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

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

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

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

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). y = -5/3x - 6 Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

Question 1 (5) 3x – 5y = 4 Find the equation of the 3x - 4 = 5y straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). 3x - 4 = 5y 5y = 3x - 4 (5) 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 y = -5/3x - 6 Straight Line Menu Back to Home

(5) Markers Comments 3x – 5y = 4 3x - 4 = 5y 5y = 3x - 4 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 5y = 3x - 4 (5) 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) m1 = 3 5 m2 = -5 3 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 y = -5/3x - 6 Back to Home

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

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

Question 2 (4) 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 Back to Home

(4) Markers Comments 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 Back to Home

STRAIGHT LINE : Question 3 X Y A B C 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

STRAIGHT LINE : Question 3 X Y A B C 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 (a) = 3/5 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

Question 3 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 X Y A B C 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 Back to Home

If no diagram is given draw a neat labelled diagram. Markers Comments 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° A B Ø ° 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° Back to Home

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

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)

Question 4 (a) mSR = -1 – (-1) 10 - 3 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) X Y P(4,-5) Q(2,3) R(10,-1) 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 Back to Home

Question 4 (b) = 4/-8 = - 1/2 (b) Midpoint of QR is (6,1) mQR = 3 – (-1) 2 - 10 = 4/-8 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 X Y P(4,-5) Q(2,3) R(10,-1) we get y – 1 = 2(x – 6) so f is y = 2x – 11 Solution to 4 (c) Begin Solution Continue Solution Markers Comments Straight Line Menu Back to Home

Question 4 (c) (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 X Y P(4,-5) Q(2,3) R(10,-1) Point of intersection is (5,-1) Begin Solution Continue Solution Markers Comments Straight Line Menu Back to Home

If no diagram is given draw a neat labelled diagram. Markers Comments 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 Q P R y x Perpendicular bisector Since it passes through (3,-1) equation of e is y = -1 Comments for 4 (b) Next Comment Straight Line Menu Back to Home

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

Markers Comments 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 ie x = 5 y = -1 y = 2x - 11 Point of intersection is (5,-1) Next Comment Straight Line Menu Back to Home

STRAIGHT LINE : Question 5 X Y G(2,-5) E(6,-3) F(12,-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

STRAIGHT LINE : Question 5 X Y G(2,-5) E(6,-3) F(12,-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

Question 5(a) mFG = -5 – (-5) 12 - 2 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) G(2,-5) Solution to 5 (b) Begin Solution Continue Solution Markers Comments Straight Line Menu Back to Home

Question 5(b) = -1/8 (X8) 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 Back to Home

Question 5(c) (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 Back to Home

mFG = -5 – (-5) 12 - 2 E F G Markers Comments 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 y x F E G (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 Back to Home

= -1/8 (X8) E F G Markers Comments To find midpoint of EG 2 + 6 -3 + (-5) 2 2 , H 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 y x F E G 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) Back to Home

intersection of the two lines solve the two equations: (c) Markers Comments 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 Back to Home

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

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

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

Question 1 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 Back to Home

Question 1 If x = 4 then: dy/dx = 1 + 16 24 16 = ¼ + 1 = 5/4 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 Begin Solution or y = 5/4x – 7 Continue Solution Markers Comments Basic Differentiation Menu Back to Home

Prepare expression for differentiation. Markers Comments 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 Back to Home

Find y coordinate at x = 4 using: If x = 4 then: Markers Comments 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 Back to Home

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

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

Question 2 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 Back to Home

Question 2 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 Back to Home

m = tan135 ° = -1 Markers Comments 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 y x 135 ° 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 Back to Home

Set m = i.e. 2x - 5 = -1 and solve for x. Markers Comments 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 Back to Home

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 EXIT Go to Main Menu

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

Question 3 Stationary points occur at x = 0 and x = p. 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 x  0  p  g(x) - 0 - 0 + (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 Back to Home

Question 3 This now gives us the following graph y = g(x) This now gives us the following graph y = g(x) r 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 Back to Home

the gradient function: Stationary points occur at x = 0 and x = p. Markers Comments 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. p y = g(x) x y a Continue Comments Next Comment Differentiation Menu Back to Home

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

+ - Markers Comments 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. p y = g(x) 2 For each interval decide if the value of 3 Draw in curve to fit information + - x y a In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates. Next Comment Differentiation Menu Back to Home

BASIC DIFFERENTIATION : Question 4 Here is part of the graph of y = x3 - 3x2 - 9x + 2. y = x3 - 3x2 - 9x + 2 Find the coordinates of the stationary points and determine their nature algebraically. Reveal answer only Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

BASIC DIFFERENTIATION : Question 4 Here is part of the graph of y = x3 - 3x2 - 9x + 2. y = x3 - 3x2 - 9x + 2 Find the coordinates of the stationary points and determine their nature algebraically. Reveal answer only (-1,7) is a maximum TP and (3,-25) is a minimum TP Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu EXIT

BASIC DIFFERENTIATION : Question 4 Here is part of the graph of y = x3 - 3x2 - 9x + 2. y = x3 - 3x2 - 9x + 2 Find the coordinates of the stationary points and determine their nature algebraically. Return to solution EXIT

Question 4 SPs occur where dy/dx = 0 ie 3x2 – 6x – 9 = 0 Here is part of the graph of y = x3 - 3x2 - 9x + 2. ie 3(x2 – 2x – 3) = 0 ie 3(x – 3)(x + 1) = 0 Find the coordinates of the stationary points and determine their nature algebraically. ie x = -1 or x = 3 Using y = x3 - 3x2 - 9x + 2 when x = -1 Continue Solution y = -1 – 3 + 9 + 2 = 7 & when x = 3 y = 27 – 27 - 27 + 2 = -25 Begin Solution So stationary points are at (-1,7) and (3,-25) Continue Solution Markers Comments Basic Differentiation Menu Back to Home

Question 4 We now consider the sign of the gradient either side of -1 and 3. Here is part of the graph of y = x3 - 3x2 - 9x + 2. x  -1  3  Find the coordinates of the stationary points and determine their nature algebraically. (x + 1) - 0 + + + (x - 3) - - - 0 + dy/dx + 0 - 0 + Back to graph Begin Solution Hence (-1,7) is a maximum TP and (3,-25) is a minimum TP Continue Solution Markers Comments Basic Differentiation Menu Back to Home

“At stationary points “ SPs occur where dy/dx = 0 Markers Comments Make the statement: “At stationary points “ SPs occur where dy/dx = 0 ie 3x2 – 6x – 9 = 0 ie 3(x2 – 2x – 3) = 0 Must attempt to find and set equal to zero ie 3(x – 3)(x + 1) = 0 ie x = -1 or x = 3 “multiply by the power and reduce the power by 1” Using y = x3 - 3x2 - 9x + 2 when x = -1 Find the value of y from y = x3 -3x2-9x+2 not from y = -1 – 3 + 9 + 2 = 7 & when x = 3 y = 27 – 27 - 27 + 2 = -25 So stationary points are at (-1,7) and (3,-25) Next Comment Differentiation Menu Continue Comments Back to Home

x -1 + 0 - Minimum requirement Markers Comments Justify the nature of each stationary point using a table of “signs” We now consider the sign of the gradient either side of -1 and 3. x -1 + 0 - x  -1  3  (x + 1) - 0 + + + Minimum requirement (x - 3) - - - 0 + dy/dx + 0 - 0 + State the nature of the stationary point i.e. Maximum T.P. Hence (-1,7) is a maximum TP and (3,-25) is a minimum TP Next Comment Differentiation Menu Back to Home

BASIC DIFFERENTIATION : Question 5 When a company launches a new product its share of the market after x months is calculated by the formula S(x) = 2 - 4 x x2 (x  2) So after 5 months the share is S(5) = 2/5 – 4/25 = 6/25 Find the maximum share of the market that the company can achieve. Reveal answer only Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu EXIT Go to Main Menu

BASIC DIFFERENTIATION : Question 5 When a company launches a new product its share of the market after x months is calculated by the formula S(x) = 2 - 4 x x2 (x  2) So after 5 months the share is S(5) = 2/5 – 4/25 = 6/25 Find the maximum share of the market that the company can achieve. Reveal answer only = 1/4 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu EXIT Go to Main Menu

= 8 - 2 x3 x2 Question 5 End points When a company launches a new product its share of the market after x months is calculated as: S(2) = 1 – 1 = 0 There is no upper limit but as x   S(x)  0. S(x) = 2 - 4 x x2 (x  2) Stationary Points S(x) = 2 - 4 x x2 = 2x-1 – 4x-2 Find the maximum share of the market that the company can achieve. So S (x) = -2x-2 + 8x-3 = 8 - 2 x3 x2 = - 2 + 8 x2 x3 Begin Solution Continue Solution Continue Solution Markers Comments Basic Differentiation Menu Back to Home

Question 5 SPs occur where S (x) = 0 When a company launches a new product its share of the market after x months is calculated as: 8 - 2 x3 x2 = 0 8 = 2 x3 x2 ( cross mult!) or S(x) = 2 - 4 x x2 (x  2) 8x2 = 2x3 8x2 - 2x3 = 0 2x2(4 – x) = 0 Find the maximum share of the market that the company can achieve. x = 0 or x = 4 In required interval NB: x  2 Go Back to Previous Continue Solution Continue Solution Markers Comments Basic Differentiation Menu Back to Home

Question 5 We now check the gradients either side of X = 4 When a company launches a new product its share of the market after x months is calculated as: x  4  S (3.9 ) = 0.00337… S(x) = 2 - 4 x x2 (x  2) S (x) + 0 - S (4.1) = -0.0029… Find the maximum share of the market that the company can achieve. Hence max TP at x = 4 So max share of market = S(4) = 2/4 – 4/16 Go Back to Previous = 1/2 – 1/4 Continue Solution = 1/4 Markers Comments Basic Differentiation Menu Back to Home

= 8 - 2 x3 x2 Markers Comments Must look for key word to spot the optimisation question i.e. End points Maximum, minimum, greatest , least etc. S(2) = 1 – 1 = 0 There is no upper limit but as x   S(x)  0. Must consider end points and stationary points. Stationary Points S(x) = 2 - 4 x x2 = 2x-1 – 4x-2 Prepare expression for differentiation. So S (x) = -2x-2 + 8x-3 = 8 - 2 x3 x2 = - 2 + 8 x2 x3 Next Comment Differentiation Menu Continue Comments Back to Home

( Note: No marks are allocated for trial and error solution.) Markers Comments Must attempt to find ( Note: No marks are allocated for trial and error solution.) SPs occur where S (x) = 0 8 - 2 x3 x2 = 0 Must attempt to find and set equal to zero 8 = 2 x3 x2 ( cross mult!) or 8x2 = 2x3 “multiply by the power and reduce the power by 1” 8x2 - 2x3 = 0 Usually easier to solve resulting equation using cross-multiplication. Take care to reject “solutions” outwith the domain. 2x2(4 – x) = 0 x = 0 or x = 4 In required interval NB: x  2 Next Comment Differentiation Menu Continue Comments Back to Home

x 4 + 0 - Minimum requirement. Markers Comments Must show a maximum value using a table of “signs”. We now check the gradients either side of X = 4 x 4 + 0 - x  4  S (3.9 ) = 0.00337… S (x) + 0 - S (4.1) = -0.0029… Minimum requirement. Hence max TP at x = 4 State clearly: Maximum T.P at x = 4 So max share of market = S(4) = 2/4 – 4/16 Next Comment = 1/2 – 1/4 Differentiation Menu = 1/4 Back to Home

HIGHER – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 1 : Recurrence Relations Please choose a question to attempt from the following: 1 2 3 Back to Unit 1 Menu EXIT

RECURRENCE RELATIONS : Question 1 A recurrence relation is defined by the formula un+1 = aun + b, where -1<a<1 and u0 = 20. If u1 = 10 and u2 = 4 then find the values of a and b. Find the limit of this recurrence relation as n  . Reveal answer only Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu EXIT Go to Main Menu

RECURRENCE RELATIONS : Question 1 A recurrence relation is defined by the formula un+1 = aun + b, where -1<a<1 and u0 = 20. If u1 = 10 and u2 = 4 then find the values of a and b. Find the limit of this recurrence relation as n  . (a) a = 0.6 Reveal answer only b = -2 Go to full solution (b) Go to Marker’s Comments L = -5 Go to Basic Differentiation Menu EXIT Go to Main Menu

Question 1 (a) Using un+1 = aun + b A recurrence relation is defined by the formula un+1 = aun + b, where -1<a<1 and u0 = 20. If u1 = 10 and u2 = 4 then find the values of a and b. (b) Find the limit of this recurrence relation as n  . we get u1 = au0 + b and u2 = au1 + b Replacing u0 by 20, u1 by 10 & u2 by 4 gives us 20a + b = 10 and 10a + b = 4 subtract  10a = 6 or a = 0.6 Continue Solution Replacing a by 0.6 in 10a + b = 4 Begin Solution gives 6 + b = 4 Continue Solution or b = -2 Markers Comments Recurrence Relations Menu Back to Home

Question 1 (b) un+1 = aun + b is now un+1 = 0.6un - 2 A recurrence relation is defined by the formula un+1 = aun + b, where -1<a<1 and u0 = 20. If u1 = 10 and u2 = 4 then find the values of a and b. (b) Find the limit of this recurrence relation as n  . This has a limit since -1<0.6<1 At this limit, L, un+1 = un = L So we now have L = 0.6 L - 2 or 0.4L = -2 or L = -2  0.4 or L = -20  4 so L = -5 Begin Solution Continue Solution Markers Comments Recurrence Relations Menu Back to Home

simultaneous equations and solve. u1 is obtained from u0 and Markers Comments Must form the two simultaneous equations and solve. u1 is obtained from u0 and u2 is obtained from u1 . A trial and error solution would only score 1 mark. (a) Using un+1 = aun + b we get u1 = au0 + b and u2 = au1 + b Replacing u0 by 20, u1 by 10 & u2 by 4 gives us 20a + b = 10 and 10a + b = 4 subtract  10a = 6 Comments for 1(b) or a = 0.6 Replacing a by 0.6 in 10a + b = 4 gives 6 + b = 4 Next Comment Recurrence Menu or b = -2 Back to Home

Must state condition for limit i.e. -1 < 0.6 < 1 Markers Comments Must state condition for limit i.e. -1 < 0.6 < 1 At limit L, state un+1 = un = L Substitute L for un+1 and un and solve for L. (b) un+1 = aun + b is now un+1 = 0.6un - 2 This has a limit since -1<0.6<1 At this limit, L, un+1 = un = L So we now have L = 0.6 L - 2 or 0.4L = -2 or L = -2  0.4 or L = -20  4 so L = -5 Next Comment Recurrence Menu Back to Home

RECURRENCE RELATIONS : Question 2 Two different recurrence relations are known to have the same limit as n  . The first is defined by the formula un+1 = -5kun + 3. The second is defined by the formula vn+1 = k2vn + 1. Find the value of k and hence this limit. Reveal answer only Go to full solution Go to Marker’s Comments Go to Recurrence Relations Menu EXIT Go to Main Menu

RECURRENCE RELATIONS : Question 2 Two different recurrence relations are known to have the same limit as n  . The first is defined by the formula un+1 = -5kun + 3. The second is defined by the formula vn+1 = k2vn + 1. Find the value of k and hence this limit. k = 1/3 Reveal answer only L = 9/8 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu EXIT Go to Main Menu

Question 2 If the limit is L then as n   we have un+1 = un = L and vn+1 = vn = L Two different recurrence relations are known to have the same limit as n  . The first is defined by the formula un+1 = -5kun + 3. The second is defined by Vn+1 = k2vn + 1. Find the value of k and hence this limit. First Sequence Second Sequence un+1 = -5kun + 3 becomes vn+1 = k2vn + 1 becomes L = k2L + 1 L = -5kL + 3 L - k2L = 1 L + 5kL = 3 L(1 - k2) = 1 L(1 + 5k) = 3 L = 1 . . (1 - k2) L = 3 . . (1 + 5k) Begin Solution Continue Solution Markers Comments Continue Solution Recurrence Relations Menu Back to Home

Question 2 L = 3 . . (1 + 5k) L = 1 . . (1 - k2) Two different recurrence relations are known to have the same limit as n  . The first is defined by the formula un+1 = -5kun + 3. The second is defined by Vn+1 = k2vn + 1. Find the value of k and hence this limit. . 3 . = . 1 . . (1 + 5k) (1 – k2) It follows that Cross multiply to get 1 + 5k = 3 – 3k2 This becomes 3k2 + 5k – 2 = 0 Or (3k – 1)(k + 2) = 0 So k = 1/3 or k = -2 Since -1<k<1 then k = 1/3 Begin Solution Continue Solution Markers Comments Continue Solution Recurrence Relations Menu Back to Home

Question 1 Since -1<k<1 then k = 1/3 Two different recurrence relations are known to have the same limit as n  . The first is defined by the formula un+1 = -5kun + 3. The second is defined by Vn+1 = k2vn + 1. Find the value of k and hence this limit. Using L = 1 . . (1 - k2) gives us L = . 1 . . (1 – 1/9) or L = 1  8/9 ie L = 9/8 Begin Solution Continue Solution Markers Comments Recurrence Relations Menu Back to Home

Since both recurrence relations have the same limit, L, find the Markers Comments Since both recurrence relations have the same limit, L, find the limit for both and set equal. If the limit is L then as n   we have un+1 = un = L and vn+1 = vn = L First Sequence Second Sequence un+1 = -5kun + 3 becomes vn+1 = k2vn + 1 becomes L = k2L + 1 L = -5kL + 3 L - k2L = 1 L + 5kL = 3 Continue Comments L(1 - k2) = 1 L(1 + 5k) = 3 L = 3 . . (1 + 5k) L = 1 . . (1 - k2) Next Comment Recurrence Menu Back to Home

Since both recurrence relations have the same limit, L, find the Markers Comments Since both recurrence relations have the same limit, L, find the limit for both and set equal. L = 3 . . (1 + 5k) L = 1 . . (1 - k2) Only one way to solve resulting equation i.e. terms to the left, form the quadratic and factorise. State clearly the condition for the recurrence relation to approach a limit. -1< k < 1. Take care to reject the “solution” which is outwith the range. . 3 . = . 1 . . (1 + 5k) (1 – k2) It follows that Cross multiply to get 1 + 5k = 3 – 3k2 This becomes 3k2 + 5k – 2 = 0 Or (3k – 1)(k + 2) = 0 So k = 1/3 or k = -2 Since -1<k<1 then k = 1/3 Next Comment Continue Solution Recurrence Menu Back to Home

Find L from either formula. Since -1<k<1 then k = 1/3 Markers Comments Find L from either formula. Since -1<k<1 then k = 1/3 Using L = 1 . . (1 - k2) gives us L = . 1 . . (1 – 1/9) or L = 1  8/9 ie L = 9/8 Next Comment Recurrence Menu Back to Home

RECURRENCE RELATIONS : Question 3 A man plants a row of fast growing trees between his own house and his neighbour’s. These trees are known grow at a rate of 1m per annum so cannot be allowed to become too high. He therefore decides to prune 30% from their height at the beginning of each year. Using the 30% pruning scheme what height should he expect the trees to grow to in the long run? The neighbour is concerned at the growth rate and asks that the trees be kept to a maximum height of 3m. What percentage should be pruned to ensure that this happens? Reveal answer only Go to full solution Go to Marker’s Comments Go to Recurrence Relations Menu EXIT

RECURRENCE RELATIONS : Question 3 A man plants a row of fast growing trees between his own house and his neighbour’s. These trees are known grow at a rate of 1m per annum so cannot be allowed to become too high. He therefore decides to prune 30% from their height at the beginning of each year. Using the 30% pruning scheme what height should he expect the trees to grow to in the long run? The neighbour is concerned at the growth rate and asks that the trees be kept to a maximum height of 3m. What percentage should be pruned to ensure that this happens? Reveal answer only Height of trees in long run is 31/3m. Go to full solution (b) 331/3% Go to Marker’s Comments Go to Recurrence Relations Menu EXIT

Question 3 (a) Removing 30% leaves 70% or 0.7 The trees are known grow at a rate of 1m per annum. He therefore decides to prune 30% from their height at the beginning of each year. If Hn is the tree height in year n then Hn+1 = 0.7Hn + 1 Since -1<0.7<1 this sequence has a limit, L. At the limit Hn+1 = Hn = L Using the 30% pruning scheme what height should he expect the trees to grow to in the long run? So L= 0.7L + 1 or 0.3 L = 1 ie L = 1  0.3 = 10  3 = 31/3 Height of trees in long run is 31/3m. Begin Solution Continue Solution Continue Solution Markers Comments Recurrence Relation Menu Back to Home

Question 3 If fraction left after pruning is a and we need the limit to be 3 The trees are known grow at a rate of 1m per annum. He therefore decides to prune 30% from their height at the beginning of each year. then we have 3 = a X 3 + 1 or 3a = 2 or a = 2/3 (b) The neighbour asks that the trees be kept to a maximum height of 3m. What percentage should be pruned to ensure that this happens? This means that the fraction pruned is 1/3 or 331/3% Begin Solution Continue Solution Markers Comments Recurrence Relation Menu Back to Home

Do some numerical work to get the “feel” for the problem. Markers Comments Do some numerical work to get the “feel” for the problem. (a) Removing 30% leaves 70% or 0.7 H0 = 1 (any value acceptable) H1 = 0.7 x1 + 1 = 1.7 H2 = 0.7 x1.7 + 1 = 2.19 etc. If Hn is the tree height in year n then Hn+1 = 0.7Hn + 1 State the recurrence relation, with the starting value. Hn+1 = 0.7 Hn + 1, H0 = 1 State the condition for the limit -1< 0.7< 1 At limit L, state Hn+1 = Hn = L Substitute L for Hn+1 and Hn and solve for L. Since -1<0.7<1 this sequence has a limit, L. At the limit Hn+1 = Hn = L So L= 0.7L + 1 or 0.3 L = 1 ie L = 1  0.3 = 10  3 Height of trees in long run is 31/3m. Next Comment Continue Solution Recurrence Menu Back to Home

Since we know the limit we are working backwards to %. L = 0.7L + 1 Markers Comments Since we know the limit we are working backwards to %. L = 0.7L + 1 New limit, L = 3 and multiplier a 3 = a x3 + 1 etc. Take care to subtract from 1 to get fraction pruned. If fraction left after pruning is a and we need the limit to be 3 then we have 3 = a X 3 + 1 or 3a = 2 or a = 2/3 This means that the fraction pruned is 1/3 or 331/3% Next Comment Recurrence Menu Back to Home

Trig Graphs & Equations HIGHER – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 1 : Trig Graphs & Equations Please choose a question to attempt from the following: 1 2 3 Back to Unit 1 Menu EXIT

TRIG GRAPHS & EQUATIONS : Question 1 /2  y = acosbx + c This diagram shows the graph of y = acosbx + c. Determine the values of a, b & c. Reveal answer only Go to full solution Go to Marker’s Comments Go to Trig Graphs & Equations Menu Go to Main Menu EXIT

TRIG GRAPHS & EQUATIONS : Question 1 /2  y = acosbx + c This diagram shows the graph of y = acosbx + c. Determine the values of a, b & c. Reveal answer only a = 3 Go to full solution b = 2 Go to Marker’s Comments c = -1 Go to Trig Graphs & Equations Menu Go to Main Menu EXIT

Question 1 a = ½(max – min) This diagram shows the graph of y = acosbx + c. Determine the values of a, b & c. = ½(2 – (-4)) = ½ X 6 = 3 /2  y = acosbx + c Period of graph =  so two complete sections between 0 & 2 ie b = 2 For 3cos(…) max = 3 & min = -3. This graph: max = 2 & min = -4. ie 1 less Begin Solution so c = -1 Continue Solution Markers Comments Trig Graphs etc. Menu Back to Home

The values chosen for a,b and c must be justified. Markers Comments The values chosen for a,b and c must be justified. Possible justification of a = 3 a = 1/2(max - min) y = cosx graph stretched by a factor of 3 etc. Possible justification of b = 2 Period of graph = 2 complete cycles in 2 2 ÷ = 2 2 complete cycles in 2 etc. Possible justification for c = -1 3cos max = 3, min = -3 This graph: max = 2, min = -4 i.e. -1 c = -1 y = 3cosx graph slide down 1 unit etc. a = ½(max – min) = ½(2 – (-4)) = ½ X 6 = 3 Period of graph =  so two complete sections between 0 & 2 ie b = 2 For 3cos(…) max = 3 & min = -3. This graph: max = 2 & min = -4. ie 1 less so c = -1 Next Comment Trig Graphs Menu Back to Home

TRIG GRAPHS & EQUATIONS : Question 2 Solve 3tan2 + 1 = 0 ( where 0 <  <  ). Reveal answer only Go to full solution Go to Marker’s Comments Go to Trig Graphs & Equations Menu Go to Main Menu EXIT

TRIG GRAPHS & EQUATIONS : Question 2 Solve 3tan2 + 1 = 0 ( where 0 <  <  ).  = 5/12 Reveal answer only  = 11/12 Go to full solution Go to Marker’s Comments Go to Trig Graphs & Equations Menu Go to Main Menu EXIT

Question 2 3tan2 + 1 = 0 3tan2 = -1 Solve 3tan2 + 1 = 0 ( where 0 <  <  ). tan2 = -1/3 Q2 or Q4  -    +  2 -  sin all tan cos tan -1(1/3) = /6 Q2: angle =  - /6 so 2 = 5/6 ie  = 5/12 1 2 3 /6 Q4: angle = 2 - /6 so 2 = 11/6 Begin Solution ie  = 11/12 Continue Solution Markers Comments tan2 repeats every /2 radians but repeat values are not in interval. Trig Graphs etc. Menu Back to Home

Solve the equation for tan . Full marks can be obtained by Markers Comments Solve the equation for tan . Full marks can be obtained by working in degrees and changing final answers back to radians. 3tan2 + 1 = 0 Use the positive value when finding tan-1. 3tan2 = -1 tan2 = -1/3 Q2 or Q4 Use the quadrant rule to find the solutions.  -    +  2 -  sin all tan cos tan -1(1/3) = /6 Must learn special angles or be able to calculate from triangles. Q2: angle =  - /6 so 2 = 5/6 1 45° 2 60° 30° Take care to reject ”solutions” outwith domain. ie  = 5/12 Q4: angle = 2 - /6 1 2 3 /6 so 2 = 11/6 ie  = 11/12 Next Comment Trig Graphs Menu tan2 repeats every /2 radians but repeat values are not in interval. Back to Home

TRIG GRAPHS & EQUATIONS : Question 3 2/3 P Q y = 2 The diagram shows a the graph of a sine function from 0 to 2/3. (a) State the equation of the graph. (b) The line y = 2 meets the graph at points P & Q. Find the coordinates of these two points. Reveal answer only Go to full solution Go to Marker’s Comments Go to Trig Graphs & Equations Menu EXIT

TRIG GRAPHS & EQUATIONS : Question 3 The diagram shows a the graph of a sine function from 0 to 2/3. y = 2 P Q 2/3 (a) State the equation of the graph. (b) The line y = 2 meets the graph at points P & Q. Find the coordinates of these two points. Graph is y = 4sin3x Reveal answer only P is (/18, 2) and Q is (5/18, 2). Go to full solution Go to Marker’s Comments Go to Trig Graphs & Equations Menu EXIT

Question 3 One complete wave from 0 to 2/3 so 3 waves from 0 to 2. The diagram shows a the graph of a sine function from 0 to 2/3. (a) State the equation of the graph. Max/min = ±4 4sin(…) 2/3 P Q y = 2 Graph is y = 4sin3x Begin Solution Continue Solution Continue Solution Markers Comments Trig Graphs etc. Menu Back to Home

Question 3 Graph is y = 4sin3x (b) The line y = 2 meets the graph at points P & Q. Find the coordinates of these two points. (b) At P & Q y = 4sin3x and y = 2 so 4sin3x = 2 or sin3x = 1/2 Q1 or Q2 2/3 P Q y = 2  -    +  2 -  sin all tan cos sin-1(1/2) = /6 Q1: angle = /6 so 3x = /6 1 2 3 /6 ie x = /18 Q2: angle =  - /6 Begin Solution so 3x = 5/6 Continue Solution ie x = 5/18 Markers Comments P is (/18, 2) and Q is (5/18, 2). Trig Graphs etc. Menu Back to Home

Identify graph is of the form y = asinbx. Markers Comments Identify graph is of the form y = asinbx. Must justify choice of a and b. Possible justification of a One complete wave from 0 to 2/3 so 3 waves from 0 to 2. Max/min = ±4 4sin(…) Max = 4, Min = -4 4sin(…) y = sinx stretched by a factor of 4 Graph is y = 4sin3x Possible justification for b Period = 3 waves from 0 to Next Comment Trig Graphs Menu Back to Home

Use the quadrant rule to find the solutions. or sin3x = 1/2 Q1 or Q2 Markers Comments At intersection y1 = y2 4sin3x = 2 Solve for sin3x Graph is y = 4sin3x (b) At P & Q y = 4sin3x and y = 2 so 4sin3x = 2 Use the quadrant rule to find the solutions. or sin3x = 1/2 Q1 or Q2  -    +  2 -  sin all tan cos sin-1(1/2) = /6 Must learn special angles or be able to calculate from triangles. Q1: angle = /6 1 45° 2 60° 30° so 3x = /6 1 2 3 /6 ie x = /18 Q2: angle =  - /6 Take care to state coordinates. so 3x = 5/6 Next Comment ie x = 5/18 Trig Graphs Menu P is (/18, 2) and Q is (5/18, 2). Back to Home

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

FUNCTIONS & GRAPHS : Question 1 This graph shows the the function y = g(x). Make a sketch of the graph of the function y = 4 – g(-x). y = g(x) -8 12 (-p,q) (u,-v) Reveal answer only Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu EXIT

FUNCTIONS & GRAPHS : Question 1 This graph shows the the function y = g(x). Make a sketch of the graph of the function y = 4 – g(-x). y = g(x) -8 12 (-p,q) (u,-v) Reveal answer only (p,-q+4) (8,4) (-12,4) (0,4) (-u,v+4) y = 4 – g(-x) Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu EXIT

Question 1 y = 4 – g(-x) = -g(-x) + 4 This graph shows the the function y = g(x). Make a sketch of the graph of the function y = 4 – g(-x). A B Reflect in X-axis C Slide 4 up Reflect in Y-axis y = g(x) -8 12 (-p,q) (u,-v) Known Points (-8,0), (-p,q), (0,0), (u,-v), (12,0) A (-8,0), (-p,-q), (0,0), (u,v), (12,0) B Begin Solution (8,0), (p,-q), (0,0), (-u,v), (-12,0) Continue Solution C Markers Comments (8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4) Functions & Graphs Menu Back to Home

Question 1 (8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4) This graph shows the the function y = g(x). Make a sketch of the graph of the function y = 4 – g(-x). Now plot points and draw curve through them. (p,-q+4) (8,4) (-12,4) (0,4) (-u,v+4) y = 4 – g(-x) y = g(x) -8 12 (-p,q) (u,-v) Begin Solution Continue Solution Markers Comments Functions & Graphs Menu Back to Home

Change order to give form: y = k.g(x) + c y = 4 – g(-x) = -g(-x) + 4 Markers Comments Change order to give form: y = k.g(x) + c y = 4 – g(-x) = -g(-x) + 4 When the function is being changed by more than one related function take each change one at a time either listing the coordinates or sketching the steps to final solution. A B Reflect in X-axis C Slide 4 up Reflect in Y-axis Known Points (-8,0), (-p,q), (0,0), (u,-v), (12,0) A (-8,0), (-p,-q), (0,0), (u,v), (12,0) B (8,0), (p,-q), (0,0), (-u,v), (-12,0) Next Comment C Functions Menu (8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4) Back to Home

Slide k units parallel to y-axis kf(x) Stretch by a factor = k -f(x) Markers Comments y = 4 – g(-x) = -g(-x) + 4 A Learn Rules: Not given on formula sheet B Reflect in X-axis C f(x) + k Slide k units parallel to y-axis kf(x) Stretch by a factor = k -f(x) Reflect in the x-axis f(x-k) Slide k units parallel to the x-axis f(-x) Reflect in y-axis Slide 4 up Reflect in Y-axis Known Points (-8,0), (-p,q), (0,0), (u,-v), (12,0) A (-8,0), (-p,-q), (0,0), (u,v), (12,0) B (8,0), (p,-q), (0,0), (-u,v), (-12,0) Next Comment C Functions Menu (8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4) Back to Home

Now plot points and draw curve through them. Markers Comments In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates. (8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4) Now plot points and draw curve through them. (p,-q+4) (8,4) (-12,4) (0,4) (-u,v+4) y = 4 – g(-x) Next Comment Functions Menu Back to Home

FUNCTIONS & GRAPHS : Question 2 y = ax (1,a) This graph shows the the function y = ax. Make sketches of the graphs of the functions (I) y = a(x+2) (II) y = 2ax - 3 Reveal answer only Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu EXIT

FUNCTIONS & GRAPHS : Question 2 ANSWER TO PART (I) This graph shows the the function y = ax. Make sketches of the graphs of the functions (I) y = a(x+2) (II) y = 2ax - 3 (-1,a) (-2,1) y = a(x+2) y = ax Reveal answer only ANSWER to PART (II) Go to full solution y = 2ax - 3 y = ax (0,-1) (1,2a-3) Go to Marker’s Comments Go to Functions & Graphs Menu EXIT

Question 2 (I) y = a(x+2) Make sketches of the graphs of the functions f(x) = ax so a(x+2) = f(x+2) y = ax (1,a) move f(x) 2 to left (0,1)(-2,1) & (1,a) (-1,a) (-1,a) (-2,1) y = a(x+2) y = ax Begin Solution Continue Solution Markers Comments Functions & Graphs Menu Back to Home

Question 2 (II) y = 2ax - 3 Make sketches of the graphs of the functions (II) y = 2ax - 3 f(x) = ax so 2ax - 3 = 2f(x) - 3 y = ax (1,a) double y-coords slide 3 down (0,1)(0,2)(0,-1) (1,a) (1,2a) (1,2a-3) y = 2ax - 3 y = ax (0,-1) (1,2a-3) Begin Solution Continue Solution Markers Comments Functions & Graphs Menu Back to Home

When the problem is given in terms of a specific function Markers Comments When the problem is given in terms of a specific function rather in terms of the general f(x), change back to f(x) eg. y = 2x, y = log3x, y = x2 + 3x, each becomes y = f(x) (I) y = a(x+2) f(x) = ax so a(x+2) = f(x+2) move f(x) 2 to left In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates. (0,1)(-2,1) & (1,a) (-1,a) (-1,a) (-2,1) y = a(x+2) y = ax Next Comment Functions Menu Back to Home

Slide k units parallel to y-axis kf(x) Stretch by a factor = k -f(x) Markers Comments (I) y = a(x+2) Learn Rules: Not given on formula sheet f(x) = ax so a(x+2) = f(x+2) f(x) + k Slide k units parallel to y-axis kf(x) Stretch by a factor = k -f(x) Reflect in the x-axis f(x-k) Slide k units parallel to the x-axis f(-x) Reflect in y-axis move f(x) 2 to left (0,1)(-2,1) & (1,a) (-1,a) (-1,a) (-2,1) y = a(x+2) y = ax Next Comment Functions Menu Back to Home

When the problem is given in terms of a specific function Markers Comments When the problem is given in terms of a specific function rather in terms of the general f(x), change back to f(x) eg. y = 2x, y = log3x, y = x2 + 3x, each becomes y = f(x) (II) y = 2ax - 3 f(x) = ax so 2ax - 3 = 2f(x) - 3 double y-coords slide 3 down In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates. (0,1)(0,2)(0,-1) (1,a) (1,2a) (1,2a-3) y = 2ax - 3 y = ax (0,-1) (1,2a-3) Next Comment Functions Menu Back to Home

Slide k units parallel to y-axis kf(x) Stretch by a factor = k -f(x) Markers Comments (II) y = 2ax - 3 Learn Rules: Not given on formula sheet f(x) = ax so 2ax - 3 = 2f(x) - 3 f(x) + k Slide k units parallel to y-axis kf(x) Stretch by a factor = k -f(x) Reflect in the x-axis f(x-k) Slide k units parallel to the x-axis f(-x) Reflect in y-axis double y-coords slide 3 down (0,1)(0,2)(0,-1) (1,a) (1,2a) (1,2a-3) y = 2ax - 3 y = ax (0,-1) (1,2a-3) Next Comment Functions Menu Back to Home

FUNCTIONS & GRAPHS : Question 3 Two functions f and g are defined on the set of real numbers by the formulae f(x) = 2x - 1 and g(x) = x2 . (a) Find formulae for (i) f(g(x)) (ii) g(f(x)) . (b) Hence show that the equation g(f(x)) = f(g(x)) has only one real solution. Reveal answer only Go to full solution Go to Marker’s Comments Go to Functions & Graphs Go to Main Menu EXIT

FUNCTIONS & GRAPHS : Question 3 Two functions f and g are defined on the set of real numbers by the formulae f(x) = 2x - 1 and g(x) = x2 . (a) Find formulae for (i) f(g(x)) (ii) g(f(x)) . (b) Hence show that the equation g(f(x)) = f(g(x)) has only one real solution. (a) (i) = 2x2 - 1 Reveal answer only Go to full solution = 4x2 – 4x + 1 (ii) Go to Marker’s Comments Go to Functions & Graphs Go to Main Menu EXIT

Question 3 (a)(i) f(g(x)) Two functions f and g are defined on the set of real numbers by the formulae f(x) = 2x - 1 and g(x) = x2 . = f(x2) = 2x2 - 1 (ii) g(f(x)) Find formulae for (i) f(g(x)) (ii) g(f(x)) . = g(2x-1) = (2x – 1)2 = 4x2 – 4x + 1 Begin Solution Continue Solution Continue Solution Markers Comments Functions & Graphs Menu Back to Home

Question 3 f(g(x)) = 2x2 - 1 Two functions f and g are defined on the set of real numbers by the formulae f(x) = 2x - 1 and g(x) = x2 . g(f(x)) = 4x2 – 4x + 1 (b) g(f(x)) = f(g(x)) 4x2 – 4x + 1 = 2x2 - 1 2x2 – 4x + 2 = 0 Hence show that the equation g(f(x)) = f(g(x)) has only one real solution. x2 – 2x + 1 = 0 (x – 1)(x – 1) = 0 x = 1 Hence only one real solution! Begin Solution Continue Solution Markers Comments Functions & Graphs Menu Back to Home

In composite function problems take at least 3 lines to Markers Comments (a) In composite function problems take at least 3 lines to answer the problem: State required composite function: f(g(x)) Replace g(x) without simplifying: f(x2) In f(x) replace each x by g(x): 2 x2 - 1 (a)(i) f(g(x)) = f(x2) = 2x2 - 1 (ii) g(f(x)) (II) State required composite function: g(f(x)) Replace f(x) without simplifying: g(2x-1) In g(x) replace each x by f(x): = g(2x-1) = (2x – 1)2 = 4x2 – 4x + 1 (2x – 1)2 Next Comment Functions Menu Back to Home

simplify and factorise. f(g(x)) = 2x2 - 1 Markers Comments (b) Only one way to solve resulting equation: Terms to the left, simplify and factorise. f(g(x)) = 2x2 - 1 g(f(x)) = 4x2 – 4x + 1 (b) g(f(x)) = f(g(x)) 4x2 – 4x + 1 = 2x2 - 1 2x2 – 4x + 2 = 0 x2 – 2x + 1 = 0 (x – 1)(x – 1) = 0 x = 1 Hence only one real solution! Next Comment Functions Menu Back to Home

FUNCTIONS & GRAPHS : Question 4 A function g is defined by the formula g(x) = . 2 (x1) (x – 1) (a) Find a formula for h(x) = g(g(x)) in its simplest form. (b) State a suitable domain for h. Reveal answer only Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu EXIT

FUNCTIONS & GRAPHS : Question 4 A function g is defined by the formula g(x) = . 2 (x1) (x – 1) (a) Find a formula for h(x) = g(g(x)) in its simplest form. (b) State a suitable domain for h. = (2x - 2) . . (3 – x) h(x) Reveal answer only Go to full solution Domain = {x  R: x  3} Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu EXIT

Question 4 = 2 - 1 = 2 = 2 (a) g(g(x)) = g( ) A function g is defined by the formula g(x) = . 2 (x1) (x – 1) = 2 . 2 . (x – 1) - 1 = 2 Find a formula for h(x) = g(g(x)) in its simplest form. 2 - (x – 1) . (x – 1) = 2 (3 - x) .(x – 1) (x - 1) .(3 – x) = 2 Begin Solution = (2x - 2) . . (3 – x) Continue Solution Markers Comments Continue Solution Functions & Graphs Menu Back to Home

Question 4 = (2x - 2) . . (3 – x) h(x) A function g is defined by the formula g(x) = . 2 (x1) (x – 1) (b) For domain 3 - x  0 Find a formula for h(x) = g(g(x)) in its simplest form. Domain = {x  R: x  3} (b) State a suitable domain for h. Begin Solution Continue Solution Markers Comments Functions & Graphs Menu Back to Home

= 2 - 1 = 2 - 1 = 2 Markers Comments (a) In composite function problems take at least 3 lines to answer the problem: State required composite function: g(g(x)) Replace g(x) without simplifying: g(2/(x-1)) In g(x) replace each x by g(x): . 2 . (x – 1) (a) g(g(x)) = g( ) = 2 . 2 . (x – 1) - 1 = 2 2 (x-1) - 1 2 - (x – 1) . (x – 1) = 2 (3 - x) .(x – 1) (x - 1) .(3 – x) = 2 = (2x - 2) . . (3 – x) Next Comment Functions Menu Back to Home

In finding a suitable domain it is often necessary to restrict R Markers Comments (b) In finding a suitable domain it is often necessary to restrict R to prevent either division by zero or the root of a negative number: In this case: 3 - x = 0 i.e. preventing division by zero. = (2x - 2) . . (3 – x) h(x) (b) For domain 3 - x  0 Domain = {x  R: x  3} Next Comment Functions Menu Back to Home

Polynomials Quadratics Integration Addition Formulae The Circle HIGHER – ADDITIONAL QUESTION BANK UNIT 2 : Polynomials Quadratics Integration Addition Formulae The Circle EXIT

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

POLYNOMIALS : Question 1 Show that x = 3 is a root of the equation x3 + 3x2 – 10x – 24 = 0. Hence find the other roots. Reveal answer only Go to full solution Go to Marker’s Comments Go to Polynomial Menu Go to Main Menu EXIT

POLYNOMIALS : Question 1 Show that x = 3 is a root of the equation x3 + 3x2 – 10x – 24 = 0. Hence find the other roots. other roots are x = -4 & x = -2 Reveal answer only Go to full solution Go to Marker’s Comments Go to Polynomial Menu Go to Main Menu EXIT

Question 1 18 Using the nested method - coefficients are 1, 3, -10, -24 Show that x = 3 is a root of the equation x3 + 3x2 – 10x – 24 = 0. 1 3 -10 -24 f(3) = 3 Hence find the other roots. 18 3 24 1 6 8 f(3) = 0 so x = 3 is a root. Also (x – 3) is a factor. Other factor: x2 + 6x + 8 or (x + 4)(x + 2) Begin Solution If (x + 4)(x + 2) = 0 then x = -4 or x = -2 Continue Solution Markers Comments Hence other roots are x = -4 & x = -2 Polynomial Menu Back to Home

State clearly in solution that f(3) = 0 x = 3 is a root Markers Comments State clearly in solution that f(3) = 0 x = 3 is a root Using the nested method - coefficients are 1, 3, -10, -24 Show completed factorisation of cubic i.e. (x - 3)(x + 4)(x + 2) = 0 1 3 -10 -24 f(3) = 3 18 3 24 1 6 8 Take care to set factorised expression = 0 f(3) = 0 so x = 3 is a root. Also (x – 3) is a factor. List all the roots of the polynomial x = 3, x = -4, x = -2 Other factor: x2 + 6x + 8 or (x + 4)(x + 2) If (x + 4)(x + 2) = 0 then x = -4 or x = -2 Next Comment Polynomial Menu Hence other roots are x = -4 & x = -2 Back to Home

POLYNOMIALS : Question 2 Given that (x + 4) is a factor of the polynomial f(x) = 3x3 + 8x2 + kx + 4 find the value of k. Hence solve the equation 3x3 + 8x2 + kx + 4 = 0 for this value of k. Reveal answer only Go to full solution Go to Marker’s Comments Go to Polynomial Menu Go to Main Menu EXIT

POLYNOMIALS : Question 2 Given that (x + 4) is a factor of the polynomial f(x) = 3x3 + 8x2 + kx + 4 find the value of k. Hence solve the equation 3x3 + 8x2 + kx + 4 = 0 for this value of k. k = -15 Reveal answer only So full solution of equation is x = -4 or x = 1/3 or x = 1 Go to full solution Go to Marker’s Comments Go to Polynomial Menu Go to Main Menu EXIT

Question 2 Since (x + 4) a factor then f(-4) = 0 . Given that (x + 4) is a factor of the polynomial f(x) = 3x3 + 8x2 + kx + 4 find the value of k. Hence solve the equation 3x3 + 8x2 + kx + 4 = 0 for this value of k. Now using the nested method - coefficients are 3, 8, k, 4 f(-4) = -4 3 8 k 4 -12 16 (-4k – 64) (-4k – 60) 3 -4 (k + 16) Since -4k – 60 = 0 then -4k = 60 Begin Solution Continue Solution so k = -15 Markers Comments Polynomial Menu Back to Home

Question 2 If k = -15 then we now have Given that (x + 4) is a factor of the polynomial f(x) = 3x3 + 8x2 + kx + 4 find the value of k. Hence solve the equation 3x3 + 8x2 + kx + 4 = 0 for this value of k. f(-4) = -4 3 8 -15 4 -12 16 -4 3 -4 1 0 Other factor is 3x2 – 4x + 1 or (3x - 1)(x – 1) If (3x - 1)(x – 1) = 0 then x = 1/3 or x = 1 Begin Solution So full solution of equation is: x = -4 or x = 1/3 or x = 1 Continue Solution Markers Comments Polynomial Menu Back to Home

The working in the nested solution can sometimes be Markers Comments The working in the nested solution can sometimes be eased by working in both directions toward the variable: Since (x + 4) a factor then f(-4) = 0 . Now using the nested method - coefficients are 3, 8, k, 4 -4 3 8 k 4 -12 16 -4 3 -4 1 0 f(-4) = -4 3 8 k 4 -12 16 (-4k – 64) (-4k – 60) 3 -4 (k + 16) k + 16 = 1 k = -15 Since -4k – 60 = 0 then -4k = 60 Next Comment so k = -15 Polynomial Menu Back to Home

Simply making f(-4) = 0 will also yield k i.e. Markers Comments Simply making f(-4) = 0 will also yield k i.e. 3(-4)3 + 8(-4)2 + k(-4) + 4 = 0 k = -15 Since (x + 4) a factor then f(-4) = 0 . Now using the nested method - coefficients are 3, 8, k, 4 f(-4) = -4 3 8 k 4 -12 16 (-4k – 64) (-4k – 60) 3 -4 (k + 16) Since -4k – 60 = 0 then -4k = 60 Next Comment so k = -15 Polynomial Menu Back to Home

Show completed factorisation of the cubic: (x + 4)(3x - 1)(x - 1) = 0 Markers Comments Show completed factorisation of the cubic: (x + 4)(3x - 1)(x - 1) = 0 If k = -15 then we now have 3 8 -15 4 f(-4) = -4 -12 16 -4 3 -4 1 0 Other factor is 3x2 – 4x + 1 or (3x - 1)(x – 1) If (3x - 1)(x – 1) = 0 So full solution of equation is: x = -4 or x = 1/3 or x = 1 Next Comment Polynomial Menu Back to Home

POLYNOMIALS : Question 3 Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of f(x). Hence express f(x) in its fully factorised form. Reveal answer only Go to full solution Go to Marker’s Comments Go to Polynomial Menu Go to Main Menu EXIT

POLYNOMIALS : Question 3 Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of f(x). Hence express f(x) in its fully factorised form. 6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2) Reveal answer only Go to full solution Go to Marker’s Comments Go to Polynomial Menu Go to Main Menu EXIT

Question 3 Using the nested method - coefficients are 6, 13, 0, -4 Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of f(x). Hence express f(x) in its fully factorised form. f(-2) = -2 6 13 0 -4 -12 -2 4 6 1 -2 f(-2) = 0 so (x + 2) is a factor Begin Solution Continue Solution Markers Comments Polynomial Menu Back to Home

Question 3 Using the nested method - coefficients are 6, 13, 0, -4 Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of f(x). Hence express f(x) in its fully factorised form. f(-2) = -2 6 13 0 -4 -12 -2 4 6 1 -2 Other factor is 6x2 + x – 2 or (3x + 2)(2x - 1) Hence 6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2) Begin Solution Continue Solution Markers Comments Polynomial Menu Back to Home

State clearly in solution that f(-2) = 0 x = -2 is a root Markers Comments State clearly in solution that f(-2) = 0 x = -2 is a root Using the nested method - coefficients are 6, 13, 0, -4 f(-2) = -2 6 13 0 -4 -12 -2 4 6 1 -2 f(-2) = 0 so (x + 2) is a factor Next Comment Polynomial Menu Back to Home

Show completed factorisation of cubic i.e. (3x + 2)(2x - 1)(x +2). Markers Comments Show completed factorisation of cubic i.e. (3x + 2)(2x - 1)(x +2). Using the nested method - coefficients are 6, 13, 0, -4 f(-2) = -2 6 13 0 -4 -12 -2 4 Can show (x + 2) is a factor by showing f(-2) = 0 but still need nested method for quadratic factor. 6 1 -2 Other factor is 6x2 + x – 2 or (3x + 2)(2x - 1) Hence 6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2) Next Comment Polynomial Menu Back to Home

POLYNOMIALS : Question 4 A busy road passes through several small villages so it is decided to build a by-pass to reduce the volume of traffic. Relative to a set of coordinate axes the road can be modelled by the curve y = -x3 + 6x2 – 3x – 10. The by-pass is a tangent to this curve at point P and rejoins the original road at Q as shown below. P Q y = -x3 + 6x2 – 3x – 10 4 bypass Find the coordinates of P and the equation of the bypass PQ. Hence find the coordinates of Q – the point where the bypass rejoins the original road. EXIT Reveal answer only Go to full solution

POLYNOMIALS : Question 4 A busy road passes through several small villages so it is decided to build a by-pass to reduce the volume of traffic. Relative to a set of coordinate axes the road can be modelled by the curve y = -x3 + 6x2 – 3x – 10. The by-pass is a tangent to this curve at point P and rejoins the original road at Q as shown below. Find the coordinates of P and the equation of the bypass PQ. Hence find the coordinates of Q – the point where the bypass rejoins the original road. P is (4,10) PQ is y = -3x + 22 Q is (-2,28) Go to full solution Go to Marker’s Comments Go to Polynomial Menu Go to Main Menu EXIT

Question 4 At point P, x = 4 so using the equation of the curve we get ….. y = -x3 + 6x2 – 3x – 10 Find the coordinates of P and the equation of the bypass PQ. y = -43 + (6 X 42) – (3 X 4) - 10 = -64 + 96 – 12 - 10 P Q y = -x3 + 6x2 – 3x – 10 4 = 10 ie P is (4,10) Gradient of tangent = gradient of curve = dy/dx = -3x2 + 12x - 3 When x = 4 then dy/dx = (-3 X 16) + (12 X 4) – 3 Begin Solution = -48 + 48 – 3 = -3 Continue Solution Markers Comments Polynomial Menu Back to Home

Question 4 P is (4,10) Find the coordinates of P dy/dx = -3 y = -x3 + 6x2 – 3x – 10 Find the coordinates of P and the equation of the bypass PQ. Now using : y – b = m(x – a) where (a,b) = (4,10) & m = -3 P Q y = -x3 + 6x2 – 3x – 10 4 We get y – 10 = -3(x – 4) or y – 10 = -3x + 12 So PQ is y = -3x + 22 Begin Solution Continue Solution Markers Comments Polynomial Menu Back to Home

Question 4 (b) The tangent & curve meet whenever y = -x3 + 6x2 – 3x – 10 y = -3x + 22 and y = -x3 + 6x2 – 3x – 10 (b)Hence find the coordinates of Q – the point where the bypass rejoins the original road. ie -3x + 22 = -x3 + 6x2 – 3x – 10 or x3 - 6x2 + 32 = 0 P Q y = -x3 + 6x2 – 3x – 10 4 PQ is y = -3x + 22 We already know that x = 4 is one solution to this so using the nested method we get ….. f(4) = 4 1 -6 0 32 4 -8 -32 Begin Solution 1 -2 -8 Continue Solution Markers Comments Other factor is x2 – 2x - 8 Polynomial Menu Back to Home

Question 4 (b) The other factor is x2 – 2x - 8 = (x – 4)(x + 2) y = -x3 + 6x2 – 3x – 10 = (x – 4)(x + 2) (b)Hence find the coordinates of Q – the point where the bypass rejoins the original road. Solving (x – 4)(x + 2) = 0 we get x = 4 or x = -2 P Q y = -x3 + 6x2 – 3x – 10 4 PQ is y = -3x + 22 It now follows that Q has an x-coordinate of -2 Using y = -3x + 22 if x = -2 then y = 6 + 22 = 28 Begin Solution Hence Q is (-2,28) Continue Solution Markers Comments Polynomial Menu Back to Home

Must use differentiation to find gradient. Learn rule: Markers Comments (a) Must use differentiation to find gradient. Learn rule: “Multiply by the power then reduce the power by 1” At point P, x = 4 so using the equation of the curve we get ….. y = -43 + (6 X 42) – (3 X 4) - 10 = -64 + 96 – 12 - 10 = 10 ie P is (4,10) Gradient of tangent = gradient of curve = dy/dx = -3x2 + 12x - 3 When x = 4 then dy/dx = (-3 X 16) + (12 X 4) – 3 Next Comment = -48 + 48 – 3 = -3 Polynomial Menu Back to Home

the point of contact (4,10)& Gradient of curve at this Markers Comments (a) P is (4,10) Use : the point of contact (4,10)& Gradient of curve at this point (m = -3) in equation y - b = m(x - a) dy/dx = -3 Now using : y – b = m(x – a) where (a,b) = (4,10) & m = -3 We get y – 10 = -3(x – 4) or y – 10 = -3x + 12 So PQ is y = -3x + 22 Next Comment Polynomial Menu Back to Home

simplify and factorise Markers Comments (b) (b) The tangent & curve meet whenever At intersection y1 = y2 y = -3x + 22 and y = -x3 + 6x2 – 3x – 10 Terms to the left, simplify and factorise ie -3x + 22 = -x3 + 6x2 – 3x – 10 or x3 - 6x2 + 32 = 0 We already know that x = 4 is one solution to this so using the nested method we get ….. f(4) = 4 1 -6 0 32 4 -8 -32 1 -2 -8 Next Comment Polynomial Menu Other factor is x2 – 2x - 8 Back to Home

Note solution x = 4 appears twice: Repeated root tangency Markers Comments (b) other factor is x2 – 2x - 8 Note solution x = 4 appears twice: Repeated root tangency = (x – 4)(x + 2) Solving (x – 4)(x + 2) = 0 we get x = 4 or x = -2 It now follows that Q has an x-coordinate of -2 Using y = -3x + 22 if x = -2 then y = 6 + 22 = 28 Hence Q is (-2,28) Next Comment Polynomial Menu Back to Home

HIGHER – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 2 : Quadratics Please choose a question to attempt from the following: 1 2 3 4 5 6 Back to Unit 2 Menu EXIT

QUADRATICS : Question 1 (a) Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b. (b) Hence, or otherwise, sketch the graph of y = f(x). Reveal answer only Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu EXIT

QUADRATICS : Question 1 (a) Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b. (b) Hence, or otherwise, sketch the graph of y = f(x). = (x – 4)2 + 5 (a) Reveal answer only y = x2 – 8x + 21 (b) Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu (4,5) EXIT

Question 1 (a) f(x) = x2 – 8x + 21 Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b. Hence, or otherwise, sketch the graph of y = f(x). = (x2 – 8x + ) + 21 16 - 16 (-82)2 = (x – 4)2 + 5 Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

Question 1 f(x) = (x – 4)2 + 5 has a minimum value of 5 when (x – 4)2 = 0 ie x = 4 Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b. Hence, or otherwise, sketch the graph of y = f(x). So the graph has a minimum turning point at (4,5). When x = 0 , y = 21 (from original formula!) so Y-intercept is (0,21). Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

Question 1 Graph looks like…. Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b. Hence, or otherwise, sketch the graph of y = f(x). y = x2 – 8x + 21 (0,21) (4,5) Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

Move towards desired form in stages: f(x) = x2 - 8x + 21 Markers Comments Move towards desired form in stages: f(x) = x2 - 8x + 21 = (x2 - 8x) + 21 = (x2 - 8x +16) + 21 - 16 (a) f(x) = x2 – 8x + 21 = (x2 – 8x + ) + 21 16 - 16 (-82)2 = (x – 4)2 + 5 Find the number to complete the perfect square and balance the expression. (a + b) 2 = a2 + 2ab + b2 = (x - 4)2 + 5 Next Comment Quadratics Menu Back to Home

The sketch can also be obtained by calculus: Markers Comments The sketch can also be obtained by calculus: f(x) = (x – 4)2 + 5 has a minimum value of 5 when (x – 4)2 = 0 ie x = 4 So the graph has a minimum turning point at (4,5). When x = 0 , y = 21 (from original formula!) so Y-intercept is (0,21). (0,21) Next Comment Quadratics Menu (4,5) Back to Home

coefficient of x2 is positive. Markers Comments Min. Turning Point (4,5), coefficient of x2 is positive. f(x) = (x – 4)2 + 5 has a minimum value of 5 when (x – 4)2 = 0 ie x = 4 So the graph has a minimum turning point at (4,5). Hence sketch. 21 (4,5) When x = 0 , y = 21 (from original formula!) so Y-intercept is (0,21). (0,21) Next Comment Quadratics Menu (4,5) Back to Home

QUADRATICS : Question 2 Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2. (b) Hence find the maximum turning point on the graph of y = f(x). Reveal answer only Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu EXIT

QUADRATICS : Question 2 Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2. (b) Hence find the maximum turning point on the graph of y = f(x). (a) = 11 - 4(x – 1)2 Reveal answer only Go to full solution Go to Marker’s Comments maximum t p is at (1,11) . (b) Go to Quadratics Menu Go to Main Menu EXIT

Question 2 (a) f(x) = 7 + 8x - 4x2 Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2. Hence find the maximum turning point on the graph of y = f(x). = - 4x2 + 8x + 7 = -4[x2 – 2x] + 7 = -4[(x2 – 2x + ) ] + 7 1 - 1 (-22)2 = -4[(x – 1)2 - 1] + 7 = -4(x – 1)2 + 4 + 7 = 11 - 4(x – 1)2 Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

Question 2 = 11 - 4(x – 1)2 Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2. Hence find the maximum turning point on the graph of y = f(x). Maximum value is 11 when (x – 1)2 = 0 ie x = 1. so maximum turning point is at (1,11) . Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

Move towards desired form in stages: f(x) = 7 + 8x - 4x2 Markers Comments Move towards desired form in stages: f(x) = 7 + 8x - 4x2 = - 4x2 + 8x + 7 = (- 4x2 + 8x) + 7 (a) f(x) = 7 + 8x - 4x2 = - 4x2 + 8x + 7 = -4[x2 – 2x] + 7 = -4[(x2 – 2x + ) ] + 7 1 - 1 (-22)2 Must reduce coefficient of x2 to 1 = -4(x2 - 2x) + 7 = -4[(x – 1)2 - 1] + 7 = -4(x – 1)2 + 4 + 7 = 11 - 4(x – 1)2 Next Comment Quadratics Menu Back to Home

Must reduce coefficient of x2 to 1 = -4(x2 - 2x) + 7 Markers Comments Must reduce coefficient of x2 to 1 = -4(x2 - 2x) + 7 (a) f(x) = 7 + 8x - 4x2 = - 4x2 + 8x + 7 = -4[x2 – 2x] + 7 Find the number to complete the perfect square and balance the expression. (a + b) 2 = a2 + 2ab + b2 = -4[(x2 – 2x + ) ] + 7 1 - 1 (-22)2 = -4[(x – 1)2 - 1] + 7 = -4[(x2 - 2x +1) -1] +7 ) = -4(x-1)2 + 7 + 4 = 11 - 4(x-1)2 Max. TP at (1,11) = -4(x – 1)2 + 4 + 7 = 11 - 4(x – 1)2 Maximum value is 11 when (x – 1)2 = 0 ie x = 1. Next Comment Quadratics Menu so maximum turning point is at (1,11) . Back to Home

QUADRATICS : Question 3 For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have equal roots. Reveal answer only Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu EXIT

QUADRATICS : Question 3 For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have equal roots. Reveal answer only k = -4 or k = 20 Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu EXIT

Question 3 Let 4x2 – kx + (k + 5) = ax2 + bx + c For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have equal roots. then a = 4, b = -k & c = (k + 5) For equal roots we need discriminant = 0 ie b2 – 4ac = 0 (-k)2 – (4 X 4 X (k + 5)) = 0 k2 – 16(k + 5) = 0 k2 – 16k - 80 = 0 (k + 4)(k – 20) = 0 k = -4 or k = 20 Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

Learn Rules relating to discriminant b2- 4ac b2- 4ac = 0 Equal roots Markers Comments Learn Rules relating to discriminant b2- 4ac b2- 4ac = 0 Equal roots b2- 4ac > 0 2 Real, distinct roots b2- 4ac < 0 No real roots b2- 4ac 0 Real roots, equal or distinct Let 4x2 – kx + (k + 5) = ax2 + bx + c then a = 4, b = -k & c = (k + 5) For equal roots we need discriminant = 0 ie b2 – 4ac = 0 (-k)2 – (4 X 4 X (k + 5)) = 0 k2 – 16(k + 5) = 0 k2 – 16k - 80 = 0 (k + 4)(k – 20) = 0 k = -4 or k = 20 Next Comment Quadratics Menu Back to Home

Must use factorisation to solve resulting quadratic. Markers Comments Must use factorisation to solve resulting quadratic. Trial and error receives no credit. Let 4x2 – kx + (k + 5) = ax2 + bx + c then a = 4, b = -k & c = (k + 5) For equal roots we need discriminant = 0 ie b2 – 4ac = 0 (-k)2 – (4 X 4 X (k + 5)) = 0 k2 – 16(k + 5) = 0 k2 – 16k - 80 = 0 (k + 4)(k – 20) = 0 k = -4 or k = 20 Next Comment Quadratics Menu Back to Home

QUADRATICS : Question 4 The equation of a parabola is f(x) = px2 + 5x – 2p . Prove that the equation f(x) = 0 always has two distinct roots. Reveal answer only Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu EXIT

QUADRATICS : Question 4 The equation of a parabola is f(x) = px2 + 5x – 2p . Prove that the equation f(x) = 0 always has two distinct roots. discriminant = b2 – 4ac Reveal answer only = 8p2 + 25 Go to full solution Since p2  0 for all values of p then 8p2 + 25 > 0. Go to Marker’s Comments The discriminant is always positive so there are always two distinct roots. Go to Quadratics Menu Go to Main Menu EXIT

Question 4 Let px2 + 5x – 2p = ax2 + bx + c The equation of a parabola is f(x) = px2 + 5x – 2p . Prove that the equation f(x) = 0 always has two distinct roots. then a = p, b = 5 & c = -2p. So discriminant = b2 – 4ac = 52 – (4 X p X (-2p)) = 25 – (-8p2) = 8p2 + 25 Since p2  0 for all values of p then 8p2 + 25 > 0. Begin Solution The discriminant is always positive so there are always two distinct roots. Continue Solution Markers Comments Quadratics Menu Back to Home

Learn Rules relating to discriminant b2- 4ac b2- 4ac = 0 Equal roots Markers Comments Learn Rules relating to discriminant b2- 4ac b2- 4ac = 0 Equal roots b2- 4ac > 0 2 Real, distinct roots b2- 4ac < 0 No real roots b2- 4ac 0 Real roots, equal or distinct Let px2 + 5x – 2p = ax2 + bx + c then a = p, b = 5 & c = -2p. So discriminant = b2 – 4ac = 52 – (4 X p X (-2p)) = 25 – (-8p2) = 8p2 + 25 Since p2  0 for all values of p then 8p2 + 25 > 0. Next Comment The discriminant is always positive so there are always two distinct roots. Quadratics Menu Back to Home

State condition you require to show true explicitly: Markers Comments State condition you require to show true explicitly: For two distinct roots b2- 4ac > 0 Let px2 + 5x – 2p = ax2 + bx + c then a = p, b = 5 & c = -2p. So discriminant = b2 – 4ac = 52 – (4 X p X (-2p)) = 25 – (-8p2) = 8p2 + 25 Since p2  0 for all values of p then 8p2 + 25 > 0. Next Comment The discriminant is always positive so there are always two distinct roots. Quadratics Menu Back to Home

algebraic logic or show the graph of 8p2 + 25 is always Markers Comments To show 8p2 + 25> 0 use algebraic logic or show the graph of 8p2 + 25 is always above the “x-axis”. Let px2 + 5x – 2p = ax2 + bx + c then a = p, b = 5 & c = -2p. So discriminant = b2 – 4ac = 52 – (4 X p X (-2p)) = 25 – (-8p2) 25 = 8p2 + 25 Min. T.P. at (0,25) hence graph always above the “x - axis.” Since p2  0 for all values of p then 8p2 + 25 > 0. Next Comment The discriminant is always positive so there are always two distinct roots. Quadratics Menu Back to Home

QUADRATICS : Question 5 Given that the roots of 3x(x + p) = 4p(x – 1) are equal then show that p = 0 or p = 48. Reveal answer only Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu EXIT

QUADRATICS : Question 5 Given that the roots of 3x(x + p) = 4p(x – 1) are equal then show that p = 0 or p = 48. For equal roots we need discriminant = 0 p(p - 48) = 0 Reveal answer only ie p = 0 or p = 48 Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu EXIT

Question 5 Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p Given that the roots of 3x(x + p) = 4p(x – 1) are equal then show that p = 0 or p = 48. 3x2 + 3px = 4px - 4p 3x2 - px + 4p = 0 Let 3x2 - px + 4p = ax2 + bx + c then a = 3, b = -p & c = 4p For equal roots we need discriminant = 0 ie b2 – 4ac = 0 (-p)2 - (4 X 3 X 4p) = 0 p2 - 48p = 0 Begin Solution p(p - 48) = 0 Continue Solution ie p = 0 or p = 48 Markers Comments Quadratics Menu Back to Home

Must put the equation into standard quadratic form before Markers Comments Must put the equation into standard quadratic form before reading off a,b and c. i.e. ax2 + bx +c = 0 Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p 3x2 - px + 4p = 0 Let 3x2 - px + 4p = ax2 + bx + c then a = 3, b = -p & c = 4p For equal roots we need discriminant = 0 ie b2 – 4ac = 0 (-p)2 - (4 X 3 X 4p) = 0 p2 - 48p = 0 Next Comment p(p - 48) = 0 Quadratics Menu ie p = 0 or p = 48 Back to Home

Learn Rules relating to discriminant b2- 4ac b2- 4ac = 0 Equal roots Markers Comments Learn Rules relating to discriminant b2- 4ac b2- 4ac = 0 Equal roots b2- 4ac > 0 2 Real, distinct roots b2- 4ac < 0 No real roots b2- 4ac 0 Real roots, equal or distinct Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p 3x2 - px + 4p = 0 Let 3x2 - px + 4p = ax2 + bx + c then a = 3, b = -p & c = 4p For equal roots we need discriminant = 0 ie b2 – 4ac = 0 (-p)2 - (4 X 3 X 4p) = 0 p2 - 48p = 0 Next Comment p(p - 48) = 0 Quadratics Menu ie p = 0 or p = 48 Back to Home

State condition you require explicitly: Markers Comments State condition you require explicitly: For two equal roots b2- 4ac = 0 Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p 3x2 - px + 4p = 0 Let 3x2 - px + 4p = ax2 + bx + c Must use factorisation to solve resulting quadratic. then a = 3, b = -p & c = 4p For equal roots we need discriminant = 0 ie b2 – 4ac = 0 (-p)2 - (4 X 3 X 4p) = 0 p2 - 48p = 0 Next Comment p(p - 48) = 0 Quadratics Menu ie p = 0 or p = 48 Back to Home

QUADRATICS : Question 6 The diagram below shows the parabola y = -2x2 + 3x + 2 and the line x + y – 4 = 0. y = -2x2 + 3x + 2 x + y – 4 = 0 Reveal answer only Go to full solution Go to Marker’s Comms Go to Quadratics Menu Go to Main Menu Prove that the line is a tangent to the curve. EXIT

QUADRATICS : Question 6 The diagram below shows the parabola y = -2x2 + 3x + 2 and the line x + y – 4 = 0. y = -2x2 + 3x + 2 x + y – 4 = 0 Reveal answer only Go to full solution Go to Marker’s Comms Go to Quadratics Menu Go to Main Menu Since the discriminant = 0 then there is only one solution to the equation so only one point of contact and it follows that the line is a tangent. Prove that the line is a tangent to the curve. EXIT

Question 6 Linear equation can be changed from x + y – 4 = 0 to y = -x + 4. The line and curve meet when y = -x + 4 and y = -2x2 + 3x + 2 . x + y – 4 = 0 So -x + 4 = -2x2 + 3x + 2 Or 2x2 - 4x + 2 = 0 Let 2x2 - 4x + 2 = ax2 + bx + c y = -2x2 + 3x + 2 Prove that the line is a tangent. then a = 2, b = -4 & c = 2. Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

Question 6 then a = 2, b = -4 & c = 2. So discriminant = b2 – 4ac = (-4)2 – (4 X 2 X 2) x + y – 4 = 0 = 16 - 16 = 0 Since the discriminant = 0 then there is only one solution to the equation so only one point of contact and it follows that the line is a tangent. y = -2x2 + 3x + 2 Prove that the line is a tangent. Begin Solution Continue Solution Markers Comments Quadratics Menu Back to Home

For intersection of line and polynomial y1 = y2 Markers Comments For intersection of line and polynomial y1 = y2 Terms to the left, simplify and factorise. Linear equation can be changed from x + y – 4 = 0 to y = -x + 4. The line and curve meet when y = -x + 4 and y = -2x2 + 3x + 2 . So -x + 4 = -2x2 + 3x + 2 Or 2x2 - 4x + 2 = 0 Let 2x2 - 4x + 2 = ax2 + bx + c then a = 2, b = -4 & c = 2. Next Comment Quadratics Menu Back to Home

“equal roots” may be used in place of the discriminant. Markers Comments Or -x + 4 = -2x2 + 3x + 2 2x2 - 4x + 2 = 0 2(x2 - 2x + 1) = 0 2(x - 1)(x - 1) = 0 x = 1 (twice) Equal roots tangency then a = 2, b = -4 & c = 2. So discriminant = b2 – 4ac = (-4)2 – (4 X 2 X 2) = 16 - 16 = 0 To prove tangency “equal roots” may be used in place of the discriminant. The statement must be made explicitly. Since the discriminant = 0 then there is only one solution to the equation so only one point of contact and it follows that the line is a tangent. Next Comment Quadratics Menu Back to Home

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

INTEGRATION : Question 1 The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k. y = x2 - 8x + 18 x = 3 x = k Reveal answer only Go to full solution Go to Marker’s Comments Show that the shaded area is given by 1/3k3 – 4k2 + 18k - 27 Go to Integration Menu Go to Main Menu EXIT

 INTEGRATION : Question 1 The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k. y = x2 - 8x + 18 x = 3 x = k Reveal answer only Go to full solution Go to Marker’s Comments Show that the shaded area is given by 1/3k3 – 4k2 + 18k - 27 Go to Integration Menu Go to Main Menu Area = (x2 - 8x + 18) dx  3 k EXIT = 1/3k3 – 4k2 + 18k – 27 as required.

[ ] [ ]  Question 1 = x3 - 8x2 + 18x Area = (x2 - 8x + 18) dx k The diagram shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k. = x3 - 8x2 + 18x [ ] 3 2 k 3 = 1/3x3 – 4x2 + 18x [ ] k 3 Show that the shaded area is given by 1/3k3 – 4k2 + 18k - 27 = (1/3k3 – 4k2 + 18k) – ((1/3 X 27) – (4 X 9) + 54) = 1/3k3 – 4k2 + 18k – 27 as required. Begin Solution Continue Solution Markers Comments Integration Menu Back to Home

[ ] [ ]  Learn result can be used to find the enclosed area shown: Markers Comments Learn result can be used to find the enclosed area shown: Area = (x2 - 8x + 18) dx  3 k = x3 - 8x2 + 18x [ ] 3 2 k 3 = 1/3x3 – 4x2 + 18x [ ] k 3 a b f(x) = (1/3k3 – 4k2 + 18k) – ((1/3 X 27) – (4 X 9) + 54) = 1/3k3 – 4k2 + 18k – 27 as required. Next Comment Integration Menu Back to Home

[ ] [ ]  Learn result for integration: “Add 1 to the power and Markers Comments Learn result for integration: “Add 1 to the power and divide by the new power.” Area = (x2 - 8x + 18) dx  3 k = x3 - 8x2 + 18x [ ] 3 2 k 3 = 1/3x3 – 4x2 + 18x [ ] k 3 = (1/3k3 – 4k2 + 18k) – ((1/3 X 27) – (4 X 9) + 54) = 1/3k3 – 4k2 + 18k – 27 as required. Next Comment Integration Menu Back to Home

INTEGRATION : Question 2 Given that dy/dx = 12x2 – 6x and the curve y = f(x) passes through the point (2,15) then find the equation of the curve y = f(x). Reveal answer only Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu EXIT

INTEGRATION : Question 2 Given that dy/dx = 12x2 – 6x and the curve y = f(x) passes through the point (2,15) then find the equation of the curve y = f(x). Equation of curve is y = 4x3 – 3x2 - 5 Reveal answer only Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu EXIT

Question 2 dy/dx = 12x2 – 6x Given that dy/dx = 12x2 – 6x and the curve y = f(x) passes through the point (2,15) then find the equation of the curve y = f(x). So = 12x3 – 6x2 + C 3 2 = 4x3 – 3x2 + C Substituting (2,15) into y = 4x3 – 3x2 + C We get 15 = (4 X 8) – (3 X 4) + C So C + 20 = 15 ie C = -5 Begin Solution Equation of curve is y = 4x3 – 3x2 - 5 Continue Solution Markers Comments Integration Menu Back to Home

Learn the result that integration undoes differentiation: i.e. given Markers Comments Learn the result that integration undoes differentiation: i.e. given dy/dx = 12x2 – 6x So = 12x3 – 6x2 + C 3 2 = 4x3 – 3x2 + C Learn result for integration: “Add 1 to the power and divide by the new power”. Substituting (2,15) into y = 4x3 – 3x2 + C We get 15 = (4 X 8) – (3 X 4) + C So C + 20 = 15 ie C = -5 Equation of curve is y = 4x3 – 3x2 - 5 Next Comment Integration Menu Back to Home

Do not forget the constant of integration!!! Markers Comments Do not forget the constant of integration!!! dy/dx = 12x2 – 6x So = 12x3 – 6x2 + C 3 2 = 4x3 – 3x2 + C Substituting (2,15) into y = 4x3 – 3x2 + C We get 15 = (4 X 8) – (3 X 4) + C So C + 20 = 15 ie C = -5 Equation of curve is y = 4x3 – 3x2 - 5 Next Comment Integration Menu Back to Home

 INTEGRATION : Question 3 Find x2 - 4 dx 2xx Reveal answer only Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu EXIT

 INTEGRATION : Question 3 Find x2 - 4 dx 2xx = xx + 4 + C 3 x Reveal answer only Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu EXIT

    Question 3 x2 - 4 dx 2xx Find x2 - 4 dx 2xx = x2 - 4 dx = 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C = 1/3x3/2 + 4x-1/2 + C = xx + 4 + C 3 x Begin Solution Continue Solution Markers Comments Integration Menu Back to Home

   Prepare expression by: 1 Dividing out the fraction. Markers Comments Prepare expression by: 1 Dividing out the fraction. 2 Applying the laws of indices. x2 - 4 2xx dx  = x2 - 4 2x3/2 2x3/2 dx  Learn result for integration: Add 1 to the power and divide by the new power. = 1/2x1/2 - 2x-3/2 dx  = 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C = 1/3x3/2 + 4x-1/2 + C Do not forget the constant of integration. = xx + 4 + C 3 x Next Comment Integration Menu Back to Home

 ( ) INTEGRATION : Question 4 Evaluate x2 - 2 2 dx x 1 2  ( ) Evaluate x2 - 2 2 dx x Reveal answer only Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu EXIT

 ( ) INTEGRATION : Question 4 = 21/5 Evaluate x2 - 2 2 dx x  ( ) Evaluate x2 - 2 2 dx x = 21/5 Reveal answer only Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu EXIT

 ( )  ( )   [ ] [ ] ( ) ( ) Question 4 = 21/5 x2 - 2 2 dx x  ( ) x2 - 2 2 dx x 1 2  ( ) Evaluate x2 - 2 2 dx x ( ) = x4 - 4x + 4 dx  x2 1 2 ( ) = x4 - 4x + 4x-2 dx  1 2 [ ] = x5 - 4x2 + 4x-1 5 2 -1 1 2 = x5 - 2x2 - 4 5 x [ ] 2 1 = (32/5 - 8 - 2) - (1/5 - 2 - 4) Begin Solution = 21/5 Continue Solution Markers Comments Integration Menu Back to Home

 ( )   [ ] [ ] ( ) ( ) Prepare expression by: Markers Comments Prepare expression by: 1 Expanding the bracket 2 Applying the laws of indices. 1 2  ( ) x2 - 2 2 dx x ( ) = x4 - 4x + 4 dx  x2 1 2 Learn result for integration: “Add 1 to the power and divide by the new power”. ( ) = x4 - 4x + 4x-2 dx  1 2 [ ] = x5 - 4x2 + 4x-1 5 2 -1 1 2 = x5 - 2x2 - 4 5 x [ ] 2 1 When applying limits show substitution clearly. = (32/5 - 8 - 2) - (1/5 - 2 - 4) = 21/5 Next Comment Integration Menu Back to Home

INTEGRATION : Question 5 The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B. y = -x2 + 8x - 10 y = x A B Reveal answer only Go to full solution Go to Marker’s Comms Go to Integration Menu Go to Main Menu (a) Find the coordinates of A and B. Hence find the shaded area between the curves. EXIT

INTEGRATION : Question 5 The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B. y = -x2 + 8x - 10 y = x A B Reveal answer only Go to full solution Go to Marker’s Comms Go to Integration Menu Go to Main Menu (a) Find the coordinates of A and B. A is (2,2) and B is (5,5) . Hence find the shaded area between the curves. = 41/2units2 EXIT

Question 5 Line & curve meet when y = x and y = -x2 + 8x - 10 . The diagram shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B. So x = -x2 + 8x - 10 or x2 - 7x + 10 = 0 (a) Find the coordinates of A and B. ie (x – 2)(x – 5) = 0 Hence find the shaded area between the curves. ie x = 2 or x = 5 Since points lie on y = x then A is (2,2) and B is (5,5) . Begin Solution Continue Solution Markers Comments Integration Menu Back to Home

  [ ] Question 5 A is (2,2) and B is (5,5) . The diagram shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B. (b) Curve is above line between limits so Shaded area = (-x2 + 8x – 10 - x) dx  2 5 = (-x2 + 7x – 10) dx  2 5 (a) Find the coordinates of A and B. = -x3 + 7x2 - 10x 3 2 [ ] 5 2 Hence find the shaded area between the curves. = (-125/3 + 175/2 – 50) – (-8/3 +14 – 20) Begin Solution = 41/2units2 Continue Solution Markers Comments Integration Menu Back to Home

At intersection of line and curve y1 = y2 Markers Comments At intersection of line and curve y1 = y2 Terms to the left, simplify and factorise. Line & curve meet when y = x and y = -x2 + 8x - 10 . So x = -x2 + 8x - 10 or x2 - 7x + 10 = 0 ie (x – 2)(x – 5) = 0 ie x = 2 or x = 5 Since points lie on y = x then A is (2,2) and B is (5,5) . Next Comment Integration Menu Back to Home

  [ ] Learn result can be used to find the enclosed area shown: y1 Markers Comments Learn result can be used to find the enclosed area shown: (b) Curve is above line between limits so Shaded area = (-x2 + 8x – 10 - x) dx  2 5 = (-x2 + 7x – 10) dx  2 5 = -x3 + 7x2 - 10x 3 2 [ ] 5 2 a b upper curve y1 area y2 lower curve = (-125/3 + 175/2 – 50) – (-8/3 +14 – 20) = 41/2units2 Next Comment Integration Menu Back to Home

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

ADDITION FORMULAE : Question 1 In triangle PQR show that the exact value of cos(a - b) is 4/5. P Q R 1 4 2 a b Reveal answer only Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

ADDITION FORMULAE : Question 1 In triangle PQR show that the exact value of cos(a - b) is 4/5. P Q R 1 4 2 a b cos(a – b) = cosacosb + sinasinb = (2/5 X 1/5 ) + (1/5 X 2/5 ) = 4/5 Reveal answer only Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

Question 1 PQ2 = 12 + 22 = 5 In triangle PQR show that the exact value of cos(a - b) is 4/5. so PQ = 5 QR2 = 42 + 22 = 20 so QR = 20 = 45 = 25 P Q R 1 4 2 a b sina = 2/25 = 1/5 & sinb = 2/5 cosa = 4/25 = 2/5 & cosb = 1/5 cos(a – b) = cosacosb + sinasinb = (2/5 X 1/5 ) + (1/5 X 2/5 ) Begin Solution = 2/5 + 2/5 Continue Solution = 4/5 Markers Comments Add Formulae Menu Back to Home

Use formula sheet to check correct expansion and relate Markers Comments Use formula sheet to check correct expansion and relate to given variables: cos(a - b) = cosacosb + sina sinb PQ2 = 12 + 22 = 5 so PQ = 5 QR2 = 42 + 22 = 20 so QR = 20 = 45 = 25 sina = 2/25 = 1/5 & sinb = 2/5 Work only with exact values when applying Pythagoras’: cosa = 4/25 = 2/5 & cosb = 1/5 2 a a = cos(a – b) = cosacosb + sinasinb = (2/5 X 1/5 ) + (1/5 X 2/5 ) = 2/5 + 2/5 Next Comment Add Form Menu = 4/5 Back to Home

ADDITION FORMULAE : Question 2 Find the exact value of cos(p + q) in the diagram below Y X O F(4,4) G(3,-1) p q Reveal answer only Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

ADDITION FORMULAE : Question 2 Find the exact value of cos(p + q) in the diagram below Y X O F(4,4) G(3,-1) p q Reveal answer only = 1/5 Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

Question 2 OF2 = 42 + 42 = 32 Find the exact value of cos(p + q) in the diagram below so OF = 32 = 162 = 42 OG2 = 32 + 12 = 10 so OG = 10 Y X O F(4,4) G(3,-1) p q sinp = 4/42 = 1/2 & sinq = 1/10 cosp = 4/42 = 1/2 & cosq = 3/10 cos(p + q) = cospcosq - sinpsinq = (1/2 X 3/10 ) - (1/2 X 1/10 ) = 3/20 - 1/20 Begin Solution = 2/20 Continue Solution = 2/4 5 Markers Comments = 2/2 5 Add Formulae Menu = 1/5 Back to Home

Use formula sheet to check correct expansion and relate Markers Comments Use formula sheet to check correct expansion and relate to given variables: cos(a + b) = cosacosb - sina sinb cos(p + q) = cospcosq - sinpsinq OF2 = 42 + 42 = 32 so OF = 32 = 162 = 42 OG2 = 32 + 12 = 10 Formula Sheet: so OG = 10 sinp = 4/42 = 1/2 & sinq = 1/10 becomes: cosp = 4/42 = 1/2 & cosq = 3/10 cos(p + q) = cospcosq - sinpsinq = (1/2 X 3/10 ) - (1/2 X 1/10 ) = 3/20 - 1/20 = 2/20 = 2/4 5 Next Comment = 2/2 5 Add Form Menu = 1/5 Back to Home

Work only with exact values when applying Pythagoras’: Markers Comments Work only with exact values when applying Pythagoras’: OF2 = 42 + 42 = 32 so OF = 32 = 162 = 42 OG2 = 32 + 12 = 10 2 a a = so OG = 10 sinp = 4/42 = 1/2 & sinq = 1/10 cosp = 4/42 = 1/2 & cosq = 3/10 cos(p + q) = cospcosq - sinpsinq = (1/2 X 3/10 ) - (1/2 X 1/10 ) = 3/20 - 1/20 = 2/20 = 2/4 5 Next Comment = 2/2 5 Add Form Menu = 1/5 Back to Home

ADDITION FORMULAE : Question 3 Solve the equation 2 - sinx° = 3cos2x° where 0<x<360. Reveal answer only Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

ADDITION FORMULAE : Question 3 Solve the equation 2 - sinx° = 3cos2x° where 0<x<360. Solution = {30, 150, 199.5, 340.5} Reveal answer only Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

Question 3 2 - sinx° = 3cos2x° 2 - sinx° = 3(1 – 2sin2x°) Solve the equation 2 - sinx° = 3cos2x° where 0<x<360. 2 - sinx° = 3(1 – 2sin2x°) 2 - sinx° = 3 – 6sin2x° Rearrange into quadratic form 6sin2x° - sinx° - 1 = 0 6s2 – s – 1 = (3s + 1)(2s – 1) (3sinx° + 1)(2sinx° - 1) = 0 sinx° = -1/3 or sinx° = 1/2 Q3 or Q4 Q1 or Q2 Begin Solution A S T C a° (180 - a)° (180 + a)° (360 - a)° Continue Solution Markers Comments Add Formulae Menu Back to Home

Question 3 sinx° = -1/3 or sinx° = 1/2 Solve the equation 2 - sinx° = 3cos2x° where 0<x<360. A S T C a° (180 - a)° (180 + a)° (360 - a)° sin-1 (1/2) = 30° Q1: x = 30 Q2: x = 180 – 30 = 150 sin-1 (1/3) = 19.5° Begin Solution Q3: x = 180 + 19.5 = 199.5 Q4: x = 360 - 19.5 = 340.5 Continue Solution Markers Comments Solution = {30, 150, 199.5, 340.5} Add Formulae Menu Back to Home

Use formula sheet to check correct expansion and relate to Markers Comments Use formula sheet to check correct expansion and relate to given variables: cos2x° =1 - 2sin2x° 2 - sinx° = 3cos2x° 2 - sinx° = 3(1 – 2sin2x°) 2 - sinx° = 3 – 6sin2x° Rearrange into quadratic form 6sin2x° - sinx° - 1 = 0 Formula must be consistent with the rest of the equation. 2 - sinx° i.e. choose formula with sinx° 6s2 – s – 1 = (3s + 1)(2s – 1) (3sinx° + 1)(2sinx° - 1) = 0 sinx° = -1/3 or sinx° = 1/2 Q3 or Q4 Q1 or Q2 A S T C a° (180 - a)° (180 + a)° (360 - a)° Next Comment Add Form Menu Back to Home

Only one way to solve the resulting quadratic: Markers Comments Only one way to solve the resulting quadratic: Terms to the left, put in standard quadratic form and factorise. 2 - sinx° = 3cos2x° 2 - sinx° = 3(1 – 2sin2x°) 2 - sinx° = 3 – 6sin2x° Rearrange into quadratic form 6sin2x° - sinx° - 1 = 0 6s2 – s – 1 = (3s + 1)(2s – 1) (3sinx° + 1)(2sinx° - 1) = 0 sinx° = -1/3 or sinx° = 1/2 Q3 or Q4 Q1 or Q2 A S T C a° (180 - a)° (180 + a)° (360 - a)° Next Comment Add Form Menu Back to Home

Take care to relate “solutions” to the given domain. Markers Comments Take care to relate “solutions” to the given domain. Since all 4 values are included. sinx° = -1/3 or sinx° = 1/2 A S T C a° (180 - a)° (180 + a)° (360 - a)° sin-1 (1/2) = 30° Q1: x = 30 Q2: x = 180 – 30 = 150 sin-1 (1/3) = 19.5° Q3: x = 180 + 19.5 = 199.5 Q4: x = 360 - 19.5 = 340.5 Next Comment Add Form Menu Solution = {30, 150, 199.5, 340.5} Back to Home

ADDITION FORMULAE : Question 4 Solve sin2 = cos  where 0 <  < 2 Reveal answer only Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

ADDITION FORMULAE : Question 4 Solve sin2 = cos  where 0 <  < 2 Soltn = {/6 , /2 , 5/6 , 3/2} Reveal answer only Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu EXIT

Question 4 sin2 = cos sin2 - cos = 0 Solve sin2 = cos  where 0 <  < 2 2sin cos - cos = 0 (common factor cos) cos(2sin - 1) = 0 cos = 0 or (2sin - 1) = 0 cos = 0 or sin = 1/2 (roller-coaster graph) Q1 or Q2  = /2 or 3/2 Begin Solution  -    +  2 -  A S T C Continue Solution Markers Comments Add Formulae Menu Back to Home

Question 4 cos = 0 or sin = 1/2 Solve sin2 = cos   = /2 or 3/2 where 0 <  < 2  = /2 or 3/2 Q1 or Q2  -    +  2 -  A S T C sin-1(1/2) = /6 Q1:  = /6 Q2:  =  - /6 = 5/6 Begin Solution Continue Solution Markers Comments Soltn = {/6 , /2 , 5/6 , 3/2} Add Formulae Menu Back to Home

Use formula sheet to check correct expansion and relate to Markers Comments Use formula sheet to check correct expansion and relate to given variables: sin2x = 2sinxcosx sin2 = cos sin2 - cos = 0 2sin cos - cos = 0 (common factor cos) Although equation is in radians possible to work in degrees and convert to radians using: cos(2sin - 1) = 0 cos = 0 or (2sin - 1) = 0 cos = 0 or sin = 1/2 (roller-coaster graph) Q1 or Q2  = /2 or 3/2  -    +  2 -  A S T C Next Comment Add Form Menu Back to Home

Only one way to solve the result: Markers Comments Only one way to solve the result: Terms to the left, put in standard quadratic form and factorise. sin2 = cos sin2 - cos = 0 2sin cos - cos = 0 (common factor cos) cos(2sin - 1) = 0 cos = 0 or (2sin - 1) = 0 cos = 0 or sin = 1/2 (roller-coaster graph) Q1 or Q2  = /2 or 3/2  -    +  2 -  A S T C Next Comment Add Form Menu Back to Home

use sketch of graph to obtain solutions: cosx = 0 Markers Comments For trig. equations sinx = 0 or 1 or cosx = 0 or 1 use sketch of graph to obtain solutions: cosx = 0 cos = 0 or sin = 1/2  = /2 or 3/2 Q1 or Q2  -    +  2 -  A S T C y x y = cos x x = , sin-1(1/2) = /6 Q1:  = /6 Q2:  =  - /6 = 5/6 Next Comment Add Form Menu Soltn = {/6 , /2 , 5/6 , 3/2} Back to Home

Take care to relate “solutions” to the given domain. Markers Comments Take care to relate “solutions” to the given domain. Since all 4 values are included cos = 0 or sin = 1/2  = /2 or 3/2 Q1 or Q2  -    +  2 -  A S T C sin-1(1/2) = /6 Q1:  = /6 Q2:  =  - /6 = 5/6 Next Comment Add Form Menu Soltn = {/6 , /2 , 5/6 , 3/2} Back to Home

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

THE CIRCLE : Question 1 The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0. The centre is at C and the winder is at W. The dial for the second hand is 1/3 the size of the face and is located half way between C and W. W C Find the coordinates of C and W. (b) Hence find the equation of the dial for the second hand. Reveal answer only Go to Marker’s Comments EXIT Go to full solution Go to Circle Menu

THE CIRCLE : Question 1 The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0. The centre is at C and the winder is at W. The dial for the second hand is 1/3 the size of the face and is located half way between C and W. W C C is (5,4) W is (5,10). Find the coordinates of C and W. (b) Hence find the equation of the dial for the second hand. (x – 5)2 + (y – 7)2 = 4 Reveal answer only Go to Marker’s Comments EXIT Go to full solution Go to Circle Menu

Question 1 Find the coordinates of C and W. The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0. The centre is at C and the winder is at W. The dial for the second hand is 1/3 the size of the face and is located half way between C and W. Large circle is x2 + y2 – 10x – 8y + 5 = 0. Comparing Coefficients x2 + y2 + 2gx + 2fy + c = 0 2g = -10, 2f = -8 and c = 5 So g = -5, f = -4 and c = 5 Centre is (-g,-f) and r = (g2 + f2 – c) C is (5,4) r = (25 + 16 – 5) r = 36 = 6 Begin Solution W is 6 units above C so W is (5,10). Continue Solution Markers Comments Circle Menu Back to Home

Question 1 (b) Hence find the equation of the dial for the second hand. The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0. The centre is at C and the winder is at W. The dial for the second hand is 1/3 the size of the face and is located half way between C and W. Radius of small circle = 6  3 = 2. Centre is midpoint of CW ie (5,7). Using (x – a)2 + (y – b)2 = r2 we get (x – 5)2 + (y – 7)2 = 4 Begin Solution Continue Solution Markers Comments Circle Menu Back to Home

Use formula sheet to check correct formulas and relate to Markers Comments Use formula sheet to check correct formulas and relate to general equation of the circle. Find the coordinates of C and W. Large circle is x2 + y2 – 10x – 8y + 5 = 0. Comparing Coefficients x2 + y2 + 2gx + 2fy + c = 0 2g = -10, 2f = -8 and c = 5 So g = -5, f = -4 and c = 5 Centre is (-g,-f) and r = (g2 + f2 – c) C is (5,4) r = (25 + 16 – 5) r = 36 = 6 Next Comment W is 6 units above C so W is (5,10). Circle Menu Back to Home

Identify centre and radius before putting values into circle equation: Markers Comments Identify centre and radius before putting values into circle equation: Centre (5,7), radius = 2 (x - a)2 + (y - b)2= r2 (x - 5)2 + (y - 7)2= 22 There is no need to expand this form of the circle equation. (b) Hence find the equation of the dial for the second hand. Radius of small circle = 6  3 = 2. Centre is midpoint of CW ie (5,7). Using (x – a)2 + (y – b)2 = r2 we get (x – 5)2 + (y – 7)2 = 4 Next Comment Circle Menu Back to Home

THE CIRCLE : Question 2 In a factory a conveyor belt passes through several rollers. Part of this mechanism is shown below. (a) The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0. belt P A(7,0) B Q The belt is a common tangent which meets the small roller at A(7,0)and the large roller at B. Find the equation of the tangent. B has an x-coordinate of 10. Find its y-coordinate. (b) Find the equation of the larger roller given that its diameter is twice that of the smaller roller, and has centre Q. Reveal answer only Go to Marker’s Comments EXIT Go to full solution Go to Circle Menu

THE CIRCLE : Question 2 In a factory a conveyor belt passes through several rollers. Part of this mechanism is shown below. (a) The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0. belt P A(7,0) B Q The belt is a common tangent which meets the small roller at A(7,0)and the large roller at B. Find the equation of the tangent. B has an x-coordinate of 10. Find its y-coordinate. y = -4x + 28 B is (10,-12) (b) Find the equation of the larger roller given that its diameter is twice that of the smaller roller, and has centre Q. (x – 18)2 + (y + 10)2 = 68 Reveal answer only Go to Marker’s Comments EXIT Go to full solution Go to Circle Menu

Question 2 = 1/4 (a) (i) Small circle is x2 + y2 – 6x + 2y - 7 = 0. The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0. Comparing coefficients x2 + y2 + 2gx + 2fy + c = 0 The belt is a common tangent which meets the small roller at A(7,0)and the large roller at B. Find the equation of the tangent. B has an x-coordinate of 10. Find its y-coordinate. 2g = -6, 2f = 2 and c = -7 So g = -3, f = 1 and c = -7 Centre is (-g,-f) and r = (g2 + f2 – c) P is (3,-1) r = (9 + 1 + 7) r = 17 Begin Solution For equation of tangent: Continue Solution Gradient of PA = 0 – (-1) 7 - 3 = 1/4 Markers Comments Circle Menu Back to Home

Question 2 = 1/4 P is (3,-1) The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0. For equation of tangent: Gradient of PA = 0 – (-1) 7 - 3 = 1/4 The belt is a common tangent which meets the small roller at A(7,0)and the large roller at B. Find the equation of the tangent. B has an x-coordinate of 10. Find its y-coordinate. Gradient of tangent = -4 (as m1m2 = -1) Using y – b = m(x – a) with (a,b) = (7,0) & m = -4 we get …. y – 0 = -4(x – 7) or y = -4x + 28 Begin Solution Continue Solution Markers Comments Circle Menu Back to Home

Question 2 (a)(ii) At B x = 10 so y = (-4 X 10) + 28 = -12. The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0. ie B is (10,-12) The belt is a common tangent which meets the small roller at A(7,0)and the large roller at B. Find the equation of the tangent. B has an x-coordinate of 10. Find its y-coordinate. Begin Solution Continue Solution Markers Comments Circle Menu Back to Home

Question 2 r = 17 P is (3,-1) The small roller has centre P Small Circle: r = 17 P is (3,-1) A(7,0) The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0. (b) From P to A is 4 along and 1 up. So from B to Q is 8 along and 2 up. Q is at (10+8, -12+2) ie (18,-10) (b) Find the equation of the larger roller given that its diameter is twice that of the smaller roller, and has centre Q. Radius of larger circle is 217 so r2 = (217)2 = 4 X 17 = 68 Using (x – a)2 + (y – b)2 = r2 we get (x – 18)2 + (y + 10)2 = 68 Begin Solution Continue Solution Markers Comments Circle Menu Back to Home

Use formula sheet to check correct formulas and relate to Markers Comments Use formula sheet to check correct formulas and relate to general equation of the circle. (a) (i) Small circle is x2 + y2 – 6x + 2y - 7 = 0. Comparing coefficients x2 + y2 + 2gx + 2fy + c = 0 2g = -6, 2f = 2 and c = -7 So g = -3, f = 1 and c = -7 Centre is (-g,-f) and r = (g2 + f2 – c) P is (3,-1) r = (9 + 1 + 7) r = 17 For equation of tangent: Next Comment Gradient of PA = 0 – (-1) 7 - 3 = 1/4 Circle Menu Back to Home

Identify centre and radius before putting values into circle equation: Markers Comments Identify centre and radius before putting values into circle equation: Centre (18,-10), radius = 2 (x - a)2 + (y - b)2= r2 (x - 18)2 + (y + 10)2= (2 ) 2 (b) From P to A is 4 along and 1 up. So from B to Q is 8 along and 2 up. Q is at (10+8, -12+2) ie (18,-10) Radius of larger circle is 217 so r2 = (217)2 = 4 X 17 = 68 There is no need to expand this form of the circle equation. If the radius is “squared out” it must be left as an exact value. Using (x – a)2 + (y – b)2 = r2 we get (x – 18)2 + (y + 10)2 = 68 Next Comment Circle Menu Back to Home

THE CIRCLE : Question 3 Part of the mechanism inside an alarm clock consists of three different sized collinear cogwheels. C D E The wheels can be represented by circles with centres C, D and E as shown. The small circle with centre C has equation x2 + (y –12)2 = 5. The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20. Find the equation of the large circle. Reveal answer only Go to Marker’s Comments EXIT Go to full solution Go to Circle Menu

THE CIRCLE : Question 3 Part of the mechanism inside an alarm clock consists of three different sized collinear cogwheels. C D E The wheels can be represented by circles with centres C, D and E as shown. The small circle with centre C has equation x2 + (y –12)2 = 5. The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20. Find the equation of the large circle. (x – 8)2 + (y – 8)2 = 45 Reveal answer only Go to Marker’s Comments EXIT Go to full solution Go to Circle Menu

Question 3 Small circle has centre (0,12) and radius 5 The small circle with centre C has equation x2 + (y –12)2 = 5. The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20. Find the equation of the large circle. Med. circle has centre (18,3) and radius 20 = 4 X 5 = 25 CE2 = (18 – 0)2 + (3 – 12)2 (Distance form!!) = 182 + (-9)2 = 324 + 81 = 405 CE = 405 = 81 X 5 = 95 Diameter of large circle = 95 - 25 - 5 = 65 Begin Solution C D E 5 35 95 25 Continue Solution Markers Comments So radius of large circle = 35 Circle Menu Back to Home

Question 3 = 4/9[( ) - ( )] 25 The small circle with centre C D E 5 35 95 25 The small circle with centre C has equation x2 + (y –12)2 = 5. The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20. Find the equation of the large circle. It follows that CD = 4/9CE = 4/9[( ) - ( )] 18 3 12 = 4/9( ) 18 -9 = ( ) 8 -4 So D is the point (0 + 8,12- 4) or (8,8) Using (x – a)2 + (y – b)2 = r2 we get (x – 8)2 + (y – 8)2 = (35)2 Begin Solution or (x – 8)2 + (y – 8)2 = 45 Continue Solution Markers Comments Circle Menu Back to Home

Use formula sheet to check correct formulas and relate to Markers Comments Use formula sheet to check correct formulas and relate to equation of the circle. Small circle has centre (0,12) and radius 5 Med. circle has centre (18,3) and radius 20 = 4 X 5 = 25 CE2 = (18 – 0)2 + (3 – 12)2 (Distance form!!) = 182 + (-9)2 = 324 + 81 = 405 CE = 405 = 81 X 5 = 95 Diameter of large circle = 95 - 25 - 5 = 65 C D E 5 35 95 25 Next Comment Circle Menu Back to Home

The section formula is used to find Centre D Markers Comments The section formula is used to find Centre D It follows that CD = 4/9CE C E D = 4/9[( ) - ( )] 18 3 12 = 4/9( ) 18 -9 So D is the point (0 + 8,12- 4) or (8,8) Using (x – a)2 + (y – b)2 = r2 we get (x – 8)2 + (y – 8)2 = (35)2 or (x – 8)2 + (y – 8)2 = 45 Next Comment Circle Menu Back to Home

THE CIRCLE : Question 4 The central driving wheels on a steam locomotive are linked by a piston rod. A B C E D The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 0 the front wheel has centre C & equation x2 + y2 – 52x – 10y + 676 = 0. If one unit is 5cm then find the size of the minimum gap between the wheels given that the gaps are equal and the wheels are identical. The central wheel has equation x2 + y2 – 28x – 10y + 196 = 0 and the midpoint of DE is M(13.5,1.5). Find the equation of DE and hence find the coordinates of D & E. EXIT Go to full solution Reveal answer only

THE CIRCLE : Question 4 The central driving wheels on a steam locomotive are linked by a piston rod. The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 0 the front wheel has centre C & equation x2 + y2 – 52x – 10y + 676 = 0. If one unit is 5cm then find the size of the minimum gap between the wheels given that the gaps are equal and the wheels are identical. Each gap = 2 units = 10cm. The central wheel has equation x2 + y2 – 28x – 10y + 196 = 0 and the midpoint of DE is M(13.5,1.5). Find the equation of DE and hence find the coordinates of D & E. Hence D is (10,2) and E is (17,1). Equation of DE x + 7y = 24 EXIT Go to full solution Reveal answer only

Question 4 (a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 . Comparing coefficients The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 0 the front wheel has centre C & equation x2 + y2 – 52x – 10y + 676 = 0 If one unit is 5cm then find the size of the minimum gap between the wheels. x2 + y2 + 2gx + 2fy + c = 0 2g = -4, 2f = -10 and c = 4 So g = -2, f = -5 and c = 4 Centre is (-g,-f) and r = (g2 + f2 – c) A is (2,5) r = (4 + 25 – 4) r = 25 = 5 Begin Solution Continue Solution Markers Comments Circle Menu Back to Home

Question 4 Front wheel is x2 + y2 – 52x – 10y + 676 = 0 (a) Front wheel is x2 + y2 – 52x – 10y + 676 = 0 Comparing coefficients The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 0 the front wheel has centre C & equation x2 + y2 – 52x – 10y + 676 = 0 If one unit is 5cm then find the size of the minimum gap between the wheels. x2 + y2 + 2gx + 2fy + c = 0 2g = -52, 2f = -10 and c = 676 So g = -26, f = -5 and c = 676 Centre is (-g,-f) r = 5 as wheels are identical C is (26,5) AC = 24 units {A is (2,5)} Both gaps = AC – 2 radii - diameter Begin Solution = 24 – 5 – 5 – 10 = 4 units Continue Solution Each gap = 2 units = 10cm. Markers Comments Circle Menu Back to Home

Question 4 Middle wheel is x2 + y2 – 28x – 10y + 196 = 0 . (b) Middle wheel is x2 + y2 – 28x – 10y + 196 = 0 . The central wheel has equation x2 + y2 – 28x – 10y + 196 = 0 and the midpoint of DE is M(13.5,1.5). Find the equation of DE and hence find the coordinates of D & E. x2 + y2 + 2gx + 2fy + c = 0 2g = -28, 2f = -10 and c = 196 So g = -14, f = -5 and c = 196 Centre is (-g,-f) B is (14,5) Equation of DE Gradient of BM = 5 – 1.5 14 – 13.5 Begin Solution = 3.5/0.5 = 7 Continue Solution Gradient of DE = -1/7 (m1m2 = -1) Markers Comments Circle Menu Back to Home

Question 4 Using y – b = m(x – a) The central wheel has equation Equation of DE Using y – b = m(x – a) The central wheel has equation x2 + y2 – 28x – 10y + 196 = 0 and the midpoint of DE is M(13.5,1.5). Find the equation of DE and hence find the coordinates of D & E. we get y – 1.5 = -1/7 (x – 13.5) ( X 7) Or 7y – 10.5 = -x + 13.5 So DE is x + 7y = 24 x + 7y = 24 can be written as x = 24 – 7y Line & circle meet when x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y. Begin Solution Substituting (24 – 7y) for x in the circle equation we get …. Continue Solution Markers Comments Circle Menu Back to Home

Question 4 (24 – 7y)2 + y2 – 28(24 – 7y) – 10y + 196 = 0 (b) (24 – 7y)2 + y2 – 28(24 – 7y) – 10y + 196 = 0 576 – 336y + 49y2 + y2 – 672 + 196y – 10y + 196 = 0 The central wheel has equation x2 + y2 – 28x – 10y + 196 = 0 and the midpoint of DE is M(13.5,1.5). Find the equation of DE and Hence find the coordinates of D & E. 50y2 – 150y + 100 = 0 (50) y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 So y = 1 or y = 2 Using x = 24 – 7y If y = 1 then x = 17 If y = 2 then x = 10 Begin Solution Hence D is (10,2) and E is (17,1). Continue Solution Markers Comments Circle Menu Back to Home

Use formula sheet to check correct formulas and relate to Markers Comments Use formula sheet to check correct formulas and relate to general equation of the circle. (a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 . Comparing coefficients x2 + y2 + 2gx + 2fy + c = 0 2g = -4, 2f = -10 and c = 4 So g = -2, f = -5 and c = 4 Centre is (-g,-f) and r = (g2 + f2 – c) A is (2,5) r = (4 + 25 – 4) r = 25 = 5 Next Comment Circle Menu Back to Home

In part b) avoid fractions when substituting into the circle equation: Markers Comments In part b) avoid fractions when substituting into the circle equation: Use x = (24 - 7y) not y = (24 -x) Equation of DE Using y – b = m(x – a) we get y – 1.5 = -1/7 (x – 13.5) Or 7y – 10.5 = -x + 13.5 So DE is x + 7y = 24 x + 7y = 24 can be written as x = 24 – 7y Line & circle meet when x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y. Substituting (24 – 7y) for x in the circle equation we get …. Next Comment Circles Menu Back to Home

Even if it appears an error has occurred continue the Markers Comments (24 – 7y)2 + y2 – 28(24 – 7y) – 10y + 196 = 0 576 – 336y + 49y2 + y2 – 672 + 196y – 10y + 196 = 0 50y2 – 150y + 100 = 0 (50) Even if it appears an error has occurred continue the solution even if it means applying the quadratic formula: y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 So y = 1 or y = 2 x = Using x = 24 – 7y If y = 1 then x = 17 If y = 2 then x = 10 Hence D is (10,2) and E is (17,1). Next Comment Circle Menu Back to Home

Wave Function Vectors Logs & Exponential Further Calculus HIGHER – ADDITIONAL QUESTION BANK UNIT 3 : Wave Function Vectors Logs & Exponential Further Calculus EXIT

HIGHER – ADDITIONAL QUESTION BANK You have chosen to study: UNIT 3 : Wave Function Please choose a question to attempt from the following: 1 2 3 Back to Unit 3 Menu EXIT

WAVE FUNCTION: Question 1 Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where k>0 and 0  a  90. Hence solve cos(x°) + 7sin (x°) = 5 for 0  x  90. Reveal answer only Go to full solution Go to Marker’s Comments Go to Wave Function Menu Go to Main Menu EXIT

WAVE FUNCTION: Question 1 Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where k>0 and 0  a  90. Hence solve cos(x°) + 7sin (x°) = 5 for 0  x  90. Hence cos(x°) + 7sin (x°) = 52cos(x° - 81.9°) Reveal answer only (a) Go to full solution Solution is {36.9} Go to Marker’s Comments (b) Go to Wave Function Menu Go to Main Menu EXIT

Question 1 Let cos(x°) + 7sin (x°) = kcos(x° - a°) Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where k>0 and 0  a  90. Hence solve cos(x°) + 7sin (x°) = 5 for 0  x  90. = k(cosx°cosa° + sinx°sina° ) = (kcosa°)cosx° + (ksina°)sinx° Comparing coefficients kcosa° = 1 & ksina° = 7 So (kcosa°)2 + (ksina°)2 = 12 + 72 or k2cos2a° + k2sin2a° = 1 + 49 or k2(cos2a° + sin2a°) = 50 or k2 = 50 (cos2a° + sin2a°) = 1 Begin Solution so k = 50 = 252 = 52 Continue Solution Markers Comments Wave Function Menu Back to Home

Question 1 so k = 50 = 252 = 52 Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where k>0 and 0  a  90. Hence solve cos(x°) + 7sin (x°) = 5 for 0  x  90. ksina° kcosa° = 7 1    so a in Q1! ksina° > 0 so a in Q1 or Q2 kcosa° > 0 so a in Q1 or Q4 tana° = 7 a° = tan-1(7) = 81.9° Hence cos(x°) + 7sin (x°) = 52cos(x° - 81.9°) Begin Solution Continue Solution Markers Comments Wave Function Menu Back to Home

Question 1 Hence cos(x°) + 7sin (x°) = 52cos(x° - 81.9°) Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where k>0 and 0  a  90. Hence solve cos(x°) + 7sin (x°) = 5 for 0  x  90. (b) If cos(x°) + 7sin (x°) = 5 then 52cos(x° - 81.9°) = 5 or cos(x° - 81.9°) = 1/2 Q1 or Q4 & cos-1(1/2) = 45° Q1: angle = 45° so x° - 81.9° = 45° so x° = 126.9°  not in range. Begin Solution Continue Solution Q4: angle = 360° - 45° so x° - 81.9° = 315° so x° = 396.9° (**) Markers Comments Wave Function Menu Back to Home

Question 1 Q4: angle = 360° - 45° so x° - 81.9° = 315° Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where k>0 and 0  a  90. Hence solve cos(x°) + 7sin (x°) = 5 for 0  x  90. (**) function repeats every 360° & 396.9° - 360° = 36.9° Solution is {36.9} Begin Solution Continue Solution Markers Comments Wave Function Menu Back to Home

Use the formula sheet for the correct expansion: kcos(x° - a°) Markers Comments Use the formula sheet for the correct expansion: kcos(x° - a°) = kcosx°cosa° + ksinx°sina° (Take care not to omit the “k” term) Let cos(x°) + 7sin (x°) = kcos(x° - a°) = k(cosx°cosa° + sinx°sina° ) = (kcosa°)cosx° + (ksina°)sinx° Comparing coefficients kcosa° = 1 & ksina° = 7 So (kcosa°)2 + (ksina°)2 = 12 + 72 or k2cos2a° + k2sin2a° = 1 + 49 or k2(cos2a° + sin2a°) = 50 or k2 = 50 (cos2a° + sin2a°) = 1 so k = 50 = 252 = 52 Next Comment Wave Func Menu Back to Home

When equating coefficients “square” and “ring” Markers Comments When equating coefficients “square” and “ring” corresponding coefficients: 1cosx° + 7sinx° = kcosx°cosa° + ksinx°sina° Let cos(x°) + 7sin (x°) = kcos(x° - a°) = k(cosx°cosa° + sinx°sina° ) = (kcosa°)cosx° + (ksina°)sinx° Comparing coefficients kcosa° = 1 & ksina° = 7 So (kcosa°)2 + (ksina°)2 = 12 + 72 or k2cos2a° + k2sin2a° = 1 + 49 or k2(cos2a° + sin2a°) = 50 or k2 = 50 (cos2a° + sin2a°) = 1 so k = 50 = 252 = 52 Next Comment Wave Func Menu Back to Home

State the resulting equations explicitly: kcosa° = 1 ksina° = 7 Markers Comments State the resulting equations explicitly: kcosa° = 1 ksina° = 7 Let cos(x°) + 7sin (x°) = kcos(x° - a°) = k(cosx°cosa° + sinx°sina° ) = (kcosa°)cosx° + (ksina°)sinx° Comparing coefficients kcosa° = 1 & ksina° = 7 So (kcosa°)2 + (ksina°)2 = 12 + 72 or k2cos2a° + k2sin2a° = 1 + 49 or k2(cos2a° + sin2a°) = 50 or k2 = 50 (cos2a° + sin2a°) = 1 so k = 50 = 252 = 52 Next Comment Wave Func Menu Back to Home

k can be found directly: Markers Comments k can be found directly: Let cos(x°) + 7sin (x°) = kcos(x° - a°) = k(cosx°cosa° + sinx°sina° ) = (kcosa°)cosx° + (ksina°)sinx° Note: k is always positive Comparing coefficients kcosa° = 1 & ksina° = 7 So (kcosa°)2 + (ksina°)2 = 12 + 72 or k2cos2a° + k2sin2a° = 1 + 49 or k2(cos2a° + sin2a°) = 50 or k2 = 50 (cos2a° + sin2a°) = 1 so k = 50 = 252 = 52 Next Comment Wave Func Menu Back to Home

WAVE FUNCTION: Question 2 Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°) where k>0 and 0 < a < 360. Reveal answer only Go to full solution Go to Marker’s Comments Go to Wave Function Menu Go to Main Menu EXIT

WAVE FUNCTION: Question 2 Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°) where k>0 and 0 < a < 360. Hence 4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°) Reveal answer only Go to full solution Go to Marker’s Comments Go to Wave Function Menu Go to Main Menu EXIT

Question 2 Let 4sin(x°)–2cos(x°) = ksin(x° + a°) Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°) where k>0 and 0 < a < 360. = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Comparing coefficients kcosa° = 4 & ksina° = -2 So (kcosa°)2 + (ksina°)2 = 42 + (-2)2 or k2cos2a° + k2sin2a° = 16 + 4 or k2(cos2a° + sin2a°) = 20 Begin Solution or k2 = 20 (cos2a° + sin2a°) = 1 Continue Solution so k = 20 = 45 = 25 Markers Comments Wave Function Menu Back to Home

Question 2 so k = 20 = 45 = 25 Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°) where k>0 and 0 < a < 360.    so a in Q4! ksina° kcosa° = -2 4 tana° = (-1/2) so Q2 or Q4 ksina° < 0 so a in Q3 or Q4 kcosa° > 0 so a in Q1 or Q4 tan-1(1/2) = 26.6° Q4: a° = 360° – 26.6° = 333.4° Begin Solution Hence 4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°) Continue Solution Markers Comments Wave Function Menu Back to Home

Use the formula sheet for the correct expansion: ksin(x° + a°) Markers Comments Use the formula sheet for the correct expansion: ksin(x° + a°) = ksinx°cosa° + kcosx°sina° (Take care not to omit the “k” term) Let 4sin(x°)–2cos(x°) = ksin(x° + a°) = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Comparing coefficients kcosa° = 4 & ksina° = -2 So (kcosa°)2 + (ksina°)2 = 42 + (-2)2 or k2cos2a° + k2sin2a° = 16 + 4 or k2(cos2a° + sin2a°) = 20 or k2 = 20 (cos2a° + sin2a°) = 1 Next Comment so k = 20 = 45 = 25 Wave Func Menu Back to Home

When equating coefficients “square” and “ring” Markers Comments When equating coefficients “square” and “ring” corresponding coefficients: 4sinx° - 2cosx° = ksinx°cosa° + kcosx°sina° Let 4sin(x°)–2cos(x°) = ksin(x° + a°) = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Comparing coefficients kcosa° = 4 & ksina° = -2 State the resulting equations explicitly: kcosa° = 4 ksina° = -2 So (kcosa°)2 + (ksina°)2 = 42 + (-2)2 or k2cos2a° + k2sin2a° = 16 + 4 or k2(cos2a° + sin2a°) = 20 or k2 = 20 (cos2a° + sin2a°) = 1 Next Comment so k = 20 = 45 = 25 Wave Func Menu Back to Home

k can be found directly: Markers Comments k can be found directly: Let 4sin(x°)–2cos(x°) = ksin(x° + a°) = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Note: k is always positive Comparing coefficients kcosa° = 4 & ksina° = -2 So (kcosa°)2 + (ksina°)2 = 42 + (-2)2 or k2cos2a° + k2sin2a° = 16 + 4 or k2(cos2a° + sin2a°) = 20 or k2 = 20 (cos2a° + sin2a°) = 1 Next Comment so k = 20 = 45 = 25 Wave Func Menu Back to Home

Use the sign of the equations to determine the correct quadrant: Markers Comments Use the sign of the equations to determine the correct quadrant: kcosa° = 4 (cos +ve) ksina° = -2 (sin -ve) so k = 20 = 45 = 25    so a in Q4! ksina° kcosa° = -2 4 tana° = (-1/2) so Q2 or Q4  cos +ve &sin -ve ksina° < 0 so a in Q3 or Q4 kcosa° > 0 so a in Q1 or Q4 tan-1(1/2) = 26.6° Q4: a° = 360° – 26.6° = 333.4° Hence 4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°) Next Comment Wave Func Menu Back to Home

WAVE FUNCTION: Question 3 The graph below is that of y = 5sin(x°) - 12cos(x°) P y = 5sin(x°) - 12cos(x°) Reveal answer only Go to full solution Go to Marker’s Comms Go to Wave Funct Menu Go to Main Menu Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0  a < 360 Hence find the coordinates of the maximum turning point at P. EXIT

WAVE FUNCTION: Question 3 The graph below is that of y = 5sin(x°) - 12cos(x°) P y = 5sin(x°) - 12cos(x°) Reveal answer only Go to full solution Go to Marker’s Comms Go to Wave Funct Menu Go to Main Menu Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0  a < 360 Hence find the coordinates of the maximum turning point at P. Hence 5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°) EXIT Max TP is at (157.4,13)

Question 3 (a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°) Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0  a < 360 Hence find the coordinates of the maximum turning point at P. = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Comparing coefficients kcosa° = 5 & ksina° = -12 So (kcosa°)2 + (ksina°)2 = 52 + (-12)2 or k2cos2a° + k2sin2a° = 25 + 144 or k2(cos2a° + sin2a°) = 169 Begin Solution or k2 = 169 (cos2a° + sin2a°) = 1 Continue Solution so k = 13 Markers Comments Wave Function Menu Back to Home

Question 3    so a in Q4! so k = 13 Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0  a < 360 Hence find the coordinates of the maximum turning point at P. ksina° kcosa° = -12 5 tana° = (-12/5) so Q2 or Q4 ksina° < 0 so a in Q3 or Q4 kcosa° > 0 so a in Q1 or Q4 tan-1(12/5) = 67.4° Q4: a° = 360° – 67.4° = 292.6° Hence 5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°) Begin Solution Continue Solution Markers Comments Wave Function Menu Back to Home

Question 3 13sin(x° + 292.6°) is 13. Hence 5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°) Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0  a < 360 Hence find the coordinates of the maximum turning point at P. The maximum value of 13sin(x° + 292.6°) is 13. Maximum on a sin graph occurs when angle = 90°. ie x + 292.6 = 90 or x = -202.6 (**) This is not in the desired range but the function repeats every 360°. Begin Solution Continue Solution Taking -202.6 + 360 = 157.4 Markers Comments Max TP is at (157.4,13) Wave Function Menu Back to Home

Use the formula sheet for the correct expansion: ksin(x° + a°) Markers Comments Use the formula sheet for the correct expansion: ksin(x° + a°) = ksinx°cosa° + kcosx°sina° (Take care not to omit the “k” term) (a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°) = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Comparing coefficients kcosa° = 5 & ksina° = -12 So (kcosa°)2 + (ksina°)2 = 52 + (-12)2 or k2cos2a° + k2sin2a° = 25 + 144 or k2(cos2a° + sin2a°) = 169 or k2 = 169 (cos2a° + sin2a°) = 1 Next Comment so k = 13 Wave Func Menu Back to Home

When equating coefficients “square” and “ring” Markers Comments When equating coefficients “square” and “ring” corresponding coefficients: 5sinx° - 12cosx° = ksinx°cosa° + kcosx°sina° (a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°) = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Comparing coefficients kcosa° = 5 & ksina° = -12 So (kcosa°)2 + (ksina°)2 = 52 + (-12)2 State the resulting equations explicitly: kcosa° = 5 ksina° = -12 or k2cos2a° + k2sin2a° = 25 + 144 or k2(cos2a° + sin2a°) = 169 or k2 = 169 (cos2a° + sin2a°) = 1 Next Comment so k = 13 Wave Func Menu Back to Home

k can be found directly: Markers Comments k can be found directly: (a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°) = k(sinx°cosa° + cosx°sina° ) = (kcosa°)sinx° + (ksina°)cosx° Comparing coefficients kcosa° = 5 & ksina° = -12 Note: k is always positive So (kcosa°)2 + (ksina°)2 = 52 + (-12)2 or k2cos2a° + k2sin2a° = 25 + 144 or k2(cos2a° + sin2a°) = 169 or k2 = 169 (cos2a° + sin2a°) = 1 Next Comment so k = 13 Wave Func Menu Back to Home

Use the sign of the equations to determine the correct quadrant: Markers Comments Use the sign of the equations to determine the correct quadrant: kcosa° = 5 (cos +ve) ksina° = -12 (sin -ve)    so a in Q4! so k = 13 ksina° kcosa° = -12 5 tana° = (-12/5) so Q2 or Q4 ksina° < 0 so a in Q3 or Q4 kcosa° > 0 so a in Q1 or Q4  cos +ve &sin -ve tan-1(12/5) = 67.4° Q4: a° = 360° – 67.4° = 292.6° Hence 5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°) Next Comment Wave Func Menu Back to Home

be found by considering rules for related functions: Markers Comments The maximum value of 13sin(x°+292.6°) can also be found by considering rules for related functions: 13sin(x° + 292.6°) slides 13sinx° graph 292.6° to the left. Maximum value of sinx° occurs at 90° The maximum value of 13sin(x° + 292.6°) is 13. Maximum on a sin graph occurs when angle = 90°. ie x + 292.6 = 90 or x = -202.6 (**) This is not in the desired range but the function repeats every 360°. Taking -202.6 + 360 = 157.4 Max TP is at (157.4,13) Next Comment Wave Func Menu Back to Home

Add 360° to bring into domain x = 157.4° 13sin(x° + 292.6°) is 13. Markers Comments Maximum value of sin(x°+292.6°) occurs at (90°- 292.6°) i.e at x = - 202.6° Add 360° to bring into domain x = 157.4° The maximum value of 13sin(x° + 292.6°) is 13. Maximum on a sin graph occurs when angle = 90°. ie x + 292.6 = 90 or x = -202.6 (**) This is not in the desired range but the function repeats every 360°. Taking -202.6 + 360 = 157.4 Max TP is at (157.4,13) Next Comment Wave Func Menu Back to Home

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

LOGS & EXPONENTIALS: Question 1 As a radioactive substance decays the amount of radioactive material remaining after t hours, At , is given by the formula At = A0e-0.161t where A0 is the original amount of material. If 400mg of material remain after 10 hours then determine how much material there was at the start. The half life of the substance is the time required for exactly half of the initial amount to decay. Find the half life to the nearest minute. Reveal answer only Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

LOGS & EXPONENTIALS: Question 1 As a radioactive substance decays the amount of radioactive material remaining after t hours, At , is given by the formula At = A0e-0.161t where A0 is the original amount of material. If 400mg of material remain after 10 hours then determine how much material there was at the start. The half life of the substance is the time required for exactly half of the initial amount to decay. Find the half life to the nearest minute. Initial amount of material = 2000mg Reveal answer only t = 4hrs 18mins Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

Question 1 (a) When t = 10, A10 = 400 so At = A0e-0.161t becomes A10 = A0e-0.161X10 If 400mg of material remain after 10 hours then determine how much material there was at the start. or 400 = A0e-1.61 and A0 = 400  e-1.61 ie A0 = 2001.12… or 2000 to 3 sfs Initial amount of material = 2000mg Begin Solution Continue Solution Markers Comments Logs & Exp Menu Back to Home

Question 1 (b) For half life At = 1/2A0 so At = A0e-0.161t becomes 1/2A0 = A0e-0.161t (b) The half life of the substance is the time required for exactly half of the initial amount to decay. Find the half life to the nearest minute. or e-0.161t = 0.5 or ln(e-0.161t ) = ln0.5 ie -0.161t = ln0.5 so t = ln0.5  (-0.161) ie t = 4.3052…hrs ( 0.3052 X 60 = 18.3..) Begin Solution Continue Solution or t = 4hrs 18mins Markers Comments Logs & Exp Menu Back to Home

The ex is found on the calculator: ex 2nd ln Markers Comments (a) When t = 10, A10 = 400 so At = A0e-0.161t The ex is found on the calculator: becomes A10 = A0e-0.161X10 2nd ln ex or 400 = A0e-1.61 and A0 = 400  e-1.61 ie A0 = 2001.12… or 2000 to 3 sfs Initial amount of material = 2000mg Next Comment Logs & Exp Menu Back to Home

The half life can be found using any real value: Markers Comments (b) (b) For half life At = 1/2A0 The half life can be found using any real value: e.g. A0 = 2000 At = 1000 so At = A0e-0.161t becomes 1/2A0 = A0e-0.161t or e-0.161t = 0.5 This results in the equation or ln(e-0.161t ) = ln0.5 At = A0e-0.161t 1000 = 2000e-0.161t etc. ie -0.161t = ln0.5 so t = ln0.5  (-0.161) ie t = 4.3052…hrs ( 0.3052 X 60 = 18.3..) Next Comment or t = 4hrs 18mins Logs & Exp Menu Back to Home

To solve an exponential equation must use logs. Markers Comments (b) (b) For half life At = 1/2A0 To solve an exponential equation must use logs. A trial and error solution will only be given minimum credit: e.g. 12 = e2x take logs. to both sides ln12 = ln (e2x) log and exponential are inverse functions ln 12 = 2x etc. so At = A0e-0.161t becomes 1/2A0 = A0e-0.161t or e-0.161t = 0.5 or ln(e-0.161t ) = ln0.5 ie -0.161t = ln0.5 so t = ln0.5  (-0.161) ie t = 4.3052…hrs ( 0.3052 X 60 = 18.3..) Next Comment or t = 4hrs 18mins Logs & Exp Menu Back to Home

LOGS & EXPONENTIALS: Question 2 log10y log10x 1 0.3 The graph illustrates the law y = k xn The line passes through (0,0.3) and (1,0). Find the values of k and n. Reveal answer only Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

LOGS & EXPONENTIALS: Question 2 log10y log10x 1 0.3 The graph illustrates the law y = k xn The line passes through (0,0.3) and (1,0). Find the values of k and n. Reveal answer only Hence k = 2 and n = 0.3 Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

Question 2 log10y log10x Intercept = (0,0.3) The graph illustrates the law y = k gradient = 0 – 0.3 1 - 0 = -0.3 xn Straight line so in form Y = mX + c with Y = log10y and X = log10x The line passes through (0,0.3) and (1,0). Find the values of k and n. log10y log10x 1 0.3 This becomes log10y = -0.3log10x + 0.3 Or log10y = -0.3log10x + log10100.3 or log10y = -0.3log10x + log102 Begin Solution or log10y = log10x-0.3 + log102 law3 Continue Solution or log10y = log102x-0.3 Markers Comments law1 Logs & Exp Menu Back to Home

Question 2 log10y log10x or log10y = log102x-0.3 law1 The graph illustrates the law y = k so y = 2x-0.3 = 2 x0.3 xn Hence k = 2 and n = 0.3 The line passes through (0,0.3) and (1,0). Find the values of k and n. log10y log10x 1 0.3 Begin Solution Continue Solution Markers Comments Logs & Exp Menu Back to Home

It is also possible to find the values of k and n by applying Markers Comments It is also possible to find the values of k and n by applying the laws of logs to the given equation and substituting two coordinates from the graph: e.g. y = kxn Take logs to both sides log y = log kxn Apply Law 1: log AB = log A + logB Intercept = (0,0.3) gradient = 0 – 0.3 1 - 0 Straight line so in form Y = mX + c with Y = log10y and X = log10x This becomes log10y = -0.3log10x + 0.3 Or log10y = -0.3log10x + log10100.3 or log10y = -0.3log10x + log102 or log10y = log10x-0.3 + log102 Next Comment or log10y = log102x-0.3 Logs & Exp Menu Back to Home

log y = logk + logxn Apply Law 3: logxn = nlogx log y = logk + nlogx Markers Comments log y = logk + logxn Apply Law 3: logxn = nlogx log y = logk + nlogx log y = nlogx + logk Intercept = (0,0.3) gradient = 0 – 0.3 1 - 0 Straight line so in form Y = mX + c with Y = log10y and X = log10x This becomes log10y = -0.3log10x + 0.3 Or log10y = -0.3log10x + log10100.3 or log10y = -0.3log10x + log102 or log10y = log10x-0.3 + log102 Next Comment or log10y = log102x-0.3 Logs & Exp Menu Back to Home

Two coordinates from the graph: Markers Comments Two coordinates from the graph: or log10y = log102x-0.3 log10y log10x 1 0.3 so y = 2x-0.3 = 2 x0.3 Hence k = 2 and n = 0.3 Next Comment Straight Line Menu Back to Home

Two coordinates from the graph: Markers Comments Two coordinates from the graph: or log10y = log102x-0.3 log y = nlogx + logk (0,0.3) 0.3 = n.0 + logk - 1 (1,0) 0 = n.1 + logk - 2 so y = 2x-0.3 = 2 x0.3 Hence k = 2 and n = 0.3 Hence solve 1 and 2 to find k and n logk = 0.3 hence k = 2, and n = - logk hence n= -0.3. Next Comment Straight Line Menu Back to Home

LOGS & EXPONENTIALS: Question 3 Solve the equation log3(5) – log3(2x + 1) = -2 Reveal answer only Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

LOGS & EXPONENTIALS: Question 3 Solve the equation log3(5) – log3(2x + 1) = -2 x = 22 Reveal answer only Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

( ) Question 3 If log3(5) – log3(2x + 1) = -2 Solve the equation then log3 5 2x + 1 ( ) = -2 law2 so 5 (2x + 1) = 3-2 or 5 (2x + 1) = 1 9 Cross mult we get 2x + 1 = 45 or 2x = 44 Begin Solution ie x = 22 Continue Solution Markers Comments Logs & Exp Menu Back to Home

( ) To solve an equation involving logs apply the laws so that the Markers Comments To solve an equation involving logs apply the laws so that the equation is reduced to log=log and the log can be removed and the equation solved: log35 - log3(2x+1) = -2 If log3(5) – log3(2x + 1) = -2 then log3 5 2x + 1 ( ) = -2 so 5 (2x + 1) = 3-2 or 5 (2x + 1) = 1 9 Cross mult Law 2: log = logA - logB A B we get 2x + 1 = 45 Must know: log33 = 1 or 2x = 44 Next Comment ie x = 22 Logs & Exp Menu Back to Home

( ) 5 log3 = -2 log33 2x+1 Law 3: log = nlogx xn 5 log3 = log33-2 2x+1 Markers Comments log3 = -2 log33 5 2x+1 If log3(5) – log3(2x + 1) = -2 Law 3: log = nlogx xn then log3 5 2x + 1 ( ) = -2 log3 = log33-2 5 2x+1 so 5 (2x + 1) = 3-2 The log terms can now be dropped from both sides of the equation and the equation solved: or 5 (2x + 1) = 1 9 Cross mult we get 2x + 1 = 45 or 2x = 44 Next Comment ie x = 22 Logs & Exp Menu Back to Home

( ) Drop log3 terms = 3-2 5 2x+1 = 5 2x+1 1 9 etc. Markers Comments If log3(5) – log3(2x + 1) = -2 = 3-2 5 2x+1 then log3 5 2x + 1 ( ) = -2 = 5 2x+1 1 9 so 5 (2x + 1) = 3-2 etc. or 5 (2x + 1) = 1 9 Cross mult we get 2x + 1 = 45 or 2x = 44 Next Comment ie x = 22 Logs & Exp Menu Back to Home

LOGS & EXPONENTIALS: Question 4 The pressure in a leaky tyre drops according to the formula Pt = P0e-kt where P0 is the initial tyre pressure and Pt is the pressure after t hours. If the tyre is inflated to 35psi but this drops to 31 psi in 30 minutes then find the value of k to 3 decimal places. By how many more psi will the pressure drop in the next 15 mins? sssss Reveal answer only Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

LOGS & EXPONENTIALS: Question 4 The pressure in a leaky tyre drops according to the formula Pt = P0e-kt where P0 is the initial tyre pressure and Pt is the pressure after t hours. If the tyre is inflated to 35psi but this drops to 31 psi in 30 minutes then find the value of k to 3 decimal places. By how many more psi will the pressure drop in the next 15 mins? sssss Reveal answer only (a) k = 0.243 to 3 dps Go to full solution 1.8psi (b) Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

Question 4 We have P0 = 35, Pt = 31 and t = 0.5 (30mins = ½ hr) If the tyre is inflated to 35psi but this drops to 31 psi in 30 minutes then find the value of k to 3 decimal places. So Pt = P0e-kt becomes 31 = 35e(-0.5k) this becomes e(-0.5k) = 31/35 or lne(-0.5k) = ln(31/35) or -0.5k = ln(31/35) or k = ln(31/35)  (-0.5) Begin Solution ie k = 0.243 to 3 dps Continue Solution Markers Comments Logs & Exp Menu Back to Home

Question 4 ie k = 0.243 to 3 dps We now have k = 0.243, P0 = 31 and t = 0.25 (15mins = 1/4hr) (b) By how many more psi will the pressure drop in the next 15 mins? using Pt = P0e-0.243t we get P0.25 = 31e(-0.243X0.25) so P0.25 = 29.2 Pressure drop in next 15 mins is 31 – 29.2 or 1.8psi Begin Solution Continue Solution Markers Comments Logs & Exp Menu Back to Home

(a) The ex is found on the calculator: ex 2nd ln Markers Comments We have P0 = 35, Pt = 31 and t = 0.5 (30mins = ½ hr) 2nd ln ex So Pt = P0e-kt becomes 31 = 35e(-0.5k) this becomes e(-0.5k) = 31/35 or lne(-0.5k) = ln(31/35) or -0.5k = ln(31/35) or k = ln(31/35)  (-0.5) ie k = 0.243 to 3 dps Next Comment Logs & Exp Menu Back to Home

To solve an exponential equation must use logs. Markers Comments To solve an exponential equation must use logs. A trial and error solution will only be given minimum credit: e.g. e(-0.5k) = 31/35 take logs. to both sides ln e(-0.5k) = ln 31/35 log and exponential are inverse functions -0.5k = ln31/35 etc. We have P0 = 35, Pt = 31 and t = 0.5 (30mins = ½ hr) So Pt = P0e-kt becomes 31 = 35e(-0.5k) this becomes e(-0.5k) = 31/35 or lne(-0.5k) = ln(31/35) or -0.5k = ln(31/35) or k = ln(31/35)  (-0.5) ie k = 0.243 to 3 dps Next Comment Logs & Exp Menu Back to Home

LOGS & EXPONENTIALS: Question 5 Given that sin = au and cos = av show that u – v = logatan. Reveal answer only Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

LOGS & EXPONENTIALS: Question 5 Given that sin = au and cos = av show that u – v = logatan. u – v = logatan Reveal answer only Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu EXIT Go to Main Menu

Question 5 Since sin = au and cos = av Given that sin = au and cos = av show that u – v = logatan. then u = logasin and v = logacos so u – v = logasin - logacos or u – v = loga(sin/cos) logx – logy = log(x/y) hence u – v = logatan Begin Solution Continue Solution Markers Comments Logs & Exp Menu Back to Home

Should be known from standard grade Markers Comments There are different routes to the same result but all involve correct application of formulas and laws of logs: Since sin = au and cos = av Should be known from standard grade then u = logasin and v = logacos so u – v = logasin - logacos sin  cos  or u – v = loga(sin/cos) e.g. tan = logx – logy = log(x/y) tan = au av hence u – v = logatan Next Comment Logs & Exp Menu Back to Home

Take logs to base a to both sides Markers Comments Take logs to base a to both sides Since sin = au and cos = av loga tan = loga au av then u = logasin and v = logacos Law2: log =logA-logB A B so u – v = logasin - logacos loga tan = loga au - loga av or u – v = loga(sin/cos) Law 3: log = nlogx xn logx – logy = log(x/y) loga tan = uloga a - vloga a hence u – v = logatan logaa=1 loga tan = u - v Next Comment Logs & Exp Menu Back to Home

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

VECTORS: Question 1 P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively. Prove that these points are collinear. (b) Given that PS = 7  PQ then find the coordinates of S. Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT Go to Main Menu

VECTORS: Question 1 P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively. Prove that these points are collinear. (b) Given that PS = 7  PQ then find the coordinates of S. (a) PQ and  PR are  Since multiples of the same Reveal answer only vector and have P as a common point then Go to full solution it follows that P, Q & R are collinear. Go to Marker’s Comments Go to Vectors Menu (b) S = (-25, 7, 16) Go to Main Menu EXIT

[ ] [ ] [ ] [ ] [ ] - - Question 1 PQ = q - p =  (a) [ ] -1 1 -2 3 - PQ = q - p =  (a) P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively. Prove that these points are collinear. (b) Given that [ ] -4 1 3 = [ ] -13 4 7 3 -2 - PR = r - p =  PS = 7  PQ then find [ ] -16 4 9 [ ] -4 1 3 = 4 = the coordinates of S. PQ and  PR are  Since multiples of the same Begin Solution vector and have P as a common point then Continue Solution it follows that P, Q & R are collinear. Markers Comments Vectors Menu Back to Home

[ ] [ ] [ ] Question 1 = 7 = (b) PS = 7  PQ [ ] -4 1 3 = 7 [ ] -28 7 21 = (b) PS = 7  PQ P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively. Prove that these points are collinear. (b) Given that S is (3,0,-5) + [ ] -28 7 21 = (3-28, 0+7, -5+21) = (-25, 7, 16) PS = 7  PQ then find the coordinates of S. Begin Solution Continue Solution Markers Comments Vectors Menu Back to Home

[ ] [ ] [ ] [ ] [ ] - - a) Must know result: Markers Comments a) [ ] -1 1 -2 3 - PQ = q - p =  (a) Must know result: Given coordinates of A and B [ ] -4 1 3 = [ ] -13 4 7 3 -2 - PR = r - p =  [ ] -16 4 9 [ ] -4 1 3 = 4 = PQ and  PR are  Since multiples of the same Next Comment vector and have P as a common point then Vectors Menu it follows that P, Q & R are collinear. Back to Home

[ ] [ ] [ ] [ ] [ ] - - Must be able to state explicitly Markers Comments Must be able to state explicitly the result for collinearity: [ ] -1 1 -2 3 - PQ = q - p =  (a) [ ] -4 1 3 = [ ] -13 4 7 3 -2 - PR = r - p =  [ ] -16 4 9 [ ] -4 1 3 = 4 = PQ and  PR are  Since multiples of the same Next Comment vector and have P as a common point then Vectors Menu it follows that P, Q & R are collinear. Back to Home

[ ] [ ] [ ] b) An alternative approach is to Markers Comments b) An alternative approach is to form a vector equation and solve it: [ ] -4 1 3 = 7 [ ] -28 7 21 = (b) PS = 7  PQ S is (3,0,-5) + [ ] -28 7 21 = (3-28, 0+7, -5+21) = (-25, 7, 16) Next Comment Vectors Menu Back to Home

[ ] [ ] [ ] b) Markers Comments = 7 = (b) PS = 7  PQ S is (3,0,-5) + [ ] -4 1 3 = 7 [ ] -28 7 21 = (b) PS = 7  PQ S is (3,0,-5) + [ ] -28 7 21 = (3-28, 0+7, -5+21) = (-25, 7, 16) Next Comment Vectors Menu Back to Home

VECTORS: Question 2 E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively. Find FE and  FG in component form. (b) Hence, or otherwise, find the size of angle EFG. Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT Go to Main Menu

[ ] [ ] VECTORS: Question 2 E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively. Find FE and  FG in component form. (b) Hence, or otherwise, find the size of angle EFG. FE  (a) [ ] -1 4 -3 = [ ] 3 -3 FG  Reveal answer only = Go to full solution so  = 73.9° Go to Marker’s Comments Go to Vectors Menu EXIT Go to Main Menu

[ ] [ ] [ ] [ ] - - Question 2 FE = e - f =  (a) [ ] 4 3 -1 5 2 - FE = e - f =  (a) E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively. Find [ ] -1 4 -3 = FE and  FG in  component form. [ ] 8 -1 5 2 - FG = g - f =  Hence, or otherwise, find the size of angle EFG. [ ] 3 -3 = Begin Solution Continue Solution Markers Comments Vectors Menu Back to Home

[ ] Question 2 . (b) Let angle EFG =  E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively. Find  E F G ie FE and  FG in  component form. [ ] -1 4 -3 3 . FE .  FG = Hence, or otherwise, find the size of angle EFG. = (-1 X 3) + (4 X 0) + (-3 X (-3)) = -3 + 0 + 9 = 6 Begin Solution |FE| = ((-1)2 + 42 + (-3)2)  = 26 Continue Solution |FG| = (32 + 02 + (-3)2)  Markers Comments = 18 Vectors Menu Back to Home

Question 2 Given that FE.FG = |FE||FG|cos  E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively. Find then cos = FE.FG  |FE||FG| = 6 26 18 FE and  FG in  so  = cos-1(6  26  18) = 73.9° component form. Hence, or otherwise, find the size of angle EFG. Begin Solution Continue Solution Markers Comments Vectors Menu Back to Home

[ ] Ensure vectors are calculated from the vertex: E F G . Markers Comments Ensure vectors are calculated from the vertex: (b) Let angle EFG =   E F G ie F G E [ ] -1 4 -3 3 . FE .  FG = = (-1 X 3) + (4 X 0) + (-3 X (-3)) = -3 + 0 + 9 = 6 |FE| = ((-1)2 + 42 + (-3)2)  Next Comment = 26 |FG| = (32 + 02 + (-3)2)  Vectors Menu = 18 Back to Home

Refer to formula sheet and relate formula to given variables: Markers Comments Refer to formula sheet and relate formula to given variables: Given that FE.FG = |FE||FG|cos  then cos = FE.FG  |FE||FG| so  = cos-1(6  26  18) = 73.9° Next Comment Vectors Menu Back to Home

[ ], [ ]and [ ]resp. VECTORS: Question 3 PQRSTUVW is a cuboid in which PQ , PS & PW are represented by the vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1. Find the components of PA & PB and hence the size of angle APB.  Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT

[ ], [ ]and [ ]resp. VECTORS: Question 3 PQRSTUVW is a cuboid in which PQ , PS & PW are represented by the vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1. Find the components of PA & PB and hence the size of angle APB.   Reveal answer only |PA| = 29 Go to full solution  |PB| = 106 Go to Marker’s Comments = 48.1° APB Go to Vectors Menu EXIT

[ ], [ ]and [ ]resp. [ ] [ ] = [ ] [ ] [ ]+ [ ]= [ ] + + Question 3 PA =  PS + SA =  PS + 1/3ST  PQRSTUVW is a cuboid in which PQ , PS & PW are represented by vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 PS + 1/3PW  = [ ] -2 4 [ ] = 3 + [ ] -2 4 3 = A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1. PB =  PQ + QV + VB    Find the components of PA & PB and hence the size of angle APB. PQ + PW + 1/2PS  = [ ] 4 2 [ ]+ 9 + [ ]= -1 [ ] 3 4 9 Begin Solution = Continue Solution Markers Comments Vectors Menu Back to Home

[ ] [ ] [ ], [ ]and [ ]resp. [ ] Question 3 . PA =  PB =  [ ] -2 4 3 [ ] 3 4 9 PB =  PQRSTUVW is a cuboid in which PQ , PS & PW are represented by vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 (b) Let angle APB =   A P B ie A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1.   Find the components of PA & PB and hence the size of angle APB. PA .  PB = [ ] -2 4 3 9 . Begin Solution = (-2 X 3) + (4 X 4) + (3 X 9) Continue Solution = -6 + 16 + 27 Markers Comments = 37 Vectors Menu Back to Home

[ ], [ ]and [ ]resp. Question 3 PA .  PB = 37 PQRSTUVW is a cuboid in which PQ , PS & PW are represented by vectors    [ ], 4 2 [ ]and -2 [ ]resp. 9 |PA| = ((-2)2 + 42 + 32)  = 29 |PB| = (32 + 42 + 92)  = 106 Given that PA.PB = |PA||PB|cos  A is 1/3 of the way up ST & B is the midpoint of UV. ie SA:AT = 1:2 & VB:BU = 1:1. then cos = PA.PB  |PA||PB| = 37 29 106   Find the components of PA & PB and hence the size of angle APB. so  = cos-1(37  29  106) Begin Solution Continue Solution = 48.1° Markers Comments Vectors Menu Back to Home

[ ] [ ] [ ] Ensure vectors are calculated from the vertex: B P A . Markers Comments Ensure vectors are calculated from the vertex: PA =  [ ] -2 4 3 [ ] 3 4 9 PB =  P A B (b) Let angle APB =   A P B ie PA .  PB = [ ] -2 4 3 9 . Next Comment = (-2 X 3) + (4 X 4) + (3 X 9) = -6 + 16 + 27 Vectors Menu = 37 Back to Home

Refer to formula sheet and relate formula to given variables: Markers Comments Refer to formula sheet and relate formula to given variables: PA .  PB = |PA| = ((-2)2 + 42 + 32)  = 29 |PB| = (32 + 42 + 92)  = 106 Given that PA.PB = |PA||PB|cos  then cos = PA.PB  |PA||PB| so  = cos-1(37  29  106) Next Comment Vectors Menu = 48.1° Back to Home

VECTORS: Question 4 b c a An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c as shown. Find b.(a + c) & comment on your answer (ii) b.(a – c) Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT Go to Main Menu

VECTORS: Question 4 b c a An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c as shown. Find b.(a + c) & comment on your answer (ii) b.(a – c) Dot product = 0 so b & (a + c) are perpendicular. (ii) b.(a – c) = b.b = 42 = 16 Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT Go to Main Menu

Question 4 NB: each angle is 60° but vectors should be “tail to tail” An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c . b c 60° 120° Find b.(a + c) & comment b.(a – c) (i) b.(a + c) = b.a + b.c = |b||a|cos1 + |b||c|cos2 = 4X4Xcos60° + 4X4Xcos120° = 16 X ½ + 16 X (-½ ) Begin Solution = 8 + (-8) Continue Solution = 0 Markers Comments Dot product = 0 so b & (a + c) are perpendicular. Vectors Menu Back to Home

Question 4 NB: a – c = a + (-c) or b . An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c . (ii) b.(a – c) = b.b = |b||b|cos0 = |b|2 Find b.(a + c) & comment b.(a – c) = 42 = 16 Begin Solution Continue Solution Markers Comments Vectors Menu Back to Home

This geometric question is based on the scalar product definition Markers Comments This geometric question is based on the scalar product definition and the distributive law: NB: each angle is 60° but vectors should be “tail to tail” b c 60° 120° and a(b+c) = ab + ac (i) b.(a + c) = b.a + b.c No other results should be applied. = |b||a|cos1 + |b||c|cos2 = 4X4Xcos60° + 4X4Xcos120° = 16 X ½ + 16 X (-½ ) = 8 + (-8) = 0 Next Comment Dot product = 0 so b & (a + c) are perpendicular. Vectors Menu Back to Home

Ensure all scalar products are calculated from the vertex: Markers Comments Ensure all scalar products are calculated from the vertex: NB: each angle is 60° but vectors should be “tail to tail” b c 60° 120° (i) b.(a + c) = b.a + b.c = |b||a|cos1 + |b||c|cos2 = 4X4Xcos60° + 4X4Xcos120° = 16 X ½ + 16 X (-½ ) = 8 + (-8) = 0 Next Comment Dot product = 0 so b & (a + c) are perpendicular. Vectors Menu Back to Home

VECTORS: Question 5 a = 2i + j – 3k, b = -i + 10k & c = -2i + j + k. (a) Find (i) 2a + b in component form (ii) | 2a + b | (b) Show that 2a + b and c are perpendicular. Reveal answer only Go to full solution Go to Marker’s Comments Go to Vectors Menu EXIT Go to Main Menu

[ ] VECTORS: Question 5 = 29 a = 2i + j – 3k, b = -i + 10k & c = -2i + j + k. (a) Find (i) 2a + b in component form (ii) | 2a + b | (b) Show that 2a + b and c are perpendicular. [ ] 3 2 4 = (a)(i) 2a + b Reveal answer only Go to full solution (ii) | 2a + b | = 29 Go to Marker’s Comments (b) Go to Vectors Menu Since the dot product is zero it follows that the vectors are perpendicular. Go to Main Menu EXIT

[ ] [ ] [ ] Question 5 + (a)(i) 2a + b = a = 2i + j – 3k, b = -i + 10k [ ] 2 1 -3 -1 10 + (a)(i) 2a + b = a = 2i + j – 3k, b = -i + 10k & c = -2i + j + k. [ ] 4 2 -6 -1 10 + = (a) Find 2a + b in component form | 2a + b | [ ] 3 2 4 = Show that 2a + b and c are perpendicular. Begin Solution Continue Solution Markers Comments Vectors Menu Back to Home

Question 5 = 29 (ii) | 2a + b | = (32 + 22 + 42) a = 2i + j – 3k, b = -i + 10k & c = -2i + j + k. = 29 (a) Find 2a + b in component form | 2a + b | Show that 2a + b and c are perpendicular. Begin Solution Continue Solution Markers Comments Vectors Menu Back to Home

[ ] [ ] . Question 5 (b) (2a + b ).c = a = 2i + j – 3k, b = -i + 10k 4 [ ] -2 1 . (b) (2a + b ).c = a = 2i + j – 3k, b = -i + 10k & c = -2i + j + k. = (3 X (-2))+(2 X 1)+(4 X 1) (a) Find 2a + b in component form | 2a + b | = -6 + 2 + 4 = 0 Since the dot product is zero it follows that the vectors are perpendicular. Show that 2a + b and c are perpendicular. Begin Solution Continue Solution Markers Comments Vectors Menu Back to Home

[ ] [ ] [ ] When a vector is given in i , j , k form change to Markers Comments When a vector is given in i , j , k form change to component form: e.g. 5i - 2j + k = [ ] 2 1 -3 -1 10 + (a)(i) 2a + b = [ ] 4 2 -6 -1 10 + = [ ] 3 2 4 = Next Comment Vectors Menu Back to Home

Must know formula for the magnitude of a vector: Markers Comments Must know formula for the magnitude of a vector: (ii) | 2a + b | = (32 + 22 + 42) = 29 Next Comment Vectors Menu Back to Home

[ ] [ ] . Must know condition for perpendicular vectors: Markers Comments Must know condition for perpendicular vectors: [ ] 3 2 4 [ ] -2 1 . (b) (2a + b ).c = = (3 X (-2))+(2 X 1)+(4 X 1) = -6 + 2 + 4 = 0 Since the dot product is zero it follows that the vectors are perpendicular. Next Comment Vectors Menu Back to Home

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

FURTHER CALCULUS : Question 1 Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx . Reveal answer only Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

FURTHER CALCULUS : Question 1 Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx . = 6cos(2x) + 2sin(4x) Reveal answer only Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

Question 1 Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx . OUTER / INNER Differentiate outer then inner Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx . y = 3sin(2x) – 1/2cos(4x) dy/dx = 3cos(2x) X 2 - (-1/2sin(4x)) X 4 = 6cos(2x) + 2sin(4x) Begin Solution Continue Solution Markers Comments Further Calc Menu Back to Home

Check formula sheet for correct result: Markers Comments Check formula sheet for correct result: 3sin(2x) 1/2cos(4x) OUTER / INNER Differentiate outer then inner y sin(ax) acos(ax) cos(ax) -asin(ax) y = 3sin(2x) – 1/2cos(4x) Relate formula to given variables dy/dx = 3cos(2x) X 2 - (-1/2sin(4x)) X 4 = 6cos(2x) + 2sin(4x) Next Comment Further Calc Menu Back to Home

When applying the “chain rule” “Peel an onion” Markers Comments When applying the “chain rule” “Peel an onion” 3sin(2x) 1/2cos(4x) OUTER / INNER Differentiate outer then inner 3sin(2x) 1/2cos(4x) OUTER / INNER Differentiate outer then inner y = 3sin(2x) – 1/2cos(4x) dy/dx = 3cos(2x) X 2 - (-1/2sin(4x)) X 4 = 6cos(2x) + 2sin(4x) e.g. 3sin - 3cos 2x 2 Next Comment Further Calc Menu Back to Home

FURTHER CALCULUS : Question 2 Given that g(x) = (6x – 5) then evaluate g´(9) . Reveal answer only Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

FURTHER CALCULUS : Question 2 Given that g(x) = (6x – 5) then evaluate g´(9) . = 3/7 Reveal answer only Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

Question 2 Given that g(x) = (6x – 5) then evaluate g´(9) . g(x) = (6x – 5) = (6x – 5)1/2 Given that g(x) = (6x – 5) then evaluate g´(9) . (6x – 5)1/2 outer / inner diff outer then inner g´(x) = 1/2(6x – 5)-1/2 X 6 = 3(6x – 5)-1/2 = 3 (6x – 5) g´(9) = 3 (6X9 – 5) Begin Solution Continue Solution = 3 49 Markers Comments Further Calc Menu = 3/7 Back to Home

Apply the laws of indices to replace the sign Markers Comments Apply the laws of indices to replace the sign g(x) = (6x – 5) = (6x – 5)1/2 (6x – 5)1/2 outer / inner diff outer then inner Apply the chain rule Learn the rule for differentiation Multiply by the power then reduce the power by 1 g´(x) = 1/2(6x – 5)-1/2 X 6 = 3(6x – 5)-1/2 = 3 (6x – 5) g´(9) = 3 (6X9 – 5) Next Comment = 3 49 Further Calc Menu = 3/7 Back to Home

Apply the laws of indices to return power to a positive value Markers Comments Apply the laws of indices to return power to a positive value then the root g(x) = (6x – 5) = (6x – 5)1/2 (6x – 5)1/2 outer / inner diff outer then inner g´(x) = 1/2(6x – 5)-1/2 X 6 Will usually work out to an exact value without need to use calculator. = 3(6x – 5)-1/2 = 3 (6x – 5) g´(9) = 3 (6X9 – 5) Next Comment = 3 49 Further Calc Menu = 3/7 Back to Home

FURTHER CALCULUS : Question 3 A curve for which dy/dx = -12sin(3x) passes through the point (/3,-2). Express y in terms of x. Reveal answer only Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

FURTHER CALCULUS : Question 3 A curve for which dy/dx = -12sin(3x) passes through the point (/3,-2). Express y in terms of x. Reveal answer only So y = 4cos(3x) + 2 Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

through the point (/3,-2). Express y in terms of x. Question 3 if dy/dx = -12sin(3x) A curve for which dy/dx = -12sin(3x) passes through the point (/3,-2). Express y in terms of x. then y =  -12sin(3x) dx = 1/3 X 12cos(3x) + C = 4cos(3x) + C At the point (/3,-2) y = 4cos(3x) + C becomes -2 = 4cos(3 X /3) + C or -2 = 4cos + C or -2 = -4 + C ie C = 2 Begin Solution Continue Solution So y = 4cos(3x) + 2 Markers Comments Further Calc Menu Back to Home

Learn the result that integration undoes differentiation: i.e. given Markers Comments Learn the result that integration undoes differentiation: i.e. given if dy/dx = -12sin(3x) then y =  -12sin(3x) dx = 1/3 X 12cos(3x) + C = 4cos(3x) + C At the point (/3,-2) y = 4cos(3x) + C becomes -2 = 4cos(3 X /3) + C or -2 = 4cos + C or -2 = -4 + C ie C = 2 Next Comment So y = 4cos(3x) + 2 Further Calc Menu Back to Home

Check formula sheet for correct result: if dy/dx = -12sin(3x) Markers Comments Check formula sheet for correct result: if dy/dx = -12sin(3x) then y =  -12sin(3x) dx sinax -cos(ax) a = 1/3 X 12cos(3x) + C + c = 4cos(3x) + C cos(ax) sin(ax) a + c At the point (/3,-2) y = 4cos(3x) + C becomes -2 = 4cos(3 X /3) + C or -2 = 4cos + C Do not forget the constant of integration. or -2 = -4 + C ie C = 2 Next Comment So y = 4cos(3x) + 2 Further Calc Menu Back to Home

 FURTHER CALCULUS : Question 4 Find the derivative of y = (2x3 + 1)2/3 where x > 0. Hence find x2 (2x3 + 1)1/3 dx  Reveal answer only Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

 FURTHER CALCULUS : Question 4 = 4x2 Find the derivative of y = (2x3 + 1)2/3 where x > 0. Hence find x2 (2x3 + 1)1/3 dx  = 1/4(2x3 + 1)2/3 + C Reveal answer only Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu EXIT

Question 4 (a) if y = (2x3 + 1)2/3 (2x3 + 1)2/3 outer inner diff outer then inner Find the derivative of y = (2x3 + 1)2/3 where x > 0. then dy/dx = 2/3 (2x3 + 1)-1/3 X 6x2 = 4x2 (2x3 + 1)1/3 Begin Solution Continue Solution Markers Comments Further Calc Menu Back to Home

    Question 4 = ¼ X (b) From (a) it follows that (b) Hence find dx  = (2x3 + 1)2/3 + C x2 (2x3 + 1)1/3 dx  now x2 (2x3 + 1)1/3 dx  4x2 (2x3 + 1)1/3 dx  = ¼ X Begin Solution = 1/4(2x3 + 1)2/3 + C Continue Solution Markers Comments Further Calc Menu Back to Home

Markers Comments . Apply the chain rule “Peel an onion” (a) if y = (2x3 + 1)2/3 (2x3 + 1)2/3 outer inner diff outer then inner (2x3 + 1)2/3 outer inner diff outer then inner then dy/dx = 2/3 (2x3 + 1)-1/3 X 6x2 = 4x2 (2x3 + 1)1/3 e.g. Next Comment Further Calc Menu Back to Home

   Learn the result that integration undoes differentiation: Markers Comments b) Learn the result that integration undoes differentiation: i.e. given (b) From (a) it follows that 4x2 (2x3 + 1)1/3 dx  = (2x3 + 1)2/3 + C now x2 (2x3 + 1)1/3 dx  4x2 (2x3 + 1)1/3 dx  = ¼ X = 1/4(2x3 + 1)2/3 + C Next Comment Further Calc Menu Back to Home