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X-STANDARD MATHEMATICS

PREPARED BY: R.RAJENDRAN. M.A., M . Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21. X-STANDARD MATHEMATICS. MODEL QUESTION. Section – A.

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X-STANDARD MATHEMATICS

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  1. PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21 X-STANDARD MATHEMATICS MODEL QUESTION

  2. Section – A Note: (i) Answer all the 15 questions(ii) Choose the correct answer in each question. Each of these questions contains four options with just one correct option(iii) Each question carries One mark 15 × 1 = 15

  3. Section – A • Let A = { 1, 3, 4, 7, 11 }, B = {–1, 1, 2, 5, 7, 9 } and f : A  B be given by f = { (1, –1), (3, 2), (4, 1), (7, 5), (11, 9) }. Then f is(A) one-one (B) onto (C) bijective (D) not a function

  4. Section – A • The common ratio of the G.P is(A) (B) 5(C) (D)

  5. Section – A • If a1 , a2 , a3 ,……are in A.P. such that then the 13th term of the A.P. is (A) (B) 0(C) 12a1 (D) 14a1

  6. Section – A • The LCM of 6x2 y, 9x2 yz,12x2 y2 z is(A) 36x2 y2 z(B) 48xy2 z2(C) 96x2 y2 z2(D) 72xy2 z

  7. Section – A • If b = a + c , then the equation ax2 + bx + c = 0 has(A) real roots (B) no roots (C) equal roots (D) no real roots

  8. Section – A • If then the order of A is (A) 2  1(B) 2  2(C) 1 2 (D) 3 2

  9. Section – A

  10. Section – A

  11. Section – A

  12. Section – A

  13. Section – A

  14. Section – A • (A) sin  + cos  (B) sin  cos (C) sin  – cos  (D) cosec  + cot 

  15. Section – A • If the total surface area of a solid hemisphere is 12 cm2 then its curved surface area is equal to(A) 6 cm2 (B) 24 cm2(C) 36 cm2 (D) 8 cm2

  16. Section – A • Mean and standard deviation of a data are 48 and 12 respectively. The coefficient of variation is(A) 42 (B) 25 (C) 28 (D) 48

  17. Section – A • If A and B are mutually exclusive events and S is the sample space such that P(A)=1/3 P(B) and S = A  B, then P(A) = (A) (B) (C) (D)

  18. Section – B Note: (i) Answer 10 questions(ii) Answer any 9 questions from the first 14 questions. Question No. 30 is Compulsory.(iii) Each question carries Two marks 10 × 2 = 20

  19. If A = {4,6,7,8,9}, B = {2,4,6} and C = {1,2,3,4,5,6}, then find A (B  C). ANSWER: B  C = {2, 4, 6}  {1, 2, 3, 4, 5, 6} = {2, 4, 6} A (B  C) = {4, 6, 7, 8, 9}  {2, 4, 6} = {2, 4, 6, 7, 8, 9}

  20. Let X = { 1, 2, 3, 4 }. Examine whether the relation g = { (3, 1), (4, 2), (2, 1) } is a function from X to X or not. Explain. X = {1, 2, 3, 4}, g = {(3,1), (4,2), (2, 1)} g is not a function. since 1 has no image. 1 2 3 4 1 2 3 4

  21. Three numbers are in the ratio 2 : 5 : 7. If 7 is subtracted from the second, the resulting numbers form an arithmetic sequence. Determine the numbers. The ratio of three numbers = 2 : 5 : 7 Let the numbers be 2x, 5x, 7x 2x, 5x – 7, 7x are in an AP 5x – 7 – 2x = 7x – (5x – 7) 3x – 7 = 2x – 5x + 7 3x – 7 = 2x + 7 3x – 2x = 7 + 7 x = 14 The numbers are 28, 70, 98

  22. If  and  are the roots of the equation 2x2 – 3x – 1 = 0, find the value of  –  if  >  , are the roots of 2x2 – 3x – 1 = 0 a = 2, b = – 3, c = – 1

  23. If then find the additive inverse of A. Additive Inverse of A = – A

  24. Find the product of the matrices, if exists

  25. The centre of a circle is at (-6, 4). If one end of a diameter of the circle is at the origin, then find the other end. Let the centre of the circle is (– 6, 4) One end = (0, 0), other end = (x, y) Midpoint of diameter = centre of the circle The other end = (– 12, 8)

  26. In ABC, DE || BC and . If AE = 3.7 cm, find EC. A Given, AE = 3.7cm E D C B In ABC, DE || BC

  27. A ladder leaning against a vertical wall, makes an angle of 60 with the ground. The foot of the ladder is 3.5 m away from the wall. Find the length of the ladder. A 60  Let AC = Length of the ladder BC = distance of the wall = 3.5m B C 3.5m In ABC, C = 60 AC = 3.5  2 = 7 Length of the ladder = 7m

  28. Prove the identity:

  29. A right circular cylinder has radius of 14 cm and height of 8 cm. Find its curved surface area. Radius = 14cm, height = 8cm Curved surface area = 2rh 2 = 2  22  2  8 = 44  16 = 704sq.cm

  30. The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume. Height = 12cm Circumference of the base = 44 2r = 44 Volume = r2h 2 = 22  7  12 = 154  12 = 1848cu.cm r = 7cm

  31. Calculate the standard deviation of the first 13 natural numbers. The standard deviation of the first ‘n’ natural numbers = The standard deviation of the first 13 natural numbers = 3 .7 14. 00 3 9 5 00 6 7 469 31

  32. Two coins are tossed together. What is the probability of getting at most one head. When two coins are tossed the sample space S = {HH, HT, TH. TT} n(S) = 4 A = {atmost one head} A = {HT, TH, TT} n(A) = 3

  33. (a) Simplify: (OR)

  34. (b) Show that the lines 2y = 4x + 3 and x + 2y = 10 are perpendicular. Slope of 2y = 4x + 3 is m1 = 2 Slope of x + 2y = 10 is m2 = - ½ m1 m2 = 2  (– ½ ) = – 1 The given lines are perpendicular.

  35. SECTION - C Note: (i) Answer 9 questions (ii) Answer any 8 questions from the first 14 questions. Question No. 45 is Compulsory. (iii) Each question carries Five marks 9 × 5 = 45

  36. Prove that A \ (B  C) = (A \ B)  (A \ C) using Venn diagram (2) A – ( BC) (3) A – B (1) BC A B A B B A C C C (4) A – C (5) (A – B)(A – C) A B B A From the diagram (2)and(5) A – (BC) = (A – B)(A – C) C C

  37. A function f: [-7, 6)R is defined as follows Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)

  38. A function f: [-7, 6)R is defined as follows Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) (i) Since -4 lies in the interval -5  x  2 f(x) = x + 5 f(-4) = – 4 + 5 = 1 Since 2 lies in the interval -5  x  2 f(x) = x + 5 f(2) = 2 + 5 = 7 2f(-4) + 3f(2) = 2(1) + 3(7) = 2 + 21 = 23

  39. Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) A function f: [-7, 6)R is defined as follows f(x) = (ii) Since -7 lies in the interval -7  x < -5 f(x) = x2 + 2x + 1 f(-7) = (–7)2 + 2(–7) + 1 = 49 – 14 + 1 = 36 Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(2) = -3 + 5 = 2 f(-7) – f(–3) = 36 – 2 = 34

  40. Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii) A function f: [-7, 6)R is defined as follows f(x) = (iii) Since -3 lies in the interval -5  x  2 f(x) = x + 5 f(-3) = -3 + 5 = 2 Since -6 lies in the interval -7  x < -5 f(x) = x2 + 2x + 1 f(-6) = (–6)2 + 2(–6) + 1 = 36 – 12 + 1 = 25 4 lies in the interval 2 < x < 6 f(x) = x – 1 f(4) = 4 – 1 = 3 Since 1 lies in the interval -5  x  2 f(x) = x + 5 f(1) = 1 + 5 = 6

  41. Find the sum of the first 40 terms of the series 12 – 22 + 32 – 42 + ……. . 40 12 – 22 + 32 – 42 + ……. . = 1 - 4 + 9 - 16 + 25 ….. to 40terms = (1 – 4) + (9 – 16) + (25 – 36) + …… to 20terms. = (-3) + (-7) + (-11) +………20terms It is an AP with a = –3, d = –4, n = 20 10 = 10(-82) = – 820

  42. Factorize the following x3 – 5x2 – 2x + 24 1 – 5 – 2 + 24 – 2 14 – 24 0 – 2 1 – 7 12 0  x + 2 is a factor Other factor = x2 – 7x + 12 = x2 – 4x – 3x + 12 = x(x – 4) – 3(x – 4) = (x – 4)(x – 3) x3 – 5x2 – 2x + 24 = (x + 2)(x – 4)(x – 3)

  43. If m – nx + 28x2 + 12x3 + 9x4 is a perfect square, then find the values of m, n 3x2 + 2x + 4 9x4 + 12x3 + 28x2 – nx + m 3x2 9x2 (-) +2x 6x2 + 12x3 + 28x2 + 12x3 + 4x2 (–) (-) – nx + m 6x2 + 4x + 4 24x2 24x2 + 16x + 16 (-) (–) (–) 0 n = -16, m = - 16

  44. If verify that (AB)T= BTAT . LHS = RHS  (AB)T= BTAT .

  45. Find the area of the quadrilateral whose vertices are(-4, -2), (-3,-5), (3,-2) and (2, 3) D(2,3) Area of the quadrilateral ABCD = ½ {(x1y2 + x2 y3 + x3y4 + x4y1) – (x2y1 + x3 y2 + x4y3 + x1y4)} A(-4,-2) C(3,-2) B(-3,-5) = ½ {(-4)(-5) + (-3)(-2) + (3)(3) + (2)(-2)} – -4 -2 -3 -5 3 -2 2 3 -4 -2 {(-4)(3) + (2)(-2) + (3)(-5) + (-3)(-2)} = ½ {(20 + 6 + 9 – 4) – (–12 – 4 – 15 + 6)} = ½ {31 – (–1)} = ½ (31 + 1) = ½ (32) = 16 sq. units

  46. The vertices of ABC are A(2, 1), B(6, –1) and C(4, 11). Find the equation of the straight line along the altitude from the vertex A. A The vertices of the triangle are A(2, 1), B(6, –1), C(4, 11) Let AD be the altitude. C B D 6(y – 1) = x – 2 6y – 6 = x – 2 x – 2 – 6y + 6 = 0 x – 6y + 4 = 0 A(2, 1), Slope m = 1/6 Required Equation of the line is y – y1 = m(x – x1)

  47. A boy is designing a diamond shaped kite, as shown in the figure where AE = 16cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be? A The vertices of the triangle are A(2, 1), B(6, –1), C(4, 11) Let AD be the altitude. C B D 6(y – 1) = x – 2 6y – 6 = x – 2 x – 2 – 6y + 6 = 0 x – 6y + 4 = 0 A(2, 1), Slope m = 1/6 Required Equation of the line is y – y1 = m(x – x1)

  48. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree. Let A be the point at which the tree is broken and let the top of the tree touch the ground at B. In ABC, BC = 30m, mB = 30 A 30 B C 30m

  49. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree. A 30 B C 30m = 10  1.732 = 17.32 = 20  1.732 = 34.64 Actual height of the tree = 17.32 + 34.64 = 51.96m

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