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5, , TERM-1, , SAMPLE PAPER, SOLVED, , MATHEMATICS, (STANDARD), , Time Allowed: 90 Minutes, , Maximum Marks: 40, , General Instructions: Same instructions as given in the Sample Paper 1., , SECTION - A, , 16 marks, , (Section A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted.), , 1. The largest number which on dividing 70 and, , 125 leaves remainders 5 and 8, respectively,, is:, , (a) 5, , (b) 13, , (c) 9, , (d) 11, , 6. In the given figure, DE || BC. The value of x is:, A, , x, , 2. If k + 1 = sec2A(1 – sin A) (1 + sin A), then the, value of k is:, , (a) 0, , (b) 1, , (c) 2, , (d) 3, , D, , 3. Find the length of diagonals of a rectangle, AOBC whose three vertices are A(0, 3),, O(0, 0) and B(5, 0)., , (a), , 23 units, , (c) 21 units, , (b) 5 units, (d), , 34 units, , 4. A wire, bent in the form of a square, encloses, , an area of 121 cm2. If the same wire is bent, in the form of a circle, then the circumference, of the circle is:, , (a) 11 cm, , (b) 22 cm, , (c) 33 cm, , (d) 44 cm, , 5. If the distance between the points A(–3, –14), and B(p, –5) is 9 units, then the value of p is:, , (a) 1, , (b) –7, , (c) –3, , (d) 5, , (a) 2, (c) 7, , x+3, , E, , 3x + 4, , 3x + 19, , B, , C, , (b) 4, (d) 11, , 7. What will be the number of the zero(s), if the, graph of a quadratic polynomial does not, intersect the x-axis?, (a) 0, (b) 1, (c) 2, (d) 3, , 8. On choosing a letter randomly from the, , letters of the word ‘ASSASSINATION’, the, probability that the letter chosen is a vowel, 6, is in the form of, . Then x is equal to:, 2x + 1, (a) 8, (b) 7, (c) 6, (d) 5, , 9. If the diameter of a wheel is 1.54 m, then the, distance covered by it in 100 revolutions is:

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(a) 143 m, (c) 484 m, , 14. Find the value of (x, y), if centroid of the, , (b) 275 m, (d) 396 cm, , 10. Two alarm clock ring their alarms at regular, intervals of 50 seconds and 48 seconds. If, they first beep together at 12 noon, at what, time will they beep again?, (a) 12 : 20 p.m., (b) 01 : 05 p.m., (c) 02 : 20 p.m., (d) 12 : 35 p.m., , 11. Two dice are thrown together. Then the, , probability that sum of the two numbers on, the dice will be multiple of 4 is:, 3, 1, (a), (b), 4, 4, 1, (c), (d) 0, 2, 12. Evaluate the zeroes of the polynomial 2x2 +, 14x + 20., (a) –2, –5, (b) 2, 5, (c) –2, 5, (d) –5, 2, , 13. Given two triangles ABC and DEF. If DABC ~, DDEF, 2AB = DE and BC = 8 cm, then find the, length of EF., D, A, , triangle with vertices (x, 0), (0, y) and, (6, 3) is (3, 4)., (a) (3, 0), (b) (6, 6), (c) (3, 9), (d) (–6, 8), , 15. In which quadrant does the mid-point of the, line segment joining the points (–1, 2) and, (3, 4) lies?, (a) I, (b) II, (c) III, (d) IV, , 16. If, , 241, 241, , then find the value of m +, =, 4000, 2m × 5n, , n, where m and n are non-negative integers., (a) 10, (b) 8, (c) 6, (d) 7, , 1, sin A, +, , if, tan A 1 + cos A, , 17. Evaluate the value of, cosec A = 2., (a) 2, (c) 1, , (b) 0, (d) –1, , 18. What is the value of k, if one zero of the, polynomial (k – 1)x2 – 10x + 3 is reciprocal of, the other?, (a) 4, (b) 5, (c) –1, (d) 0, , 19. Calculate the value of x, if LCM (x, 18) = 36, and HCF (x, 18) = 2., (a) 4, (c) 2, , B, , C, , (a) 10 cm, (c) 8 cm, , E, , F, , (b) 12 cm, (d) 16 cm, , 20. The value of, , (b) 8, (d) 6, , sin3q + cos3q, sinq + cosq, , (a) sin q cos q, (c) cot q, , + sin q cos q is:, , (b) tan q, (d) 1, , SECTION - B, , 16 marks, , (Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted.), , 21. Rajesh and Mahesh are playing a game. In, , this game, each player throws two dice and, note down the numbers on the dice. By the, rules of the game, Mahesh needs to get two, numbers such that their product is a perfect, square, in order to win the game. What is the, probability that Mahesh will win the game?, 1, 2, (a), (b), 9, 9, 1, 2, (c), (d), 3, 7, , 22. In the given figure, if AB = BC = CD = 7 cm,, then the perimeter of shaded region is:, , (a) 21 cm, (c) 35 cm, , (b) 42 cm, (d) 66 cm, , 23. What is the value of x, if DADE ~ DACB, ∠DEC, = 105° and ∠ECB = 65°?

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y, , A, , l₂, , x, l₁, , E, D, , 105, , º, 65º, C, , B, , (a) 45°, , (b) 60°, , (c) 13°, , (d) 40°, , x', , x, , O, , 24. Two triangles are similar and their areas, , are 121 cm2 and 64 cm2 respectively. If, the median of the first triangle is 12.1 cm,, calculate the measure of corresponding, median of the other triangle., , (a) 6.4 cm, , (b) 8.8 cm, , (c) 9.6 cm, , (d) 7.6 cm, , y', , (a) Infinite, , (b) Two, , (c) Unique, , (d) No solution, , 2, 3, BC and area of DABC = 81 cm2, then the area, of DDAE is:, , 30. Consider a DABC, where DE || BC. If DE =, , 1 1, + , if a and b are the, α β, zeroes of the polynomial x2 + x + 1., , (a) 24 cm2, , (b) 16 cm2, , (a) 1, , (b) 0, , (c) 36 cm2, , (d) 32 cm2, , (c) –1, , (d) 2, , 25. Find the value of, , 26. On choosing a number x from the numbers 1,, 2, 3 and a number y from the numbers 1, 4,, 9, the probability of P(xy < 9) is:, 5, 1, (a), (b), 9, 9, (c), , 4, 9, , (d), , 27. Find the value of n if a = 23 × 3, b = 2 × 3 × 5,, c = 3n × 5 and LCM (a, b, c) = 23 × 32 × 5., (b) 2, , (c) 3, , (d) 4, , angled at B with AB = 4 cm. What is the, length of AC?, (a) 2 cm, , (b) 2 2 cm, , (c) 4 cm, , (d) 4 2 cm, , 32. Amit and Prem are very good cricketers and also, , 3, 9, , (a) 1, , 31. ABC is an isosceles triangle, which is right, , represented their school team at district and, even state levels. One day, after their match,, they measured the height of the wickets and, found it to be 28 inches. They marked a point, P on the ground as shown in the figure below:, , 28. From the following figure, the value of sin A, cos A + sin C cos C is:, , C, , 3, , A, , B, 4, , (a), (c), , 12, 5, 7, 12, , (b), (d), , 24, 25, 7, 24, , 29. The number of solutions of the pair of linear, equations, shown in the graph is:, , 3, If cot P = , the length of PQ is:, 4, (a) 3 in, (b) 7 in, (c) 21 in, (d) 35 in, , 33. Co-prime numbers is a set of numbers which, have 1 as their ....................... .

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37. How many zeroes will be there for the, , (a) only factor, , (b) LCM, , (c) HCF, , (d) The two factors, , 34. If one of the zeroes of the polynomial, , f(x) = x2 – 7x – 8 is – 1, then find the other, zero., , (a) 7, , (b) 1, , (c) 8, , (d) 5, , of radius 30 cm, subtends an angle q at the, centre O. The value of q is:, , (a) 30°, , (b) 37°, , (c) 45°, , (d) 52°, , 36. Evaluate sin2, (a) 1, (c) –, , 1, 3, , 38. Calculate the minimum number by which, 8 should be multipled so as to get a, rational number., , 35. An arc of length of length 19 cm of a circle, , ≠ 0 and, , polynomial f(x) = (x – 2)2 + 4?, (a) 0, (b) 1, (c) 2, (d) 3, , – cos2 , if 3 tan, is an acute angle., 1, (b), 3, (d) – 1, , = 3 sin ,, , (a), , 2, , (b), , 3, , (c), , 5, , (d), , 6, , 39. Find the value of x if, , 4 – sin2 45°, = 3.5., cot x. tan 60°, , (a) 0°, (c) 30°, , (b) 15°, (d) 60°, , 40. If DABC ~ DPQR, then evaluate the length of, AC, if perimeter of DABC = 20 cm, perimeter, of DPQR = 40 cm and PR = 8 cm., (a) 4 cm, (b) 6 cm, (c) 10 cm, (d) 3 cm, , SECTION - C, , 8 marks, , (Case Study Based Questions.), (Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted.), Q. 41-45 are baded on Case Study–1, Case Study–1:, Sam went for an outing with his friends.They, went to dominos to enjoy the delicious pizza. He, was enjoying the pizza with his friends and share, with them by slicing it. During slicing the pizza, he, noticed that the pair of linear equations formed. (i.e.,, straight lines), Let these pair of linear equations be y – 2x = 1 and, 5y – x = 14., , 42. At what point, does the linear equation, y – 2x = 1 intersect the y-axis?, , (a) (0, 1), ,  1 , (b)  − , 0 ,  2 , ,  14 , (c)  0,, , , 5 , , (d) (0, –14), , 43. The system of linear equations 2x – 3y + 6 =, 0 and 2x + 3y – 18 = 0:, , (a) has infinitely many solutions, (b) has no solution, (c) has a unique solution, (d) May or may not have a solution, , 44. For what value(s) of k, the system of linear, equations 2x – ky + 3 = 0 and 3x + 2y – 1 = 0, has no solution?, , (a) –6, 4, (b), 3, , (b) 6, (d) –, , 4, 3, , 45. If a pair of linear equations in two variables, 41. What is the point of intersection of the lines, represented by the equations y – 2x = 1 and, 5y – x = 14?, (a) (1, 3), (b) (6, 4), (c) (–2, 3), (d) (–4, 2), , is inconsistent, then the lines represented by, two equations are:, , (a) parallel, (b) intersecting, (c) always coincident, (d) Intersecting or coincident

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46. Find the ratio in which C divides the line, , Q. 46-50 are baded on Case Study–2, Case Study–2:, Shyla is a very talented lady. She is always, interested in doing something creative in her free, time after the household work. She embroidered a, leaf by knitting on her table cloth. Her son trace the, design on a coordinate plane as shown below., , joining W and E., (a) 5 : 4, (c) 2 : 5, , (b) 5 : 3, (d) 1 : 1, , 47. What is the ratio in which x-axis divides the, line joining the points P and D?, (a) 1 : 1, (b) 4 : 5, (c) 2 : 1, (d) 8 : 3, , 48. What is the ratio in which y-axis divides the, line joining the points L and U?, (a) 1 : 4, (b) 7 : 9, (c) 4 : 7, (d) 9 : 2, , 49. What is the distance of point K from the, origin?, (a) 3 units, (c) 7 units, , (b) 5 units, (d) 10 units, , 50. From the given points, the mid-point of which, doesn't lie on y-axis?, (a) U and G, (b) P and L, (c) Q and K, , (d) U and F, , SOLUTION, SAMPLE PAPER - 5, SECTION - A, 1. (b) 13, Explanation:, Required number = HCF (70 – 5, 125 – 8), = HCF (65, 117), ·.· , 65 = 5 × 13, 2, , , and , 117 = 3 × 13, .·. Required number = 13., , , ⇒ , , k+1 =1, , , ⇒ , , k =0, , 3. (d) 34 units, Explanation: Length of diagonal = AB =, (5 – 0) 2 + (0 – 3) 2 =, , 25 + 9 = 34, , A(0, 3), , C, , O(0, 0), , B(5, 0), , 2. (a) 0, Explanation: We have,, , k + 1 = sec2 A(1 – sin A) (1 + sin A), = sec2 A(1 – sin2 A), [·.· (a – b)(a + b) = a2 – b2], = sec2 A . cos2 A, [·.· sin2 q + cos2 q = 1], =, , 1, cos2 A, , × cos2 A, ·.· sec q = 1 , cos q , , , Concept Applied, ,  Diagonals of a rectangle are equal in length.

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4. (d) 44 cm, , .·., , Explanation: Let a be the side of the square., a2 = 121 cm2, , Then, , .·. , , a=, , [Given], , 121 = 11, , Perimeter of square = 4a = 4 × 11 = 44 cm, , , = Circumference of the circle, , 5. (c) –3, , Explanation:, ·.·, , AB = 9 units, , .·. Using distance formula,, , ( p + 3)2 + ( −5 + 14 )2 = 9, , ⇒, (p + 3)2 + (9)2 = 81 [squaring both sides], , , , ⇒, (p + 3)2 + 81 = 81, , ⇒, , (p + 3)2 = 0, , , ⇒, , p+3=0, , ⇒ p = –3, , 6. (a) 2, Explanation: DE || BC, \ By Thale's theorem,, AD, AE, , =, BD, CE, x, x+3, ⇒, =, 3x + 4, 3x + 19, ⇒, ⇒, ⇒, , x(3x + 19) = (3x + 4)(x + 3), 3x2 + 19x = 3x2 + 13x + 12, 6x = 12 ⇒ x = 2, , 7. (a) 0, , = 100 × Circumference of wheel, = 100 × 2pr, = 100 × 2 ×, = 484 m, , 8. (c) 6, Explanation: There are 13 letters in the word, ‘ASSASSINATION’., , \ Total number of outcomes = 13, There are 6 vowels in the word ‘ASSASSINATION’., 6, , \ Required probability =, 13, But given that,, 6, 6, , =, 13, 2x + 1, , ⇒ , 2x + 1 = 13, 2x = 12, x=6, , 9. (c) 484 m, , Explanation: Diameter of wheel = 1.54 m, 1.54, .·. Radius of wheel =, m, 2, Now, Distance covered by it in one revolution, = Circumference of the wheel, , 22, 7, , ×, , 1.54, 2, , 10. (a) 12 : 20 p.m., Explanation: We have,, 50 = 2 × 52, , 48 = 24 × 3, Time after which they beep together, = LCM (50, 48), = 24 × 3 × 52, = 1200 s, or 20 min, Since, the two clocks first beep together at 12, noon, so next they will beep together at 12, noon + 20 min i.e., 12 : 20 pm., 1, 4, , Explanation: Let E be the event of getting the, sum of two numbers as a multiple of 4., , 11. (b), , , i.e., E = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4),, (5, 3), (6, 2), (6, 6)}, , \ , n(E) = 9, , Here, total number of events, n(S) = 36, n (E), 9, , \ Required probability =, =, n (S), 36, =, , , Explanation: If the graph of a quadratic, polynomial does not intersect the x-axis, then, the number of zero(s) is zero., , , ⇒ , , ⇒ , , Distance covered by it in 100 revolutions, , 12. (a) –2, –5, , 1, 4, , , Explanation: Let, , p(x) = 2x2 + 14x + 20, = 2(x2 + 7x + 10), = 2(x2 + 5x + 2x + 10), , , [by splitting the middle term], , = 2[x(x + 5) + 2 (x + 5)], = 2(x + 2) (x + 5), To determine the zeroes, Put p(x) = 0, Þ, , 2(x + 2) (x + 5) = 0, , \ , x = –2 and x = –5, Hence, the zeroes of the given polynomial are, –2 and –5., , 13. (d) 16 cm, Explanation: Q, , \ , ⇒ , , DABC ~ DDEF[Given], BC, AB, =, EF, DE, AB, 8, =, [Q DE = 2AB], EF, 2 AB

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⇒ , , 8, 1, =, EF, 2, , , \ , , EF = 16 cm, , Caution,  L earning of trigonometric values for some standard, angles is needed., , 18. (a) 4, , 14. (c) (3, 9), , Explanation: Since, (3, 4) is the centroid of a, triangle with vertices (x, 0), (0, y) and (6, 3)., x+0+6, 0+y+3, , \, 3 =, and 4 =, 3, 3, , ⇒, x = 3 and y = 9, , \ (x, y) = (3, 9), , 15. (a) I, , , Explanation: Let one of the zeroes of the, polynomial be a., Then another zero is, Now,, , Explanation:, Mid-point of line segment, –1 + 3 2 + 4, n = (1, 3), = d, ,, 2, 2, \ (1, 3) lies in quadrant I., , 16. (b) 8, Explanation: Given, , 3, 1, =, (, k, − 1), α, 1=, , \, , 3, k −1, , , Þ, , k–1=3, , , Þ, , k=4, , 19. (a) 4, , 241, 241, = m, 2 × 5n, 4000, , 241, , a., , 1, ., α, , Explanation: We have,, LCM(x, 18) × HCF(x, 18) = x × 18, , 241, , Þ, , 36 × 2 = 18x, , On comparing, we have m = 5, n = 3, , Þ, , x=, , .·., , \, , x=4, , , Þ, , 25 × 53, , =, , 2m × 5n, , m+n =5+3=8, , 17. (a) 2, , Explanation: Given: cosec A = 2 ⇒ A = 30°, sin 30°, sin A, 1, 1, , \, +, =, +, tan A, tan 30°, 1 + cos 30°, 1 + cos A, 1, 1, 2, =, +, 1, 3, 1+, 3, 2, 1, 2, = 3 +, 2+ 3, 2, 2– 3, 1, = 3 +, ×, 2– 3, 2+ 3, 2– 3, 4–3, , =, , 3 +, , =, , 3 +2– 3 =2, , 36 × 2, 18, , 20. (d) 1, Explanation: We have,, sin3 q + cos3 q, , + sin q cos q, sin q + cos q, =, , (sin q + cos q)(sin2 q + cos2 q – sin q cos q), sin q + cos q, + sin q cos q, [·.· a3 + b3 = (a + b)(a2 + b2 – ab)], , = (sin2 q + cos2 q – sin qcos q) + sin qcos q, = (1 – sin q cos q) + sin q cos q, [·.· sin2 q + cos2 q = 1], = 1, , SECTION - B, 2, 9, Explanation: Number of possible outcomes on, throwing two dice = 36, Clearly, Mahesh will win when he get the, product of numbers as a perfect square i.e., when, he will get (1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4),, (5, 5), (6, 6)., , 21. (b), , , \ Number of favourable outcomes = 8, , \ P(getting a product of perfect square), =, , 8, 2, =, 36 9, , , \ Probability that Mahesh will win the game is, , 2, 9

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22. (d) 66 cm, Explanation: Perimeter of shaded region =, Length of semi-circular arc APB + Length of, semi-circular arc ARC + Length of semi-circular, arc BSD + Length of semi-circular arc CQD,  AB ,  AC ,  BD ,  CD , = π , +π , +π , +π , , , ,  2 ,  2 ,  2 ,  2 , , 5, 9, Explanation: Total number of possible cases =, 3 × 3 = 9., , 26. (a), , , \ Favourable cases = {(1, 1), (1, 4), (2, 1),, (2, 4), (3, 1)}, 5, P(xy < 9) =, 9, , , \ , , 27. (b) 2, , , , 7,  14 ,  14 , 7, = π   +π   +π   +π  , 2,  2 ,  2 , 2, , Explanation: We have,, , , , = p(3.5 + 7 + 7 + 3.5), , , , a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5, , \, , LCM (a, b, c) = 23 × 3n × 5, , 22, , , , =, , × 21, , , , = 66 cm, , 7, , , Þ, , 23 × 3n × 5 = 23 × 32 × 5, , , Þ, , n =2, , 28. (b), , 23. (d) 40°, Explanation:, , Q DADE ~ DACB[Given], , \ ∠ACB = ∠ADE = 65°, , 24, , 25, , Explanation:, theorem,, , In, , using, , DABC,, , Pythagoras, , AC2 = AB2 + BC2, , , , Also, ∠AED = 180° – 105° = 75° = ∠ABC, , = 42 + 32, , In DADE,, , = 16 + 9 = 25, , , ∠ADE + ∠AED + ∠DAE = 180°, , .·., , , ⇒ 65° + 75° + ∠DAE = 180°, , Now,, , , ⇒ , , ∠DAE = 180° – 140° = 40°, , 24. (b) 8.8 cm, Explanation: Let the corresponding median of, the other triangle be x cm., , \ , ,  12.1, 121, = ,  x , 64, , 2, , [Q The ratio of the areas of two similar triangles, , , is equal to the square of the ratio of their, , , , corresponding medians], , , ⇒ , , ⇒ , , ⇒ , , 11, = 12.1, 8, x, 12.1 × 8, x=, 11, x = 8.8, , , \ Corresponding median of the other triangle, , AC =, , sin A cos A + sin C cos C =, , 25 = 5, BC, , ×, , AC, , = 2 ×, = 2 ×, , AB, AC, , BC, AC, 3, 5, , ×, , ×, 4, 5, , +, , AB BC, ×, AC AC, , AB, AC, =, , 24, 25, , 29. (d) No solution, , Explanation: In the graph, the pair of linear, equations represent parallel lines. Since,, parallel lines never intersect, so they have no, solution., , 30. (c) 36 cm2, Explanation: Here, DE =, And DE || BC, , 2, BC, 3, , A, , is 8.8 cm., , 25. (c) –1, Explanation: a and b are the zeroes of the, polynomial x2 + x + 1., , \ , and , Now, , , D, , a + b = –1, ab = 1, 1 1, α +β, + =, = –1, αβ, α β, , B, , , \, , , \, , E, , C, , DADE ~ DABC , (by AA similarity), AD, AE, DE, =, =, AB, AC, BC

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Also,, , , 2, , 2, , 2, ar (3 ADE), DE, =, = d n, 3, ar (3 ABC), BC2, 4, 9, ar (TADE), 4, =, 81, 9, , =, ⇒ , , 35. (b) 37°, Explanation: Radius of circle = 30 cm, θ, Length of an arc AB =, × 2πr, 360°, where, q is the angle subtended by the arc AB, at the centre of circle., , ar(DDAE) = 36 cm2, , ⇒ , , 31. (d) 4 2 cm, , O, , , Explanation: Since DABC is an isosceles,, then,, \, , AB = BC., AB = BC = 4 cm, C, , A, , 19 × 7 × 180, =q, 22 × 30, 36.27° = q, q - 37°, , ⇒, , A, , θ, 22, × 2×, ×30, 360, 7, , 19 =, , \, , B, , B, , ⇒, ⇒, , 1, 3, Explanation: We have,, , 3 tan q = 3 sin q, , = (4)2 + (4)2, , ⇒ , , 3, , = 16 + 16, , ⇒ , , Using Pythagoras theorem, we have, , , , ⇒ , , , AC2 = AB2 + BC2, , AC =, , 32 = 4 2 cm, , , Explanation: It is given that QR = 28 in and, 3, cot P = ., 4, Base, We know that,, cot P =, Perpendicular, , , 33. (c) HCF, , PQ, PQ, =, =, ., 28, QR, , PQ, 3, 28 × 3, =, ⇒ PQ =, = 21 in., 28, 4, 4, , Explanation: Co-prime numbers have only 1, as their common factor., , 34. (c) 8, Explanation : We have,, , , f(x) = x2 – 7x – 8, , Now, sum of the zeroes = 7, Since, one of the zero is – 1., \ Other zero = 7 – (– 1) = 7 + 1 = 8, , sin θ, = 3 sin q, cos θ, 3, =3, cos θ, , AC2 = 32, , 32. (c) 21 in, , Therefore,, , 36. (b), , 3, = cos q, 3, 1, cos q =, 3, , ⇒ , \ , , ...(i), , Now, sin2 q – cos2 q = 1 – cos2 q – cos2 q, = 1 – 2 cos2 q, = 1 – 2 × e, = 1 –, , 1 2, o [Using (i)], 3, , 2, 1, =, 3, 3, , 37. (a) 0, , Explanation: The given polynomial is, f(x) = (x – 2)2 + 4, , , For zeroes, put, Þ , Þ , , f(x) = 0, , (x – 2)2 + 4 = 0, (x – 2)2 = – 4, , Which is not possible,, Hence, this polynomial has no zeroes., , 38. (a), , 2, , Explanation: The smallest number will be, Because,, , 2., , 8 × 2 = 16 = 4 , which is rational.

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39. (d) 60°, , , ⇒, , Explanation: We have,, , , , ⇒, , 4 – sin2 45°, = 3.5, cot x. tan 60°, , , ⇒, , , ⇒, , ⇒, , 1, = cot 60°, 3, , x = 60°, , 40. (a) 4 cm, , 1 2, o, 4–e, 2, = 3.5, cot x 3, 1, 4–, 2, = 3.5, 3 cot x, 3.5, = 3.5, 3 cot x, , , ⇒, , cot x =, , , Explanation: Since, DABC ~ DPQR,, , \ , , AC, Perimeter of ∆ABC, =, PR, Perimeter of ∆PQR, , , ⇒ , , AC, 20, =, 8, 40, , , ⇒ , , AC =, , 3 cot x = 1, , 20 × 8, = 4 cm, 40, , SECTION - C, 41. (a) (1, 3), , 45. (a) parallel, , Explanation: From the graph, it is clear that, lines are intersecting at a point (1, 3)., \ Point of intersection of lines is (1, 3)., , 46. (b) 5 : 3, , 42. (a) (0, 1), Explanation: From the figure, it is clear that, the equation y – 2x = 1 is intersecting y-axis, at (0, 1)., , 43. (c) has a unique solution, Explanation: We have system of linear, equations, , and, \, , 2x – 3y + 6 = 0, 2x + 3y – 18 = 0, a1, b1, 3, 2, =, = 1,, = − = –1, a2 2, b2, 3, a1, , ∵, , a2, , ¹, , 4, 3, Explanation: Given system of linear equations, 2x – ky + 3 = 0 and 3x + 2y – 1 = 0, For no solution, we have,, b1, a1, c, ¹ 1, , =, b2, a2, c2, , Þ , , Since, WC = 5 units and CE = 3 units, , \ C divides the line joining W and E in the ratio, 5 : 3., , 47. (c) 2 : 1, , Explanation: Clearly, the coordinates of P and, D are (–4, 8) and (5, –4) respectively., Let the x-axis divides the join of P and D at, point (x, y) in the ratio of k : 1., k:1, P, (–4, 8), , b2, , 44. (d) –, , Þ, , , Explanation: Clearly, the coordinates of W, C, and E are (–4, –3), (1, –3) and (4, –3), respectively., , b1, , \ It has a unique solution., , Þ, , , Explanation: When two lines are parallel then, the pair of the linear equations is inconsistent, i.e., no solution., , 2, k ¹ 3, = −, 3, 2, −1, , 4, 3, , D, (5, –4), , –4k + 8, k+1, But (x, y) lies on x-axis, therefore y = 0, Then, , , y=, , , ⇒ –4k + 8 = 0 ⇒ 4k = 8 ⇒ k = 2, Thus, the required ratio is 2 : 1., , 48. (c) 4 : 7, , Explanation: Clearly, the coordinates of L and, U are (4, 9) and (–7, 2) respectively., Let the y-axis divides the join of L and U at the, point (x, y) in the ratio k : 1., k:1, , 2, k, = −, 3, 2, k = −, , (x, y), , L, (4, 9), , , , Then,, , (x, y), , x=, , –7k + 4, k+1, , U, (–7, 2)

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But (x, y) lies on y-axis, therefore x = 0, , , ⇒ , , –7k + 4, =0, k+1, , , ⇒ , , 7k = 4, , 4, 7, Thus, the required ratio is 4 : 7., , ⇒ , , k=, , G are (–7, 2) and (7, 2) respectively, therefore, their mid-point is (0, 2), which lies on y-axis., Also, the coordinates of P and L are (–4, 8) and, (4, 9) respectively, therefore their mid-point is, 17, c 0,, m , which also lies on y-axis., 2, And, the coordinates of Q and K are (–3, 3) and, , Explanation: Coordinates of K are (3, 4),, therefore its distance from origin, , (3, 4) respectively, therefore, thier mid-point is, 7, c0, m which also lies on y-axis., 2, , 32 + 42 = 9 + 16 = 25 = 5 units, , Lastly, the coordinates of U and F are (–7, 2) and, , 49. (b) 5 units, , =, , 50. (d) U and F, Explanation: Clearly the coordinates of U and, , (9, 1) respectively, therefore, their mid-point is,  3, 1, 2  which doesn't lie on y-axis.