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X-STANDARD MATHEMATICS

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1 X-STANDARD MATHEMATICS
PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21 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) 12a (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) (B) 25 (C) (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. 1 2 3 4 1 2 3 4 X = {1, 2, 3, 4}, g = {(3,1), (4,2), (2, 1)} g is not a function. since 1 has no image.

21 Three numbers are in the ratio 2 : 5 : 7
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)
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.
Given, AE = 3.7cm E D C B In ABC, DE || BC

27 Let AC = Length of the ladder BC = distance of the wall = 3.5m
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 Let AC = Length of the ladder BC = distance of the wall = 3.5m 60  B C 3.5m In ABC, C = 60 AC = 3.5  2 = 7 Length of the ladder = 7m

28 Prove the identity:

29 Curved surface area = 2rh
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 Circumference of the base = 44 2r = 44 Volume = r2h
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 3 14. 00 9 6 7 5 00 469 31

32 Two coins are tossed together
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
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(2) = 2 + 5 = 7 2f(-4) + 3f(2) = 2(1) + 3(7) = = 23

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

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

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

43 Factorize the following x3 – 5x2 – 2x + 24
1 – 5 – – 2 – 2 14 – 24 1 – 7 12  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)

44 3x2 + 2x + 4 3x2 9x4 + 12x3 + 28x2 – nx + m 9x2 6x2 +2x + 12x3 + 28x2
If m – nx + 28x2 + 12x3 + 9x4 is a perfect square, then find the values of m, n 3x2 + 2x + 4 3x2 9x x x2 – nx + m 9x2 (-) 6x2 +2x + 12x x2 + 12x x2 (–) (-) 6x2 + 4x + 4 24x2 – nx m 24x x (-) (–) (–) n = -16, m = - 16

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

46 Find the area of the quadrilateral whose vertices are (-4, -2), (-3,-5), (3,-2) and (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)} – 3 -2 2 3 {(-4)(3) + (2)(-2) + (3)(-5) + (-3)(-2)} = ½ {( – 4) – (–12 – 4 – )} = ½ {31 – (–1)} = ½ (31 + 1) = ½ (32) = 16 sq. units

47 Required Equation of the line is y – y1 = m(x – x1)
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)

48 Required Equation of the line is y – y1 = m(x – x1)
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)

49 A vertical tree is broken by the wind
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

50 Actual height of the tree = 17.32 + 34.64 = 51.96m
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 = = 51.96m

51 Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere. Cone: Radius = 12cm, Height = 48cm Volume of the sphere = volume of the cone 12 r3 = 12  12  12 r = 12cm

52 Calculate the standard deviation of the following data.
X 3 8 13 18 23 f 7 10 15 x f d = x – 3 d2 fd fd2 3 – 3 = 0 8 – 3 = 5 13 – 3 = 10 18 – 3 = 15 23 – 3 = 20 25 100 225 400 50 150 160 510 250 1500 2250 3200 7200

53 Standard deviation 6 .3 6 39.96 36 12 3 3 96 369 27

54 The probability that a new car will get an award for its design is 0
The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards. Let A be the event that a new car will get an award for its design B be the event that a new car will get an award for efficient use of fuel. P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15 P(atleast one of the two awards) = P(AB)

55 The probability that a new car will get an award for its design is 0
The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards. Let A be the event that a new car will get an award for its design B be the event that a new car will get an award for efficient use of fuel. P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15 P(atleast one of the two awards) = P(AB) P(AB) = P(A) + P(B) – P(AB) = – 0.15 = 0.60 – 0.15 = 0.45

56 The probability that a new car will get an award for its design is 0
The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards. P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15 P(getting only one of the awards) = P (only A or only B) = [P(A) – P(A B)] + [P(B) – P(A  B)] = (0.25 – 0.15) + (0.35 – 0.15) = = 0.30

57 The sum of three consecutive terms in an AP is – 6 and their product is 90. Find the numbers.
Let a – d, a, a + d be three numbers in an AP Sum = –6 a – d + a + a + d = – 6 3a = –6 a = – 6/3 a = –2 Product = 90 (a – d)  a  (a + d) = 90 (a2 – d2)a = 90 {(–2)2 – d2 }(–2) = 90 4 – d2 = – 45 – d2 = –45 –4 d2 = 49 d = 7 The given numbers are – 2, 5, 12 (or) –2, –9, –16

58 A Cylindrical jar of diameter 14cm and depth 20cm is half-full of water lead shots of same size are dropped into the jar and the level of water raises by 2.8cm. Find the diameter of each lead shots. Cylindrical jar: Diameter = 14cm Radius = 7cm Height of water level raises = 2.8cm Volume of 300 lead shots = volume of raised water

59 A Cylindrical jar of diameter 14cm and depth 20cm is half-full of water lead shots of same size are dropped into the jar and the level of water raises by 2.8cm. Find the diameter of each lead shots. 100 7 100

60 SECTION - D Note: 2 ×10 = 20 marks
(i) This section contains Two questions, each with two alternatives. (ii) Answer both the questions choosing either of the alternatives. (iii) Each question carries Ten marks

61 Draw two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also measure the lengths of the tangents.

62 Draw two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also measure the lengths of the tangents. Length of the tangent segment A 8 6 10 M O P 8 B

63 (a) Draw the graph of x2 – x – 8 = 0 and hence find the roots of x2 – 2x – 15 = 0.
In x axis 1cm = 1 unit y axis 1cm = 2 units. x x -x –8 –8 –8 –8 –8 – –8 –8 –8 y

64 Solution set = {– 3, 5}

65 (b) A cyclist travels from a place A to a place B along the same route at a uniform speed on different days. The following table gives the speed of his travel and the corresponding time he took to cover the distance. Speed in km/hr 2 4 6 10 12 Time in hrs 60 30 20

66 Draw the speed-time graph and use it to find
Speed in km/hr 2 4 6 10 12 Time in hrs 60 30 20 Draw the speed-time graph and use it to find (i) the number of hours he will take if he travels at a speed of 5 km / hr (ii) the speed with which he should travel if he has to cover the distance in 40 hrs.

67 If x = 5, y = 24 Time in hr If y = 40, y = 2.4 Speed in km/hr


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