MATHEMATICS-24-09- 11th (J-Batch) Final SOLUTION

REVIEW TEST-3 Class : XI (J-Batch) Time : 100 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain 14 questions. All questions are compulsory. 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. 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(J-BATCH) MATHEMATICS REVIEW TEST-3 Q.1 If tan , tan  are the roots of x2 – px + q = 0 and cot , cot  are the roots of x2 – rx + s = 0 then find the value of rs in terms of p and q. [4] p [Sol. x2 – px + q = 0 tan  tan .....(1) ; x2 – rx + s = 0 [Ans. cot cot q2 ] ....(2) hence roots of 2nd are reciprocal of (1)  put x  1 x in (2) 1 r x2 – x + s = 0 sx2 – rx + 1 = 0 (3) comparing (1) and (3) 1 p s = r = q 1 p p s = q ; r = q hence rs = q2 ] Q.2 Let P(x) = ax2 + bx + 8 is a quadratic polynomial. If the minimum value of P(x) is 6 when x = 2, find the values of a and b. [4] [Ans. a = 1/2, b = – 2] [Sol. P(x) = ax2 + bx + 8 (1) P(2) = 4a + 2b + 8 = 6 (2) b  – 2a = 2;  4a = – b from (2), we get – b + 2b = – 2   4a = – (– 2)  ]   1  Q.3 Let P = 102n1  then find log n1   1 1 1 0.01 (P). [4] [Ans. –1 ] [Sol. P = 101 ·10 2 ·1022 ·10 23 ............ 1 1  1  1 ......  1 1  1 P = 10 2 22 23 = 10 2 = 102 = 100  log0.01(P) = – 1] Q.4 Prove the identity sec8A 1 sec 4A 1 = tan 8A tan 2A . [4] [Sol. LHS = 1 cos8A  cos 4A 1 cos 4A cos8A 2sin 2 4A·cos 4A = 1 cos2 2A·cos8A sin 4A·sin 8A = 2sin 2 2A·cos8A tan 8A·2 sin 2A·cos 2A = 2 sin2 2A = tan 8A tan 2A Hence proved.] Q.5 Find the general solution set of the equation logtan x(2 + 4 cos2x) = 2. [4] [Sol. 2 + 4 cos2x = tan2x 3 + 4 cos2x = sec2x 4 cos4x + 3 cos2x – 1 = 0 let cos2x = t 4t2 + 3t – 1 = 0 (4t – 1)(t + 1) = 0 t = 1/4 or t = – 1 cos2x = 1 or cos2x = – 1 (not possible) 4  cos2x = cos2  3  x = n + 3 , n  I] sin   sin 3  sin 5 ......... sin 17  Q.6 Find the value of cos  cos 3  cos 5   cos17 when  = 24 . [4] [Sol. sin   sin 3  sin 5   sin 17 cos   cos 3  cos 5   cos17  where  = 24 [Ans. 1] Nr = 1 2sin  [2 sin  (sin  + sin 3 + + sin 17)] = 1 2sin 2   cos 2  cos 4  cos16  cos18 2 sin  1 ––  1cos 2   2 sin 2 9 = 2 sin  [1 – cos 18] = 2 sin  Dr = = 1 [2 sin  (cos  + cos 3 + + cos 17)] 2 sin  1 2sin  [sin 2 + sin 0 + sin 4 – sin 2 + sin 18 – sin 16)] = 1 [sin 18 ] = 2 sin  2 sin 9 cos 9 2 sin  Nr  Dr sin 2 9 = sin  sin  × sin 9 cos9 9 = tan 9 = tan 24 3 = tan 8 = 1 Ans. ] Q.7(a) Sum the following series to infinity 1 1·4·7 1 + 4·7·10 1 + 7·10·13 + ........... (b) Sum the following series upto n-terms. 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + ............. [3+3] 1 1 [Ans. (a) 24 ; (b) Sn = 5 n(n + 1)(n + 2)(n + 3)(n + 4)] 1 [Sol.(a) Tn = [1 (n 1)3][1 3n][4  3n] 1 = (3n  2)(2n  1)(3n  4) T = 1  1  1  ] n 6 (3n  2)(3n  1) (3n 1)(3n  4)  T = 1  1  1  1 6 1·4 4·7  T = 1  1  1  2 6  4·7 7 ·10 ⁝ ⁝ ⁝ T = 1  1  1  n 6 (3n  2)(3n  1) (3n 1)(3n  4)  ——————————————————  S = T = 1  1  1  =  1  1  n  n as n   6 1·4 (3n 1)(3n  4)   24 6(3n 1)(3n  4)   S = 1 24 Ans. (b) 1 · 2 · 3 · 4 + 2 · 3 · 4 · 5 + 3 · 4 · 5 · 6 + ............. 1 Tn = 5 n(n + 1)(n + 2)(n + 3) [(n + 4) – (n – 1)]  T1 1 = 5 [1 · 2 · 3 · 4 · 5 – 0] 1 T2 = 5 [2 · 3 · 4 · 5 · 6 – 1 · 2 · 3 · 4 · 5] ⁝ ⁝ ⁝ 1 Tn = 5 [n(n + 1)(n + 2)(n + 3)(n + 4) – (n – 1)n(n + 1)(n + 2)(n + 3)]  Sn = Tn = 1 5 [n(n + 1)(n + 2)(n + 3)(n + 4)] Q.8 The equation cos2x – sin x + a = 0 has roots when x  (0, /2) find 'a'. [6] [Ans. a  (–1, 1)] [Sol. 1 – sin2x – sin x + a = 0 sin2x + sin x – (a + 1) = 0 let sin x = t  t2 + t – (a + 1) = 0, t  (0, 1) 1 1 4(a  1) t = 2 t =  t = now 0 < 1 1 1 4a  5 2 4a  5 2 4a  5 2 (reject – ve sign) < 1 0 < – 1 + < 2 or 1 < 4a  5 < 3 1 < 4a + 5 < 9 – 4 < 4a < 4  – 1 < a < 1  a  (–1, 1) Ans. ] Q.9s&p A, B and C are distinct positive integers, less than or equal to 10. The arithmetic mean of A and B is 9. The geometric mean of A and C is 6 . Find the harmonic mean of B and C. [6] [Ans. 9 9 ] 19 [Sol. A + B = 18 .....(1) [13th 24-9-2006] AC = 72 (2) There are only two possibilities A = 10 and B = 8 or A = 8 and B = 10 If A = 10 then from (2) C is not an integer. Hence A = 8 and B = 10; C = 9  H.M. between B and C = 2·10·9 10  9 180 = 19 Ans. ] Q.10 Express cos 5x in terms of cos x and hence find general solution of the equation cos 5x = 16 cos5x. [6]  [Hint: Ans. x = (2n + 1) 2 or x = n ±  , where cos 5x = 16 cos5x – 20 cos3x + 5 cos x] 3 Q.11 If x is real and 4y2 + 4xy + x + 6 = 0, then find the complete set of values of x for which y is real. [6] [Ans. x  (– , – 2]  [3, )] [13th 24-9-2006] [Sol. 4y2 + 4xy + x + 6 = 0 y  R  D  0 16x2 – 16(x + 6)  0  x2 – x – 6  0  (x – 3)(x + 2)  0  x  (– , – 2]  [3, ) Ans. ] Q.12 Find the sum of all the integral solutions of the inequality [6] 2 log3x – 4 logx27  5. [Ans. 3320] [Sol. Let log3x = y  x = 3y (1)  2 log3x – 12 logx3  5 2y – 12 y  2y2  5y 12 y (2y  3)(y  4)  0  y  0  y   ,  3   0, 4  2  3 using (1), –  < log3x  – 2 0  x  3–3/2 or 1 < x  81 81·82 hence integral solutions are S = 2 + 3 + 4 + ....... + 81 = 2 1 = 81 · 41 – 1 = 3321 – 1 = 3320 Ans. ] 1 tan  1 tan   1 tan   Q.13 If  +  +  =   , show that         sin   sin   sin  1 = . 21/Ex-1, ph-1 2 1 tan   1 tan   1 tan   2 2 2 cos   cos  cos        sin   sin   sin   sin(    ) [7] [Sol. RHS = cos  cos  cos   cos(    ) [T/S, Q.21, Ex-1, Ph-1] 2 sin    cos      2 sin    cos     2   2   2   2   2    =                 2 cos cos   2 cos     2 cos         2         2  sin     cos      cos     2   2    2   2  =   ·      cos     cos     cos     2               sin    sin     = 2 tan            2   4 2  2 cos    cos            2   1 tan    1 tan    1 tan   tan     tan     tan      2   2   2  =    4 2      4 2   4  ;     1 tan      1 tan 2       1 tan 2   = RHS]  2  Q.14 In any  ABC prove that (a) c2 = (a – b)2cos2 C + (a + b)2sin2 C . 2 2 (b) a3cos(B – C) + b3cos(C – A) + c3cos(A – B) = 3 abc. [4 + 4] [Sol: (a) RHS (a – b)2cos2 C + (a + b)2sin2 C 2 2 (a2 + b2 – 2ab) cos2 C + (a2 + b2 + 2ab) sin2 C a cos  sin 2 2 2   b cos  sin 2 2 2   2abcos 2 a2 + b2– 2ab cosC = c2 = LHS (using cosine law) (b) T1 = a3cos(B – C) T2 = b3cos(C – A) T3 = c3cos(A – B) T1 = a2·a cos(B – C) = a2 ·ksin A cos(B – C) [using sine law] = a2 · k ·sin(B + C) cos(B – C) [A + B + C =   B + C =  – A  sin (B + C) = sin A] T = a2 k 1 2 [sin 2B + sin 2C] = a [2k sinB cosB + 2k sinC cosC] 2 = a2 [b cosB + c cosC] [using sine law] lly T2 = b2[a cosA + c cosC] & T3 = c2 [a cosA + b cosB] now T1 + T2 + T3 = a2[b cosB+c cosC] + b2[a cosA+c cosC] + c2[a cosA + b cosB] = ab [a cosB + b cosA] + ac [a cosC+c cosA] + bc[b cosC + c cosB] = abc + acb + bca [using projection rule] = 3abc = RHS ]