MATHEMATICS-27-08- 11th (J-Batch)

REVIEW TEST-2 MATHEMATICS Class : XI (J-Batch) Time : 100 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain 17 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-2 Q.1QE(TN) Find the set of values of 'a' for which the quadratic polynomial (a + 4)x2 – 2ax + 2a – 6 < 0  x  R. [3] [Ans. (– , – 6)] Q.2 Solve the inequality by using method of interval, x +1  x + 5 . [3] QE(TN) x 1 x +1 [Ans. (– , – 1)  (1, 3] ] Q.34(dpp-22) Find the minimum vertical distance between the graphs of y = 2 + sin x and y = cos x. [3]  x    [Ans. 2 – ] [Hint: dmin = min(2 + sin x – cos x) = min[2 + sin    ] = 2 –  ] [12 & 13 25-09-2005] Q.4 Solve: d  3 cos x  cos3 x  when x = 18°. [3] (new) [Sol. y = –   dx  4 cos 3x 4 dy dx =   3 sin 3x 4 [Ans. 3 ( 16 +1)] dy = 3 sin 54 = 3  ( 5 +1)  3 = +1) Ans.] dx x=18 4  4  4   16 Q.57(dpp-23)33/1 If p, q are the roots of the quadratic equation x2 + 2bx + c = 0, prove that 2 log ( + y  q ) = log 2 + log  y + b + y2 + 2by + c  . [4] [Sol. x2 + 2bx + c = 0 ; p + q = 2b  pq = c  ....(1) TPT 2 log ( + y  q ) = log 2 + log  y + b + y2 + 2by + c   LHS = log(y  p + y  q + 2 y  p y  q ) = log 2y  (p + q) + 2 y2  (p + q)y + pq   = log 2y + 2b + 2 y2 + 2by + c   = log 2 + log y + b + y2 + 2by + c  = RHS ] Q.6QE(TN) Find the maximum and minimum value of y = x2 +14x + 9 x2 + 2x + 3  x  R. [4] [Ans. Maximum = 4 ; minimum = – 5] 7 Q.72log(New) Suppose that a and b are positive real numbers such that log27a + log9b = 2 and log27b + log9a = 3 . Find the value of the ab. [4] [Ans. 243] 7 [Sol. log27a + log9b = 2 2 ; log27b + log9a = 3 1 1 7 3 log3a + 2 log3b = 2 1 1 2 3 log3b + 2 log3a = 3 adding the equation 1 1 7 2 25 3 log3(ab) + 2 log3(ab) = 2 + 3 = 6 5 25 log (ab) =  log (ab) = 5  ab = 35 = 243 Ans. ] 6 3 6 3 Q.8s&p(New) Given sin2y = sin x · sin z where x, y, z are in an A.P. Find all possible values of the common difference of the A.P. and evaluate the sum of all the common differences which lie in the interval (0, 315). [4] [Ans. 5050] [Sol. Let x = a – d; y = a and z = a + d given sin2a = sin(a – d) · sin (a + d)  sin2a = sin(a – d) · sin(a + d) = sin2a – sin2d  sin2d = 0  d = n, n  I  sum = (1 + 2 + 3 + + 100) = 5050] tan 8 Q.910(DPP-5) Prove that tan  = (1 + sec2) (1 + sec4) (1 + sec8). [4] [Sol. RHS = 1+ cos 2 cos 2  1+ cos 4 cos 4  1+ cos 8 cos 8 2 cos2  2 cos2 2 2 cos2 4 = cos 2 cos 4 cos 8 [8 cos cos 2cos 4]cos  cos 8 = sin 8 cos  sin   ] cos 8 Q.10 Find the exact value of tan2  + tan2 3 + tan2 5 + tan2 7 . [4] 9(dpp-9)50/06 16 16 16 16 [Sol. Let  16 =  tan2 + tan23 + tan25 + tan27 [Ans : 28] = (tan2  + cot2  ) + (tan23 + cot23) [ Note that tan7 = tan(8 – ) = cot and tan5 = tan(8 –3) = cot3 ] = (cot – tan)2 + (cot3 – tan3)2 + 4 = 4 [cot22 + cot26] + 4 = 4 [ cot22 + tan22 ] + 4 = 4 [ (cot2 – tan2)2 + 2] + 4 = 4 (cot2 – tan2)2 + 12 = 4 . 4 cot2 4 + 12 = 16 × 12 + 12 = 28 Ans ] Q.11 48/6(New) 89 Evaluate  1 . [5] [Sol. S = n=11+ (tan n) 1 + 1 + 1 + ........ + 1 + 1 1+ (tan1)2 reversing the sum 1+ (tan 2)2 1+ (tan 3)2 1+ (tan 88)2 1+ (tan 89)2 S = 1 + 1 + .............................. + 1 + 1 1+ (cot1)2 1+ (cot 2)2 1+ (cot 88)2 1+ (cot 89)2 89 1 1 89 1 + (tan n)2 2S = + n=11+ (tan n) 1+ (cot n) 89 =  1 = 1 + 1 + ....... + 1 = 89 n=1  S = 44.5 Ans. ] n=1 1+ (tan n) 1+ (tan n)2 Q.12QE(TN) Find the value of k for which one root of the equation of x2 – (k + 1)x + k2 + k–8=0 exceed 2 and other is smaller than 2. [5] [Ans. k  (–2, 3)] [Sol. Since a > 0 f (0) < 0  f (2) < 0  4 – 2(k + 1) + k2 + k – 8 < 0 k2 – K + 6 < 0 (k + 2)(k – 3) < 0  k  (–2, 3) ] *Q.134s&p(New) Let an be the nth term of an arithmetic progression. Let Sn be the sum of the first n terms of the arithmetic progression with a1 = 1 and a3 = 3a8. Find the largest possible value of Sn. [5] 100 2 [Sol. From a3 = 3a8 we obtain 1 + 2d = 3(1 + 7d)  d = – 19 . [Ans. 19 ] n  2  (n 1) 2  n (20  n)n Then Sn = 2  19  = 19 [19 – (n – 1)] = 19 . now consider 20n – n2 = – [n2 – 20n] = – [(n – 10)2 – 100] 100  (n 10)2  Sn = 19 100 now, Sn will be maximum if n = 10 and (Sn)max = 19 Ans. ] Q.14 (a) If A+B+C =  & sin A + C  = k sin C , then find the value of tan A ·tan B 8(Dpp-7)44/ph-1 2 2 2 2  x2 + x  in terms of k. *(b)QE(TN) Solve the inequality, log0.5 log6  x + 4  < 0. [2 + 4]  [Ans. (a) k  1 ; (b) (– 4, – 3)  (8, )] k + 1 Q.15(New) 25(Ex-1, ph-2) Let f (x) = sin6x + cos6x + k(sin4x + cos4x) for some real number k. Determine (a) all real numbers k for which f (x) is constant for all values of x. (b) all real numbers k for which there exists a real number 'c' such that f (c) = 0. [2 + 4] [Ans. (a) – 3 ; (b) k  1,  1  ] [T/S, Q.25, Ex-1, Ph-2] 2  2 [Sol.(a) f (x) = (sin2x + cos2x)3 – 3 sin2cos2x(sin2x + cos2x) + k[(sin2 + cos2x)2 – 2sin2x cos2x] = 1 – 3sin2x cos2x + k(1 – 2 sin2x cos2x) f (x) = (k + 1) – sin2x cos2x (2k + 3) (1) for f (x) to be independent of x 3 k = – 2 Ans. (b) f (c) = (k + 1) – sin2c cos2c (2k + 3) = 0  sin2c cos2c = k +1 2k + 3  1 (sin22c) = 4 k +1 2k + 3  sin22c = 4(k +1) 2k + 3 but 0  sin22c  1  0  4(k +1) 2k + 3  1 solving 4(k + 1) 2k + 3  0; (k +1) 2k + 3  0 3 hence k  – 1 or k < – 2 again solving 4(k +1) 2k + 3  1; 4k + 4 2k + 3 – 1  0; 4k + 4  2k  3 2k + 3  0 2k +1 2k + 3 Hence k  1,  1   0 Ans. ]  2  Q.16QE(TN) Given the product p of sines of the angles of a triangle & product q of their cosines, find the cubic equation, whose coefficients are functions of p & q & whose roots are the tangents of the angles of the triangle. [6] [Ans: qx3  px2 + (1 + q) x  p = 0] [REE’92, 6] [Sol. Given sinA sinB sinC = p ; cosA cosB cosC = q Hence tanA tanB tanC = tanA + tanB + tanC = p/q Hence equation of cubic is x3  p x2 + d tan A tan Bix  p = 0 ...(i) q q now tan A tan B= sin A sin Bcos C + sin Bsin C cos A + sin C sin A cos B cos A cos BcosC We know that A + B + C =  cos(A+B+C) = –1 cos(A+B) cosC – sin(A+B) sinC = –1 ( cosA cosB – sinA sin B) cosC – sinC (sinA cosB + cosA sinB) = –1 1+ cosA cosB cosC= sinA sinB cosC + sinB sinC cosA + sinC sinA cosB dividing by cosA cosB cosC 1+ q = tan A tan B q Hence (i) becomes qx3 – px2 + (1 + q)x – p = 0 Ans.] Q.17QE(TN) If each pair of the equations x2 + p1x + q1 = 0 x2 + p2 x + q2 = 0 x2 + p3x + q3 = 0 has exactly one root in common then show that (p1 + p2 + p3)2 = 4(p1p2 + p2p3 + p3p1 – q1 – q2 – q3). [6] [Sol. Let x2 + p1x + q1 = 0 x2 + p2x + q2 = 0 x2 + p3x + q3 = 0 Now | ( + ) – ( + ) |2 = |  –  |2 p2 + p2  2p p = p2  4q 1 2 1 2 3 3 or p2 + p2  p2 = 2p1p2 – 4q3 1 2 3 |||ly p2 + p2  p2 = 2p2p3 – 4q1 2 3 1 and p2 + p2  p2 = 2p3p1 – 4q2 3 1 2 adding p2 + p2 + p2 = 2 p1 p2 – 4 q1 1 2 3 (p1 + p2 + p3)2 = 4[p1 p2  q1 ] ]