MATHEMATICS-09-07- 11th (PQRS) SOLUTION

REVIEW TEST-2 Class : XI (PQRS) Time : 100 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain 16 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(PQRS) MATHEMATICS SOLUTIONS REVIEW TEST-2 2log21/ 4 x  3log27 (x2 +1)3  2x dy Q.1 Let y = 74 log49 x – x 1 and dx = ax + b, find the value of a and b. [4] [Sol. y = x4  (x2 + 2x +1) 2 = x2 + x + 1 x  x 1 dy  dx = 2x + 1 = ax + b hence a = 2 and b = 1 Ans. ] Q.2 Show that cos2A + cos2(A + B) + 2 cosA cos(180° + B) · cos(360° + A + B) is independent of A. Hence find its value when B = 810°. [4] [Ans. 1] [Sol. cos2A + cos2(A + B) – [2 cosA · cosB · cos (A + B)] cos2A + cos2(A + B) – [ {cos(A + B) + cos(A – B) } cos (A + B) ] cos2A + cos2(A + B) – cos2(A + B) – (cos2A – sin2B) = sin2B which is independent of A now, sin2(810°) = sin2(720° + 90°) = sin290° = 1 Ans. ] Q.312 QE Find the product of the roots of the equation, | x2 | + | x | – 6 = 0. [Sol. | x |2 + | x | – 6 = 0. let | x | = t t2 + t – 6 = 0 (t + 3)(t – 2) = 0  t = – 3 or t = 2 but | x | = – 3 is not possible. hence | x | = 2  x = 2 or – 2. Hence product of roots = – 4 Ans. ] [Ans. – 4] [4] Q.419 QE One root of mx2 – 10x + 3 = 0 is two third of the other root. Find the sum of the roots. [4] [Ans. 5/4] 2 [Hint:  + 3 = 10  5 = 10 m 3 m 2 3   = 6 ; m 9 also  · 3 = m  22 = m 2· 36 m2 = 9  m 24 m = 3  m = 8  sum = 10 = 10 = 5 Ans. ] m 8 4 Q.57 trig Suppose x and y are real numbers such that tan x + tan y = 42 and cot x + cot y = 49. Find the value of tan(x + y). [4] [Ans. 294] [Sol. tan x + tan y = 42 and cot x + cot y = 49 tan x + tan y tan(x + y) = 1 tan x tan y now, cot x + cot y = 49  1 + 1 = 49  tan y + tan x = 49 tan x + tan y tan x tan y 42 6 tan x ·tan y tan x · tan y = 49 = 49 = 7 tan (x + y) = = = 294 Ans. ] Q.621 QE Find the solution set of k so that y = kx is secant to the curve y = x2 + k. [4] [Ans. k  (– , 0)  (4, )] [Sol. put y = kx in y = x2 + k kx = x2 + k = 0 x2 – kx + k = 0 for line to be secand, D > 0 k2 – 4k > 0 k(k – 4) > 0 hence k > 4 or k < 0 k  (– , 0)  (4, ) Ans. ] Q.7 A quadratic polynomial p(x) has 1 + and 1 – as roots and it satisfies p(1) = 2. Find the quadratic polynomial. [4] [Sol. sum of the roots = 2 product of the roots = – 4 ]  let p(x) = a(x2 – 2x – 4) p(1) = 2  2 = a(12 – 2 · 1 – 4)  a = – 2/5  p (x) = – 2 (x2 – 2x – 4) Ans. ] 5 Q.8 Solve the equation 0.5log x (x2 x) log 4 3 . [4] 1 (x2 x) log 2 (1 2) log x log 2 [Sol. x = 3 3  xlogx (x2 x) = 2  x2 – x = 2 x2 – x – 2 = 0 (x – 2)(x + 1) = 0  x = 2 or x = – 1 (rejected)  x = 2 Ans. ]  Q.9 Find the sum of the series, cos 2n +1 3 + cos 2n +1 5 + cos 2n +1 + upto n terms. Do not use any direct formula of summation. [5]  [Sol. Let  = 2n +1 S = cos  + cos 3 + cos 5 + cos (2n – 1) (2 sin ) S = 2 sin  [cos  + cos 3 + cos 5 + cos (2n – 1)] T1 = sin 2 – 0 T2 = sin 4 – sin 2 T3 = sin 6 – sin 4 Tn = sin 2n – sin 2(n – 1) ——————————— (2 sin ) S = sin2n sin 2n S = 2n +1 = 1 2 sin  2 2n +1 Ans. ] Q.10 Find the minimum and maximum value of f (x, y) = 7x2 + 4xy + 3y2 subjected to x2 + y2 = 1. [5] [Sol. Let x = cos  and y = sin  y = f () = 7 cos2 + 4 sin  cos  + 3 sin2 = 3 + 2 sin 2 + 2(1 + cos 2) = 5 + 2(sin 2 + cos 2) but –  (sin 2 + cos 2)   ymax = 5 + 2 and ymin = 5 – 2 ] Q.11 Find the minimum & maximum value of (sin x – cos x – 1) (sin x + cos x – 1)  x  R. [5] [Ans. maximum = 4; minimum = – 1/2] [Hint. y = (sin x – 1)2 – cos2x = (sin x – 1)2 – (1 – sin2x) = 2sin2x – 2 sin x  y = 2 (sin2x – sin x) = 2 sin x   1 2   1  –  4 hence y = 2  9  1  = 4 when sin x = – 1 max  4 4 y = 2 0  1  = – 1 when sin x = 1 ] min  4 2 2 Q.12 Given that log2a = s, log4b = s2 and logc2 (8) = 2 s3 +1 a 2b5 . Write log2 c4 as a function of 's' (a, b, c > 0, c  1). [5] [Sol. Given log2a = s (1) log2b = 2s2 (2) s3 +1 log8c2 = 2 log c 2 s3 +1 ....(3)  3log 2 = 2  4 log2c = 3(s3 + 1) (4) to find 2 log2a + 5 log2b – 4 log2c  2s + 10s2 – 3(s3 + 1) ] Q.13 Find the range of the expression y = [Sol. x2y – 4xy – 5y = x2 – 2x – 8 x2  2x  8 x2  4x  5 , for all permissible value of x. [5] (y – 1)x2 + 2x(1 – 2y) + 8 – 5y = 0  x  R hence D  0 4(1 – 2y)2 – 4(y – 1)(8 – 5y)  0 (4y2 – 4y + 1) – (13y – 8 – 5y2)  0 9y2 – 17y + 9  0 (1) since coefficient of y2 > 0 and D = 289 – 324 < 0 Hence (1) is always true Hence range of y is (– , ) ] Q.14 Find whether a triangle ABC can exists with the tangents of its interior angle satisfying, tan A = x, tan B = x + 1 and tan C = 1 – x for some real value of x. Justify your assertion with adequate reasoning. [6] [Sol. In a triangle tan A = tan A (to be proved) x + x + 1 + 1 – x = x(1 + x)(1 – x) 2 + x = x – x3 x3 = – 2 x = – 21/3 Hence tanA = x < 0 and tanB = x + 1 = 1 – 21/3 < 0 Hence A and B both are obtuse. Which is not possible in a triangle. Hence no such triangle can exist. ] Q.15 Solve the equation, 5 sin x + 5 – 5 = 2 sin2x + 2 sin x 1 2 sin2 x if x  (0, ). [6]  1   1   1 2  [Sol. 5 sin x +  – 5 = 2 sin 2 x +  = 2 sin x +  1  2 sin x   1 4 sin 2 x   2 sin x   Let sin x + 2 sin x = t 5t – 5 = 2(t2 – 1)  2t2 – 5t + 3 = 0  (2t – 3)(t – 1) = 0 t = 1 or t = 3/2 t = 1, 2 sin2x – 2 sin x + 1 = 0 D < 0 no solution if t = 3/2, 2 sin2x – 3 sin x + 1 = 0  sin x = 1 or sin = 1/2  x =  2  5 or 6 , 6    x   , ,  5  ]  Q.16 Find the value of x, y, z satisfying the equations log2x + log4y + log4z = 2 log9x + log3y + log9z = 2 and log16x + log16y + log4z = 2. [6] [Sol. from (1) log2(x2yz) = 4  x2yz = 24 (1) |||ly y2zx = 34 (2) z2xy = 44 (3) (1) × (2) × (3)  x4y4z4 = (2 · 3 · 4)4  xyz = 24 16 2 from (1)  x = 24 = 3 81 27 from (2)  y = 24 = 8 256 32 from (3)  z = 24 = 3 ]