MATHEMATICS-24-12- 11th (PQRS & J) SOLUTION

REVIEW TEST-7 Class : XI (P,Q,R,S & J) PAPER CODE : A Time : 2 hour Max. Marks : 130 INSTRUCTIONS 1. The question paper contains 00 pages and 3-parts. Part-A contains 20 objective question, Part-B contains 3 "Match the Column" questions and Part-C contains 4 "Subjective" questions. 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. PART-A (i) Q.1 to Q.15 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.16 to Q.20 have one or more than one correct alternative(s) 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.3 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 NEGATIVE marking. 0.5 Marks will be deducted for each wrong match. Marks will be awarded only if all the correct alternative(s) is/are selected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions and carry 9 marks each. There is NO NEGATIVE marking. Marks will be awarded only if all the correct bubbles are filled in your OMR answer sheet. 2. Indicate the correct answer for each question by filling appropriate bubble in your OMR 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. 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 bubbles is P Q R S (A) (B) (C) (D) PART-C Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column will be treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimal places. e.g. 86 should be filled as 0086.00 XI(PQRS) MATHEMATICS REVIEW TEST-7 PART-A Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking and 1 mark will be deducted for each wrong answer. Q. 1ph-3 A triangle with sides a = 15, b = 28 and c = 41. The length of the altitude from the vertex B on the side AC is (A) 6 (B) 7 (C*) 9 (D) 16 [Sol. Note that C must be largest cos C = a 2 + b2  c2 2ab (15)2 + (28)2  (41)2 = 2·15·28 4 = – 5 . Hence sin C = 3/5 now h = a sin(180 – C) = a sin C 3 = 15 · 5 = 9 Ans. ] Q.2 If sides a, b and c of triangle ABC satisfy a3 + b3 + c3  C  = c2 then tan  4  has the value equal to (A) – 1 (B*) 2 – a + b + c (C) 1/   (D) 2 + [Hint: a3 + b3 + c/ 3 = (a + b)c2 + c/ 3 a2 + b2 – ab = c2 a 2 + b2  c2 1 2ab = 2 1 Hence, cos C = 2  C = 60° C  4 = 15°  C  So, tan  4  = 2 – Ans.]   Q.3ph-3 In a triangle ABC, ABC = 45°, point D is on BC so that 2BD = CD and  DAB = 75°. ACB equals (A*) 15° (B) 60° (C) 30° (D) 75° [Hint: Using m-n theorem 3 cot 120° = 2 cot 45° – 1 · cot  cot  = 2 + 3  = 15° Ans. ] Q.4 The first term of an infinite geometric series is 2 and its sum be denoted by S. If | S – 2 | < 1/10 then the true set of the range of common ratio of the series is  1 , 1   1 , 1  (A) 10 5  (B)   2  – {0}   1 , 1   1 , 1  (C)  19 20  – {0} (D*)  19 21  – {0}     [Hint: 2  2 1 r < 1  10 1 < 1  20 < 1  20 1 < 20 ] Q.5ph-2 Number of solution satisfying the equation, tan22x = 2 tan 2x · tan 3x + 1 in [0, 2] is (A*) 0 (B) 1 (C) 2 (D) 4 [Sol. 2 tan 2x · tan 3x = tan22x – 1 2 tan 2x  1  tan 2 2x = – cot 3x   + 3x  tan 4x = tan     4x = n +   + 3x  x = n + which is not in domain 2  no solution ] cos   5 2   sin  cos 5  sin   Q.6ph-1 Numerical value of sin + cos  + 12  12 4  , is 12  12 4    1 (A) 2 (B*) 3 2 (C) 2 + (D) 1 + 3 2 [Sol. cos 15°(sin 75° + cos 45°) + sin 15°(cos 75° – sin 45°) sin(75° + 15°) + cos(45° + 15°) 1 3 1 + 2 = 2 ] Q.7cir Two circles both touching the coordinate axes and pass through the point (6, 3). The radii of the two circles are the roots of the equation (A) t2 – 12t + 20 = 0 (B) t2 – 15t + 36 (C*) t2 – 18t + 45 = 0 (D) t2 – 14t + 48 = 0 [Hint: r1 = 15; r2 = 3]  1   1  Q.8qe Let 'a' and 'b'are the roots of the equation x2 – mx + 2 = 0. Suppose that a + b  and  b + a  are the     roots of the equation x2 – px + q = 0. If p = 2q then the value of m is equal to (A) 4 (B*) 6 (C) 8 (D) 9 [Sol. a + b = m; ab = 2 1 (a + b) + a 1 + b = p a + b (a + b) + ab = p  p = m + m = 3m 2 2 ....(1) a + 1  b + 1  1 also     b   a  = q  ab + ab + 2 = q  2 + 1 2 + 2 = q  q = 9 2 ....(2) now from (1) and (2) p = 2q 3m 2 = 9  m = 6 Ans. ] 1 Q.9 The value of the determinant x y 0 1 1 1 x x 1+ x  y depends on (A) only x (B) only y (C) both x and y (D*) neither x nor y [Sol. C1  C1 + C3 0 D = 1 1+ x 0 1 1 1 x x 1+ x  y = –1[x – 1 – x] = 1 Ans.] Q.10 The sum of all the positive integers greater than 1 and less than 1000, which leave a remainder of one when divided by 2, 3, 4, 5 and 6, is (A*) 8176 (B) 7936 (C) 8167 (D) none [Sol. Any number that leaves a remainder of 1 when divided by 2, 3, 4, 5 and 6 must exactly 1 more than a number that is divisible by all 5 of these. The smallest being 60. Hence the series n 61 + 121 + 181 + ....... + 961 now 961 = 61 + n  1 60  n = 16  S = 16 (61 + 961) = 8(1022) = 8176 Ans. ] 2 Direction for Q.11 and Q.12 (2 questions together) Consider the digits 1, 2, 2, 3, 3, 3 and answer the following Q.11 If all the 6 digit numbers using these digits only are formed and arranged in ascending order of their magnitude then 29th number will be (A) 213332 (B) 233321 (C*) 233312 (D) none 5! [Sol. Starting with 1 Starting with 2 Hence 30th number is 233321  29th number is 233312  C ] 3!2! = 10 5! 3! = 20 Q.12 Let M denotes the number of six digit numbers using only the given digits if not all the 2's are together and N denotes the corresponding figure if no 3's are together then M – N equals (A) 16 (B*) 28 (C) 54 (D) 36 [Sol. M = Total – When all 2's together = 6! 2!3! – 5! = 60 – 20 = 40 3! N = When no 3's are together = M – N = 40 – 12 = 28 Ans.] 3! × 4 C = 12 | 1 | 2 | 2 | 2! Q.13 Number of selections that can be made of 6 letters from the word "COMMITTEE" is (A) 20 (B) 17 (C) 34 (D*) 35 [Sol. 2 alike + 2 other alike + 2 other alike = 1 M's = 2; T's = 2; E's = 2 2 alike + 2 other alike + 2 different = 3C2 · 4C2 = 18 C's = 1; 2 alike + 4 different = 3C1 · 5C1 = 15 All 6 different = 1 —— = 35 Ans. ] O's = 1; I's = 1 Q.14 A circle of radius r touches the lines given by the equation 4x2 – 4xy + y2 – 18x + 9y – 36 = 0. Area of the circle in square units is (A) 45  (B) 75  (C) 45/2 (D*) 45/4 [Hint: Line pair is 2x – y + 3 = 0 and 2x – y – 12 = 0 d =  A = r2 = 2r  r = 3 5 2  D ] Q.15 If the maximum and minimum value of the expression x + 2 2x2 + 3x + 6 (x  R) are M and m respectively then the value of 1  1 M m equals to (A) – 13 (B) – 10 (C) 10 (D*) 16 x + 2 [Sol. y = 2x2 + 3x + 6  2x2y + 3xy + 6y = x + 2  2x2y + x(3y – 1) + 2(3y – 1) = 0, x  R  D  0  (3y – 1)2 – 16y(3y – 1)  0  9y2 – 6y – 48y2 + 16y + 1  0  – 39y2 + 10y + 1  0  39y2 – 10y – 1  0  39y2 – 13y + 3y – 1  0  (3y – 1)(13y + 1)  0  maximum value (M) = 1 and minimum value (m) = – 1 3 13  1  1 M m = 3 + 13 = 16 Ans. ] Select the correct alternatives. (more than one are correct) [5 × 5 = 25] There is NO NEGATIVE marking. Q.16 If sin (x + 20°) = 2 sin x cos 40° where x   0,   then which of the following hold good    2  x x (A*) sec 2 =  (B*) cot 2 = (2 + ) (C) tan 4x = 3 (D) cosec 4x = 2 [Sol. sin x cos 20° + cos x sin 20° = 2 sin x cos 40° sin 20° cos x = sin x (2 cos 40° – cos 20°) tan x = sin 20 2 cos 40 – cos 20 = sin 20 sin 20 cos 40+ cos 40 – cos 20 sin 20 sin 20 = cos 40+ 2 sin 30sin (10) = sin 50 – sin10 = 2 cos 30sin 20 tan x =  x = 30°  A and B ] Q.17 If the vertices of an equilateral triangle ABC are (1, 1); (–1, –1) and (a, b) then (A*) a2 + b2 must be equals to 6 (B*) a + b must be equals to zero (C) a + b can be equal to 2 (D*) length of its median is [Hint: (a, b)  ( 3,  3) or (a, b)  ( 3, 3)  a2 + b2 = 6 & a + b = 0 ] Q.18 The sides of a right triangle T1 are 20, x and hypotenuse y. The sides of another right triangle T2 are 30, x – 5 and hypotenuse y + 5. If P1 and P2 are the radii of the circles inscribed and 1 and 2 are the areas of the triangles T1 and T2 respectively then which of the following hold good? (A) 61 = 52 (B*) 81 = 72 (C*) P1 = P2 (D) 2P1 = P2 [Sol. x2 + 400 = y2 ...(1) & 900 + (x – 5)2 = (y + 5)2  900 + x2 – 10x = y2 + 10y  900 + x2 – y2 = 10(x + y)  500 = 10(x + y)  x + y = 50 ...(2) From (1) & (2) y – x = 8 ...(3) From (2) & (3) So, y = 29 & x = 21  1 = 210 & 2 = 240  2 = 8  8 = 7 1 7 1 2 Also, P1 = 1 = S 210 35 = 6 & P2 = 2 = S 240 40 = 6 ] Q.19 ABCD is a quadrilateral co-ordinates of whose vertices are A(1, 0), B(–1, 0), C(3,4) and D(–3, 4) then (A) The diagonals of the quadrilateral are equal but not at right angle (B*) Area of the quadrilateral is 16 (C*) Circle passing through any three points of this quadrilateral also passes through the fourth point (D*) The quadrilateral ABCD is an equilateral trapezium Q.20 Let A  (1, 2); B  (3, 4) and C  (x, y) be any point satisfying (x – 1)(x – 3) + (y – 2)(y – 4) = 0 then which of the following hold good? (A*) Maximum possible area of the triangle ABC is 2 square units (B*) Maximum number of positions of C in the XY plane for the area of the triangle ABC to be unity, is 4 (C*) Least radius of the circle passing through A and B is 2 (D*) If 'O' is the origin then the orthocentre as well as circumcentre of the triangle OAB lies outside this triangle PART-B MATCH THE COLUMN [3 × 8 = 24] There is NEGATIVE marking. 0.5 Marks will be deducted for each wrong match. 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 mayhave the matching with the same entries of column-II and one entryof column-I mayhave one or more than one matching with entries of column-II. Q.1 Column I Column II (A) Number of increasing permutations of m symbols (P) nm are there from the n set numbers {a1, a2, , an} where the order among the numbers is given by a1 < a2 < a3 <  an–1 < an is (B) There are m men and n monkeys. Number of ways (Q) mCn in which every monkey has a master, if a man can have any number of monkeys (C) Number of ways in which n red balls and (m – 1) green (R) nCm balls can be arranged in a line, so that no two red balls are together, is (balls of the same colour are alike) (D) Number of ways in which 'm' different toys can be (S) mn distributed in 'n' children if every child may receive any number of toys, is [Ans. (A)  R; (B)  S; (C)  Q; (D)  P] Q.2 (A) Column I Column II If the lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 (P) Arithmetic Progression and cx + 4y + 1 = 0 passes through the same point, then a, b, c are in (B) Let a, b, c be distinct non-negative numbers. (Q) Geometric Progression If the lines ax + ay + c = 0, x + 1 = 0 and cx + cy + b = 0 passes through the same point, then a, b, c are in (C) If the lines ax + amy + 1 = 0, bx + (m + 1)by + 1 = 0 (R) Harmonic Progression and cx + (m + 2)cy + 1 = 0, where m  0 are concurrent then a, b, c are in (D) If the roots of the equation (S) None x2 – 2(a + b)x + a(a + 2b + c) = 0 [Hint: (D) be equal then a, b, c are in [Ans. (A)  P; (B)  S; (C)  R; (D)  Q] Roots equal  D = 0  4(a + b)2 = 4a(a + 2b + c) a2 + b2 + 2ab = a2 + 2ab + ac  b2 = ac  a, b, c are in G.P.  (Q)] Q.3 Column I Column II n n C (A) Lim  2 equals (P) 0 n n=1 2n (B) Let the roots of f (x) = 0 are 2, 3, 5, 7 and 9 (Q) 1 and the roots of g (x) = 0 are – 1, 3, 5, 7 and 8. f (x) Number of solutions of the equation g(x) = 0 is (C) Let y = sin3 x cos x cos3 x + sin x  where 0 < x < 2 , (R) 3/2 then the minimum value of y is (D) A circle passes through vertex D of the square ABCD, (S) 2 and is tangent to the sides AB and BC. If AB = 1, the radius of the circle can be expressed as p + q , then p + q has the value equal to [Ans. (A)  S; (B)  S; (C)  Q; (D)  Q ] n n C [Hint: (A) S = Lim  2 n n=1 2n S = 1 + 3 + 6 + 10 + ........ .....(1) 4 8 16 32 S = + 1 + 3 + 6 2 8 16 32 + ........ .....(2) ——————————— S = 1 + 2 + 3 + 4 + ........ .....(3) [From (1) – (2)] 2 4 8 16 32 S = + 1 + 2 + 3 4 8 16 32 +........ .....(4) ———————————— S = 1 + 2 + 1 + 1 +........ [From (3) – (4)] 4 4  S = 8 16 (1 4) = 32 1 · 2 = 1  S = 2 Ans. ] 4 1 (1 2) 4 1 2 (B) Uncommon roots when g (x)  0 and f (x) is zero are 2, 9 which are the solutions.] (C) y = 1 2 sin 2 x cos2 x sin x cos x sin 2 x + cos2 x = sin x cos x – 2 sin x cos x = (tan x + cot x) – sin 2x Now, (tan x + cot x) is minimum at x = /4 and sin 2x is maximum at x = /4  ymin occurs at x = /4 and ymin = 1 ] (D) r2 = 2(1 – r)2 r2 – 4r + 2 = 0 r = 2 – ] PART-C SUBJECTIVE: [4 × 9 = 36] There is NO NEGATIVE marking.  1 33 Q.1 If x1 and x2 are the two solutions of the equation 3log2 x 12xlog16 9 = log   , then find the value of 3 3  x2 + x2 . [Ans. x = 4 or x = 16; x2 + x2 = 272 ] 1 2 1 33 1 2 1 2  1  [Sol. 3log2 x 12·xlog16 9 = log   or xlog2 3 12·xlog4 3 = log   3 3  3  33  or let  log 3 1 log 3 x 12·x xlog 2 3 = y  =  log3 333 = – 27  3  y – 12 + 27 = 0  ( y  9) (  3) = 0  y = 81 or y = 9 hence xlog2 3 = 34 or 3log2 x = 34  Ans. or So, 3log2 x = 32 x2 + x2 = 272 = 32  Ans. ] 1 2 Q.2cir A circle with center in the first quadrant is tangent to y = x + 10, y = x – 6, and the y-axis. Let (h, k) be the center of the circle. If the value of (h + k) = a + b where is a surd, find the value of a + b. [Sol. Given h > 0, k > 0 [Ans. h + k = 2 + 8 ;  a + b = 10] 2r = distance between the two parallel lines x – y + 10 = 0 and x – y – 6 = 0 2r = but = 8  r = 4 h = r = 4 h  k +10 now r = 2 = 4  (0, 0) and (h, k) lies on the same side h – k + 10 > 0  h – k + 10 = 8  k = 2 + 4 2  h + k = 2 + 8  a + b = 10 Ans. ] Q.3 Suppose that there are 5 red points and 4 blue points on a circle. Find the number of convex polygons whose vertices are among the 9 points and having at least one blue vertex. [Ans. 450] [Sol. 9 ; [T/S, Q.15, Ex-2, prob in prob form] Remarks: When 6 or more vertices are taken then at least one blue has to be taken. n (A) = (84 – 10) + (126 – 5) + (126 – 1) + 84 + 36 + 3 + 1 = 74 + 121 + 125 + 130 = 450 Ans.] Q.4 Triangle ABC lies in the Cartesian plane and has an area of 70 sq. units. The coordinates of B and C are (12, 19) and (23, 20) respectively and the coordinates of Aare (p, q). The line containing the median to the side BC has slope –5. Find the largest possible value of (p + q). [Ans. 47] [Sol. = – 5  From slope 39 – 2q = –5(35 – 2p) 39 – 2q = –175 + 10p i.e. Also, p ± 12 23 q 1 19 1 20 1 = 140  From Area i.e. 11q – p = 337 11q – p = 57 Also 51p4+ q2 =41307 solving 51p4+ q2 =41307 solving  p = 15 & q = 32 So, p + q = 47 p = 20 & q = 7 p + q = 27 Hence, largest possible value = 47 Ans. ]