Let Maths take you Further…

Name: Let Maths take you Further…
Uploaded: 2017-08-22T02:52:43+00:00
Duration: PTM50S32
Channel: Joslyn Peevy
Description: Let Maths take you Further…

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

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!

ALGEBRA

Key Topics

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

Quick Quiz

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

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

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

Worked Exam Questions

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

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.

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

Questions for you to try now...

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

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

Answers to selected questions

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

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

Solutions Question 2 The formula for r is given by: A) B) C) D)
E) None of these. C1(MEI) 6th June 2006

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

More practice for you.

C1(AQA) Jan 2006 Question 1

C1(Edexcel) Jan 2006 Question 1

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

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

COORDINATE GEOMETRY

Key Topics

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

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

Quick Quiz

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

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

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

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

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)

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

Question for you to try (part 1)
Question 3 (Part One) C1(MEI) 7th June 2007

Question for you to try (part 2)
Question 3 (Part Two) C1(MEI) 7th June 2007

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

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

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

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)

C1(AQA) Jun 2006 Question 7

C1(Edexcel) Jun 2006 Question 10

CURVE SKETCHING (AND INDICES)

Key Topics

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

Quick Quiz

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]

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

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)

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)

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

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

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

Solution to Question 1 (ii)
Solution to (ii) is: A) B) C) D) E) Don’t know C1(MEI) June 2006

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)

CALCULUS (NON-MEI)

Key Topics

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

Quick Quiz

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.

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

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’.

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

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

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

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

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

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

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

POLYNOMIALS (MEI)

Key Topics

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

Quick Quiz

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

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

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

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

C1(MEI) 16th January 2006

C1(MEI) 6th June 2006

C1(MEI) 7th June 2007

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

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

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

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.

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

C1(MEI) 6th June 2006 Question 8

C1(MEI) 16th January 2007 Question 4

C1(MEI) January 2008 Question 6

C1(MEI) January 2008 Question 7

C1(MEI) June 2008 Question 3

EXAM PRACTICE

That’s all folks!!!

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