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HIGHER – ADDITIONAL QUESTION BANK

Practice and master straight line equations with a question bank covering gradients, parallel lines, and triangles in unit 1. Improve your skills and understanding of straight line functions and graphs.

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HIGHER – ADDITIONAL QUESTION BANK

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  1. HIGHER – ADDITIONAL QUESTION BANK Please decide which Unit you would like to revise: UNIT 1 UNIT 2 UNIT 3 Straight Line Functions & Graphs Trig Graphs & Equations Basic Differentiation Recurrence Relations Polynomials Quadratic Functions Integration Addition Formulae The Circle Vectors Further Calculus Exponential / Logarithmic Functions The Wave Function EXIT

  2. HIGHER – ADDITIONAL QUESTION BANK Straight Line Trig Graphs & Equations UNIT 1 : Functions & Graphs Basic Differentiation Recurrence Relations EXIT

  3. HIGHER – ADDITIONAL QUESTION BANK You have chosen to study: Straight Line UNIT 1 : Please choose a question to attempt from the following: 1 2 3 4 5 Back to Unit 1 Menu EXIT

  4. STRAIGHT LINE : Question 1 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  5. y = -5/3x - 6 STRAIGHT LINE : Question 1 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  6. Question 1 y = -5/3x - 6 Back to Home 3x – 5y = 4 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). 3x - 4 = 5y (5) 5y = 3x - 4 y = 3/5x - 4/5 Using y = mx + c , gradient of line is 3/5 So required gradient = -5/3 , ( m1m2 = -1) We now have (a,b) = (-6,4) & m =-5/3 . Using y – b = m(x – a) We get y – 4 = -5/3 (x – (-6)) y – 4 = -5/3 (x + 6) Begin Solution y – 4 = -5/3x - 10 Continue Solution Markers Comments Straight Line Menu

  7. Markers Comments -5 3 m1 = m2 = 3 5 y = -5/3x - 6 Back to Home • An attempt must be made to put the original equation into the form • y = mx + c to read off the gradient. 3x – 5y = 4 3x - 4 = 5y (5) 5y = 3x - 4 • State the gradient clearly. • State the condition for perpendicular lines m1 m2 = -1. y = 3/5x - 4/5 Using y = mx + c , gradient of line is 3/5 • When finding m2 simply invert and change the sign on m1 So required gradient = -5/3 , ( m1m2 = -1) We now have (a,b) = (-6,4) & m =-5/3 . Using y – b = m(x – a) • Use the y - b = m(x - a) form to obtain the equation of the line. We get y – 4 = -5/3 (x – (-6)) y – 4 = -5/3 (x + 6) Next Comment y – 4 = -5/3x - 10 Straight Line Menu

  8. STRAIGHT LINE : Question 2 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3). Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  9. STRAIGHT LINE : Question 2 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3). y = -2x + 7 Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  10. Question 2 Back to Home 8x + 4y – 7 = 0 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3). 4y = -8x + 7 (4) y = -2x + 7/4 Using y = mx + c , gradient of line is -2 So required gradient = -2 as parallel lines have equal gradients. We now have (a,b) = (5,-3) & m = -2. Using y – b = m(x – a) We get y – (-3) = -2(x – 5) Begin Solution y + 3 = -2x + 10 Continue Solution Markers Comments y = -2x + 7 Straight Line Menu

  11. Markers Comments Back to Home • An attempt must be made to • put the original equation into • the form y = mx + c to • read off the gradient. 8x + 4y – 7 = 0 4y = -8x + 7 (4) y = -2x + 7/4 • State the gradient clearly. Using y = mx + c , gradient of line is -2 • State the condition for • parallel lines m1 = m2 So required gradient = -2 as parallel lines have equal gradients. • Use the y - b = m(x - a) form • to obtain the equation of • the line. We now have (a,b) = (5,-3) & m = -2. Using y – b = m(x – a) We get y – (-3) = -2(x – 5) y + 3 = -2x + 10 Next Comment Straight Line Menu y = -2x + 7

  12. Y C X A B STRAIGHT LINE : Question 3 In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). (a) Find the gradients of AC and BC. (b) Hence find the size of ACB. Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  13. Y C X A B STRAIGHT LINE : Question 3 In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). (a) Find the gradients of AC and BC. (b) Hence find the size of ACB. mAC = 3/5 (a) Reveal answer only mBC = - 3 Go to full solution Go to Marker’s Comments (b) = 77.4° Go to Straight Line Menu Go to Main Menu EXIT

  14. Question 3 Y C X B A Back to Home • Using the gradient formula: In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). mAC = 3 – 0 7 - 2 = 3/5 • Find the gradients of AC • and BC. mBC = 3 – 0 7 - 8 = - 3 (b) Hence find the size of ACB. (b) Using tan = gradient If tan = 3/5 then CAB = 31.0° If tan = -3 then CBX = (180-71.6)° = 108.4 o so ABC = 71.6° Begin Solution Hence : ACB = 180°– 31.0° – 71.6° Continue Solution Markers Comments = 77.4° Straight Line Menu

  15. Markers Comments B Ø ° A Back to Home • If no diagram is given draw a • neat labelled diagram. • In calculating gradients state • the gradient formula. • Using the gradient formula: mAC = 3 – 0 7 - 2 = 3/5 • Must use the result that the • gradient of the line is equal • to the tangent of the angle • the line makes with the • positive direction of the x-axis. • Not given on the formula sheet. mBC = 3 – 0 7 - 8 = - 3 (b) Using tan = gradient If tan = 3/5 then CAB = 31.0° mAB = tanØ ° Ø ° = tan-1 mAB If tan = -3 then CBX = (180-71.6)° = 108.4 o so ABC = 71.6° Hence : ACB = 180°– 31.0° – 71.6° Next Comment Straight Line Menu = 77.4°

  16. STRAIGHT LINE : Question 4 Y In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Q(2,3) Find (a) the equation of the line e, the median from R of triangle PQR. X R(10,-1) P(4,-5) (b) the equation of the line f, the perpendicular bisector of QR. Reveal answer only (c) The coordinates of the point of intersection of lines e & f. Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  17. STRAIGHT LINE : Question 4 Y In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Q(2,3) Find (a) the equation of the line e, the median from R of triangle PQR. X R(10,-1) P(4,-5) (b) the equation of the line f, the perpendicular bisector of QR. Reveal answer only (c) The coordinates of the point of intersection of lines e & f. Go to full solution Go to Marker’s Comments (a) y = -1 Go to Straight Line Menu (b) y = 2x – 11 Go to Main Menu EXIT (5,-1) (c)

  18. Question 4 (a) Y Q(2,3) X R(10,-1) P(4,-5) Back to Home • Midpoint of PQ is (3,-1): let’s call this S In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Using the gradient formula m = y2 – y1 x2 – x1 Find (a) the equation of the line e, the median from R of triangle PQR. mSR = -1 – (-1) 10 - 3 = 0 (ie line is horizontal) Since it passes through (3,-1) equation of e is y = -1 Solution to 4 (b) Begin Solution Continue Solution Markers Comments Straight Line Menu

  19. Question 4 (b) (b) Midpoint of QR is (6,1) Y Q(2,3) mQR = 3 – (-1) 2 - 10 = 4/-8 X R(10,-1) P(4,-5) so f is y = 2x – 11 Back to Home In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). = - 1/2 Find required gradient = 2 (m1m2 = -1) (b) the equation of the line f, the perpendicular bisector of QR. Using y – b = m(x – a) with (a,b) = (6,1) & m = 2 we get y – 1 = 2(x – 6) Solution to 4 (c) Begin Solution Continue Solution Markers Comments Straight Line Menu

  20. Question 4 (c) Y Q(2,3) X R(10,-1) P(4,-5) Back to Home (c) e & f meet when y = -1 & y = 2x -11 In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). so 2x – 11 = -1 Find ie 2x = 10 (c) The coordinates of the point of intersection of lines e & f. ie x = 5 Point of intersection is (5,-1) Begin Solution Continue Solution Markers Comments Straight Line Menu

  21. Markers Comments y Q x R P Back to Home • If no diagram is given draw a neat labelled diagram. • Midpoint of PQ is (3,-1): let’s call this S Using the gradient formula m = y2 – y1 x2 – x1 • Sketch the median and the • perpendicular bisector mSR = -1 – (-1) 10 - 3 (ie line is horizontal) median Perpendicular bisector Since it passes through (3,-1) equation of e is y = -1 Next Comment Comments for 4 (b) Straight Line Menu

  22. Markers Comments (b) Midpoint of QR is (6,1) mQR = 3 – (-1) 2 - 10 = 4/-8 required gradient = 2 (m1m2 = -1) , 2 + 10 3 + (-1) 2 2 Using y – b = m(x – a) with (a,b) = (6,1) & m = 2 y we get y – 1 = 2(x – 6) Q so f is y = 2x – 11 x R P Back to Home • To find midpoint of QR • Look for special cases: = - 1/2 Horizontal lines in the form y = k Vertical lines in the form x = k Next Comment Comments for 4 (c) Straight Line Menu

  23. Markers Comments Back to Home • To find the point of intersection of the two lines solve the two equations: (c) e & f meet when y = -1 & y = 2x -11 so 2x – 11 = -1 ie 2x = 10 y = -1 y = 2x - 11 ie x = 5 Point of intersection is (5,-1) Next Comment Straight Line Menu

  24. Y X E(6,-3) F(12,-5) G(2,-5) STRAIGHT LINE : Question 5 In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find (a) the equation of the altitude from vertex E. (b) the equation of the median from vertex F. (c) The point of intersection of the altitude and median. Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  25. Y X E(6,-3) F(12,-5) G(2,-5) STRAIGHT LINE : Question 5 In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find (a) the equation of the altitude from vertex E. (b) the equation of the median from vertex F. (c) The point of intersection of the altitude and median. Reveal answer only Go to full solution (a) x = 6 Go to Marker’s Comments (b) x + 8y + 28 = 0 Go to Straight Line Menu (c) Go to Main Menu (6,-4.25) EXIT

  26. Question 5(a) Back to Home • Using the gradient formula In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). mFG = -5 – (-5) 12 - 2 = 0 Find (a) the equation of the altitude from vertex E. (ie line is horizontal so altitude is vertical) Y Altitude is vertical line through (6,-3) ie x = 6 X E(6,-3) F(12,-5) Solution to 5 (b) G(2,-5) Begin Solution Continue Solution Markers Comments Straight Line Menu

  27. Question 5(b) Back to Home • Midpoint of EG is (4,-4)- let’s call this H In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). mFH = -5 – (-4) 12 - 4 = -1/8 Find Using y – b = m(x – a) with (a,b) = (4,-4) & m =-1/8 (b) the equation of the median from vertex F. Y we get y – (-4) = -1/8(x – 4) (X8) X E(6,-3) or 8y + 32 = -x + 4 F(12,-5) G(2,-5) Median is x + 8y + 28 = 0 Begin Solution Solution to 5 (c) Continue Solution Markers Comments Straight Line Menu

  28. Question 5(c) Back to Home (c) Lines meet when x = 6 & x + 8y + 28 = 0 In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). put x =6 in 2nd equation 8y + 34 = 0 Find ie 8y = -34 (c) The point of intersection of the altitude and median. ie y = -4.25 Y Point of intersection is (6,-4.25) X E(6,-3) F(12,-5) G(2,-5) Begin Solution Continue Solution Markers Comments Straight Line Menu

  29. Markers Comments y E x F G Back to Home • If no diagram is given draw a • neat labelled diagram. • Sketch the altitude and • the median. • Using the gradient formula mFG = -5 – (-5) 12 - 2 = 0 (ie line is horizontal so altitude is vertical) Altitude is vertical line through (6,-3) ie x = 6 median Comments for 5 (b) altitude Next Comment Straight Line Menu

  30. Markers Comments • To find midpoint of EG 2 + 6 -3 + (-5) 2 2 , H y E x F G Back to Home • Midpoint of EG is (4,-4)- call this H • Look for special cases: mFH = -5 – (-4) 12 - 4 = -1/8 Horizontal lines in the form y = k Vertical lines in the form x = k Using y – b = m(x – a) with (a,b) = (4,-4) & m =-1/8 we get y – (-4) = -1/8(x – 4) (X8) or 8y + 32 = -x + 4 Median is x + 8y + 28 = 0 Next Comment Straight Line Menu Comments for 5 (c)

  31. Markers Comments Back to Home • To find the point of • intersection of the two lines • solve the two equations: (c) Lines meet when x = 6 & x + 8y + 28 = 0 x = 6 x + 8y = -28 put x =6 in 2nd equation 8y + 34 = 0 ie 8y = -34 ie y = -4.25 Point of intersection is (6,-4.25) Next Comment Straight Line Menu

  32. HIGHER – ADDITIONAL QUESTION BANK You have chosen to study: Basic Differentiation UNIT 1 : Please choose a question to attempt from the following: 1 2 3 4 5 Back to Unit 1 Menu EXIT

  33. BASIC DIFFERENTIATION : Question 1 Find the equation of the tangent to the curve (x>0) at the point where x = 4. Reveal answer only Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

  34. BASIC DIFFERENTIATION : Question 1 Find the equation of the tangent to the curve (x>0) at the point where x = 4. Reveal answer only y = 5/4x – 7 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

  35. Question 1 Back to Home NB:a tangent is a line so we need a point of contact and a gradient. Find the equation of the tangent to the curve y = x – 16 x (x>0) at the point where x = 4. Point If x = 4 then y = 4 – 16 4 = 2 – 4 = -2 so (a,b) = (4,-2) Gradient: y = x – 16 x = x1/2 – 16x -1 Continue Solution dy/dx = 1/2x-1/2 + 16x-2 = 1 + 16 2x x2 If x = 4 then: Begin Solution dy/dx = 1 + 16 24 16 Continue Solution = ¼ + 1 = 5/4 Markers Comments Basic Differentiation Menu

  36. Question 1 Back to Home If x = 4 then: Find the equation of the tangent to the curve y = x – 16 x (x>0) at the point where x = 4. dy/dx = 1 + 16 24 16 = ¼ + 1 = 5/4 Gradient of tangent = gradient of curve so m = 5/4 . We now use y – b = m(x – a) this gives us y – (-2) = 5/4(x – 4) Back to Previous or y + 2 = 5/4x – 5 or y = 5/4x – 7 Begin Solution Continue Solution Markers Comments Basic Differentiation Menu

  37. Markers Comments Back to Home • Prepare expression for differentiation. NB:a tangent is a line so we need a point of contact and a gradient. Point • Find gradient of the tangent • using rule: If x = 4 then y = 4 – 16 4 = 2 – 4 = -2 “multiply by the power and reduce the power by 1” so (a,b) = (4,-2) Gradient: y = x – 16 x = x1/2 – 16x -1 • Find gradient = at x = 4. dy/dx = 1/2x-1/2 + 16x-2 = 1 + 16 2x x2 Continue Comments If x = 4 then: dy/dx = 1 + 16 24 16 Next Comment Differentiation Menu = ¼ + 1 = 5/4

  38. Markers Comments Back to Home • Find y coordinate at x = 4 using: If x = 4 then: dy/dx = 1 + 16 24 16 = ¼ + 1 = 5/4 • Use m = 5/4 and (4,-2) in • y - b = m(x - a) Gradient of tangent = gradient of curve so m = 5/4 . We now use y – b = m(x – a) this gives us y – (-2) = 5/4(x – 4) or y + 2 = 5/4x – 5 or y = 5/4x – 7 Next Comment Differentiation Menu

  39. BASIC DIFFERENTIATION : Question 2 Find the coordinates of the point on the curve y = x2 – 5x + 10 where the tangent to the curve makes an angle of 135° with the positive direction of the X-axis. Reveal answer only Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

  40. BASIC DIFFERENTIATION : Question 2 Find the coordinates of the point on the curve y = x2 – 5x + 10 where the tangent to the curve makes an angle of 135° with the positive direction of the X-axis. (2,4) Reveal answer only Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

  41. Question 2 Back to Home NB: gradient of line = gradient of curve Find the coordinates of the point on the curve y = x2 – 5x + 10 where the tangent to the curve makes an angle of 135° with the positive direction of the X-axis. Line Using gradient = tan we get gradient of line = tan135° = -tan45° = -1 Curve Gradient of curve = dy/dx = 2x - 5 Continue Solution It now follows that 2x – 5 = -1 Begin Solution Or 2x = 4 Continue Solution Or x = 2 Markers Comments Basic Differentiation Menu

  42. Question 2 Back to Home Using y = x2 – 5x + 10 with x = 2 Find the coordinates of the point on the curve y = x2 – 5x + 10 where the tangent to the curve makes an angle of 135° with the positive direction of the X-axis. we get y = 22 – (5 X 2) + 10 ie y = 4 So required point is (2,4) Back to Previous Begin Solution Continue Solution Markers Comments Basic Differentiation Menu

  43. Markers Comments y 135 ° x Back to Home • Find gradient of the tangent • using rule: NB: gradient of line = gradient of curve “multiply by the power and reduce the power by 1” Line Using gradient = tan • Must use the result that the gradient • of the line is also equal to the tangent • of the angle the line makes with the • positive direction of the x- axis. • Not given on the formula sheet. we get gradient of line = tan135° = -tan45° = -1 Curve Gradient of curve = dy/dx = 2x - 5 m = tan135 ° = -1 It now follows that 2x – 5 = -1 Or 2x = 4 Next Comment Or x = 2 Differentiation Menu Continue Comments

  44. Markers Comments • Set m = • i.e. 2x - 5 = -1 • and solve for x. Back to Home It now follows that 2x – 5 = -1 Or 2x = 4 Or x = 2 Using y = x2 – 5x + 10 with x = 2 we get y = 22 – (5 X 2) + 10 ie y = 4 So required point is (2,4) Next Comment Differentiation Menu

  45. BASIC DIFFERENTIATION : Question 3 y = g(x) The graph of y = g(x) is shown here. There is a point of inflection at the origin, a minimum turning point at (p,q) and the graph also cuts the X-axis at r. r Make a sketch of the graph of y = g(x). (p,q) Reveal answer only Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

  46. BASIC DIFFERENTIATION : Question 3 y = g(x) The graph of y = g(x) is shown here. There is a point of inflection at the origin, a minimum turning point at (p,q) and the graph also cuts the X-axis at r. r Make a sketch of the graph of y = g(x). (p,q) Reveal answer only y = g(x) Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

  47. Question 3 x  0  p  g(x) - 0 - 0 + Back to Home y = g(x) Stationary points occur at x = 0 and x = p. (We can ignore r.) We now consider the sign of the gradient either side of 0 and p: r (p,q) Make a sketch of the graph of y = g(x). new y-values Begin Solution Click for graph Continue Solution Markers Comments Basic Differentiation Menu

  48. Question 3 Back to Home y = g(x) This now gives us the following graph y = g(x) r 0 p (p,q) Make a sketch of the graph of y = g(x). Begin Solution Return to Nature Table Continue Solution Markers Comments Basic Differentiation Menu

  49. Markers Comments y = g(x) y 0 p x a Back to Home To sketch the graph of the gradient function: Stationary points occur at x = 0 and x = p. 1 Mark the stationary points on the x axis i.e. Continue Comments Next Comment Differentiation Menu

  50. Markers Comments y a x  0  p  g(x) - 0 - 0 + Back to Home To sketch the graph of the gradient function: Stationary points occur at x = 0 and x = p. 1 Mark the stationary points on the x axis i.e. (We can ignore r.) We now consider the sign of the gradient either side of 0 and p: 2 For each interval decide if the value of + x new y-values - - Next Comment Differentiation Menu Continue Comments

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