Download presentation

1
**Let Maths take you Further…**

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

2
**Outline of the Day 10:00 11:00 Algebra 11:00 11:15 Break**

11:15 12:15 Co-ordinate Geometry 12:15 1:00pm Lunch 1:00 2:00 Curve Sketching and Indices 2:00 3:00 Calculus 3:00pm Home time!

3
ALGEBRA

4
Key Topics

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

8
Quick Quiz

9
Question 1 The expression (2x-5)(x+3) is equivalent to: A) 2x2 + x - 15 B) 2x2 - x - 15 C) 2x2 + 11x - 15 D) 2x2 - 2x - 15 E) Don’t know (2x-5)(x+3) = 2x2 +6x – 5x -15 = 2x2 +x – 15

10
**Question 2 b2 – 4ac = 52 – 4 x 2 x (-1) = 25 + 8 =33**

The discriminant of the quadratic equation 2x2 +5x-1=0 is: A) 17 B) 33 C) 27 D) -3 E) I don’t know b2 – 4ac = 52 – 4 x 2 x (-1) = =33

11
**Question 3 3 x (2) 9x –3 y =15 (3) + x + 3y = 5 (1) 10x = 20 x=2**

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
Worked Exam Questions

13
Worked Example Write the expression x2 -8x – 29 in the form (x+a)2 + b, where a and b are constants. Hence find the roots of the equation x2 -8x – 29 = 0. Express the roots in the form c±d√5 where c and d are constants to be determined. Take away 16 since the -4 in the bracket will give us an extra 16. So a=-4 and b=-45. Don’t forget the plus and minus. A very common error through out A-level! We must complete the question C1(Edexcel) Jun 2005, Q3

14
**Worked Example Solve the simultaneous equations x – 2y = 1, 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! C1(Edexcel) Jun 2005, Question 5.

15
**6x+3>5-2x and 2x2 -7x > >-3.**

Worked Example a) Find the set of values of x for which 6x+3>5-2x. b) Find the set of values of x for which 2x2 -7x > >-3. c) Hence, or otherwise, find the set of values of x for which 6x+3>5-2x and 2x2 -7x > >-3. C1(Edexcel) Jun 2005, Q6 3 3

16
**Questions for you to try now...**

17
Question for you to try Question 1 C1(MEI) 16th January 2006

18
Question for you to try Question 2 C1(MEI) 6th June 2006

19
Question for you to try Question 3 C1(MEI) 6th June 2006

20
**Answers to selected questions**

21
**Solutions A) a =30, b = 2; B) a = 120, b = 2; C) a = 30, b = 5;**

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. C1(MEI) 16th January 2006

22
Worked Solution Question 1 C1(MEI) 16th January 2006 a=30 and b=2

23
**Solutions Question 2 The formula for r is given by: A) B) C) D)**

E) None of these. C1(MEI) 6th June 2006

24
**Solutions A) -3<x<1 B) -3>x>1 C) -3<x or x>1**

Question 3 The set of values for x is: A) -3<x<1 B) -3>x>1 C) -3<x or x>1 D) -3>x or x>1 E) None of these. C1(MEI) 6th June 2006

25
More practice for you.

26
C1(AQA) Jan 2006 Question 1

27
C1(AQA) Jan 2007 Question 3

28
C1(AQA) Jan 2007 Question 7

29
C1(Edexcel) Jan 2006 Question 1

30
C1(Edexcel) Jan 2006 Question 5 C1(Edexcel) Jan 2006

31
C1(Edexcel) Jun 2006 Question 2 C1(Edexcel) Jun 2006

32
C1(Edexcel) Jun 2006 Question 6 C1(Edexcel) Jun 2006

33
C1(Edexcel) Jun 2006 Question 8 C1(Edexcel) Jun 2006

34
C1(Edexcel) Jan 2007 Question 2 C1(Edexcel) Jan 2007

35
C1(Edexcel) Jan 2007 Question 5 C1(Edexcel) Jan 2007

36
C1(Edexcel) Jun 2007 Question 1 C1(Edexcel) Jun 2007

37
C1(Edexcel) Jun 2007 Question 6 C1(Edexcel) Jun 2007

38
C1(Edexcel) Jun 2007 Question 7 C1(Edexcel) Jun 2007

39
C1(Edexcel) Jan 2008 Question 2 C1(Edexcel) Jan 2008

40
C1(Edexcel) Jan 2008 Question 3 C1(Edexcel) Jan 2008

41
C1(Edexcel) Jan 2008 Question 8 C1(Edexcel) Jan 2008

42
COORDINATE GEOMETRY

43
Key Topics

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 = y2 – y1**

change in x x2 – x1 y = mx + c m = gradient c = y intercept

46
**Distance between two points**

Equation of a circle: Centre: (a, b) Radius: r

47
Quick Quiz

48
**Question 1 A straight line has equation 10y = 3x + 15.**

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 C) The gradient is 15 and the y-intercept is 3 D) The gradient is 1.5 and the y-intercept is 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
Worked Exam Questions

52
**Worked Example y x O A C B(3,4) NOT TO SCALE**

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. Gradient of perpendicular = -1 / (gradient of original) 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. C1(MEI) 16th January 2006

53
**Worked Example A circle has equation (x-2)2 + y2 = 45.**

State the centre and radius of this circle. The circle intersects the line with equation x + y = 5 at two points, A and B. Find algebraically the coordinates of A and B. Compute the distance between A and B to 2 decimal places. Centre of circle is (2,0) and radius is √45 Equation of line implies: x = 5-y. Using this in the equation of the circle gives: (5-y-2) 2 + y2 = 45 (3-y) 2 + y2 = 45 9-6y+y2 +y2 =45 2 y2 -6y + 9 =45 2 y2 -6y -36 =0 y2 -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 = r2 c) Draw a diagram: Distance = Watch the minus signs C1(MEI) 16th January 2006 (1,6) (8,-3)

54
**Questions for you to try now...**

55
Question for you to try Question 1 C1(MEI) 6th June 2006

56
Question for you to try Question 2 C1 (MEI) June 2006

57
**Question for you to try (part 1)**

Question 3 (Part One) C1(MEI) 7th June 2007

58
**Question for you to try (part 2)**

Question 3 (Part Two) C1(MEI) 7th June 2007

59
**Answers to selected questions**

60
**Solution to 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. C1(MEI) 6th June 2006

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 + 13. (This is the same as the last line since we have 2y = -3x + 26, so 3x + 2y =26. C1(MEI) 6th June 2006

62
**Solution to Question 2 The radius of the circle in (ii) is: A) √45**

B) ½ √45 C) √85 D) ½ √85 E) Don’t know. C1 (MEI) June 2006

63
**Solution to Question 2 Gradient = change in y/change in x D(x,y)**

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. 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 C1 (MEI) June 2006 D(x,y) (6,4.5) B(5,0)

64
More practice for you.

65
C1(AQA) Jan 2006 Question 2

66
C1(AQA) Jan 2006 Question 5

67
C1(AQA) Jun 2006 Question 7

68
C1(AQA) Jan 2007 Question 4

69
C1(AQA) Jan 2007 Question 2

70
C1(AQA) Jun 2007 Question 1

71
C1(AQA) Jun 2007 Question 5

72
C1(AQA) Jan 2008 Question 1

73
C1(AQA) Jan 2008 Question 4

74
C1(Edexcel) Jan 2006 Question 3

75
C1(Edexcel) Jun 2006 Question 10

76
C1(Edexcel) Jun 2006 Question 11

77
C1(Edexcel) Jun 2007 Question 10

78
C1(Edexcel) Jun 2007 Question 11

79
C1(Edexcel) Jan 2008 Question 4

80
**CURVE SKETCHING (AND INDICES)**

81
Key Topics

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

85
Quick Quiz

86
Question 1 The vertex of the quadratic graph y=(x-2)2 - 3 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 x2 -2x-3 can be written in the form: A) (x+1)2 - 4 B) (x-1)2 - 4 C) (x-1)2 - 3 D) (x-1)2 - 2 E) Don’t know x2 -2x-3 =(x-1) =(x-1)2 - 4 The -1 is present to correct for the +1 we get when multiplying out (x-1)2

88
Question 3 The graph of y=x2 -2x-1 has a minimum point at: A) (1,-1) B) (-1,-1) C) (-1,-2) D) (1,-2) E) Don’t know y = x2 -2x-1 = (x-1) = (x-1)2 - 2 So minimum point is (1,-2)

89
Worked Exam Questions

90
**Worked Exam Question i) Write x2 -2x - 2 in the form (x-p)2 + q.**

ii) State the coordinates of the minimum point on the graph y= x2 -2x - 2 . iii) Find the coordinates of the points where the graph of y= x2 -2x - 2 crosses the axes and sketch the graph. iv) Show that the graphs of y= x2 -2x - 2 and y= x2 +4x - 5 intersect only once. Find the x coordinate of the point of intersection. x2 -2x – 2 = (x-1) = (x-1)2 -3. So p=1 and q =-3. (x-1)2 has its smallest value when x=1, y value at this point is -3. So minimum point is (1,-3). Graph crosses x-axis when y=0. x2 -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) C1(MEI) 16th January 2006 y x (1-√3,0) (1+√3,0) (0,-2) (1,-3)

91
**Worked Exam Question i) Write x2 -2x - 2 in the form (x-p)2 + q.**

ii) State the coordinates of the minimum point on the graph y= x2 -2x - 2 . iii) Find the coordinates of the points where the graph of y= x2 -2x - 2 crosses the axes and sketch the graph. iv) Show that the graphs of y= x2 -2x - 2 and y= x2 +4x - 5 intersect only once. Find the x coordinate of the point of intersection. Intersection when x2 -2x – 2 = x2 +4x – 5. So -2x-2 = 4x -5, giving 6x – 3 =0. So x= ½. Intersection when x=½. C1(MEI) 16th January 2006

92
**Questions for you to try now...**

93
Question for you to try Question 1 C1(MEI) June 2006

94
Question for you to try Question 2 C1(MEI) 16th January 2007

95
Question for you to try Question 3 C1(MEI) 7th June 2007

96
Question for you to try Question 4 C1(MEI) 16th January 2006

97
**Answers to selected questions**

98
**Solution to Question 1(i)**

Solution to (i) is: A) 4/27 B) 8/81 C) 4/3 D) 4/81 E) Don’t know C1(MEI) June 2006

99
**Solution to Question 1 (ii)**

Solution to (ii) is: A) B) C) D) E) Don’t know C1(MEI) June 2006

100
**Solution to Question 3 Question 3 x (1.5,0) (-4.5,0) (-3,9) (0,27)**

The graph of y=4x2 -24x + 27 is: A) B) C) D) E) Don’t know x (4.5,0) (1.5,0) (3,-9) (0,27) C1(MEI) June 2006 (3,9) x (-1.5,0) (-4.5,0) (-3,-9) (0,27) x (1.5,0) (4.5,0) (0,-27)

101
More practice for you.

102
C1(AQA) Jan 2006 Question 3

103
C1(AQA) Jan 2007 Question 1

104
C1(AQA) Jun 2007 Question 3

105
C1(AQA) Jan 2008 Question 5

106
C1(Edexcel) Jun 2006 Question 3

107
C1(Edexcel) Jun 2006 Question 9

108
C1(Edexcel) Jan 2007 Question 10

109
CALCULUS (NON-MEI)

110
Key Topics

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

114
Quick Quiz

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

120
**Worked Example i) If y = x² – 3x + 1 then dy/dx = 2x -3.**

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. C2 (MEI) Chapter Assessment Differentiation Q4

121
**Worked Example i) If y = 2x3 -3x2 – 8x + 9, then dy/dx = 6x2 – 6x -8.**

A curve has equation y = 2x3 -3x2 – 8x + 9. Find the equation of the tangent to the curve at the point P (2, -3). Find the coordinates of the point Q at which the tangent is parallel to the tangent at P. i) If y = 2x3 -3x2 – 8x + 9, then dy/dx = 6x2 – 6x -8. When x = 2, dy/dx = 6x22 – 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 = 6x2 – 6x -8 = 4 So 6x2 – 6x -12 =0, so x2 – 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). C2 (MEI) Chapter Assessment Differentiation Q5

122
**Questions for you to try now...**

123
**Questions for you to try.**

A curve has equation y = 10 – 3x7. Find dy/dx Find an equation for the tangent to the curve at the point where x=2. Determine whether y is increasing or decreasing when x = -3. Based on C1(AQA) Jun 2006

124
**Questions for you to try.**

A curve has equation y=x x2 + 29x Find dy/dx Hence find the coordinates of the points on the curve where dy/dx=0. Based on C1(AQA) Jun 2006

125
**Answers to selected questions**

126
**Questions for you to try.**

A curve has equation y = 10 – 3x7. Find dy/dx Find an equation for the tangent to the curve at the point where x=2. 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 = -21x6 + c D) y = 21x6 + c E) Don’t know Based on C1(AQA) Jun 2006

127
**Questions for you to try.**

A curve has equation y = 10 – 3x7. Find dy/dx Find an equation for the tangent to the curve at the point where x=2. Determine whether y is increasing or decreasing when x = -3. dy/dx = -21x6 When x = 2, dy/dx = - 21 x 26 = 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 When x=-3, dy/dx = -21 x (-3) 6 = < 0. So y is decreasing. Based on C1(AQA) Jun 2006

128
**Questions for you to try.**

A curve has equation y=x x2 + 29x Find dy/dx 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 3x2 +88x + 29 = 0. So 3x2 +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)2 + 29(-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)2 + 29(-29) = So second point has coordinate (-29,11774) Based on C1(AQA) Jun 2006

129
More practice for you.

130
C1(AQA) Jan 2006 Question 7

131
C1(AQA) Jan 2007 Question 5

132
C1(AQA) Jun 2007 Question 4

133
C1(AQA) Jan 2008 Question 2

134
C1(Edexcel) Jan 2006 Question 9

135
C1(Edexcel) Jan 2006 Question 10

136
C1(Edexcel) Jun 2006 Question 5

137
C1(Edexcel) Jan 2007 Question 1

138
C1(Edexcel) Jan 2007 Question 8

139
C1(Edexcel) Jun 2007 Question 3

140
C1(Edexcel) Jan 2008 Question 5

141
POLYNOMIALS (MEI)

142
Key Topics

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

146
Quick Quiz

147
**Question 1 Which of the following is a factor of x³ + x² + 2x + 8 x+2**

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 f(-2) = 0, so x+2 is a factor. f(2) =24, so x-2 is not a factor. f(-1) = 6, so x+1 is not a factor. 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
Worked Exam Questions

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. C1(MEI) 16th January 2006

155
Worked Example When x3 + 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) = x3 + 3x +k , then f(1) = x1 +k = 6. So 4+k =6, so k=2. C1(MEI) 16th January 2006

156
**Questions for you to try now...**

157
**Questions for you to try.**

C1(MEI) 16th January 2006

158
**Questions for you to try.**

C1(MEI) 6th June 2006

159
**Questions for you to try.**

C1(MEI) 7th June 2007

160
**Answers to selected questions**

161
**Solution to Question 1 Question 1**

i) We need the discriminant to be greater than or equal to zero. b2 -4ac = 52 -4x1xk =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 =0 implies (2x+5)(2x+5)=0, so x=-5/2. C1(MEI) 16th January 2006

162
**Solution to Question 2 Question 2 f(x) = x3 + ax2 +7**

Put x=-2, to get f(-2) = (-2)3 + a(-2)2 +7 = 0 So a +7 =0. Thus 4a=1, a = ¼. C1(MEI) 6th June 2006

163
Solution to Question 3 Question 3 C1(MEI) 7th June 2007

164
**Solution to Question 3 ctd**

C1(MEI) 7th June 2007 -2c=10, c=-5 a=2 -2a + b = -1 -4 + b =-1, b =3 ax2 bx c , c=-5 x a = 2 bx2 -2 -2ax2 -2c=10 So x=1 and x=-5/2 are other roots of the equation.

165
**Solution to Question 3 ctd**

C1(MEI) 7th June 2007 (3,0) (0,-12) y=-22

166
More practice for you.

167
C1(MEI) 6th June 2006 Question 8

168
C1(MEI) 6th June 2006 Question 12

169
C1(MEI) 16th January 2007 Question 4

170
C1(MEI) 16th January 2007 Question 5

171
C1(MEI) 16th January 2007 Question 8

172
C1(MEI) 7th June 2007 Question 4

173
C1(MEI) 7th June 2007 Question 6

174
C1(MEI) January 2008 Question 6

175
C1(MEI) January 2008 Question 7

176
C1(MEI) June 2008 Question 3

177
C1(MEI) June 2008 Question 8

178
C1(MEI) June 2008 Question 11

179
EXAM PRACTICE

181
That’s all folks!!!

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google