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Further Mathematics Support Programme www.furthermaths.org.uk Core 1 Revision Day Let Maths take you Further… Outline of the Day 10:00 11:00 11:00 11:15 11:15 12:15 Algebra Break Co-ordinate Geometry 12:15 1:00pm Lunch 1:00 2:00 Curve Sketching and Indices Calculus 2:00 3:00 3:00pm Home time! 2 ALGEBRA 3 4 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 5 6 7 8 Question 1 The expression (2x-5)(x+3) is equivalent to: A) 2x2 + x - 15 B) 2x2 - x - 15 (2x-5)(x+3) 2 C) 2x + 11x - 15 2 +6x – 5x -15 = 2x D) 2x2 - 2x - 15 = 2x2 +x – 15 E) Don’t know Question 2 The discriminant of the quadratic equation 2x2 +5x-1=0 is: A) 17 b2 – 4ac B) 33 = 52 – 4 x 2 x (-1) C) 27 = 25 + 8 D) -3 E) I don’t know =33 Question 3 Consider the simultaneous equations: x + 3y = 5 3x – y =5 The correct value of x for the solution is: 3 x (2) 9x –3 y =15 A) x=1 B) x= -1 + x + 3y = 5 C) x=2 10x = 20 D) x= -2 x=2 E) I don’t know (3) (1) 12 Worked Example a) Write the expression x2 -8x – 29 in the form (x+a)2 + b, where a and b are constants. b) 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. a) x2 8x 29 ( x 4)2 16 29 Take away 16 since the -4 in the bracket will give us an extra 16. ( x 4)2 45 So a=-4 and b=-45. b) x2 8x 29 0 ( x 4)2 45 0 ( x 4)2 45 We must complete the question Don’t forget the plus and minus. A very common error through out Alevel! x 4 45 x 4 45 x 4 5 9 4 3 5 x 4 3 5 so c=4 and d=3 13 Worked Example Solve the simultaneous equations 1) x – 2y = 1, 2) x2 + y2 = 29. Label the equations 1 and 2 Equation 1 does not have any squared terms, so it is easier to expression x in terms of y Equation 1 x 1 2 y Using this in Equation 2 gives (1+2y)2 y 2 29 1 4 y 4 y 2 y 2 29 0 5 y 2 4 y 28 0 You must know how to solve quadratic equations with ease! You could use the formula if you wanted! (5 y 14)( y 2) 0 y 2, y 14 5 When y 14 , x=1+2 14 = 23 14 5 5 5 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. a)6 x 3 5 2 x 8 x 2 x 1 4 b)2x 2 7 x 3 2x 2 7 x 3 0 (2 x 1)(x 3) 0 Draw a quick sketch to help you. The function changes signs at x=3, x= 1 . 2 ½ So we need x< 1 or x>3 2 3 c) Draw a number line ¼ ½ Since both inequalities must be true 3 we look for where there is red and blue 15 1 x 1 4 2 x3 16 Question for you to try Question 1 17 Question for you to try Question 2 18 Question for you to try Question 3 19 20 Solutions Question 1 For part (i) your values of a and b are: A) B) C) D) E) a =30, b = 2; a = 120, b = 2; a = 30, b = 5; a = 120, b = 5; None of these. 21 Worked Solution Question 1 i)5 8 4 50 5 4 2 4 25 2 5 2 2 4 5 2 10 2 20 2 30 2 a=30 and b=2 (6 3) 6 3 3 6 3 3 2 3 1 36 3 33 11 11 6 3 6 3 (6 3) 1 2 p ,q 11 11 (6 3) (6 3) 36 6 3 6 3 ( 3)2 36 3 ii) 3 3 22 Solutions Question 2 The formula for r is given by: A) r V 3h B) r 3V h C) r 3V h D) r Vh 3 E) None of these. 23 Solutions 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. 24 25 Question 1 C1(AQA) Jan 2006 26 Question 3 C1(AQA) Jan 2007 27 Question 7 C1(AQA) Jan 2007 28 Question 1 C1(Edexcel) Jan 2006 29 Question 5 C1(Edexcel) Jan 2006 30 Question 2 C1(Edexcel) Jun 2006 31 Question 6 C1(Edexcel) Jun 2006 32 Question 8 C1(Edexcel) Jun 2006 33 Question 2 C1(Edexcel) Jan 2007 34 Question 5 C1(Edexcel) Jan 2007 35 Question 1 C1(Edexcel) Jun 2007 36 Question 6 C1(Edexcel) Jun 2007 37 Question 7 C1(Edexcel) Jun 2007 38 Question 2 C1(Edexcel) Jan 2008 39 Question 3 C1(Edexcel) Jan 2008 40 Question 8 C1(Edexcel) Jan 2008 41 COORDINATE GEOMETRY 42 43 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 44 Gradient = change in y = y2 – y1 change in x x2 – x1 y = mx + c m = gradient c = y intercept Distance between two points Equation of a circle: Centre: (a, b) Radius: r 47 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 15 C) The gradient is 15 and the y-intercept is 3 y =and 3/10 + 15/10 is 0.3 D) The gradient is 1.5 thexy-intercept E) Don’t know y = 0.31 x + 1.5 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 (4 1) (7 5) 9 4 13 D) isosceles (4 5) (7 2) 1 25 26 E) Don’t know (5 1) (2 5) 16 9 25 5 2 2 2 2 2 2 The sides are all different lengths 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 The equation represents a D) The point (−3,−1) lies on the circle circle with centre E) Don’t know (-3, 1) and radius 2. So the statement is incorrect 51 Worked Example y NOT TO SCALE B(3,4) O A C x 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 line AB is 5 So gradient of line BC is -1/5. Gradient of perpendicular = -1 / (gradient of original) 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 =52-1/5 x + 23/5. So x = 23. Worked Example A circle has equation (x-2)2 + y2 = 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 + 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: (1,6) Watch the minus signs (8,-3) Distance = (8 1) 2 (6 (3)) 2 72 92 53 49 81 130 11.40 54 Question for you to try Question 1 55 Question for you to try Question 2 56 Question for you to try (part 1) Question 3 (Part One) 57 Question for you to try (part 2) Question 3 (Part Two) 58 59 Solution to Question 1 Question 1 The equation of the line is: A) B) C) D) E) 3x + 2y = 26 3x + 2y = 13 -3x + 2y = 26 -3x + 2y = 13. Don’t know. 60 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 * 2 + 2 * 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 = 10 + 3 = 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. 61 Question 2 Solution to Question 2 The radius of the circle in (ii) is: A) B) C) D) E) √45 ½ √45 √85 ½ √85 Don’t know. 62 Solution to Question 2 i) ii) Gradient of AB =(8-0)/(9-5) = 8/4 =2 Gradient of BC = (1-0)/(3-5) = 1/(-2) = -½ Gradient = change in y/change in x 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 = 9 3 , 8 1 6,4.5 2 iii) 2 AC = √ ((9-3)2 + (8-1)2 ) = √ (62 + 72 ) = √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 = 1 + 81/4 = 85/4. So B lies on the circle. Let (x,y) be coordinates of D. D(x,y) Midpoint of BD is centre of circle (6,4.5). So 5 x 0 y , 6,4.5 5 x 12, y 9, x 7, y 9. 2 (6,4.5) 2 So coordinates of D are (7,9). 63 B(5,0) 64 Question 2 C1(AQA) Jan 2006 65 Question 5 C1(AQA) Jan 2006 66 Question 7 C1(AQA) Jun 2006 67 Question 4 C1(AQA) Jan 2007 68 Question 2 C1(AQA) Jan 2007 69 Question 1 C1(AQA) Jun 2007 70 Question 5 C1(AQA) Jun 2007 71 Question 1 C1(AQA) Jan 2008 72 Question 4 C1(AQA) Jan 2008 73 Question 3 C1(Edexcel) Jan 2006 74 Question 10 C1(Edexcel) Jun 2006 75 Question 11 C1(Edexcel) Jun 2006 76 Question 10 C1(Edexcel) Jun 2007 77 Question 11 C1(Edexcel) Jun 2007 78 Question 4 C1(Edexcel) Jan 2008 79 CURVE SKETCHING (AND INDICES) 80 81 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 82 85 Question 1 The vertex of the quadratic graph y=(x-2)2 - 3 is : A) Minimum (2,-3) B) Minimum (-2,3) The graph has a minimum C) Maximum (2,-3) point, since the coefficient of D) Maximum (-2,-3) x² is positive. E) Don’t know The smallest possible value of (x-2)2 is 0, when x = 2. [When x = 2 y = -3] 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 2 -2x-3 2 -1 -3 =(x-1)2 - 4 2 x =(x-1) D) (x-1) - 2 E) Don’t know The -1 is present to correct for the +1 we get when multiplying out (x-1)2 Question 3 The graph of y=x2 -2x-1 has a minimum point at: y = x2 -2x-1 2 -1 -1 = (x-1) A) (1,-1) = (x-1)2 - 2 B) (-1,-1) So minimum point is (1,-2) C) (-1,-2) D) (1,-2) E) Don’t know 89 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)2 -1 -2 = (x-1)2 -3. So p=1 and q =-3. ii) (x-1)2 has its smallest value when x=1, y value at this point is -3. y So minimum point is (1,-3). iii) 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. (1-√3,0) Coordinates are (1 +√3,0) and (1 -√3,0) Graph crosses y-axis when x=0. (0,-2) So coordinates are (0,-2) 90 i) x (1+√3,0) (1,-3) 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. iv) Intersection when x2 -2x – 2 = x2 +4x – 5. So -2x-2 = 4x -5, giving 6x – 3 =0. So x= ½. Intersection when x=½. 91 92 Question for you to try Question 1 93 Question for you to try Question 2 94 Question for you to try Question 3 95 Question for you to try Question 4 96 97 Solution to Question 1(i) Question 1 Solution to (i) is: A) B) C) D) E) 4/27 8/81 4/3 4/81 Don’t know 98 Solution to Question 1 (ii) Question 1 Solution to (ii) is: B) 3a10 b 8 c2 3 C) 3a10b8c 2 A) 8 3 b D) c2 E)99 Don’t know Question 3 Solution to Question 3 (-3,9) The graph of y=4x2 -24x + 27 is: A) C) (0,27) (0,27) (-4.5,0) x (1.5,0) x (1.5,0) (4.5,0) (3,-9) B) (3,9) (0,27) D) x (-4.5,0) (-1.5,0) (-3,-9) (1.5,0) (0,-27) 100 E) Don’t know (4.5,0) x 101 Question 3 C1(AQA) Jan 2006 102 Question 1 C1(AQA) Jan 2007 103 C1(AQA) Jun 2007 Question 3 104 C1(AQA) Jan 2008 Question 5 105 C1(Edexcel) Jun 2006 Question 3 106 C1(Edexcel) Jun 2006 Question 9 107 C1(Edexcel) Jan 2007 Question 10 108 CALCULUS (NON-MEI) 109 110 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 111 114 Question 1 The gradient of the curve y=3x2 – 4 at the point (2,8) is : A) 12 B) 6x 2–4 If y=3x C) 48 then dy/dx = 6x D) 8 x=2 E) Don’t know dy/dx = 6x2 = 12 So gradient of curve at (2,8) is 12. Question 2 If 3t 3 23t t2 3 2t 2 3 x x 2 t 1 2 2t 2t 2 dx 3 1 .5 dt 2 then A) B) C) D) E) dy/dx =1.5 dy/dx = 3/2 - t dx/dt = 1.5 dt/dx = 1.5 Don’t know Question 3 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 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. Worked Example A curve has equation y = 2x3 -3x2 – 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 = 2x3 -3x2 – 8x + 9, then dy/dx = 6x2 – 6x -8. When x = 2, dy/dx = 6x22 – 6x2 -8. = 24 – 12 - 8 =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 = -2 -3 +8 +9 =12. So coordinates of Q are (-1,12). 122 Questions for you to try. Question 1 A curve has equation y = 10 – 3x7. 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. 123 Questions for you to try. Question 2 A curve has equation y=x3 + 44x2 + 29x i) Find dy/dx ii) Hence find the coordinates of the points on the curve where dy/dx=0. 124 125 Questions for you to try. Question 1 A curve has equation y = 10 – 3x7. 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) B) C) D) E) y = -1344x + 2314. y = 1344x - 3062. y = -21x6 + c y = 21x6 + c Don’t know 126 Questions for you to try. Question 1 A curve has equation y = 10 – 3x7. 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 = -21x6 ii) When x = 2, dy/dx = - 21 x 26 = -1344. So y = -1344 x + c. When x = 2, y = 10 – 3 (2) 7 = - 374. So (2,-374) is point on the curve. Using this point in y = -1344 x + c gives -374 = -1344 x 2 + c. So c = 2314. Equation of tangent is y = -1344x + 2314. iii. When x=-3, dy/dx = -21 x (-3) 6 = -15309 < 0. So y is decreasing. 127 Questions for you to try. Question 2 A curve has equation y=x3 + 44x2 + 29x i) Find dy/dx ii) Hence find the coordinates of the points on the curve where dy/dx=0. i) dy/dx = 3x2 + 88x + 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)3 + 44(-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)3 + 44(-29)2 + 29(-29) = 11774. So second point has coordinate (-29,11774)128 129 C1(AQA) Jan 2006 Question 7 130 C1(AQA) Jan 2007 Question 5 131 C1(AQA) Jun 2007 Question 4 132 C1(AQA) Jan 2008 Question 2 133 Question 9 C1(Edexcel) Jan 2006 134 C1(Edexcel) Jan 2006 Question 10 135 C1(Edexcel) Jun 2006 Question 5 136 C1(Edexcel) Jan 2007 Question 1 137 C1(Edexcel) Jan 2007 Question 8 138 C1(Edexcel) Jun 2007 Question 3 139 C1(Edexcel) Jan 2008 Question 5 140 POLYNOMIALS (MEI) 141 142 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 143 146 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 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. 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 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. Question 3 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 Example Find the binomial expansion of (3+x)4, writing each term as simply as possible. Binomial Expansion (a b) n a n nan1b n(n 1) n2 2 n(n 1) 2 n1 n a b ab b 2! (n 1)! In our example, a=3, b = x and n=4. (3 x) 4 34 4 341 x 4 3 4 2 2 4 3 2 4 3 3 3 x 3 x x4 2! 3! 81 108x 54x 2 12x3 x 4 154 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) = 13 + 3x1 +k = 6. So 4+k =6, so k=2. 155 156 Questions for you to try. Question 1 157 Questions for you to try. Question 2 158 Questions for you to try. Question 3 159 160 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) 4x2 +20 x + 25 = (2x+5)(2x+5) So 4x2 +20 x + 25 =0 implies (2x+5)(2x+5)=0, so x=-5/2. 161 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 -8 + 4a +7 =0. Thus 4a=1, a = ¼. 162 Question 3 Solution to Question 3 i)(x 3)(2x 2 5x 4) 2x3 5x2 4x 6x 2 15x 12 2 x3 x 2 11x 12 Computethediscriminant of quadratic T hediscriminant of 2 x 2 5x 4 is 52 4 2 4 25 32 7 0 T hediscriminant is negativeso no real rootsof thequadratic equation Henceonlyone real rootof thecubic equationf(x) 0. 163 Solution to Question 3 ctd ii)Put x 2 into f ( x) 22 toget 2(2)3 (2)2 11(2) 12 22 0 So x 2 is root of f ( x) 22. Computingf(x) 22 gives 2x3 x 2 11x 12 22 2x3 x 2 11x 10 ax2 bx x a=2 bx2 -2 -2ax2 c , c=-5 -2c=10 So 2x3 x 2 11x 10 ( x 2)(2x 2 3x 5) ( x 2)(2 x 5)(x 1) So x=1 and x=-5/2 are other roots of the equation. 164 -2c=10, c=-5 a=2 -2a + b = -1 -4 + b =-1, b =3 Solution to Question 3 ctd We havef(x) 22 ( x 2)(2 x 5)(x 1) So f(x) ( x 2)(2 x 5)(x 1) 22 So graph of f(x)is graph of ( x 2)(2 x 5)(x 1) shift down by 22 units. (3,0) (0,-12) 165 y=-22 166 C1(MEI) 6th June 2006 Question 8 167 C1(MEI) 6th June 2006 Question 12 168 C1(MEI) 16th January 2007 Question 4 169 C1(MEI) 16th January 2007 Question 5 170 C1(MEI) 16th January 2007 Question 8 171 C1(MEI) 7th June 2007 Question 4 172 C1(MEI) 7th June 2007 Question 6 173 Question 6 C1(MEI) January 2008 174 Question 7 C1(MEI) January 2008 175 Question 3 C1(MEI) June 2008 176 Question 8 C1(MEI) June 2008 177 Question 11 C1(MEI) June 2008 178 EXAM PRACTICE 179 That’s all folks!!! 181