratic inequality sin2x + a cos x + a2  1 + cos x true for every x  R. [5] [Ans. a  (– , – 2)] [Sol. or 1 – cos2x + a cos x + a2 – 1 – cos x  0 cos2x + (1 – a)cos x – a2  0 put cos x = t t2 + (1 – a)t – a2  0  t  [– 1, 1] f (t) = t2 + (1 – a)t – a2 f (1)  0 and f (–1)  0 f (1)  0  (a + 2)(a – 1)  0 and f (–1)  0  a (a – 1)  0 hence a  (– , –2]  [1, ) since a < 0  final answer is a  (– , – 2] ] Q.10 Solve for x, log x2 2 log (x2 2) – 5log2 x2 1 . [5] [Ans. x = {–2, 2}] [Sol. 5log2 x2  3log2 (x2 2) = – 5log2 x2 1 5log2 x2  3(log2 x2 1) = 3(1+log2 x2 )  5log2 x2 1 let log x2 = t 5t – 3(t – 1) = 3(t + 1) – 5(t – 1) 5t – 3 3 = 3 · 3t – 5 5  6   10   3   5  5t  5  = 3t  3   5t  5  = 3t  3          5(t – 2) = 3(t – 2)  5 t2   3  = 1    t – 2 = 0  t = 2 log2 x2 = 2  x2 = 4  x = ± 2  x = {–2, 2} Ans. ] Q.11 In a triangle ABC if a2 + b2 = 101c2 then find the value of cot C cot A + cot B . [5] [Ans. 50] [Sol. Using c2 = a2 + b2 – 2ab cos C [T/S Q.18 Ex-I, Ph-1]  c2 + 2ab cos C = 101 c2 (using a2 + b2 = 101c2)  100 c2 = 2ab cos C  ab cos C = 50 c2 (1) also cot C cot A + cot B = cos C·sin A·sin B sin Csin(A + B) = cos C sin A·sin B sin 2 C ab = cos C c2  cot C  cot A + cot B = cos C · ab c2 (2) from (1) ab cos C = 50 c2 cot C cot A + cot B = 50 Ans. ] 1 1 + log (sin x) 1 + log (cosx) Q.12 Solve the equation for x, 52 + 52 5 =152 15 [5] [Sol. + 5(5log5 sin x ) = 15(15log15 cos x ) [Ans. x = 2n + 6 , n  I] [Q.2, Ex-1, ph-2] (1 + sin x) =  1 3 (cos x)  1 + sin x =  cos x  1 + sin x – cos x = 0 1 + 2  sin x   cos x = 0 2   x     x    1  7  1 + 2 sin    = 0  sin  3   3  = – 2 = sin    6   x –  7 = or x –  11 = 3 6 7 2 x = 6 + 6 9 3 3 6 11 2 or x = 6 + 6 13  x = 6 = 2 or x = 6 = 2 + 6  x = 2n +  , n  I Ans. ; rejecting x = 3 ; log cos x not defined ] 6 2 15  2 Q.13 Evaluate the sum  n . [5] n=1 6 42 [Hint: Use method of difference twice ] [Ans. S = 125 ] 12 [Sol. S = 6 S 22 + 62 12 32 + 63 22 42 + 64 32 + ................. [T/S, Q.24, Ex-I] 6 = + 62 + 63 + 64 + ................. – – – – ———————————————— 5S = 1 + 3 + 5 + 7 + ...................... 6 16 4 642 4 643 4 4642 4 4 4 4 4 43 = S1 say let S1 = 1 + 3 6 62 + 5 + 7 63 64 +...................... S1 = 6 + 1 + 3 + 5 62 63 64 + ...................... – – – – ———————————————— 5S1 = 6 1 + 2 + 2 + 2 6 62 63 64  1  +...................... 5S1 = 1 + 2 36   5S1 = 1  1 + 2  6   5S1 = 1 + 1 6 6 1   6  6 6 36 5  6 6 15 5S1 = 6 7  S = 7 30 1 25 5S  6 = S1 = 7  S = 25 42 125 Ans. ] Q.1429/1 Suppose that P(x) is a quadratic polynomial such that P(0) = cos340°, P(1) = (cos 40°)(sin240°) and P(2) = 0. Find the value of P(3). [8] [Ans. – 1/2] [Sol. Let P(x) = ax2 + bx + c P(0) = c = cos340° = m3 ....(1) where m = cos 40° P(1) = a + b + c = cos 40° sin240° = m(1 – m2) (2) P(2) = 4a + 2b + c = 0 (3) now P(3) = 9a + 3b + c = 3(4a + 2b + c) – 3(a + b + c) + c = 3P(2) – 3P(1) + P(0)  P(3) = 0 – 3{m(1 – m2)} + m3 = 4m3 – 3m = 4 cos340° – 3 cos 40° 1 = cos 120° = – 2 Ans. ] Q.15 If l, m, n are 3 numbers in G.P. prove that the first term of an A.P. whose lth, mth, nth terms are in H.P. is to the common difference as (m + 1) to 1. [8] [Sol. Given l, m, n  G.P.  m2 = l · n and 1 , 1 , l Tm 1  A.P. Tn where Tl , Tm , Tn are lth, mth and nth terms of A.P. 2 1 1  = + m l Tn 2 1 1 a + (m 1)d = a + (l 1)d + a + (n 1)d  a(2m – n – l) + d(lm + l + n + mn – 2ln – 2m) = 0 a  d = – (lm + l + n + mn  2ln  2m) 2m  n  l = – (lm + l + n + mn  2m2  2m) 2m  n  l – (l(m + 1) + n(m + 1)  2m(m +1)) = 2m  n  l = a (m +1)(2m  n  l) (2m  n  l) d = m + 1. Hence proved. ]