HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs.

HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: 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

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

HIGHER – ADDITIONAL QUESTION BANK UNIT 1 : Straight Line You have chosen to study: Please choose a question to attempt from the following: 12345 EXIT Back to Unit 1 Menu

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

Markers Comments Begin Solution Continue Solution 3x – 5y = 4 3x - 4 = 5y 5y = 3x - 4 (5)(5) y = 3 / 5 x - 4 / 5 Using y = mx + c, gradient of line is 3 / 5 So required gradient = -5 / 3, ( m 1 m 2 = -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) y – 4 = -5 / 3 x - 10 y = -5 / 3 x - 6 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). Straight Line Menu Back to Home

3x – 5y = 4 3x - 4 = 5y 5y = 3x - 4 (5)(5) y = 3 / 5 x - 4 / 5 Using y = mx + c, gradient of line is 3 / 5 So required gradient = -5 / 3, ( m 1 m 2 = -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) y – 4 = -5 / 3 x - 10 y = -5 / 3 x - 6 Markers Comments Straight Line Menu Back to Home An attempt must be made to put the original equation into the form y = mx + c to read off the gradient. State the gradient clearly. State the condition for perpendicular lines m 1 m 2 = -1. When finding m 2 simply invert and change the sign on m 1 m 1 = 3535 m 2 = -5 3 Use the y - b = m(x - a) form to obtain the equation of the line. Next Comment

Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu Reveal answer only 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).

Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu Reveal answer only 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

Markers Comments Begin Solution Continue Solution Question 2 Straight Line Menu Back to Home 8x + 4y – 7 = 0 4y = -8x + 7 (4)(4) y = -2x + 7 / 4 y = -2x + 7 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) y + 3 = -2x + 10 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).

Markers Comments Straight Line Menu Back to Home Next Comment An attempt must be made to put the original equation into the form y = mx + c to read off the gradient. State the gradient clearly. State the condition for parallel lines m 1 = m 2 Use the y - b = m(x - a) form to obtain the equation of the line. 8x + 4y – 7 = 0 4y = -8x + 7 (4)(4) y = -2x + 7 / 4 y = -2x + 7 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) y + 3 = -2x + 10

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

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

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

Markers Comments Straight Line Menu Back to Home Next Comment (a)Using the gradient formula: m AC = 3 – 0 7 - 2 m BC = 3 – 0 7 - 8 = - 3 (b) Using tan  = gradient = 3 / 5 If tan  = 3 / 5 then CAB = 31.0° then CBX = (180-71.6)° = 108.4 o Hence : ACB = 180° – 31.0° – 71.6° = 77.4° If tan  = -3 If no diagram is given draw a neat labelled diagram. In calculating gradients state the gradient formula. 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. A B Ø ° m AB = tanØ ° Ø ° = tan -1 m AB so ABC = 71.6°

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

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

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

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

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

Markers Comments Straight Line Menu Back to Home Next Comment If no diagram is given draw a neat labelled diagram. Q P R y x median Perpendicular bisector (a)Midpoint of PQ is (3,-1): let’s call this S Using the gradient formula m = y 2 – y 1 x 2 – x 1 m SR = -1 – (-1) 10 - 3 Since it passes through (3,-1) equation of e is y = -1 (ie line is horizontal) Comments for 4 (b) Sketch the median and the perpendicular bisector

Markers Comments Straight Line Menu Back to Home Next Comment Q P R y x (b) Midpoint of QR is (6,1) m QR = 3 – (-1) 2 - 10 = 4 / -8 required gradient = 2 (m 1 m 2 = -1) Using y – b = m(x – a) with (a,b) = (6,1) & m = 2 we get y – 1 = 2(x – 6) so f is y = 2x – 11 = - 1 / 2 To find midpoint of QR 2 + 103 + (-1) 2, Look for special cases: Horizontal lines in the form y = k Vertical lines in the form x = k Comments for 4 (c)

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

STRAIGHT LINE : Question 5 Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu Reveal answer only EXIT 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. X Y G(2,-5) E(6,-3) F(12,-5)

STRAIGHT LINE : Question 5 Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu Reveal answer only EXIT 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. X Y G(2,-5) E(6,-3) F(12,-5) x = 6 (a) x + 8y + 28 = 0 (b) (6,-4.25) (c)

Markers Comments Begin Solution Continue Solution Question 5(a) Straight Line Menu Back to Home 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. X Y G(2,-5) E(6,-3) F(12,-5) (a)Using the gradient formula m FG = -5 – (-5) 12 - 2 = 0 (ie line is horizontal so altitude is vertical) Altitude is vertical line through (6,-3) ie x = 6 Solution to 5 (b)

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

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

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

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

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

HIGHER – ADDITIONAL QUESTION BANK UNIT 1 : Basic Differentiation You have chosen to study: Please choose a question to attempt from the following: 12345 EXIT Back to Unit 1 Menu

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

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

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

Markers Comments Begin Solution Continue Solution Question 1 Basic Differentiation Menu Back to Home Find the equation of the tangent to the curve y =  x – 16 x (x>0) at the point where x = 4. If x = 4 then: 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 usy – (-2) = 5 / 4 (x – 4) or y + 2 = 5 / 4 x – 5 or y = 5 / 4 x – 7 Back to Previous

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

Markers Comments Differentiation Menu Back to Home Next Comment If x = 4 then: = 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 usy – (-2) = 5 / 4 (x – 4) or y + 2 = 5 / 4 x – 5 dy / dx or y = 5 / 4 x – 7 Find y coordinate at x = 4 using: Use m = 5/4 and (4,-2) in y - b = m(x - a)

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

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

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

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

Markers Comments Differentiation Menu Back to Home Next Comment NB: gradient of line = gradient of curve Line Using gradient = tan  we get gradient of line = tan135° = -tan45° = -1 Curve Gradient of curve = dy / dx = 2x - 5 It now follows that 2x – 5 = -1 Or 2x = 4 Or x = 2 Find gradient of the tangent using rule: “multiply by the power and reduce the power by 1” 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. m = tan 135 ° = -1 y x 135 ° Continue Comments

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

BASIC DIFFERENTIATION : Question 3 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu Reveal answer only EXIT 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. Make a sketch of the graph of y = g(x). y = g(x) (p,q) r

BASIC DIFFERENTIATION : Question 3 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu Reveal answer only EXIT 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. Make a sketch of the graph of y = g(x). y = g(x) (p,q) r y = g(x)

Markers Comments Begin Solution Continue Solution Question 3 Basic Differentiation Menu Back to Home 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: new y-values x  0  p  g(x) - 0 - 0 + Click for graph y = g(x) (p,q) r Make a sketch of the graph of y = g(x).

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

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

Markers Comments Differentiation Menu Back to Home Next Comment 2 For each interval decide if the value of To sketch the graph of the gradient function: 1 Mark the stationary points on the x axis i.e. 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: new y-values x  0  p  g(x) - 0 - 0 + Continue Comments + -- x y a

Markers Comments Differentiation Menu Back to Home Next Comment To sketch the graph of the gradient function: 1 Mark the stationary points on the x axis i.e. 0 p y = g(x) Stationary points occur at x = 0 and x = p. 3 Draw in curve to fit information 2 For each interval decide if the value of In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates. + -- x y a

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

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

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

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

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

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

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

BASIC DIFFERENTIATION : Question 5 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu Reveal answer only EXIT 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 x 2 (x  2) So after 5 months the share isS(5) = 2 / 5 – 4 / 25 = 6 / 25 Find the maximum share of the market that the company can achieve.

BASIC DIFFERENTIATION : Question 5 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu Reveal answer only EXIT 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 x 2 (x  2) So after 5 months the share isS(5) = 2 / 5 – 4 / 25 = 6 / 25 Find the maximum share of the market that the company can achieve. = 1 / 4

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

Markers Comments Continue Solution Question 5 Basic Differentiation Menu Back to Home S(x) = 2 - 4 x x 2 (x  2) Find the maximum share of the market that the company can achieve. When a company launches a new product its share of the market after x months is calculated as: SPs occur where S (x) = 0 8 - 2 x 3 x 2 or 8 = 2 x 3 x 2 ( cross mult!) 8x 2 = 2x 3 8x 2 - 2x 3 = 0 2x 2 (4 – x) = 0 x = 0 or x = 4 NB: x  2 In required interval Continue Solution Go Back to Previous = 0

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

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

Markers Comments Differentiation Menu Back to Home Next Comment Must attempt to find ( Note: No marks are allocated for trial and error solution.) SPs occur where S (x) = 0 8 - 2 x 3 x 2 or 8 = 2 x 3 x 2 ( cross mult!) 8x 2 = 2x 3 = 0 8x 2 - 2x 3 = 0 2x 2 (4 – x) = 0 x = 0 or x = 4 NB: x  2 In required interval “multiply by the power and reduce the power by 1” Must attempt to find and set equal to zero Usually easier to solve resulting equation using cross-multiplication. Take care to reject “solutions” outwith the domain. Continue Comments

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

HIGHER – ADDITIONAL QUESTION BANK UNIT 1 : Recurrence Relations You have chosen to study: Please choose a question to attempt from the following: 123 EXIT Back to Unit 1 Menu

RECURRENCE RELATIONS : Question 1 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu Reveal answer only EXIT A recurrence relation is defined by the formula u n+1 = au n + b, where -1<a<1 and u 0 = 20. (a)If u 1 = 10 and u 2 = 4 then find the values of a and b. (b)Find the limit of this recurrence relation as n  .

RECURRENCE RELATIONS : Question 1 Go to full solution Go to Marker’s Comments Go to Basic Differentiation Menu Go to Main Menu Reveal answer only EXIT A recurrence relation is defined by the formula u n+1 = au n + b, where -1<a<1 and u 0 = 20. (a)If u 1 = 10 and u 2 = 4 then find the values of a and b. (b)Find the limit of this recurrence relation as n  . a = 0.6 b = -2 (a) L = -5 (b)

Markers Comments Begin Solution Continue Solution Question 1 Recurrence Relations Menu Back to Home A recurrence relation is defined by the formula u n+1 = au n + b, where -1<a<1 and u 0 = 20. (a)If u 1 = 10 and u 2 = 4 then find the values of a and b. (b) Find the limit of this recurrence relation as n  . Using u n+1 = au n + b we get u 1 = au 0 + b and u 2 = au 1 + b Replacing u 0 by 20, u 1 by 10 & u 2 by 4 gives us20a + b = 10 and10a + b = 4 subtract  10a = 6 or (a) Replacing a by 0.6 in 10a + b = 4 gives 6 + b = 4 or b = -2 a = 0.6 Continue Solution

Markers Comments Begin Solution Continue Solution Question 1 Recurrence Relations Menu Back to Home A recurrence relation is defined by the formula u n+1 = au n + b, where -1<a<1 and u 0 = 20. (a)If u 1 = 10 and u 2 = 4 then find the values of a and b. (b) Find the limit of this recurrence relation as n  . (b) L = -5 u n+1 = au n + b is now u n+1 = 0.6u n - 2 This has a limit since -1<0.6<1 At this limit, L, u n+1 = u n = L So we now have L = 0.6 L - 2 or 0.4L = -2 or L = -2  0.4 or L = -20  4 so

Markers Comments Recurrence Menu Back to Home Next Comment Using u n+1 = au n + b we get u 1 = au 0 + b and u 2 = au 1 + b Replacing u 0 by 20, u 1 by 10 & u 2 by 4 gives us20a + b = 10 and10a + b = 4 subtract  10a = 6 or (a) Replacing a by 0.6 in 10a + b = 4 gives 6 + b = 4 or b = -2 a = 0.6 Must form the two simultaneous equations and solve. u 1 is obtained from u 0 and u 2 is obtained from u 1. A trial and error solution would only score 1 mark. Comments for 1(b)

Markers Comments Recurrence Menu Back to Home Next Comment (b) u n+1 = au n + b is now u n+1 = 0.6u n - 2 This has a limit since -1<0.6<1 At this limit, L, u n+1 = u n = 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 Must state condition for limit i.e. -1 < 0.6 < 1 At limit L, state u n+1 = u n = L Substitute L for u n+1 and u n and solve for L.

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

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

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

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

Markers Comments Begin Solution Continue Solution Question 1 Recurrence Relations Menu Back to Home Two different recurrence relations are known to have the same limit as n  . The first is defined by the formula u n+1 = -5ku n + 3. The second is defined by Vn+1 = k 2 v n + 1. Find the value of k and hence this limit. Since -1<k<1 then k = 1 / 3 UsingL = 1.. (1 - k 2 ) gives us L =. 1.. (1 – 1 / 9 ) or L = 1  8 / 9 ie L = 9 / 8

Markers Comments Recurrence Menu Back to Home Next Comment If the limit is L then as n   we have u n+1 = u n = L and v n+1 = v n = L First Sequence u n+1 = -5ku n + 3 becomes L + 5kL = 3 L(1 + 5k) = 3 L = 3.. (1 + 5k) L = -5kL + 3 Second Sequence v n+1 = k 2 v n + 1 becomes L = k 2 L + 1 L - k 2 L = 1 L(1 - k 2 ) = 1 L = 1.. (1 - k 2 ) Since both recurrence relations have the same limit, L, find the limit for both and set equal. Continue Comments

Markers Comments Recurrence Menu Back to Home Next Comment L = 1.. (1 - k 2 ) It follows that L = 3.. (1 + 5k).3.=. 1.. (1 + 5k) (1 – k 2 ) Cross multiply to get 1 + 5k = 3 – 3k 2 This becomes 3k 2 + 5k – 2 = 0 Or (3k – 1)(k + 2) = 0 So k = 1 / 3 or k = -2 Since -1<k<1 then k = 1 / 3 Since both recurrence relations have the same limit, L, find the limit for both and set equal. 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. Continue Solution

Markers Comments Recurrence Menu Back to Home Next Comment Since -1<k<1 then k = 1 / 3 UsingL = 1.. (1 - k 2 ) gives us L =. 1.. (1 – 1 / 9 ) or L = 1  8 / 9 ie L = 9 / 8 Find L from either formula.

RECURRENCE RELATIONS : Question 3 Go to full solution Go to Marker’s Comments Go to Recurrence Relations Menu Reveal answer only EXIT 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. (a)Using the 30% pruning scheme what height should he expect the trees to grow to in the long run? (b)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?

RECURRENCE RELATIONS : Question 3 Go to full solution Go to Marker’s Comments Go to Recurrence Relations Menu Reveal answer only EXIT 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. (a)Using the 30% pruning scheme what height should he expect the trees to grow to in the long run? (b)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? Height of trees in long run is 3 1 / 3 m. 33 1 / 3 % (b)

Markers Comments Begin Solution Continue Solution Question 3 Recurrence Relation Menu Back to Home 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. (a)Using the 30% pruning scheme what height should he expect the trees to grow to in the long run? (a) Removing 30% leaves 70% or 0.7 If H n is the tree height in year n then H n+1 = 0.7H n + 1 Since -1<0.7<1 this sequence has a limit, L. At the limit H n+1 = H n = L SoL= 0.7L + 1 or 0.3 L = 1 ieL = 1  0.3= 10  3 = 3 1 / 3 Height of trees in long run is 3 1 / 3 m. Continue Solution

Markers Comments Begin Solution Continue Solution Question 3 Recurrence Relation Menu Back to Home 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. (b)If fraction left after pruning is a and we need the limit to be 3 then we have3 = a X 3 + 1 or3a = 2 or a = 2 / 3 This means that the fraction pruned is 1 / 3 or 33 1 / 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?

Markers Comments Recurrence Menu Back to Home Next Comment (a) Removing 30% leaves 70% or 0.7 If H n is the tree height in year n then H n+1 = 0.7H n + 1 Since -1<0.7<1 this sequence has a limit, L. At the limit H n+1 = H n = L SoL= 0.7L + 1 or 0.3 L = 1 ieL = 1  0.3= 10  3 Height of trees in long run is 3 1 / 3 m. Do some numerical work to get the “feel” for the problem. H 0 = 1(any value acceptable) H 1 = 0.7 x1 + 1 = 1.7 H 2 = 0.7 x1.7 + 1 = 2.19 etc. State the recurrence relation, with the starting value. H n+1 = 0.7 H n + 1, H 0 = 1 State the condition for the limit -1< 0.7< 1 At limit L, state H n+1 = H n = L Substitute L for H n+1 and H n and solve for L. Continue Solution

Markers Comments Recurrence Menu Back to Home Next Comment (b)If fraction left after pruning is a and we need the limit to be 3 then we have3 = a X 3 + 1 or3a = 2 or a = 2 / 3 This means that the fraction pruned is 1 / 3 or 33 1 / 3 % 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.

HIGHER – ADDITIONAL QUESTION BANK UNIT 1 : Trig Graphs & Equations You have chosen to study: Please choose a question to attempt from the following: 123 EXIT Back to Unit 1 Menu

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

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

Markers Comments Begin Solution Continue Solution Question 1 Trig Graphs etc. Menu Back to Home This diagram shows the graph of y = acosbx + c. Determine the values of a, b & c. /2/2  y = acosbx + c a = ½(max – min) = ½(2 – (-4)) = 3 = ½ X 6 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

Markers Comments Trig Graphs Menu Back to Home Next Comment a = ½(max – min) = ½(2 – (-4)) = ½ X 6 Period of graph =  so two complete sections between 0 & 2  For 3cos(…) max = 3 & min = -3. This graph: max = 2 & min = -4. ie 1 less so c = -1 ie b = 2 = 3 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.

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

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

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

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

TRIG GRAPHS & EQUATIONS : Question 3 Go to full solution Go to Marker’s Comments Go to Trig Graphs & Equations Menu Reveal answer only EXIT The diagram shows a the graph of a sine function from 0 to 2  / 3. 2  / 3 P Q y = 2 (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.

TRIG GRAPHS & EQUATIONS : Question 3 Go to full solution Go to Marker’s Comments Go to Trig Graphs & Equations Menu Reveal answer only EXIT The diagram shows a the graph of a sine function from 0 to 2  / 3. 2  / 3 P Q y = 2 (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 P is (  / 18, 2) and Q is ( 5  / 18, 2).

Markers Comments Begin Solution Continue Solution Question 3 Trig Graphs etc. Menu Back to Home 2  / 3 PQPQ y = 2 (a) State the equation of the graph. The diagram shows a the graph of a sine function from 0 to 2  / 3. (a)One complete wave from 0 to 2  / 3 so 3 waves from 0 to 2 . Max/min = ±44sin(…) Graph is y = 4sin3x Continue Solution

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

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

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

HIGHER – ADDITIONAL QUESTION BANK UNIT 1 : Functions & Graphs You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 1 Menu

FUNCTIONS & GRAPHS : Question 1 Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu Reveal answer only EXIT 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) -812 (-p,q) (u,-v)

FUNCTIONS & GRAPHS : Question 1 Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu Reveal answer only EXIT 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) -812 (-p,q) (u,-v) (p,-q+4) (8,4) (-12,4) (0,4) (-u,v+4) y = 4 – g(-x)

Markers Comments Begin Solution Continue Solution Question 1 Functions & Graphs Menu Back to Home 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) -812 (-p,q) (u,-v) y = 4 – g(-x) = -g(-x) + 4 Reflect in X-axis Reflect in Y- axis Slide 4 up A B C Known Points (-8,0), (-p,q), (0,0), (u,-v), (12,0) (-8,0), (-p,-q), (0,0), (u,v), (12,0) A B (8,0), (p,-q), (0,0), (-u,v), (-12,0) C (8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)

Markers Comments Begin Solution Continue Solution Question 1 Functions & Graphs Menu Back to Home 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) -812 (-p,q) (u,-v) (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)

Markers Comments Functions Menu Back to Home Next Comment y = 4 – g(-x) = -g(-x) + 4 Reflect in X-axis Reflect in Y- axis Slide 4 up A B C Known Points (-8,0), (-p,q), (0,0), (u,-v), (12,0) (-8,0), (-p,-q), (0,0), (u,v), (12,0) A B (8,0), (p,-q), (0,0), (-u,v), (-12,0) C (8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4) Change order to give form: y = k.g(x) + c 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.

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

Markers Comments Functions Menu Back to Home Next Comment 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)

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

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

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

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

Markers Comments Functions Menu Back to Home Next Comment (I) y = a (x+2) f(x) = a x so a (x+2) = f(x+2) move f(x) 2 to left (0,1)  (-2,1) & (1,a)  (-1,a) (-1,a) (-2,1) y = a (x+2) y = a x 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 = 2 x, y = log 3 x, y = x 2 + 3x, each becomes y = f(x) In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates.

Markers Comments Functions Menu Back to Home Next Comment (I) y = a (x+2) f(x) = a x so a (x+2) = f(x+2) move f(x) 2 to left (0,1)  (-2,1) & (1,a)  (-1,a) (-1,a) (-2,1) y = a (x+2) y = a x Learn Rules: Not given on formula sheet f(x) + kSlide 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

Markers Comments Functions Menu Back to Home Next Comment (II) y = 2a x - 3 f(x) = a x so 2a x - 3 = 2f(x) - 3 double y-coordsslide 3 down (0,1)  (0,2)  (0,-1) (1,a)  (1,2a)  (1,2a-3) y = 2a x - 3 y = a x (0,-1) (1,2a-3) 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 = 2 x, y = log 3 x, y = x 2 + 3x, each becomes y = f(x) In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates.

Markers Comments Functions Menu Back to Home Next Comment Learn Rules: Not given on formula sheet f(x) + kSlide 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 (II) y = 2a x - 3 f(x) = a x so 2a x - 3 = 2f(x) - 3 double y-coordsslide 3 down (0,1)  (0,2)  (0,-1) (1,a)  (1,2a)  (1,2a-3) y = 2a x - 3 y = a x (0,-1) (1,2a-3)

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

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

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

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

Markers Comments Functions Menu Back to Home Next Comment (a)(i) f(g(x)) = f(x 2 ) = 2x 2 - 1 (ii) g(f(x)) = g(2x-1) = (2x – 1) 2 = 4x 2 – 4x + 1 (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(x 2 ) In f(x) replace each x by g(x): 2 x 2 - 1 (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): (2x – 1) 2

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

FUNCTIONS & GRAPHS : Question 4 Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu Reveal answer only EXIT 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.

FUNCTIONS & GRAPHS : Question 4 Go to full solution Go to Marker’s Comments Go to Functions & Graphs Menu Go to Main Menu Reveal answer only EXIT 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. h(x) = (2x - 2).. (3 – x) Domain = {x  R: x  3}

Markers Comments Begin Solution Continue Solution Question 4 Functions & Graphs Menu Back to Home Continue Solution (a) g(g(x)) = g ( ). 2. (x – 1) = 2. 2. (x – 1) - 1 = 2 2 - (x – 1). (x – 1) = 2 (3 - x).(x – 1) = 2 ( x - 1).(3 – x) = (2x - 2).. (3 – x) 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.

Markers Comments Begin Solution Continue Solution Question 4 Functions & Graphs Menu Back to Home h(x) = (2x - 2).. (3 – x) 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. (b) For domain 3 - x  0 Domain = {x  R: x  3}

Markers Comments Functions Menu Back to Home Next Comment (a) g(g(x)) = g ( ). 2. (x – 1) = 2. 2. (x – 1) - 1 = 2 2 - (x – 1). (x – 1) = 2 (3 - x).(x – 1) = 2 ( x - 1).(3 – x) = (2x - 2).. (3 – x) (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) 2 - 1

Markers Comments Functions Menu Back to Home Next Comment h(x) = (2x - 2).. (3 – x) (b) For domain 3 - x  0 Domain = {x  R: x  3} (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.

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

HIGHER – ADDITIONAL QUESTION BANK UNIT 2 : Polynomials You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 2 Menu

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

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

Markers Comments Begin Solution Continue Solution Question 1 Polynomial Menu Back to Home Show that x = 3 is a root of the equation x 3 + 3x 2 – 10x – 24 = 0. Hence find the other roots. Using the nested method - coefficients are 1, 3, -10, -24 f(3) = 3 13-10-24 3 18 24 16 8 0 f(3) = 0 so x = 3 is a root. Also (x – 3) is a factor. Other factor: x 2 + 6x + 8 or (x + 4)(x + 2) If (x + 4)(x + 2) = 0then x = -4 or x = -2 Hence other roots are x = -4 & x = -2

Markers Comments Polynomial Menu Back to Home Next Comment Using the nested method - coefficients are 1, 3, -10, -24 f(3) = 3 13-10-24 3 18 24 16 8 0 Other factor: x 2 + 6x + 8 or (x + 4)(x + 2) If (x + 4)(x + 2) = 0then x = -4 or x = -2 Hence other roots are x = -4 & x = -2 f(3) = 0 so x = 3 is a root. Also (x – 3) is a factor. State clearly in solution that f(3) = 0 x = 3 is a root Show completed factorisation of cubic i.e. (x - 3)(x + 4)(x + 2) = 0 Take care to set factorised expression = 0 List all the roots of the polynomial x = 3, x = -4, x = -2

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

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

Markers Comments Begin Solution Continue Solution Question 2 Polynomial Menu Back to Home Given that (x + 4) is a factor of the polynomial f(x) = 3x 3 + 8x 2 + kx + 4 find the value of k. Hence solve the equation 3x 3 + 8x 2 + kx + 4 = 0 for this value of k. Since (x + 4) a factor then f(-4) = 0. Now using the nested method - coefficients are 3, 8, k, 4 f(-4) = -4 38k 4 -1216(-4k – 64) 3-4(k + 16) (-4k – 60 ) Since -4k – 60 = 0 then -4k = 60 so k = -15

Markers Comments Begin Solution Continue Solution Question 2 Polynomial Menu Back to Home Given that (x + 4) is a factor of the polynomial f(x) = 3x 3 + 8x 2 + kx + 4 find the value of k. Hence solve the equation 3x 3 + 8x 2 + kx + 4 = 0 for this value of k. If k = -15 then we now have f(-4) = -438-15 4 -1216-4 Other factor is 3x 2 – 4x + 1 or (3x - 1)(x – 1) 3-4 1 0 If (3x - 1)(x – 1) = 0then x = 1 / 3 or x = 1 So full solution of equation is: x = -4 or x = 1 / 3 or x = 1

Markers Comments Polynomial Menu Back to Home Next Comment Since (x + 4) a factor then f(-4) = 0. Now using the nested method - coefficients are 3, 8, k, 4 f(-4) = -4 38k 4 -1216(-4k – 64) 3-4(k + 16) (-4k – 60 ) Since -4k – 60 = 0 then -4k = 60 so k = -15 The working in the nested solution can sometimes be eased by working in both directions toward the variable: -438k4 -12 16 -4 3-410 k + 16 = 1 k = -15

Markers Comments Polynomial Menu Back to Home Next Comment Since (x + 4) a factor then f(-4) = 0. Now using the nested method - coefficients are 3, 8, k, 4 f(-4) = -4 38k 4 -1216(-4k – 64) 3-4(k + 16) (-4k – 60 ) Since -4k – 60 = 0 then -4k = 60 so k = -15 Simply making f(-4) = 0 will also yield k i.e. 3(-4) 3 + 8(-4) 2 + k(-4) + 4 = 0 k = -15

Markers Comments Polynomial Menu Back to Home Next Comment If k = -15 then we now have f(-4) = -4 -1216-4 Other factor is 3x 2 – 4x + 1 or (3x - 1)(x – 1) 3-4 1 0 If (3x - 1)(x – 1) = 0 So full solution of equation is: x = -4 or x = 1 / 3 or x = 1 38-15 4 Show completed factorisation of the cubic: (x + 4)(3x - 1)(x - 1) = 0

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

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

Markers Comments Begin Solution Continue Solution Question 3 Polynomial Menu Back to Home Given that f(x) = 6x 3 + 13x 2 - 4 show that (x + 2) is a factor of f(x). Hence express f(x) in its fully factorised form. Using the nested method - coefficients are 6, 13, 0, -4 f(-2) = -2 613 0 -4 6 -12 1 -2 4 0 f(-2) = 0 so (x + 2) is a factor

Markers Comments Begin Solution Continue Solution Question 3 Polynomial Menu Back to Home Given that f(x) = 6x 3 + 13x 2 - 4 show that (x + 2) is a factor of f(x). Hence express f(x) in its fully factorised form. Using the nested method - coefficients are 6, 13, 0, -4 f(-2) = -2 613 0 -4 6 -12 1 -2 4 0 Other factor is 6x 2 + x – 2 or (3x + 2)(2x - 1) Hence 6x 3 + 13x 2 - 4 = (3x + 2)(2x - 1)(x + 2)

Markers Comments Polynomial Menu Back to Home Next Comment Using the nested method - coefficients are 6, 13, 0, -4 f(-2) = -2 613 0 -4 6 -12 1 -2 4 0 f(-2) = 0 so (x + 2) is a factor State clearly in solution that f(-2) = 0 x = -2 is a root

Markers Comments Polynomial Menu Back to Home Next Comment Using the nested method - coefficients are 6, 13, 0, -4 f(-2) = -2 613 0 -4 6 -12 1 -2 4 0 Show completed factorisation of cubic i.e. (3x + 2)(2x - 1)(x +2). Other factor is 6x 2 + x – 2 or (3x + 2)(2x - 1) Hence 6x 3 + 13x 2 - 4 = (3x + 2)(2x - 1)(x + 2) Can show (x + 2) is a factor by showing f(-2) = 0 but still need nested method for quadratic factor.

(a)Find the coordinates of P and the equation of the bypass PQ. (b)Hence find the coordinates of Q – the point where the bypass rejoins the original road. POLYNOMIALS : Question 4 Go to full solutionReveal answer only EXIT 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 = -x 3 + 6x 2 – 3x – 10. The by-pass is a tangent to this curve at point P and rejoins the original road at Q as shown below. bypass P Q y = -x 3 + 6x 2 – 3x – 10 4

(a)Find the coordinates of P and the equation of the bypass PQ. (b)Hence find the coordinates of Q – the point where the bypass rejoins the original road. POLYNOMIALS : Question 4 Go to full solution EXIT 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 = -x 3 + 6x 2 – 3x – 10. The by-pass is a tangent to this curve at point P and rejoins the original road at Q as shown below. Go to Marker’s Comments Go to Polynomial MenuGo to Main Menu P is (4,10)PQ is y = -3x + 22 Q is (-2,28)

Markers Comments Begin Solution Continue Solution Question 4 Polynomial Menu Back to Home y = -x 3 + 6x 2 – 3x – 10 P Q 4 (a)Find the coordinates of P and the equation of the bypass PQ. (a)At point P, x = 4 so using the equation of the curve we get ….. y = -4 3 + (6 X 4 2 ) – (3 X 4) - 10 = -64 + 96 – 12 - 10 = 10 ieP is (4,10) Gradient of tangent = gradient of curve = dy / dx = -3x 2 + 12x - 3 When x = 4 then dy / dx = (-3 X 16) + (12 X 4) – 3 = -48 + 48 – 3 = -3

Markers Comments Begin Solution Continue Solution Question 4 Polynomial Menu Back to Home y = -x 3 + 6x 2 – 3x – 10 P Q 4 (a)Find the coordinates of P and the equation of the bypass PQ. P is (4,10) dy/dx =-3 Now using :y – b = m(x – a) where (a,b) = (4,10) & m = -3 We gety – 10 = -3(x – 4) ory – 10 = -3x + 12 So PQ is y = -3x + 22

Markers Comments Begin Solution Continue Solution Question 4 Polynomial Menu Back to Home y = -x 3 + 6x 2 – 3x – 10 P Q 4 (b) The tangent & curve meet whenever y = -3x + 22 and y = -x 3 + 6x 2 – 3x – 10 ie -3x + 22 = -x 3 + 6x 2 – 3x – 10 or x 3 - 6x 2 + 32 = 0 (b)Hence find the coordinates of Q – the point where the bypass rejoins the original road. We already know that x = 4 is one solution to this so using the nested method we get ….. f(4) = 41-6032 4-8-32 1-2-80 Other factor is x 2 – 2x - 8 PQ is y = -3x + 22

Markers Comments Begin Solution Continue Solution Question 4 Polynomial Menu Back to Home y = -x 3 + 6x 2 – 3x – 10 P Q 4 (b) The (b)Hence find the coordinates of Q – the point where the bypass rejoins the original road. other factor is x 2 – 2x - 8 = (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) PQ is y = -3x + 22

Markers Comments Polynomial Menu Back to Home Next Comment Must use differentiation to find gradient. Learn rule: “Multiply by the power then reduce the power by 1” (a) (a)At point P, x = 4 so using the equation of the curve we get ….. y = -4 3 + (6 X 4 2 ) – (3 X 4) - 10 = -64 + 96 – 12 - 10 = 10 ieP is (4,10) Gradient of tangent = gradient of curve = dy / dx = -3x 2 + 12x - 3 When x = 4 then dy / dx = (-3 X 16) + (12 X 4) – 3 = -48 + 48 – 3 = -3

Markers Comments Polynomial Menu Back to Home Next Comment (a) P is (4,10) dy/dx =-3 Now using :y – b = m(x – a) where (a,b) = (4,10) & m = -3 We gety – 10 = -3(x – 4) ory – 10 = -3x + 12 So PQ is y = -3x + 22 Use : 1.the point of contact (4,10)& 2.Gradient of curve at this point (m = -3) in equation y - b = m(x - a)

Markers Comments Polynomial Menu Back to Home Next Comment (b) (b) The tangent & curve meet whenever y = -3x + 22 and y = -x 3 + 6x 2 – 3x – 10 ie -3x + 22 = -x 3 + 6x 2 – 3x – 10 or x 3 - 6x 2 + 32 = 0 We already know that x = 4 is one solution to this so using the nested method we get ….. f(4) = 41-6032 4-8-32 1-2-80 Other factor is x 2 – 2x - 8 At intersectiony 1 = y 2 Terms to the left, simplify and factorise

Markers Comments Polynomial Menu Back to Home Next Comment (b) other factor is x 2 – 2x - 8 = (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) Note solution x = 4 appears twice: Repeated root tangency

HIGHER – ADDITIONAL QUESTION BANK UNIT 2 : Quadratics You have chosen to study: Please choose a question to attempt from the following: 12345 EXIT Back to Unit 2 Menu 6

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

QUADRATICS : Question 1 Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu Reveal answer only EXIT (a) Express f(x) = x 2 – 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) y = x 2 – 8x + 21 (4,5) (b)

Markers Comments Begin Solution Continue Solution Question 1 Quadratics Menu Back to Home (a)Express f(x) = x 2 – 8x + 21 in the form (x – a) 2 + b. (b)Hence, or otherwise, sketch the graph of y = f(x). (a) f(x) = x 2 – 8x + 21 = (x 2 – 8x + ) + 2116 - 16 (-8  2) 2 = (x – 4) 2 + 5

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

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

Markers Comments Quadratics Menu Back to Home Next Comment (a) f(x) = x 2 – 8x + 21 = (x 2 – 8x + ) + 2116 - 16 (-8  2) 2 = (x – 4) 2 + 5 Move towards desired form in stages: f(x) = x 2 - 8x + 21 = (x 2 - 8x) + 21 = (x 2 - 8x +16) + 21 - 16 Find the number to complete the perfect square and balance the expression. (a + b) 2 = a 2 + 2ab + b 2 = (x - 4) 2 + 5

Markers Comments Quadratics Menu Back to Home Next Comment (b)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). (4,5) (0,21) The sketch can also be obtained by calculus:

Markers Comments Quadratics Menu Back to Home Next Comment (b)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). (4,5) (0,21) Min. Turning Point (4,5), coefficient of x 2 is positive. Hence sketch. 21 (4,5)

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

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

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

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

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

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

QUADRATICS : Question 3 Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu Reveal answer only EXIT For what value(s) of k does the equation 4x 2 – kx + (k + 5) = 0 have equal roots.

QUADRATICS : Question 3 Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu Reveal answer only EXIT For what value(s) of k does the equation 4x 2 – kx + (k + 5) = 0 have equal roots. k = -4 or k = 20

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

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

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

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

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

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

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

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

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

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

QUADRATICS : Question 5 Go to full solution Go to Marker’s Comments Go to Quadratics Menu Go to Main Menu Reveal answer only EXIT 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 iep = 0 or p = 48

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

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

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

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

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

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

Markers Comments Begin Solution Continue Solution Question 6 Quadratics Menu Back to Home y = -2x 2 + 3x + 2 x + y – 4 = 0 Prove that the line is a tangent. 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 = -2x 2 + 3x + 2. So -x + 4 = -2x 2 + 3x + 2 Or 2x 2 - 4x + 2 = 0 Let 2x 2 - 4x + 2 = ax 2 + bx + c then a = 2, b = -4 & c = 2.

Markers Comments Begin Solution Continue Solution Question 6 Quadratics Menu Back to Home y = -2x 2 + 3x + 2 x + y – 4 = 0 Prove that the line is a tangent. then a = 2, b = -4 & c = 2. So discriminant = b 2 – 4ac = (-4) 2 – (4 X 2 X 2) = 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.

Markers Comments Quadratics Menu Back to Home Next Comment 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 = -2x 2 + 3x + 2. So -x + 4 = -2x 2 + 3x + 2 Or 2x 2 - 4x + 2 = 0 Let 2x 2 - 4x + 2 = ax 2 + bx + c then a = 2, b = -4 & c = 2. For intersection of line and polynomial y 1 = y 2 Terms to the left, simplify and factorise.

Markers Comments Quadratics Menu Back to Home Next Comment Or -x + 4 = -2x 2 + 3x + 2 2x 2 - 4x + 2 = 0 2(x 2 - 2x + 1) = 0 2(x - 1)(x - 1) = 0 x = 1 (twice) Equal roots tangency then a = 2, b = -4 & c = 2. So discriminant = b 2 – 4ac = (-4) 2 – (4 X 2 X 2) = 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. To prove tangency To prove tangency “equal roots” may be used in “equal roots” may be used in place of the discriminant. place of the discriminant. The statement must be made The statement must be made explicitly. explicitly.

HIGHER – ADDITIONAL QUESTION BANK UNIT 2 : Integration You have chosen to study: Please choose a question to attempt from the following: 12345 EXIT Back to Unit 2 Menu

y = x 2 - 8x + 18 x = 3 x = k Show that the shaded area is given by 1 / 3 k 3 – 4k 2 + 18k - 27 INTEGRATION : Question 1 Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu Reveal answer only EXIT The diagram below shows the curve y = x 2 - 8x + 18 and the lines x = 3 and x = k.

y = x 2 - 8x + 18 x = 3 x = k Show that the shaded area is given by 1 / 3 k 3 – 4k 2 + 18k - 27 INTEGRATION : Question 1 Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu Reveal answer only EXIT Area = (x 2 - 8x + 18) dx  3 k = 1 / 3 k 3 – 4k 2 + 18k – 27 as required.

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

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

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

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

INTEGRATION : Question 2 Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu Reveal answer only EXIT Given that dy / dx = 12x 2 – 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 = 4x 3 – 3x 2 - 5

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

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

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

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

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

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

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

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

INTEGRATION : Question 4 Go to full solution Go to Marker’s Comments Go to Integration Menu Go to Main Menu Reveal answer only EXIT 1 2  ( ) Evaluatex 2 - 2 2 dx x = 2 1 / 5

Markers Comments Begin Solution Continue Solution Question 4 Integration Menu Back to Home 1 2  ( ) Evaluatex 2 - 2 2 dx x = 2 1 / 5 1 2  ( ) x 2 - 2 2 dx x ( ) = x 4 - 4x + 4 dx  x2x2 1 2 ( ) = x 4 - 4x + 4x -2 dx  1 2 [ ] = x 5 - 4x 2 + 4x -1 5 2 -1 1 2 = x 5 - 2x 2 - 4 5 x [ ] 2 1 = ( 32 / 5 - 8 - 2) - ( 1 / 5 - 2 - 4)

Markers Comments Integration Menu Back to Home Next Comment = 2 1 / 5 1 2  ( ) x 2 - 2 2 dx x ( ) = x 4 - 4x + 4 dx  x2x2 1 2 ( ) = x 4 - 4x + 4x -2 dx  1 2 [ ] = x 5 - 4x 2 + 4x -1 5 2 -1 1 2 = x 5 - 2x 2 - 4 5 x [ ] 2 1 = ( 32 / 5 - 8 - 2) - ( 1 / 5 - 2 - 4) Prepare expression by: 1 Expanding the bracket 2 Applying the laws of indices. Learn result for integration: “Add 1 to the power and divide by the new power”. When applying limits show substitution clearly.

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

(a) Find the coordinates of A and B. (b)Hence find the shaded area between the curves. y = -x 2 + 8x - 10 y = x A B INTEGRATION : Question 5 Go to full solution Go to Marker’s Comms Go to Integration Menu Go to Main Menu Reveal answer only EXIT The diagram below shows the parabola y = -x 2 + 8x - 10 and the line y = x. They meet at the points A and B. A is (2,2) and B is (5,5). = 4 1 / 2 units 2

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

Markers Comments Begin Solution Continue Solution Question 5 Integration Menu Back to Home The diagram shows the parabola y = -x 2 + 8x - 10 and the line y = x. They meet at the points A and B. (a) Find the coordinates of A and B. (b)Hence find the shaded area between the curves. A is (2,2) and B is (5,5). (b) Curve is above line between limits so Shaded area = (-x 2 + 8x – 10 - x) dx  2 5 = (-x 2 + 7x – 10) dx  2 5 = -x 3 + 7x 2 - 10x 3 2 [] 5 2 = ( -125 / 3 + 175 / 2 – 50) – ( -8 / 3 +14 – 20) = 4 1 / 2 units 2

Markers Comments Integration Menu Back to Home Next Comment (a)Line & curve meet when y = x and y = -x 2 + 8x - 10. Sox = -x 2 + 8x - 10 orx 2 - 7x + 10 = 0 ie(x – 2)(x – 5) = 0 iex = 2 or x = 5 Since points lie on y = x then A is (2,2) and B is (5,5). At intersection of line and curve y 1 = y 2 Terms to the left, simplify and factorise.

Markers Comments Integration Menu Back to Home Next Comment (b) Curve is above line between limits so Shaded area = (-x 2 + 8x – 10 - x) dx  2 5 = (-x 2 + 7x – 10) dx  2 5 = -x 3 + 7x 2 - 10x 3 2 [] 5 2 = ( -125 / 3 + 175 / 2 – 50) – ( -8 / 3 +14 – 20) = 4 1 / 2 units 2 Learn result can be used to find the enclosed area shown: a b upper curve y1y1 area y2y2 lower curve

HIGHER – ADDITIONAL QUESTION BANK UNIT 2 : Addition Formulae You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 2 Menu

ADDITION FORMULAE : Question 1 Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu Reveal answer only EXIT In triangle PQR show that the exact value of cos(a - b) is 4 / 5. P Q R 14 2 ab

ADDITION FORMULAE : Question 1 Go to full solution Go to Marker’s Comments Go to Add Formulae Menu Go to Main Menu Reveal answer only EXIT In triangle PQR show that the exact value of cos(a - b) is 4 / 5. P Q R 14 2 ab cos(a – b) = cosacosb + sinasinb = ( 2 /  5 X 1 /  5 ) + ( 1 /  5 X 2 /  5 ) = 4 / 5

Markers Comments Begin Solution Continue Solution Question 1 Add Formulae Menu Back to Home In triangle PQR show that the exact value of cos(a - b) is 4 / 5. P Q R 14 2 ab PQ 2 = 1 2 + 2 2 = 5 so PQ =  5 QR 2 = 4 2 + 2 2 = 20 so QR =  20 =  4  5 = 2  5 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 ) = 2 / 5 + 2 / 5 = 4 / 5

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

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

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

Markers Comments Begin Solution Continue Solution Question 2 Add Formulae Menu Back to Home Find the exact value of cos(p + q) in the diagram below Y X O F(4,4) G(3,-1) p q OF 2 = 4 2 + 4 2 = 32 so OF =  32 =  16  2 = 4  2 OG 2 = 3 2 + 1 2 = 10 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 = 2 / 2  5 = 1 /  5

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

Markers Comments Add Form Menu Back to Home Next Comment OF 2 = 4 2 + 4 2 = 32 so OF =  32 =  16  2 = 4  2 OG 2 = 3 2 + 1 2 = 10 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 = 2 / 2  5 = 1 /  5 2 a a = Work only with exact values when applying Pythagoras’:

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

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

Markers Comments Begin Solution Continue Solution Question 3 Add Formulae Menu Back to Home Solve the equation 2 - sinx° = 3cos2x° where 0<x<360. 2 - sinx° = 3cos2x° 2 - sinx° = 3(1 – 2sin 2 x°) 2 - sinx° = 3 – 6sin 2 x° Rearrange into quadratic form 6sin 2 x° - sinx° - 1 = 0 (3sinx° + 1)(2sinx° - 1) = 0 6s 2 – s – 1 = (3s + 1)(2s – 1) sinx° = -1 / 3 or sinx° = 1 / 2 Q3 or Q4Q1 or Q2 AS TC a°a°(180 - a)° (180 + a)°(360 - a)°

Markers Comments Begin Solution Continue Solution Question 3 Add Formulae Menu Back to Home Solve the equation 2 - sinx° = 3cos2x° where 0<x<360. sinx° = -1 / 3 or sinx° = 1 / 2 AS TC a°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 Solution = {30, 150, 199.5, 340.5}

Markers Comments Add Form Menu Back to Home Next Comment 2 - sinx° = 3cos2x° 2 - sinx° = 3(1 – 2sin 2 x°) 2 - sinx° = 3 – 6sin 2 x° Rearrange into quadratic form 6sin 2 x° - sinx° - 1 = 0 (3sinx° + 1)(2sinx° - 1) = 0 6s 2 – s – 1 = (3s + 1)(2s – 1) sinx° = -1 / 3 or sinx° = 1 / 2 Q3 or Q4Q1 or Q2 AS TC a°a°(180 - a)° (180 + a)°(360 - a)° Use formula sheet to check correct expansion and relateto given variables: cos2x° =1 - 2sin 2 x° Formula must be consistent with the rest of the equation. 2 - sinx° i.e. choose formula with sinx°

Markers Comments Add Form Menu Back to Home Next Comment 2 - sinx° = 3cos2x° 2 - sinx° = 3(1 – 2sin 2 x°) 2 - sinx° = 3 – 6sin 2 x° Rearrange into quadratic form 6sin 2 x° - sinx° - 1 = 0 (3sinx° + 1)(2sinx° - 1) = 0 6s 2 – s – 1 = (3s + 1)(2s – 1) sinx° = -1 / 3 or sinx° = 1 / 2 Q3 or Q4Q1 or Q2 AS TC a°a°(180 - a)° (180 + a)°(360 - a)° Only one way to solve the resulting quadratic: Terms to the left, put in standard quadratic form and factorise.

Take care to relate “solutions” to the given domain. Since all 4 values are included. Markers Comments Add Form Menu Back to Home Next Comment sinx° = -1 / 3 or sinx° = 1 / 2 AS TC a°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 Solution = {30, 150, 199.5, 340.5}

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

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

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

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

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

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

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

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

HIGHER – ADDITIONAL QUESTION BANK UNIT 2 : The Circle You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 2 Menu

(a)Find the coordinates of C and W. (b) Hence find the equation of the dial for the second hand. W C THE CIRCLE : Question 1 Go to full solution Go to Marker’s Comments Go to Circle Menu Reveal answer only EXIT The face of a stopwatch can be modelled by the circle with equation x 2 + y 2 – 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.

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

Markers Comments Begin Solution Continue Solution Question 1 Circle Menu Back to Home The face of a stopwatch can be modelled by the circle with equation x 2 + y 2 – 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. (a)Find the coordinates of C and W. Large circle is x 2 + y 2 – 10x – 8y + 5 = 0. x 2 + y 2 + 2gx + 2fy + c = 0 Comparing Coefficients 2g = -10, 2f = -8 and c = 5 So g = -5, f = -4 and c = 5 Centre is (-g,-f) and r =  (g 2 + f 2 – c) C is (5,4) r =  (25 + 16 – 5) r =  36 = 6 W is 6 units above C so W is (5,10).

Markers Comments Begin Solution Continue Solution Question 1 Circle Menu Back to Home The face of a stopwatch can be modelled by the circle with equation x 2 + y 2 – 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. (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 = r 2 we get (x – 5) 2 + (y – 7) 2 = 4

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

Markers Comments Circle Menu Back to Home Next Comment (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 = r 2 we get (x – 5) 2 + (y – 7) 2 = 4 Identify centre and radius before putting values into circle equation: Centre (5,7), radius = 2 (x - a) 2 + (y - b) 2 = r 2 (x - 5) 2 + (y - 7) 2 = 2 2 There is no need to expand this form of the circle equation.

THE CIRCLE : Question 2 Go to full solution Go to Marker’s Comments Go to Circle Menu Reveal answer only EXIT 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 x 2 + y 2 – 6x + 2y - 7 = 0. (i)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. (ii)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. belt P A(7,0) B Q

THE CIRCLE : Question 2 Go to full solution Go to Marker’s Comments Go to Circle Menu Reveal answer only EXIT 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 x 2 + y 2 – 6x + 2y - 7 = 0. (i)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. (ii)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. belt P A(7,0) B Q y = -4x + 28 B is (10,-12) (x – 18) 2 + (y + 10) 2 = 68

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

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

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

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

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

Markers Comments Circle Menu Back to Home Next Comment Identify centre and radius before putting values into circle equation: Centre (18,-10), radius = 2 (x - a) 2 + (y - b) 2 = r 2 (x - 18) 2 + (y + 10) 2 = (2 ) 2 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. (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 r 2 = (2  17) 2 = 4 X 17 = 68 Using (x – a) 2 + (y – b) 2 = r 2 we get (x – 18) 2 + (y + 10) 2 = 68

THE CIRCLE : Question 3 Go to full solution Go to Marker’s Comments Go to Circle Menu Reveal answer only EXIT 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 x 2 + (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.

THE CIRCLE : Question 3 Go to full solution Go to Marker’s Comments Go to Circle Menu Reveal answer only EXIT 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 x 2 + (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

Markers Comments Begin Solution Continue Solution Question 3 Circle Menu Back to Home Small circle has centre (0,12) and radius  5 The small circle with centre C has equation x 2 + (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 CE 2 = (18 – 0) 2 + (3 – 12) 2 = 18 2 + (-9) 2 = 324 + 81 = 405 (Distance form!!) CE =  405 =  81 X  5 = 9  5 Diameter of large circle = 9  5 - 2  5 -  5 = 6  5 So radius of large circle = 3  5 CDE 553535 9595 25 25

Markers Comments Begin Solution Continue Solution Question 3 Circle Menu Back to Home The small circle with centre C has equation x 2 + (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. CDE 553535 9595 25 25 It follows that CD = 4 / 9 CE = 4 / 9 [( ) - ( )] 18 3 0 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 = r 2 we get (x – 8) 2 + (y – 8) 2 = (3  5) 2 or (x – 8) 2 + (y – 8) 2 = 45

Markers Comments Circle Menu Back to Home Next Comment Small circle has centre (0,12) and radius  5 Med. circle has centre (18,3) and radius  20 =  4 X  5 = 2  5 CE 2 = (18 – 0) 2 + (3 – 12) 2 = 18 2 + (-9) 2 = 324 + 81 = 405 (Distance form!!) CE =  405 =  81 X  5 = 9  5 Diameter of large circle = 9  5 - 2  5 -  5 = 6  5 CDE 553535 9595 25 25 Use formula sheet to check correct formulas and relate to equation of the circle.

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

(b)The central wheel has equation x 2 + y 2 – 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. THE CIRCLE : Question 4 Go to full solutionReveal answer only EXIT The central driving wheels on a steam locomotive are linked by a piston rod. AB C E D (a)The rear wheel has centre A & equation x 2 + y 2 – 4x – 10y + 4 = 0 the front wheel has centre C & equation x 2 + y 2 – 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.

(b)The central wheel has equation x 2 + y 2 – 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. THE CIRCLE : Question 4 Go to full solutionReveal answer only EXIT The central driving wheels on a steam locomotive are linked by a piston rod. (a)The rear wheel has centre A & equation x 2 + y 2 – 4x – 10y + 4 = 0 the front wheel has centre C & equation x 2 + y 2 – 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. Hence D is (10,2) and E is (17,1). Equation of DE x + 7y = 24

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

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

Markers Comments Begin Solution Continue Solution Question 4 Circle Menu Back to Home (b) The central wheel has equation x 2 + y 2 – 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. Middle wheel is x 2 + y 2 – 28x – 10y + 196 = 0. x 2 + y 2 + 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) Gradient of BM = 5 – 1.5 14 – 13.5 Equation of DE = 3.5 / 0.5 = 7 Gradient of DE = - 1 / 7 (m 1 m 2 = -1)

Markers Comments Begin Solution Continue Solution Question 4 Circle Menu Back to Home (b) The central wheel has equation x 2 + y 2 – 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. Equation of DE Usingy – b = m(x – a) we get y – 1.5 = - 1 / 7 (x – 13.5)( X 7) Or 7y – 10.5 = -x + 13.5 So DE isx + 7y = 24 Line & circle meet when x 2 + y 2 – 28x – 10y + 196 = 0 & x = 24 – 7y. x + 7y = 24 can be written as x = 24 – 7y Substituting (24 – 7y) for x in the circle equation we get ….

Markers Comments Begin Solution Continue Solution Question 4 Circle Menu Back to Home (b) The central wheel has equation x 2 + y 2 – 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. (24 – 7y) 2 + y 2 – 28(24 – 7y) – 10y + 196 = 0 576 – 336y + 49y 2 + y 2 – 672 + 196y – 10y + 196 = 0 50y 2 – 150y + 100 = 0(  50) y 2 – 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 Hence D is (10,2) and E is (17,1).

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

Markers Comments Circles Menu Back to Home Next Comment Equation of DE Usingy – b = m(x – a) we get y – 1.5 = - 1 / 7 (x – 13.5) Or 7y – 10.5 = -x + 13.5 So DE isx + 7y = 24 Line & circle meet when x 2 + y 2 – 28x – 10y + 196 = 0 & x = 24 – 7y. x + 7y = 24 can be written as x = 24 – 7y Substituting (24 – 7y) for x in the circle equation we get …. In part b) avoid fractions when substituting into the circle equation: Use x = (24 - 7y) not y = (24 -x)

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

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

HIGHER – ADDITIONAL QUESTION BANK UNIT 3 : Wave Function You have chosen to study: Please choose a question to attempt from the following: 123 EXIT Back to Unit 3 Menu

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

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

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

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

Markers Comments Begin Solution Continue Solution Question 1 Wave Function Menu Back to Home (a)Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where k>0 and 0  a  90. (b)Hence solve cos(x°) + 7sin (x°) = 5 for 0  x  90. Hence cos(x°) + 7sin (x°) = 5  2cos(x° - 81.9°) (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. Q4: angle = 360° - 45° so x° - 81.9° = 315° so x° = 396.9° (**)

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

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

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

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

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

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

WAVE FUNCTION: Question 2 Go to full solution Go to Marker’s Comments Go to Wave Function Menu Go to Main Menu Reveal answer only EXIT 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°)

Markers Comments Begin Solution Continue Solution Question 2 Wave Function Menu Back to Home 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 = 4 2 + (-2) 2 ork 2 cos 2 a° + k 2 sin 2 a° = 16 + 4 or k 2 (cos 2 a° + sin 2 a°) = 20 or k 2 = 20 (cos 2 a° + sin 2 a°) = 1 so k =  20 =  4  5 = 2  5

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

Markers Comments Wave Func Menu Back to Home Next Comment 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 = 4 2 + (-2) 2 ork 2 cos 2 a° + k 2 sin 2 a° = 16 + 4 or k 2 (cos 2 a° + sin 2 a°) = 20 or k 2 = 20 (cos 2 a° + sin 2 a°) = 1 so k =  20 =  4  5 = 2  5

Markers Comments Wave Func Menu Back to Home Next Comment 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 = 4 2 + (-2) 2 ork 2 cos 2 a° + k 2 sin 2 a° = 16 + 4 or k 2 (cos 2 a° + sin 2 a°) = 20 or k 2 = 20 (cos 2 a° + sin 2 a°) = 1 so k =  20 =  4  5 = 2  5 When equating coefficients “square” and “ring” corresponding coefficients: 4sinx° - 2cosx° = ksinx°cosa° + kcosx°sina° State the resulting equations explicitly: kcosa° = 4 ksina° = -2

Markers Comments Wave Func Menu Back to Home Next Comment 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 = 4 2 + (-2) 2 ork 2 cos 2 a° + k 2 sin 2 a° = 16 + 4 or k 2 (cos 2 a° + sin 2 a°) = 20 or k 2 = 20 (cos 2 a° + sin 2 a°) = 1 so k =  20 =  4  5 = 2  5 k can be found directly: Note: k is always positive

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

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

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

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

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

Markers Comments Begin Solution Continue Solution Question 3 Wave Function Menu Back to Home (a)Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0  a < 360 (b)Hence find the coordinates of the maximum turning point at P. Hence 5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°) (b)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)

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

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

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

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

Markers Comments Wave Func Menu Back to Home Next Comment (b)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) 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°

Markers Comments Wave Func Menu Back to Home Next Comment (b)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) Maximum value of sin(x°+292.6°) occurs at (90°- 292.6°) i.e at x = - 202.6° Add 360° to bring into domainx = 157.4°

HIGHER – ADDITIONAL QUESTION BANK UNIT 3 : Logs & Exponentials You have chosen to study: Please choose a question to attempt from the following: 12345 EXIT Back to Unit 3 Menu

LOGS & EXPONENTIALS: Question 1 Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu Go to Main Menu Reveal answer only EXIT As a radioactive substance decays the amount of radioactive material remaining after t hours, A t, is given by the formula A t = A 0 e -0.161t where A 0 is the original amount of material. (a)If 400mg of material remain after 10 hours then determine how much material there was at the start. (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.

LOGS & EXPONENTIALS: Question 1 Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu Go to Main Menu Reveal answer only EXIT As a radioactive substance decays the amount of radioactive material remaining after t hours, A t, is given by the formula A t = A 0 e -0.161t where A 0 is the original amount of material. (a)If 400mg of material remain after 10 hours then determine how much material there was at the start. (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. Initial amount of material = 2000mg t = 4hrs 18mins

Markers Comments Begin Solution Continue Solution Question 1 Logs & Exp Menu Back to Home (a) When t = 10, A 10 = 400 so A t = A 0 e -0.161t (a)If 400mg of material remain after 10 hours then determine how much material there was at the start. becomes A 10 = A 0 e -0.161X10 or 400 = A 0 e -1.61 and A 0 = 400  e -1.61 ie A 0 = 2001.12… or 2000 to 3 sfs Initial amount of material = 2000mg

Markers Comments Begin Solution Continue Solution Question 1 Logs & Exp Menu Back to Home (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. (b) For half life A t = 1 / 2 A 0 so A t = A 0 e -0.161t becomes 1 / 2 A 0 = A 0 e -0.161t or e -0.161t = 0.5 orln(e -0.161t ) = ln0.5 ie-0.161t = ln0.5 sot = ln0.5  (-0.161) iet = 4.3052…hrs ort = 4hrs 18mins ( 0.3052 X 60 = 18.3..)

Markers Comments Logs & Exp Menu Back to Home Next Comment (a) When t = 10, A 10 = 400 so A t = A 0 e -0.161t becomes A 10 = A 0 e -0.161X10 or 400 = A 0 e -1.61 and A 0 = 400  e -1.61 ie A 0 = 2001.12… or 2000 to 3 sfs Initial amount of material = 2000mg The e x is found on the calculator: 2nd ln exex

Markers Comments Logs & Exp Menu Back to Home Next Comment (b) For half life A t = 1 / 2 A 0 so A t = A 0 e -0.161t becomes 1 / 2 A 0 = A 0 e -0.161t or e -0.161t = 0.5 orln(e -0.161t ) = ln0.5 ie-0.161t = ln0.5 sot = ln0.5  (-0.161) iet = 4.3052…hrs ort = 4hrs 18mins ( 0.3052 X 60 = 18.3..) (b) The half life can be found using any real value: e.g. A 0 = 2000 A t = 1000 This results in the equation A t = A 0 e -0.161t 1000 = 2000e -0.161t etc.

Markers Comments Logs & Exp Menu Back to Home Next Comment (b) For half life A t = 1 / 2 A 0 so A t = A 0 e -0.161t becomes 1 / 2 A 0 = A 0 e -0.161t or e -0.161t = 0.5 orln(e -0.161t ) = ln0.5 ie-0.161t = ln0.5 sot = ln0.5  (-0.161) iet = 4.3052…hrs ort = 4hrs 18mins ( 0.3052 X 60 = 18.3..) (b) To solve an exponential equation must use logs. A trial and error solution will only be given minimum credit: e.g.12 = e 2x take logs. to both sides ln12 = ln (e 2x ) log and exponential are inverse functions ln 12 = 2x etc.

LOGS & EXPONENTIALS: Question 2 Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu Go to Main Menu Reveal answer only EXIT log 10 y log 10 x 1 0.3 The graph illustrates the law y = k xnxn The line passes through (0,0.3) and (1,0). Find the values of k and n.

LOGS & EXPONENTIALS: Question 2 Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu Go to Main Menu Reveal answer only EXIT log 10 y log 10 x 1 0.3 The graph illustrates the law y = k xnxn The line passes through (0,0.3) and (1,0). Find the values of k and n. Hencek = 2 and n = 0.3

Markers Comments Begin Solution Continue Solution Question 2 Logs & Exp Menu Back to Home Intercept = (0,0.3) gradient = 0 – 0.3 1 - 0 Straight line so in formY = mX + c with Y = log 10 y and X = log 10 x This becomeslog 10 y = -0.3log 10 x + 0.3 Or log 10 y = -0.3log 10 x + log 10 10 0.3 = -0.3 or log 10 y = -0.3log 10 x + log 10 2 or log 10 y = log 10 x -0.3 + log 10 2 or log 10 y = log 10 2x -0.3 law3 law1 The graph illustrates the law y = k xnxn The line passes through (0,0.3) and (1,0). Find the values of k and n. log 10 y log 10 x 1 0.3

Markers Comments Begin Solution Continue Solution Question 2 Logs & Exp Menu Back to Home or log 10 y = log 10 2x -0.3 law1 The graph illustrates the law y = k xnxn The line passes through (0,0.3) and (1,0). Find the values of k and n. log 10 y log 10 x 1 0.3 soy = 2x -0.3 = 2 x 0.3 Hencek = 2 and n = 0.3

Markers Comments Logs & Exp Menu Back to Home Next Comment Intercept = (0,0.3) gradient = 0 – 0.3 1 - 0 Straight line so in formY = mX + c with Y = log 10 y and X = log 10 x This becomeslog 10 y = -0.3log 10 x + 0.3 Or log 10 y = -0.3log 10 x + log 10 10 0.3 or log 10 y = -0.3log 10 x + log 10 2 or log 10 y = log 10 x -0.3 + log 10 2 or log 10 y = log 10 2x -0.3 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 = kx n Take logs to both sides log y = log kx n Apply Law 1: log AB = log A + logB

Markers Comments Logs & Exp Menu Back to Home Next Comment Intercept = (0,0.3) gradient = 0 – 0.3 1 - 0 Straight line so in formY = mX + c with Y = log 10 y and X = log 10 x This becomeslog 10 y = -0.3log 10 x + 0.3 Or log 10 y = -0.3log 10 x + log 10 10 0.3 or log 10 y = -0.3log 10 x + log 10 2 or log 10 y = log 10 x -0.3 + log 10 2 or log 10 y = log 10 2x -0.3 log y = logk + logx n Apply Law 3: logx n = nlogx log y = logk + nlogx log y = nlogx + logk

Markers Comments Straight Line Menu Back to Home Next Comment or log 10 y = log 10 2x -0.3 soy = 2x -0.3 = 2 x 0.3 Hencek = 2 and n = 0.3 Two coordinates from the graph: log 10 y log 10 x 1 0.3

Markers Comments Straight Line Menu Back to Home Next Comment or log 10 y = log 10 2x -0.3 soy = 2x -0.3 = 2 x 0.3 Hencek = 2 and n = 0.3 Two coordinates from the graph: log y = nlogx + logk (0,0.3) 0.3 = n.0 + logk- 1 (1,0) 0 = n.1 + logk - 2 Hence solve 1 and 2 to find k and n logk = 0.3 hence k = 2, and n = - logk hence n= -0.3.

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

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

Markers Comments Begin Solution Continue Solution Question 3 Logs & Exp Menu Back to Home Iflog 3 (5) – log 3 (2x + 1) = -2 Solve the equation log 3 (5) – log 3 (2x + 1) = -2 then log 3 5 2x + 1 ( ) = -2 law2 so5 (2x + 1) = 3 -2 or5 (2x + 1) = 1 9 Cross mult we get 2x + 1 = 45 or 2x = 44 ie x = 22

Markers Comments Logs & Exp Menu Back to Home Next Comment Iflog 3 (5) – log 3 (2x + 1) = -2 then log 3 5 2x + 1 ( ) = -2 so5 (2x + 1) = 3 -2 or5 (2x + 1) = 1 9 Cross mult we get 2x + 1 = 45 or 2x = 44 ie x = 22 To solve an equation involving logs apply the lawsso that the equation is reduced to log=log and the logcan be removed and the equation solved: log 3 5 - log 3 (2x+1) = -2 Law 2:log = logA - logB ABAB Must know: log 3 3 = 1

Markers Comments Logs & Exp Menu Back to Home Next Comment Iflog 3 (5) – log 3 (2x + 1) = -2 then log 3 5 2x + 1 ( ) = -2 so5 (2x + 1) = 3 -2 or5 (2x + 1) = 1 9 Cross mult we get 2x + 1 = 45 or 2x = 44 ie x = 22 log 3 = -2 log 3 3 5 2x+1 log 3 = log 3 3 -2 5 2x+1 Law 3:log = nlogx xnxn The log terms can now be dropped from both sides of the equation and the equation solved:

Markers Comments Logs & Exp Menu Back to Home Next Comment Iflog 3 (5) – log 3 (2x + 1) = -2 then log 3 5 2x + 1 ( ) = -2 so5 (2x + 1) = 3 -2 or5 (2x + 1) = 1 9 Cross mult we get 2x + 1 = 45 or 2x = 44 ie x = 22 Drop log 3 terms = 3 -2 5 2x+1 etc. = 5 2x+1 1919

LOGS & EXPONENTIALS: Question 4 Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu Go to Main Menu Reveal answer only EXIT The pressure in a leaky tyre drops according to the formula P t = P 0 e -kt where P 0 is the initial tyre pressure and P t is the pressure after t hours. (a)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. (b)By how many more psi will the pressure drop in the next 15 mins? sssss

LOGS & EXPONENTIALS: Question 4 Go to full solution Go to Marker’s Comments Go to Logs & Exp Menu Go to Main Menu Reveal answer only EXIT The pressure in a leaky tyre drops according to the formula P t = P 0 e -kt where P 0 is the initial tyre pressure and P t is the pressure after t hours. (a)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. (b)By how many more psi will the pressure drop in the next 15 mins? sssss 1.8psi (a) k = 0.243to 3 dps (b)

Markers Comments Begin Solution Continue Solution Question 4 Logs & Exp Menu Back to Home (a)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. (a)We have P 0 = 35, P t = 31 and t = 0.5 (30mins = ½ hr) SoP t = P 0 e -kt becomes31 = 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 = 0.243to 3 dps

Markers Comments Begin Solution Continue Solution Question 4 Logs & Exp Menu Back to Home ie k = 0.243to 3 dps (b) By how many more psi will the pressure drop in the next 15 mins? (b)We now have k = 0.243, P 0 = 31 and t = 0.25 (15mins = 1 / 4 hr) using P t = P 0 e -0.243t we getP 0.25 = 31e (-0.243X0.25) soP 0.25 = 29.2 Pressure drop in next 15 mins is 31 – 29.2 or 1.8psi

Markers Comments Logs & Exp Menu Back to Home Next Comment (a)We have P 0 = 35, P t = 31 and t = 0.5 (30mins = ½ hr) SoP t = P 0 e -kt becomes31 = 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 = 0.243to 3 dps The e x is found on the calculator: 2nd ln exex (a)

Markers Comments Logs & Exp Menu Back to Home Next Comment 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 = ln 31 / 35 etc. (a)We have P 0 = 35, P t = 31 and t = 0.5 (30mins = ½ hr) SoP t = P 0 e -kt becomes31 = 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 = 0.243to 3 dps

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

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

Markers Comments Begin Solution Continue Solution Question 5 Logs & Exp Menu Back to Home Given that sin  = a u and cos  = a v show that u – v = log a tan . Since sin  = a u and cos  = a v thenu = log a sin  and v = log a cos  sou – v = log a sin  - log a cos  oru – v = log a ( sin  / cos  ) henceu – v = log a tan  logx – logy = log( x / y )

Markers Comments Logs & Exp Menu Back to Home Next Comment Since sin  = a u and cos  = a v thenu = log a sin  and v = log a cos  sou – v = log a sin  - log a cos  oru – v = log a ( sin  / cos  ) henceu – v = log a tan  logx – logy = log( x / y ) There are different routes to the same result but all involve correct application of formulas and laws of logs: e.g.tan  = sin  cos  Should be known from standard grade tan  = auavauav

Markers Comments Logs & Exp Menu Back to Home Next Comment Since sin  = a u and cos  = a v thenu = log a sin  and v = log a cos  sou – v = log a sin  - log a cos  oru – v = log a ( sin  / cos  ) henceu – v = log a tan  logx – logy = log( x / y ) Take logs to base a to both sides log a tan  = log a auavauav log a tan  = log a a u - log a a v Law2: log =logA-logB ABAB log a tan  = ulog a a - vlog a a Law 3:log = nlogx xnxn log a a=1 log a tan  = u - v

HIGHER – ADDITIONAL QUESTION BANK UNIT 3 : Vectors You have chosen to study: Please choose a question to attempt from the following: 12345 EXIT Back to Unit 3 Menu

VECTORS: Question 1 Go to full solution Go to Marker’s Comments Go to Vectors Menu Go to Main Menu Reveal answer only EXIT (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 PS = 7  PQ then find  the coordinates of S.

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

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

Markers Comments Begin Solution Continue Solution Question 1 Vectors Menu Back to Home (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 PS = 7  PQ then find  the coordinates of S. (b)PS = 7  PQ  [ ] -4 1 3 = 7 [ ] -28 7 21 =S is (3,0,-5) + [ ] -28 7 21 = (3-28, 0+7, -5+21) = (-25, 7, 16)

Markers Comments Vectors Menu Back to Home Next Comment Must know result: Given coordinates of A and B a) PQ = q - p =  (a) [ ] 1 -2 [ ] 3 0 -2 - [ ] -4 1 3 = PR = r - p =  [ ] -13 4 7 [ ] 3 0 -2 - [ ] -16 4 9 [ ] -4 1 3 = 4 = Since PQ and  PR are  common point then vector and have P as a it follows that P, Q & R are collinear. multiples of the same

Markers Comments Vectors Menu Back to Home Next Comment PQ = q - p =  (a) [ ] 1 -2 [ ] 3 0 -2 - [ ] -4 1 3 = PR = r - p =  [ ] -13 4 7 [ ] 3 0 -2 - [ ] -16 4 9 [ ] -4 1 3 = 4 = Since PQ and  PR are  common point then vector and have P as a it follows that P, Q & R are collinear. multiples of the same Must be able to state explicitly the result for collinearity:

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

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

VECTORS: Question 2 Go to full solution Go to Marker’s Comments Go to Vectors Menu Go to Main Menu Reveal answer only EXIT (a)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.

VECTORS: Question 2 Go to full solution Go to Marker’s Comments Go to Vectors Menu Go to Main Menu Reveal answer only EXIT (a)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) [ ] 4 -3 = FG  [ ] 3 0 -3 = so  = 73.9°

Markers Comments Begin Solution Continue Solution Question 2 Vectors Menu Back to Home FE = e - f =  (a) [ ] 4 3 [ ] 5 2 - [ ] 4 -3 = FG = g - f =  [ ] 8 [ ] 5 2 - [ ] 3 0 -3 = (a)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.

Markers Comments Begin Solution Continue Solution Question 2 Vectors Menu Back to Home (a)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. (b) Let angle EFG =   E F G ie FE.  FG =  [ ] 4 -3 [ ] 3 0 -3. = (-1 X 3) + (4 X 0) + (-3 X (-3)) = -3 + 0 + 9 = 6 |FE| =  ((-1) 2 + 4 2 + (-3) 2 )  |FG| =  (3 2 + 0 2 + (-3) 2 )  =  26 =  18

Markers Comments Begin Solution Continue Solution Question 2 Vectors Menu Back to Home (a)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. Given that FE.FG = |FE||FG|cos   then cos  = FE.FG  |FE||FG| = 6  26  18 so  = cos -1 (6   26   18) = 73.9°

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

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

VECTORS: Question 3 Go to full solution Go to Marker’s Comments Go to Vectors Menu Reveal answer only EXIT P Q R S T U V W A B PQRSTUVW is a cuboid in which PQ, PS & PW are represented by the vectors    [ ], 4 2 0 [ ] and -2 4 0 [ ] resp. 0 0 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.  

VECTORS: Question 3 Go to full solution Go to Marker’s Comments Go to Vectors Menu Reveal answer only EXIT P Q R S T U V W A B PQRSTUVW is a cuboid in which PQ, PS & PW are represented by the vectors    [ ], 4 2 0 [ ] and -2 4 0 [ ] resp. 0 0 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.   |PA|  =  29 |PB|  =  106 = 48.1° APB

Markers Comments Begin Solution Continue Solution Question 3 Vectors Menu Back to Home PA =  PS + SA =  PS + 1 / 3 ST  PS + 1 / 3 PW  = [ ] -2 4 0 [ ] = 0 0 3 + [ ] -2 4 3 = PB =  PQ + QV + VB  PQ + PW + 1 / 2 PS   = [ ] 4 2 0 [ ] + 0 0 9 + [ ] = 2 0 [ ] 3 4 9 = PQRSTUVW is a cuboid in which PQ, PS & PW are represented by vectors    [ ], 4 2 0 [ ] and -2 4 0 [ ] resp. 0 0 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.  

Markers Comments Begin Solution Continue Solution Question 3 Vectors Menu Back to Home PA =  [ ] -2 4 3 PB =  [ ] 3 4 9 PQRSTUVW is a cuboid in which PQ, PS & PW are represented by vectors    [ ], 4 2 0 [ ] and -2 4 0 [ ] resp. 0 0 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.   (b) Let angle APB =   A P B ie PA.  PB =  [ ] -2 4 3 [ ] 3 4 9. = (-2 X 3) + (4 X 4) + (3 X 9) = -6 + 16 + 27 = 37

Markers Comments Begin Solution Continue Solution Question 3 Vectors Menu Back to Home PQRSTUVW is a cuboid in which PQ, PS & PW are represented by vectors    [ ], 4 2 0 [ ] and -2 4 0 [ ] resp. 0 0 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.   PA.  PB =  37 |PA| =  ((-2) 2 + 4 2 + 3 2 )  =  29 |PB| =  (3 2 + 4 2 + 9 2 )  =  106 Given that PA.PB = |PA||PB|cos   then cos  = PA.PB  |PA||PB| = 37  29  106 so  = cos -1 (37   29   106) = 48.1°

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

Markers Comments Vectors Menu Back to Home Next Comment PA.  PB =  |PA| =  ((-2) 2 + 4 2 + 3 2 )  =  29 |PB| =  (3 2 + 4 2 + 9 2 )  =  106 Given that PA.PB = |PA||PB|cos   then cos  = PA.PB  |PA||PB| so  = cos -1 (37   29   106) = 48.1° Refer to formula sheet and relate formula to given variables:

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

VECTORS: Question 4 Go to full solution Go to Marker’s Comments Go to Vectors Menu Go to Main Menu Reveal answer only EXIT An equilateral triangle has sides 4 units long which are represented by the vectors a, b & c as shown. b c a Find (i)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 = 4 2 = 16

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

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

Markers Comments Vectors Menu Back to Home Next Comment NB: each angle is 60° but vectors should be “tail to tail” b 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 Dot product = 0 so b & (a + c) are perpendicular. This geometric question is based on the scalar product definition and the distributive law: anda(b+c) = ab + ac No other results should be applied.

Markers Comments Vectors Menu Back to Home Next Comment NB: each angle is 60° but vectors should be “tail to tail” b 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 Dot product = 0 so b & (a + c) are perpendicular. Ensure all scalar products are calculated from the vertex:

VECTORS: Question 5 Go to full solution Go to Marker’s Comments Go to Vectors Menu Go to Main Menu Reveal answer only EXIT 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.

VECTORS: Question 5 Go to full solution Go to Marker’s Comments Go to Vectors Menu Go to Main Menu Reveal answer only EXIT 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. (a)(i)2a + b [ ] 3 2 4 = =  29 (ii) | 2a + b | Since the dot product is zero it follows that the vectors are perpendicular. (b)

Markers Comments Begin Solution Continue Solution Question 5 Vectors Menu Back to Home 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. (a)(i)2a + b = [ ] 2 1 -3 [ ] 0 10 + 2 [ ] 4 2 -6 [ ] 0 10 += [ ] 3 2 4 =

Markers Comments Begin Solution Continue Solution Question 5 Vectors Menu Back to Home 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. (ii) | 2a + b | =  (3 2 + 2 2 + 4 2 ) =  29

Markers Comments Begin Solution Continue Solution Question 5 Vectors Menu Back to Home 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. (b) (2a + b ).c = [ ] 3 2 4 -2 1 1. = (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.

Markers Comments Vectors Menu Back to Home Next Comment (a)(i)2a + b = [ ] 2 1 -3 [ ] 0 10 + 2 [ ] 4 2 -6 [ ] 0 10 += [ ] 3 2 4 = When a vector is given in i, j, k form change to component form: e.g.5i - 2j + k =

Markers Comments Vectors Menu Back to Home Next Comment (ii) | 2a + b | =  (3 2 + 2 2 + 4 2 ) =  29 Must know formula for the magnitude of a vector:

Markers Comments Vectors Menu Back to Home Next Comment (b) (2a + b ).c = [ ] 3 2 4 -2 1 1. = (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. Must know condition for perpendicular vectors:

HIGHER – ADDITIONAL QUESTION BANK UNIT 3 : Further Calculus You have chosen to study: Please choose a question to attempt from the following: 1234 EXIT Back to Unit 3 Menu

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

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

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

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

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

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

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

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

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

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

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

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

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

Markers Comments Further Calc Menu Back to Home Next Comment if dy / dx = -12sin(3x) theny =  -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 Soy = 4cos(3x) + 2 Learn the result that integration undoes differentiation: i.e.given

Markers Comments Further Calc Menu Back to Home Next Comment if dy / dx = -12sin(3x) theny =  -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 Soy = 4cos(3x) + 2 Check formula sheet for correct result: sinax -cos(ax) a cos(ax) sin(ax) a + c Do not forget the constant of integration.

FURTHER CALCULUS : Question 4 Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu Reveal answer only EXIT (a)Find the derivative of y = (2x 3 + 1) 2 / 3 where x > 0. (b)Hence find x2x2 (2x 3 + 1) 1 / 3 dx 

FURTHER CALCULUS : Question 4 Go to full solution Go to Marker’s Comments Go to Further Calculus Menu Go to Main Menu Reveal answer only EXIT (a)Find the derivative of y = (2x 3 + 1) 2 / 3 where x > 0. (b)Hence find x2x2 (2x 3 + 1) 1 / 3 dx  = 4x 2 (2x 3 + 1) 1 / 3 = 1 / 4 (2x 3 + 1) 2 / 3 + C

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

Markers Comments Begin Solution Continue Solution Question 4 Further Calc Menu Back to Home (b) Hence find x2x2 (2x 3 + 1) 1 / 3 dx  (b) From (a) it follows that 4x 2 (2x 3 + 1) 1 / 3 dx  = (2x 3 + 1) 2 / 3 + C 4x 2 (2x 3 + 1) 1 / 3 dx  now x2x2 (2x 3 + 1) 1 / 3 dx  = ¼ X = 1 / 4 (2x 3 + 1) 2 / 3 + C

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

Markers Comments Further Calc Menu Back to Home Next Comment (b) From (a) it follows that 4x 2 (2x 3 + 1) 1 / 3 dx  = (2x 3 + 1) 2 / 3 + C 4x 2 (2x 3 + 1) 1 / 3 dx  now x2x2 (2x 3 + 1) 1 / 3 dx  = ¼ X = 1 / 4 (2x 3 + 1) 2 / 3 + C Learn the result that integration undoes differentiation: i.e.given b)

HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs.

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HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs.

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Presentation on theme: "HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs."— Presentation transcript:

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