MATHEMATICS-15-10- 11th (PQRS) SOLUTION

REVIEW TEST-5 Class : XI (P,Q,R,S) PAPER CODE : A Time : 3 hour Max. Marks : 228 INSTRUCTIONS 1. The question paper contain 2-parts. Part-A contains 54 objective question, Part-B contains 2 "Match the Column" questions. All questions are compulsory. PART-A (i) Q.1 to Q.45 have only one correct alternative and carry 3 marks each. There is NEGATIVE marking and 1 mark will be deducted for each wrong answer. (ii) Q.46 to Q.54 have More than one are correct alternative and carry 5 marks each. There is NO NEGATIVE marking. Marks will be awarded only if all the correct alternatives are selected. PART-B (iii) Q.1 to Q.9 are "Match the Column" type which may have one or more than one matching options and carry 8 marks for each question. 2 marks will be awarded for each correct match within a question. There is NO NEGATIVE marking. 2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet. 3. Use only HB pencil for darkening the bubble. 4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only. USEFUL DATA Atomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14, Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Pt = 195, Useful constant : g = 9.8 m/sec, h = 6.626 × 10–34 Js PART-A For example if only 'B' choice is correct then, the correct method for filling the bubble is A B C D For example if only 'B & D' choices are correct then, the correct method for filling the bubble is A B C D the wrong method for filling the bubble are The answer of the questions in wrong or any other manner will be treated as wrong. PART-B For example if Correct match for (A) is P, Q; for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling the bubble is P Q R S (A) (B) (C) (D) XI(PQRS) MATHEMATICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [45×3 = 135] There is NEGATIVE marking and 1 mark will be deducted for each wrong answer. Q.1s&p Consider the sequence 8A + 2B, 6A + B, 4A, 2A – B, Which term of this sequence will have a coefficient of A which is twice the coefficient of B? (A) 10th (B) 14th (C) 17th (D*) none [Sol. coefficient of A in nth term = 8 + (n – 1)(– 2) [11th, 12th and 13th] = 10 – 2n coefficient of B in nth term = 2 + (n – 1)(– 1) = 3 – n 10 – 2n = 2(3 – n)  10 = 6 which is absurd  none ] Q.2 s&p Find the sum of the infinite series 1 + 1 + 1 + 9 18 30 1 + 1 45 63 + ..... 1 (A*) 3 1 (B) 4 1 (C) 5 2 (D) 3 [Hint: T = 1 1 + 1 + 1 + 1 + 1 +  n 3 3 6 10 15 ........ 21  2  1 = 3  2·3 + 1 3·4 + 1 4·5 + 1 5·6  +........  hence T 2 1 using method of diff; T = 2  1 1  = n  S = 2 ·1 n 1 = Ans. ] 3 (n +1)(n + 2) 3  n +1 n + 2  n 3 2 3 Q.3ph-2 Number of degrees in the smallest positive angle x such that 8 sin x cos5x – 8 sin5x cos x = 1, is (A) 5° (B*) 7.5° (C) 10° (D) 15° [Sol. 2 sin x cos x[4 cos4x – 4 sin4x] = 1 (sin 2x)[2(cos2x + sin2x)] [2 cos2x – 2 sin2x] = 1 (sin 2x) 2 · 2 cos 2x = 1 2 sin 4x = 1 1 sin 4x = 2  4x = 30°  x = 7.5° Ans. ] Q.4log There exist positive integers A, B and C with no common factors greater than 1, such that A log2005 + B log2002 = C. The sum A + B + C equals (A) 5 (B*) 6 (C) 7 (D) 8 [Sol. A log2005 + B log2002 = C A log 5 + Blog 2 = C log 200 log 200 A log 5 + B log 2 = C log 200 = C log(52 23) = 2C log 5 + 3 C log 2 hence, A = 2C B = 3C for no common factor greater than 1, C = 1  A = 2; B = 3  A + B + C = 6 Ans. ] Q.5ph-3 A triangle with sides 5, 12 and 13 has both inscribed and circumscribed circles. The distance between the centres of these circles is (A) 2 (B) 5 2 (C) (D*) 65 2  [Sol. r = s = (12·5)·2 2(5 +12 +13) 60 = 30 = 2  5 2  d2 =  2  2 + (4)2   1 = 4 + 16 = 65 Ans. ] 2 Q.6QE The graph of a certain cubic polynomial is as shown. If the polynomial can be written in the form f (x) = x3 + ax2 + bx + c, then (A) c = 0 (B) c < 0 (C*) c > 0 (D) c = – 1 [Sol. From graph we see that this cubic has 2 + ve and one – ve root  product of roots is – ve and product of roots – c  – c is + ve ] Q.7ph-3 In an isosceles triangle ABC, AB = AC, BAC = 108° and AD trisects BAC and BD > DC. The ratio BD is DC (A*) 5 +1 2 3 (B) 2 (C) 1 x (D) 2 [Sol. In  ABC ( y = ?) x + y sin108 x = sin 36 x + y x 1 + y x sin 108 = sin 36 sin 72 = sin 36 ....(1) = 2 cos 36° y 5 + 1 x = 2 x – 1 = 5 1 2 5 +1 y = = 2 Ans. ] Q.8QE If a function f (x) = ax3 + bx2 + cx + d where a, b, c and d are integers and a > 0 is such that sin    18  = 0. Then the smallest possible value of f (1) is (A) 1 (B) 2 (C*) 3 (D) 4 [Sol. sin  = sin 10°, sin 30° = 18 1 [for 11th, 12th and 13th] 2 also sin 30° = 3 sin 10° – 4 sin310° 1 = 3 sin 10° – 4 sin310° 2 8 sin310° + 0 sin210° – 6 sin 10° + 1 = 0 (1) given, f (sin 10°) = 0 a sin310° + b sin210° + c sin 10° + d = 0 (2) comparing (1) and (2) a = 8, b = 0, c = – 6, d = 1 hence f (1) = a + b + c + d f (1) = 3 Ans. ] Q.9ph-3 The sides of a triangle are 6 and 8 and the angle  between these sides varies such that 0° <  < 90°. The length of 3rd side x is (A) 2 < x < 14 (B) 0 < x < 10 (C*) 2 < x < 10 (D) 0 < x < 14 [Sol. x2 = b2 + c2 – 2bc cos  x2 = 64 + 36 – 96 cos  = 100 – 96 cos  and   (0, 90°) cos   (0, 1)  x2|max = 100 – 0  x = 10 x2|min = 100 – 96 = 4  x = 2  2 < x < 10 Ans. ] Q.10s&p The sequence a1, a2, a3, satisfies a1 = 19, a9 = 99, and for all n  3, an is the arithmetic mean of the first n – 1 terms. Then a2 is equal to (A*) 179 (B) 99 (C) 79 (D) 59 [Sol. n  3, a3 = a1 + a 2 2 ....(1) a4 = (a1 + a 2 ) + a3 3 = 2a3 + a3 3  a4 = a3 a5 = (a1 + a2 + a3 + a 4 ) 4 = 3a 4 + a 4 4 = a4 a3 = a4 = a5 = = a9 = 99 put in equation (1) 99 = 19 + a 2 2  a2 = 179 Ans. ] Q.11s&p If b is the arithmetic mean between a and x; b is the geometric mean between 'a' and y; 'b' is the harmonic mean between a and z, (a, b, x, y, z > 0) then the value of xyz is (A) a3 (B) b3 (C) [Sol. 2b = x + a (1) b2 = ay (2) b3(2a  b) 2b  a (D*) b3(2b  a) 2a  b 2az b = a + z ....(3) b2 x = 2b – a; y = a 2 1 1 and b = a + z  z = ab 2a  b b2  xyz = (2b – a) a ab · 2a  b = b3(2b  a) 2a  b Ans. ] Q.12st.line Given A(0, 0), ABCD is a rhombus of side 5 units where the slope of AB is 2 and the slope of AD is 1/2. The sum of abscissa and ordinate of the point C is (A) 4 5 (B) 5 (C*) 6 (D) 8 [Sol. tan  = 2 tan  = 1/2 coordinates of B (5 cos , 5 sin )  5 , 10   5 5  and coordinates of D (5 cos , 5 sin )  10 , 5    15 , 15  middle point of AC is M:    5 2 5   coordinates of C (3 5, 3 5 ) ] Q.13st.line A circle of finite radius with points (–2, –2), (1, 4) and (k, 2006) can exist for (A) no value of k (B) exactly one value of k (C) exactly two values of k (D*) infinite values of k 6 [Sol. If 3 2006  4  k 1  k  1002 hence infinite number of value of k given by R – 1002 ] Q.14st.line If a  ABC is formed by 3 staright lines u = 2x + y – 3 = 0; v = x – y = 0 and w = x – 2 = 0 then for k = – 1 the line u + kv = 0 passes through its (A) incentre (B*) centroid (C) orthocentre (D) circumcentre [Hint: 2x + y – 3 – x + y = 0 x + 2y = 3 (1)  5 , 2  centroid is    which satisfies (1)   line passes through centroid. ] Q. 5general If a, b and c are numbers for which the equation identity, then a + b + c equals x2 +10x  36 a x(x  3)2 = x b + x  3 + c (x  3)2 is an (A*) 2 (B) 3 (C) 10 (D) 8 [Sol. x2 +10x  36 x(x  3)2 a(x  3)2 + b(x  3)x + cx = x(x  3)2 hence x2 + 10x – 36 = a(x – 3)2 + b(x – 3)x + cx put x = 0; – 36 = 9a  a = – 4 x2 + 10x – 36 = x2(–4 + b) + x(24 – 3b + c) + (–36) comparing coefficients also, – 4 + b = 1  b = 5 24 – 15 + c = 10  9 + c = 10  c = 1 a = – 4; b = 5; c = 1 i.e. a + b + c = 2 Ans. ] Select the correct alternatives. (More than one are correct) [9 × 5 = 45] Q.16 If the quadratic equation ax2 + bx + c = 0 (a > 0) has sec2 and cosec2 as its roots then which of the following must hold good? (A*) b + c = 0 (B*) b2 – 4ac  0 (C*) c  4a (D) 4a + b  0 Q.17 The equation | x  2 |10x2 1 = | x  2 |3x has (A) 3 integral solutions (B*) 4 real solutions (C*) 1 prime solution (D*) no irrational solution [Sol. x – 2  0  x  2 Case-I: x – 2 > 0  x > 2 (10x2 – 1) log(x – 2) = 3x log (x – 2) log(x – 2)(10x2 – 3x – 1) = 0 x – 2 = 1 or 10x2 – 5x + 2x – 1 = 0  (5x + 1)(2x – 1) = 0 x = 3 x = – 1/5 (reject), x = 1/2 (reject) Case-II: x – 2 < 0  x < 2 (10x2 – 1) log(2 – x) = 3x log(2 – x) log(2 – x)(10x2 – 3x – 1) = 0 2 – x = 1 or x = – 1/5, x = 1/2  1 x = 1  x   , 1  , 1, 3 ]  5 2  Q.18 Which of the following statements hold good? (A*) If M is the maximum and m is the minimum value of y = 3 sin2x + 3 sin x · cos x + 7 cos2x then the mean of M and m is 5.  (B) The value of cosec 18 –  sec 18 is a rational which is not integral. (C*) If x lies in the third quadrant, then the expression + 4 cos2    x  is independent of x. (D) There are exactly 2 values of  in [0, 2] which satisfy 4 cos2  2 [Sol. (A)829/ph-1 3 sin2 x + 3 sin x . cos x + 7 cos2 x     cos   1 = 0. = 3(1 cos 2x) + 3sin 2x + 7(1+ cos 2x) 2 2 2 3sin 2x + 4 cos 2x y = 2 + 5 ymax = 5/2 + 5 = 7.5 ymin = –5/2+5 = 2.5 2  1 cos   3 sin   1  2 18 2 18  (B)84/ph-1 sin  /18  cos  /18   = sin( 9) 2 [Quiz] 4sin  cos   cos  sin   =  6 18 6 sin  9 2 18 = 4 Ans +     x  (C)824/ph-1 4 cos 4 2         + 21+ cos 2  x  1+1 2 cos 2x  + 2(1+ sin x)   + 2 + 2sin x 2 | sin x | + 2 + 2 sin x = – 2 sin x + 2 + 2 sin x = 2 Ans (D)25/ph-1 4 cos2 – 2 cos – 1 = 0 cos = 2 2  8 +16 = 8 2  6 4 cos = 6 + 2   =  ; 2   = 23 4 12 cos =  = – 4 12 12 cos = cos(–5/12) ; cos(+5/12)  = 7/12 ; 17/12] PART-B MATCH THE COLUMN [6 × 8 = 48] INSTRUCTIONS: Column-I and column-II contains four entries each. Entries of column-I are to be matched with some entries of column-II. One or more than one entries of column-I may have the matching with the same entries of column-II and one entry of column-I may have one or more than one matching with entries of column-II. Column-I Column-II Q.1 (A) Area of the triangle formed by the straight lines (P) 1 x + 2y – 5 = 0, 2x + y – 7 = 0 and x – y + 1 = 0 in square units is equal to (Q) 3/4 (B) Abscissa of the orthocentre of the triangle whose vertices are the points (–2, –1); (6, – 1) and (2, 5) (R) 2 (C) Variable line 3x( + 1) + 4y( – 1) – 3( – 1) = 0 for different values of  are concurrent at the (S) 3/2 point (a, b). The sum (a + b) is (D) The equation ax2 + 3xy – 2y2 – 5x + 5y + c = 0 represents two straight lines perpendicular to each other, then | a + c | equals [Ans. (A) P; (B) R; (C) Q; (D) P] [Sol. (A) r = 3 0 + 4  3 + 5 0 3 + 4 + 5 12 = 12 = 1  Inradius is 1 Ans. (D) a = 2, c = – 3 ] Column-I Column-II If the perimeter of the parallelogram is 2( a + b ) where a, b  N then (a + b) equals [Ans. (A) S; (B) R; (C) S; (D) R] [Hint: (A) b + c = a + d = 2 · 10  a + b + c + d = 40 (B) (ar2)2 = a2 + a2r2 where a = 2  r4 = 1 + r2 r4 – r2 – 1 = 0 let r2 = t t2 – t – 1 = 0 t = 1 5 2 r2 = 1+ 5 2  1+ 5 2 1+ 5  ( 1 5 2 reject)  hypotenuse is 2 ×    = 1 +  comparing with a + a = 1, b = 5  a2 + b2 = 1 + 25 = 26 Ans. (C) a, ar, ar2  G.P. | r | < 1 a + ar + ar2 = 70  10ar = 4a + 4ar2 10r = 4 + 4r2 2r2 – 5r + 2 = 0 2r2 – 4r – r + 2 = 0 (2r – 1)(r – 2) = 0  r = 2 (reject) or r = 1/2 for r = 1/2 a a a + 2 + 4 3a = 70 7a a + 4 = 70  4 = 70  a = 40  series is 40, 20, 10  first term of G.P. is 40 Ans. (D) Using cosine rule  1  a2 = 9 + 4 – 2 · 2 · 3 ·   2  = 13 + 6 = 19   a2 = 19  a =  1  |||ly b2 = 9 + 4 – 2 · 2 · 3 ·  2    b2 = 7  b =  P = 2( + 7 )  a + b = 26 Ans. ]