Download presentation

Presentation is loading. Please wait.

Published byJoslyn Peevy Modified about 1 year ago

1
Core 1 Revision Day Let Maths take you Further… Further Mathematics Support Programme

2
2 Outline of the Day 10:00 11:00AlgebraAlgebra 11:0011:15Break 11:15 12:15Co-ordinate GeometryCo-ordinate Geometry 12:151:00pmLunch 1:002:00Curve Sketching and IndicesCurve Sketching and Indices 2:003:00CalculusCalculus 3:00pmHome time!

3
3 ALGEBRA

4
4

5
5 For AS-core you should know: How to solve quadratic equations by factorising, completing the square and “the formula”. The significance of the discriminant of a quadratic equation. How to solve simultaneous equations (including one linear one quadratic). How to solve linear and quadratic inequalities. QUICK QUIZ

6
6

7
7

8
8

9
Question 1 The expression (2x-5)(x+3) is equivalent to: A) 2x 2 + x - 15 B) 2x 2 - x - 15 C) 2x x - 15 D) 2x 2 - 2x - 15 E) Don’t know (2x-5)(x+3) = 2x 2 +6x – 5x -15 = 2x 2 +x – 15

10
Question 2 The discriminant of the quadratic equation 2x 2 +5x-1=0 is: A) 17 B) 33 C) 27 D) -3 E) I don’t know b 2 – 4ac = 5 2 – 4 x 2 x (-1) = =33

11
Question 3 Consider the simultaneous equations: x + 3y = 5 3x – y =5 The correct value of x for the solution is: A) x=1 B) x= -1 C) x=2 D) x= -2 E) I don’t know 3 x (2) 9x –3 y =15(3) + x + 3y = 5(1) 10x = 20 x=2

12
12

13
a)Write the expression x 2 -8x – 29 in the form (x+a) 2 + b, where a and b are constants. b)Hence find the roots of the equation x 2 -8x – 29 = 0. Express the roots in the form c±d√5 where c and d are constants to be determined. 13 Worked Example Take away 16 since the -4 in the bracket will give us an extra 16. We must complete the question Don’t forget the plus and minus. A very common error through out A- level! So a=-4 and b=-45.

14
Solve the simultaneous equations x – 2y = 1, x 2 + y 2 = Worked Example 1) 2) Equation 1 does not have any squared terms, so it is easier to expression x in terms of y You must know how to solve quadratic equations with ease! You could use the formula if you wanted!

15
15 Worked Example 3½ ½3¼ a) Find the set of values of x for which 6x+3>5-2x. b) Find the set of values of x for which 2x 2 -7x > >-3. c) Hence, or otherwise, find the set of values of x for which 6x+3>5-2x and 2x 2 -7x > >-3.

16
16

17
17 Question for you to try Question 1

18
18 Question for you to try Question 2

19
19 Question for you to try Question 3

20
20

21
21 Solutions Question 1 For part (i) your values of a and b are: A) a =30, b = 2; B) a = 120, b = 2; C) a = 30, b = 5; D) a = 120, b = 5; E) None of these.

22
22 Worked Solution Question 1 a=30 and b=2

23
23 Solutions Question 2 The formula for r is given by: A) B) C) D) E) None of these.

24
24 Solutions Question 3 The set of values for x is: A) -3

25
25

26
26 C1(AQA) Jan 2006 Question 1

27
27 C1(AQA) Jan 2007 Question 3

28
28 C1(AQA) Jan 2007 Question 7

29
29 C1(Edexcel) Jan 2006 Question 1

30
30 C1(Edexcel) Jan 2006 Question 5

31
31 C1(Edexcel) Jun 2006 Question 2

32
32 C1(Edexcel) Jun 2006 Question 6

33
33 C1(Edexcel) Jun 2006 Question 8

34
34 C1(Edexcel) Jan 2007 Question 2

35
35 C1(Edexcel) Jan 2007 Question 5

36
36 C1(Edexcel) Jun 2007 Question 1

37
37 C1(Edexcel) Jun 2007 Question 6

38
38 C1(Edexcel) Jun 2007 Question 7

39
39 C1(Edexcel) Jan 2008 Question 2

40
40 C1(Edexcel) Jan 2008 Question 3

41
41 C1(Edexcel) Jan 2008 Question 8

42
42 COORDINATE GEOMETRY

43
43

44
44 For AS-core you should know: How to calculate and interpret the equation of a straight line. How to calculate the distance between two points, the midpoint of two points and the gradient of the straight line joining two points. Relationships between the gradients of parallel and perpendicular lines. How to calculate the point of intersection of two lines. Calculating equations of circles and how to interpret them. Circle Properties. QUICK QUIZ

45
Gradient = change in y = y 2 – y 1 change in x x 2 – x 1 y = mx + c m = gradient c = y intercept

46
Distance between two points Equation of a circle: Centre: (a, b) Radius: r

47
47

48
Question 1 A straight line has equation 10y = 3x Which of the following is true? A) The gradient is 0.3 and the y-intercept is 1.5 B) The gradient is 3 and the y-intercept is 15 C) The gradient is 15 and the y-intercept is 3 D) The gradient is 1.5 and the y-intercept is 0.3 E) Don’t know y = 3/10 x + 15/10 y = 0.31 x + 1.5

49
Question 2 A is the point (1, 5), B is the point (4, 7) and C is the point (5, 2). Triangle ABC is A) right-angled B) scalene with no right angle C) equilateral D) isosceles E) Don’t know The sides are all different lengths

50
Question 3 A circle has the equation (x + 3)² + (y − 1)² = 4. Which of the following statements is false? A) The y coordinate of the centre is −1 B) The radius of the circle is 2 C) The x coordinate of the centre is −3 D) The point (−3,−1) lies on the circle E) Don’t know The equation represents a circle with centre (-3, 1) and radius 2. So the statement is incorrect

51
51

52
52 Worked Example The line AB has equation y=5x-11 and passes through the point B(3,4) as shown above. The line BC is perpendicular to AB and cuts the x-axis at C. Find the equation of the line BC and the x-coodinate of C. y x O A C B(3,4) NOT TO SCALE Gradient of line AB is 5 So gradient of line BC is -1/5. Equation of BC is: y = -1/5 x + c Using B(3,4) we get: 4 = -1/5 * 3 + c. So c = 4 + 3/5 = 23/5 So Equation of BC is y = -1/5 x + 23/5 X coordinate of C is given when y = 0. So 0 = -1/5 x + 23/5. So x = 23. Gradient of perpendicular = -1 / (gradient of original)

53
53 Worked Example A circle has equation (x-2) 2 + y 2 = 45. a)State the centre and radius of this circle. b)The circle intersects the line with equation x + y = 5 at two points, A and B. Find algebraically the coordinates of A and B. c)Compute the distance between A and B to 2 decimal places. a)Centre of circle is (2,0) and radius is √45 b)Equation of line implies: x = 5-y. Using this in the equation of the circle gives: (5-y-2) 2 + y 2 = 45 (3-y) 2 + y 2 = y+y 2 +y 2 =45 2 y 2 -6y + 9 =45 2 y 2 -6y -36 =0 y 2 -3y -18 =0 (y-6)(y+3)=0 So y = 6 or y = -3. When y=6, x = 5 – 6 =1. When y=-3, x = 5-(-3) = 8. So coordinates are (1,6) and (8,-3) “State” means you should be able to write down the answer. Equation of circle with centre (a,b) and radius r is (x-a) 2 + (y-b) 2 = r 2 c) Draw a diagram: Distance = (8,-3) (1,6) Watch the minus signs

54
54

55
55 Question for you to try Question 1

56
56 Question for you to try Question 2

57
57 Question for you to try (part 1) Question 3 (Part One)

58
58 Question for you to try (part 2) Question 3 (Part Two)

59
59

60
60 Solution to Question 1 Question 1 The equation of the line is: A) 3x + 2y = 26 B) 3x + 2y = 13 C) -3x + 2y = 26 D) -3x + 2y = 13. E) Don’t know.

61
61 Solution to Question 1 Question 1 Any line parallel to 3x + 2y = 6 must have the same gradient, but a different intercept. (Short method): So any line parallel to given line has the form 3x + 2y = c (constant) If the line goes through (2,10) then 3 * * 10 = c, so c =26. Hence equation is 3x + 2y = 26. (Long method): Rearrange equation to get y = 3 – (3/2) x. Gradient is –(3/2). So new line must have the equation y = -(3/2) x + c Use the point (2,10) to get 10 = -(3/2) * 2 + c. So c = = 13. Thus y = (-3/2)x (This is the same as the last line since we have 2y = -3x + 26, so 3x + 2y =26.

62
62 Solution to Question 2 Question 2 The radius of the circle in (ii) is: A) √45 B) ½ √45 C) √85 D) ½ √85 E) Don’t know.

63
63 Solution to Question 2 i)Gradient of AB =(8-0)/(9-5) = 8/4 =2 Gradient of BC = (1-0)/(3-5) = 1/(-2) = -½ Product of gradients = 2 x (-½) = -1, so perpendicular. ii)If AC is diameter then midpoint of AC is centre of the circle. Midpoint of AC = AC = √ ((9-3) 2 + (8-1) 2 ) = √ ( ) = √85 So diameter is √85 and hence radius is ½√85 So equation of the circle is (x-6) 2 + (y-4.5) 2 = (½√85) 2 =85/4 So equation is (x-6) 2 + (y-4.5) 2 =85/4 Coordinates of B(5,0) give (5-6) 2 + (0-4.5) 2 = /4 = 85/4. So B lies on the circle. iii)Let (x,y) be coordinates of D. Midpoint of BD is centre of circle (6,4.5). So So coordinates of D are (7,9). Gradient = change in y/change in x B(5,0) D(x,y) (6,4.5)

64
64

65
65 C1(AQA) Jan 2006 Question 2

66
66 C1(AQA) Jan 2006 Question 5

67
67 C1(AQA) Jun 2006 Question 7

68
68 C1(AQA) Jan 2007 Question 4

69
69 C1(AQA) Jan 2007 Question 2

70
70 C1(AQA) Jun 2007 Question 1

71
71 C1(AQA) Jun 2007 Question 5

72
72 C1(AQA) Jan 2008 Question 1

73
73 C1(AQA) Jan 2008 Question 4

74
74 C1(Edexcel) Jan 2006 Question 3

75
75 C1(Edexcel) Jun 2006 Question 10

76
76 C1(Edexcel) Jun 2006 Question 11

77
77 C1(Edexcel) Jun 2007 Question 10

78
78 C1(Edexcel) Jun 2007 Question 11

79
79 C1(Edexcel) Jan 2008 Question 4

80
80 CURVE SKETCHING (AND INDICES)

81
81

82
82 For AS-core you should know: How to sketch the graph of a quadratic given in completed square form. The effect of a translation of a curve. The effect of a stretch of a curve. QUICK QUIZ

83

84

85
85

86
Question 1 The vertex of the quadratic graph y=(x-2) is : A) Minimum (2,-3) B) Minimum (-2,3) C) Maximum (2,-3) D) Maximum (-2,-3) E) Don’t know The graph has a minimum point, since the coefficient of x² is positive. The smallest possible value of (x-2) 2 is 0, when x = 2. [When x = 2 y = -3]

87
Question 2 The quadratic expression x 2 -2x-3 can be written in the form: A) (x+1) B) (x-1) C) (x-1) D) (x-1) E) Don’t know x 2 -2x-3 =(x-1) =(x-1) The -1 is present to correct for the +1 we get when multiplying out (x-1) 2

88
Question 3 The graph of y=x 2 -2x-1 has a minimum point at: A) (1,-1) B) (-1,-1) C) (-1,-2) D) (1,-2) E) Don’t know y = x 2 -2x-1 = (x-1) = (x-1) So minimum point is (1,-2)

89
89

90
90 Worked Exam Question i) Write x 2 -2x - 2 in the form (x-p) 2 + q. ii) State the coordinates of the minimum point on the graph y= x 2 -2x - 2. iii) Find the coordinates of the points where the graph of y= x 2 -2x - 2 crosses the axes and sketch the graph. iv) Show that the graphs of y= x 2 -2x - 2 and y= x 2 +4x - 5 intersect only once. Find the x coordinate of the point of intersection. i)x 2 -2x – 2 = (x-1) = (x-1) So p=1 and q =-3. ii)(x-1) 2 has its smallest value when x=1, y value at this point is -3. So minimum point is (1,-3). iii)Graph crosses x-axis when y=0. x 2 -2x – 2 =0 implies (x-1) 2 -3 =0. So (x-1) = ±√3. So x= 1 ±√3. Coordinates are (1 +√3,0) and (1 -√3,0) Graph crosses y-axis when x=0. So coordinates are (0,-2) x y (1+√3,0)(1-√3,0) (1,-3) (0,-2)

91
91 Worked Exam Question i) Write x 2 -2x - 2 in the form (x-p) 2 + q. ii) State the coordinates of the minimum point on the graph y= x 2 -2x - 2. iii) Find the coordinates of the points where the graph of y= x 2 -2x - 2 crosses the axes and sketch the graph. iv) Show that the graphs of y= x 2 -2x - 2 and y= x 2 +4x - 5 intersect only once. Find the x coordinate of the point of intersection. iv)Intersection when x 2 -2x – 2 = x 2 +4x – 5. So -2x-2 = 4x -5, giving 6x – 3 =0. So x= ½. Intersection when x=½.

92
92

93
93 Question for you to try Question 1

94
94 Question for you to try Question 2

95
95 Question for you to try Question 3

96
96 Question for you to try Question 4

97
97

98
98 Solution to Question 1(i) Question 1 Solution to (i) is: A) 4/27 B) 8/81 C) 4/3 D) 4/81 E) Don’t know

99
99 Solution to Question 1 (ii) Question 1 Solution to (ii) is: A) B) C) D) E) Don’t know

100
100 Solution to Question 3 Question 3 The graph of y=4x 2 -24x + 27 is: A) B) x (-1.5,0) (-4.5,0) (-3,-9) (0,27) C) D) E) Don’t know x (1.5,0) (-4.5,0) (-3,9) (0,27) x (4.5,0) (1.5,0) (3,-9) (0,27) x (4.5,0)(1.5,0) (3,9) (0,-27)

101
101

102
102 C1(AQA) Jan 2006 Question 3

103
103 C1(AQA) Jan 2007 Question 1

104
104 C1(AQA) Jun 2007 Question 3

105
105 C1(AQA) Jan 2008 Question 5

106
106 C1(Edexcel) Jun 2006 Question 3

107
107 C1(Edexcel) Jun 2006 Question 9

108
108 C1(Edexcel) Jan 2007 Question 10

109
109 CALCULUS (NON-MEI)

110
110

111
111 For AS-core you should know: How the derivative of a function is used to find the gradient of its curve at a given point. What is meant by a chord and how to calculate the gradient of a chord. How the gradient of a chord can be used to approximate the gradient of a tangent. How to differentiate integer powers of x and rational powers of x. What is meant by a stationary point of a function and how differentiation is used to find them. Using differentiation to find lines which are tangential to and normal to a curve. QUICK QUIZ

112

113

114
114

115
Question 1 The gradient of the curve y=3x 2 – 4 at the point (2,8) is : A) 12 B) 6x C) 48 D) 8 E) Don’t know If y=3x 2 – 4 then dy/dx = 6x x=2 dy/dx = 6x2 = 12 So gradient of curve at (2,8) is 12.

116
Question 2 If then A) dy/dx =1.5 B) dy/dx = 3/2 - t C) dx/dt = 1.5 D) dt/dx = 1.5 E) Don’t know

117
Question 3

118
Solution to Question 3 The correct answer is (B) POINT A: the gradient is positive (sloping upwards from left to right) when x = 0. Hence, the graph of the derivative crosses the y-axis at a positive value of y. POINT’S B: the gradient is zero, this means that the graph of the derivative must cross the x-axis at the points labelled B ’. The original curve looks like the graph of a cubic, so we would expect the graph of its derivative to be a quadratic graph (a parabola), passing through the points labelled A ’ and B’.

119
119

120
Worked Example A curve has equation y = x² – 3x + 1. i)Find the equation of the tangent to the curve at the point where x = 1. ii)Find the equation of the normal to the curve at the point where x = 3. i) If y = x² – 3x + 1 then dy/dx = 2x -3. When x=1, gradient is given by dy/dx evaluated at x=1: gradient = 2x1 -3 = -1. Equation of a straight line y=mx +c So y = -x + c When x = 1, y = 1² – 3x1 + 1 = -1. So line passes through (1,-1). So -1 = -1 + c, so c= 0. Equation of tangent is y=-x. ii) When x=3, gradient is given by dy/dx: gradient = 2 x 3 – 3 = 3. Gradient of normal = -1/gradient of tangent. So gradient of normal is -1/3. When x=3, y = 3² – 3x3 + 1 = 1. So line passes though (3,1). Equation of line is y = -1/3 x + c Using the point (3,1) gives: 1 = (-1/3) x 3 + c. So x = -2. Equation of normal is y = -1/3 x -2.

121
Worked Example A curve has equation y = 2x 3 -3x 2 – 8x + 9. i)Find the equation of the tangent to the curve at the point P (2, -3). ii)Find the coordinates of the point Q at which the tangent is parallel to the tangent at P. i) If y = 2x 3 -3x 2 – 8x + 9, then dy/dx = 6x 2 – 6x -8. When x = 2, dy/dx = 6x2 2 – 6x2 -8. = 24 – =4. Tangent has gradient 4 and passes through (2,-3). Using y = mx + c we have y = 4x +c. Using the point (2,-3) we have -3 = 4 x 2 + c, so c = -11. Hence equation of tangent is y = 4x -11. ii) Tangent at P has gradient 4. Tangent is parallel when gradient is the same. So dy/dx = 6x 2 – 6x -8 = 4 So 6x 2 – 6x -12 =0, so x 2 – x - 2 =0. Thus (x-2)(x+1) = 0, which implies x=2 or x=-1. WE NEED THE COORDINATES P is where x=2, so Q is the point where x=-1. When x = -1, y = 2(-1) 3 -3(-1) 2 – 8(-1) + 9 = =12. So coordinates of Q are (-1,12).

122
122

123
123 Questions for you to try. Question 1 A curve has equation y = 10 – 3x 7. i)Find dy/dx ii)Find an equation for the tangent to the curve at the point where x=2. iii)Determine whether y is increasing or decreasing when x = -3.

124
124 Questions for you to try. Question 2 A curve has equation y=x x x i)Find dy/dx ii)Hence find the coordinates of the points on the curve where dy/dx=0.

125
125

126
126 Questions for you to try. Question 1 A curve has equation y = 10 – 3x 7. i)Find dy/dx ii)Find an equation for the tangent to the curve at the point where x=2. iii)Determine whether y is increasing or decreasing when x = -3. The equation of the tangent in part (ii) is: A) y = -1344x B) y = 1344x C) y = -21x 6 + c D) y = 21x 6 + c E) Don’t know

127
127 Questions for you to try. Question 1 A curve has equation y = 10 – 3x 7. i)Find dy/dx ii)Find an equation for the tangent to the curve at the point where x=2. iii)Determine whether y is increasing or decreasing when x = -3. i)dy/dx = -21x 6 ii)When x = 2, dy/dx = - 21 x 2 6 = So y = x + c. When x = 2, y = 10 – 3 (2) 7 = So (2,-374) is point on the curve. Using this point in y = x + c gives -374 = x 2 + c. So c = Equation of tangent is y = -1344x iii.When x=-3, dy/dx = -21 x (-3) 6 = < 0. So y is decreasing.

128
128 Questions for you to try. Question 2 A curve has equation y=x x x i)Find dy/dx ii)Hence find the coordinates of the points on the curve where dy/dx=0. i) dy/dx = 3x x + 29 ii) dy/dx = 0 when 3x 2 +88x + 29 = 0. So 3x 2 +88x + 29 = 0 implies (3x+1)(x+29) = 0, so x=-29 or x =-1/3. We need the coordinates. When x = -1/3, y = (-1/3) (-1/3) (-1/3) = 1/9 + 44/9 -29/3 = -14/3. So one point has coordinate (-1/3, -14/3). When x = -29, y = (-29) (-29) (-29) = So second point has coordinate (-29,11774)

129
129

130
130 C1(AQA) Jan 2006 Question 7

131
131 C1(AQA) Jan 2007 Question 5

132
132 C1(AQA) Jun 2007 Question 4

133
133 C1(AQA) Jan 2008 Question 2

134
134 C1(Edexcel) Jan 2006 Question 9

135
135 C1(Edexcel) Jan 2006 Question 10

136
136 C1(Edexcel) Jun 2006 Question 5

137
137 C1(Edexcel) Jan 2007 Question 1

138
138 C1(Edexcel) Jan 2007 Question 8

139
139 C1(Edexcel) Jun 2007 Question 3

140
140 C1(Edexcel) Jan 2008 Question 5

141
141 POLYNOMIALS (MEI)

142
142

143
143 For AS-core you should know: How to add, subtract and multiply polynomials. How to use the factor theorem. How to use the remainder theorem. The curve of a polynomial of order n has at most (n – 1) stationary points. How to find binomial coefficients. The binomial expansion of (a + b) n. QUICK QUIZ

144

145

146
146

147
Question 1 Which of the following is a factor of x³ + x² + 2x + 8 A)x+2 B)x-2 C)x+1 D)x-1 E)Don’t know

148
Solution to Question 1 The solution is (A). If (x-a) is a factor of f(x), then f(a)=0. If f(x) = x³ + x² + 2x + 8 then A)f(-2) = 0, so x+2 is a factor. B)f(2) =24, so x-2 is not a factor. C)f(-1) = 6, so x+1 is not a factor. D)f(1) = 12, so x-1 is not a factor.

149
Question 2 If x-2 is a factor of f(x)=3x³ – 5x² + ax + 2, then the value of a is: A) 21 B) 3 C)-21 D)-3 E) Don’t know

150
Solution to Question 2 The correct answer is (d). If x-a is a factor of f(x), then f(a)=0 If f(x)=3x³ – 5x² + ax + 2, then f(2) = 3(2)³ – 5(2)² + a(2) + 2 = 2a+6. Since 2 is a factor f(2)=0, so 2a+6 =0, so a=-3.

151
Question 3

152
Solution to Question 3 The correct answer is C The graph of y=(x-a)(x-b)(x+c) crosses the x-axis at (a, 0), (b, 0) and (-c, 0). Since two of the intersections are with the positive x-axis and one with the negative x-axis, the graph must be either A or C. Since y is positive for large positive values of x, the correct graph is C

153
153

154
154 Worked Example Find the binomial expansion of (3+x) 4, writing each term as simply as possible. Binomial Expansion In our example, a=3, b = x and n=4.

155
155 Worked Example When x 3 + 3x +k is divided by x-1, the remainder is 6. Find the value of k. Remainder Theorem: If f(x) is divided by x-a, then the remainder is f(a) If f(x) = x 3 + 3x +k, then f(1) = x1 +k = 6. So 4+k =6, so k=2.

156
156

157
157 Questions for you to try. Question 1

158
158 Questions for you to try. Question 2

159
159 Questions for you to try. Question 3

160
160

161
161 Solution to Question 1 Question 1 i) We need the discriminant to be greater than or equal to zero. b 2 -4ac = x1xk =25-4k. For one or more real roots we need 25-4k≥0. So 25/4 ≥ k. ii) 4x x + 25 = (2x+5)(2x+5) So 4x x + 25 =0 implies (2x+5)(2x+5)=0, so x=-5/2.

162
162 Solution to Question 2 Question 2 f(x) = x 3 + ax 2 +7 Put x=-2, to get f(-2) = (-2) 3 + a(-2) 2 +7 = 0 So a +7 =0. Thus 4a=1, a = ¼.

163
163 Solution to Question 3 Question 3

164
164 Solution to Question 3 ctd ax 2 bxc, c=-5 xa = 2bx ax 2 -2c=10 -2c=10, c=-5 a=2 -2a + b = b =-1, b =3 So x=1 and x=-5/2 are other roots of the equation.

165
165 Solution to Question 3 ctd y=-22 (0,-12) (3,0)

166
166

167
167 C1(MEI) 6 th June 2006 Question 8

168
168 C1(MEI) 6 th June 2006 Question 12

169
169 C1(MEI) 16 th January 2007 Question 4

170
170 C1(MEI) 16 th January 2007 Question 5

171
171 C1(MEI) 16 th January 2007 Question 8

172
172 C1(MEI) 7 th June 2007 Question 4

173
173 C1(MEI) 7 th June 2007 Question 6

174
174 C1(MEI) January 2008 Question 6

175
175 C1(MEI) January 2008 Question 7

176
176 C1(MEI) June 2008 Question 3

177
177 C1(MEI) June 2008 Question 8

178
178 C1(MEI) June 2008 Question 11

179
179 EXAM PRACTICE

180

181
181 That’s all folks!!!

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google