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Core 1 Revision Day Let Maths take you Further… Further Mathematics Support Programme www.furthermaths.org.uk.

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Presentation on theme: "Core 1 Revision Day Let Maths take you Further… Further Mathematics Support Programme www.furthermaths.org.uk."— Presentation transcript:

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) -3x>1 C) -3 1 D) -3>x or x>1 E) None of these.

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

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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

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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

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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!!!


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