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- 1. HIGHER PRELIM REVISION UNIT 1
- 2. <ul><li>Find the equation of the straight line through the </li></ul><ul><li>points A(-1,5) and B(3,1) </li></ul>(b) Find the angle which AB makes with the positive direction of the x -axis. EQUATION OF LINE BETWEEN POINTS HOW TO DO: 1. Find gradient between points using 2. Sub this gradient and EITHER point into y - b = m(x - a) 3. To find angle use gradient found in 1. and
- 3. Equation of line through (3 , 1) is: Worked Solution: From part (a) m = -1 Exact Values in non-calc papers Tan is –ve in Q2 & 4 – we want a 0<a<180. A(-1,5) and B(3,1)
- 4. Homework : Heinemann, p.11, Ex 1G, Q2
- 5. PERPENDICULAR BISECTORS HOW TO DO: 1. Find the midpoint between the points. 3. To find perpendicular gradient : “ Flip the fraction change the sign ” 4. Find equation by using this gradient and midpoint in y – b = m(x - a). Find the equation of the perpendicular bisector of the line joining the points (-1, -2) and (3, 8). 2. Find gradient between points using
- 6. Worked Solution: Gradient of perpendicular: Equation of line through (1, 3) is: 5y + 2x = 17 X 5 (-1, -2) and (3, 8).
- 7. Homework : Heinemann, p.13, Ex 1 I , Q1
- 8. STRAIGHT LINES FORMING TRIANGLES WITH AXIS
- 9. STRAIGHT LINES FORMING TRIANGLES WITH AXIS HOW TO DO: 1. Make sure EACH line is in form y = mx + c 2. Read off gradient (number in front of x). 3. To find angles use gradient found in 2. and 4. Use these angles to fill in missing angles in triangle.
- 10. Solution: y = 2x + 4 Must be in form y = mx + c m = 2 Using: x + y = 13 m = -1 63.4 0 45 0 135 0 AS angles in triangle = 180 0 , acute angle is 71.6 0
- 11. Homework : Same question using lines x + y = 15 & 2y = 3x + 2
- 12. Composite Functions f(x) = 5x + 1 and g(x) = 2x 2 - 4 (a) If B = 11, find the corresponding values of A & C. (b) Find a formula for h(x)
- 13. Solution to (a): Composite Functions From diagram : B = f(A) = 5A + 1 So if B = 11 : 11 = 5A + 1 10 = 5A A = 2 From diagram : C = g(B) = 2B 2 - 4 So if B = 11 : C = 2 x 11 2 – 4 C = 2 x 121 – 4 C = 238 f(x) = 5x + 1 and g(x) = 2x 2 - 4
- 14. Solution to (b): Composite Functions From diagram : h is g applied to f h(x) = 2(f) 2 – 4 f(x) = 5x + 1 and g(x) = 2x 2 - 4 h(x) = 2(5x + 1) 2 – 4 h(x) = 2(25x 2 + 10x + 1) – 4 h(x) = 50x 2 + 20x + 2 – 4 h(x) = 50x 2 + 20x – 2
- 15. Homework : Heinemann, p.26, Ex 2C, Q3 & 5(a) – (d)
- 16. 2004 – Paper 1 – Question 4 (a) 2 Marks (b) 2 Marks Sketching Graphs
- 17. Sketching Graphs HOW TO DO: 1. Set up a table for the KEY POINTS of f(x). <ul><li>Remember basic rule; If transformation is in bracket </li></ul><ul><li>with x the x co-ordinate is changed. Otherwise </li></ul><ul><li>only y coordinate changes. </li></ul>4. Plot points in order left to right and join up. <ul><li>Add another row underneath and for each point </li></ul><ul><li>write down new co-ordinate after transformation. </li></ul>
- 18. Solution to (a): Learn rules for transforming graphs and identify key points on graphs : intercepts, roots, SP’s etc. 2004 – Paper 1 – Question 4 From graph key points are: For y = -g(x) reflect in x-axis or change sign of y coordinate Now sketch these points. Rules for transforming
- 19. Graph compressed horizontally No change Divide by k. Graph stretched vertically Multiply by k No change Graph reflected in y-axis. No change Change sign Graph reflected in x-axis Change sign No change Graph moves left (+) or right (-) by a. No change Graph moves up or down by a. No change Comment Y-coord X-coord Graph
- 20. continue 2004 – Paper 1 – Question 4
- 21. Solution to (b): Learn rules for transforming graphs and identify key points on graphs : intercepts, roots, SP’s etc. 2004 – Paper 1 – Question 4 From graph key points are: For y = -g(x) reflect in x-axis or change sign of y coordinate Now sketch these points. Rules for transforming For y =-g(x) + 3 add 3 to y-coordinates of -g(x).
- 22. 2004 – Paper 1 – Question 4
- 23. Homework : Heinemann, p.48, Ex 3P, Q4
- 24. Solving TRIG equations with one trig function HOW TO DO: 1. Check interval. 3. Solve for 2x using quadrant rules. 4. ADD Period (360 0 ) to each answer. At this stage you should have 2 pairs for each multiple of x. Solve 2. Rearrange to make x subject. 5. Which of answers fall in given interval?
- 25. Worked Solution: 180 0 Solution is x = {15 0 , 75 0 }
- 26. Exact Values 1 Und. 0 tan 0 1 cos 1 0 sin 0 90 0 60 0 45 0 30 0 0 0
- 27. All +ve Cos +ve Tan +ve Sin +ve - + 2 - 0 Quadrant Rules
- 28. Homework : Heinemann, p.63, Ex 4H, Q1(a), (b), (c)
- 29. 2002 (Winter)– Paper 2 – Question 3 <ul><li>Calculate the limit as of the sequence defined </li></ul><ul><li>by </li></ul>(b) Determine the least value for n for which u n is greater than half of this limit and the corresponding value of u n . 3 Marks 2 Marks Limits of Recurrence Relations
- 30. HOW TO DO: 1. Establish m and c. <ul><li>Remember basic rule; </li></ul>4. If asked to find when relation reaches a particular “ target value” keep applying recurrence relation to each successive answer until target is reached. <ul><li>Make statement </li></ul>Limits of Recurrence Relations M is % or fraction C is “addy on bit.” Limit exists as
- 31. Solution: <ul><li>For limit need m and c. </li></ul>m = 0.9 c = + 10 Limit of Recurrence Relation: 2002 (Winter)– Paper 2 – Question 3 As 0.1 = 1/10 flip fraction So if this was 0.4 x by 10 then divide by 4 MUST WRITE THIS Limit exists as
- 32. Solution: (b) We want to find first term when sequence exceeds 50: Exceeds 50 at u 7 and 2002 (Winter)– Paper 2 – Question 3
- 33. Homework : Heinemann, p.78, Ex 5H, Q2
- 34. Recurrence Relations with limits
- 35. HOW TO DO: 1. Establish m and c and use them to make RR. 4. Remember basic rule; 2. If asked to find when relation reaches a particular “ target value” keep applying recurrence relation to each successive answer until target is reached. 3. “Will this be safe in the long run” questions are about limits. Make statement Limits of Recurrence Relations M is % or fraction C is “addy on bit.” Limit exists as
- 36. Recurrence Relations with limits Solution to (a): m = 100 – 25% = 75% = 0.75 c = +1 g per week TARGET = 2g Week 0 : 0g Week 1 : 0.75 x 0 + 1 = 1g Week 2 : 0.75 x 1 + 1 = 1.75 g Week 3 : 0.75 x 1.75 + 1 = 2.3125 g So it takes 3 weeks for bioforce to be effective.
- 37. Recurrence Relations with limits Solution to (b): (b) (ii) m = 0.75 c = +1 = 4g As the level of bioforce tends to 4g in the long term, and this is less than 5g, it is safe to continue this feeding programme.
- 38. Homework : Heinemann, p.79, Ex 5H, Q8
- 39. 1Mark (a) The point (-1, t) lies on the graph of Find the value of t. 3 Marks Increasing / Decreasing Functions (b) Show that the function is increasing at this point.
- 40. Increasing / Decreasing Functions HOW TO DO: <ul><li>To find a missing coordinate substitute the value </li></ul><ul><li>you know into the given function. </li></ul>2. To show a function is increasing / decreasing: - Differentiate (dy/dx) - substitute in x coordinate to dy/dx - positive values of dy/dx means increasing function - negative values of dy/dx means decreasing
- 41. Solution to (a): To find missing coordinates substitute known values into function. Increasing / Decreasing Functions If x = -1: So t = 1
- 42. Solution to (b): To find whether function is increasing/decreasing find dy/dx and then sub in x coordinate. Increasing / Decreasing Functions If x = -1: AS f ‘ (-1) is positive the function is increasing at that point.
- 43. Homework : Heinemann, p.104, Ex 6L, Q1
- 44. Maximums and Minimums The cost of painting the local community centre has been calculated as where x is the surface area of the walls to be painted. <ul><li>Calculate the surface area which will minimise the </li></ul><ul><li>cost. </li></ul>(b) Calculate the minimum cost of painting the community centre.
- 45. HOW TO DO: 1. Prepare the function for differentiation. <ul><li>Find SP’s and MAKE STATEMENT: </li></ul><ul><li>at SP’s dy/dx = 0 . </li></ul>4. Justify maximum/minimum from Nature Table. <ul><li>Find dy/dx. </li></ul>Maximums and Minimums 5. To find actual max/min substitute x value found in 3. INTO ORIGINAL FUNCTION .
- 46. Maximums and Minimums Solution to (a): At max/min x 3x 2 Multiply by denominator
- 47. Maximums and Minimums Solution to (a): At max/min x 3x 2 Multiply by denominator x = 60.6 Surface area which minimises costs
- 48. -ve Maximums and Minimums Solution to (a): +ve Slope 0 60.6 X
- 49. Maximums and Minimums Solution to (b): x = 60.6 Surface area which minimises costs Minimum cost is £121.10 ORIGINAL FUNCTION
- 50. Homework : Heinemann, p.117, Ex 6S, Q17 &18

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