PLC Papers. Created For: - PDF Free Download

PLC Papers Created For:

Algebra and proof 2 Grade 8 Objective: Use algebra to construct proofs Question 1 a) If n is a positive integer explain why the expression 2n + 1 is always an odd number. b) Use algebra to prove that the product of two odd numbers is also odd. (1) Question 2 (4) a) If x > 3 and prove that F > 1 b) Explain what happens if x = 3 (4) (1) Total /10

Composite Functions 2 Grade 7 Objective: notation Interpret the succession of two functions as a composite function including the correct Question 1. The functions f and g are such that ( ) = 4 ( ) = 2 + 1 Write in its simplest form the function (a) ( ). (2) (b) ( ). (2) (Total 4 marks)

Question 2. The functions p and r are such that ( ) = 2 2 ( ) = 4-3 (a) Write in its simplest form, the function ( ) (b) Calculate the value of. (2) (1) (c) Calculate the value of. (2) ( 1). (2) (Total 6 marks) Total /10

Expand the product of two or more binomials 2 Grade 7 Objective: Expand the product of two or more binomials Question 1. (a) Show that ( 2)( + 1)(2 3)= 2 3 5 2 3 + 9. (3) (b) Show that (3 1)( + 5)(3 1) = 9 3 + 39 2 29 + 5. (3) (Total 6 marks)

Question 2. Work out (2 1) 2 (3 4 ) 3 3 + 7. (Total 4 marks) Total /10

Exponential Graphs 2 Grade 8 Objective: Recognise, sketch, and interpret graphs of exponential functions Question 1 For each of the 3 statements, indicate whether it is TRUE or FALSE. You explain your answers. (a) The graph =2 passes through the point (2, 2). (b) The graph =5 passes through the point (-1, 0.2). (c) The graph =10 passes through the point (3, 100).

Question 2 Here are the equations of six different graphs: 5 +2 8=0 =5 = 5 =( +5) 2 1 =5 3 +12 =5 Match one of the equations to each of the following graphs: (3)

Question 3 (a) Complete the table of values for y = 2 x x 2 1 0 1 2 y (2) (b) On the grid, draw the graph of y = 2 x (2) (Total 4 marks) TOTAL /10

Factorising difficult quadratic expressions 2 Grade 7 Objective: Factorise a quadratic expression of the form ax 2 + bx + c Question 1 Factorise x 2-17x+30 (2 Marks) Question 2 Factorise 5x 2 +14x+8 (3 Marks) Question 3 Factorise 9x 2-25 (2 Marks) Question 4 Factorise 5x 2-8x-4 (3 Marks) Total marks / 10

Geometric Sequences 2 Grade 7 Objective: Recognise and use geometric sequences (r n, where n is an integer and r can be a surd) Question 1. Find the 5 th and 6 th terms of the sequences below. (a) 0.3, 0.9, 2.7, 8.1,,, (1) (b) -5, 1, -0.2, 0.04,,, (1) Question 2. (Total 2 marks) (a) Write down the first four terms of the geometric sequence with nth term 3 n.... (2) (b) State the term-to-term rule of the sequence.... (1) Question 3. In this geometric sequence, the first term is 3 and the term-to-term rule is multiply by 3. Continue the sequence for three more terms. 5, 5, 5 5,,,,

Question 4. Work out the missing terms in this geometric sequence., 3 8, 1 1 8, 3 3 8, (Total 2 marks) Total /10

Gradients and area under a graph 2 Grade 8 Objective: Calculate or estimate the gradient of a graph and the area under a graph Question 1 A straight line has been drawn on a grid. Calculate the gradient of the line. (2) (Total 2 marks) Question 2 Work out the gradient of the line 5 3 =20 (2) (Total 2 marks)

Question 3 The graph of = 3 +3 2 2 1 is drawn on the grid below. Calculate an estimate to the gradient of the curve at the point Q(-1, 3). (3)

Question 4 The scatter graph shows the cost of cars in a used car showroom. (a) Draw a line of best fit and calculate the gradient of this line. (2) (b) Give an interpretation of this gradient. (1) TOTAL /10

Quadratic equations (completing the square) 2 Grade 8 Objective: Solve quadratic equations by completing the square. Question 1. Rewrite 2 + 6 + 7 in the form ( + ) 2 (Total 1 mark) Question 2. Solve 2 10 + 9 = 0 by completing the square. (Total 2 marks)

Question 3. Solve 2 8 12 = 0 by completing the square. Leave your answers in surd form. Question 4. Solve 4 2 + 28 24 = 0 by completing the square. Give your answers to 3 significant figures. (Total 4 marks) TOTAL /10

Quadratic equations (needing re-arrangements) 2 Grade 7 Objective: Solve quadratic equations that need rearrangement Question 1. Solve 6(18+4 ) +2 = 12 (Total 2 marks) Question 2. Show that 10 6 = 1 1 2 can be written as 2 9 + 18 = 0 and hence calculate the two solutions. (Total 2 marks)

Question 3. Solve 21 5 = 4 +2 +1 Question 4. Solve 42 8 = 5 +3 +1 TOTAL /10

Quadratic equations (quadratic formula) 2 Grade 7 Objective: Solve quadratic equations by using the quadratic equation formula. Question 1. Solve 3 2 5 + 1 = 0. Give your answer to 3 significant figures. = or = Question 2. Solve 1 2 2 + 3 19 = 0. Give your answer to 2 decimal places. = or =

Question 3. a) Solve 2 3 2 = 0 Give your answer to 2 decimal places. = or = (3) b) Write down the solutions, correct to 2 decimal places, of 4 2 12 8 = 0 = or = (1) (Total 4 marks) TOTAL /10

Represent quadratic inequalities 2 Grade 7 Objective: Represent the solution to a quadratic inequality on a number line, using set notation and on a graph Question 1. a) Solve 2 + 11x + 24 > 0 Represent your solution on a number line. b) Write the integer answers for part a) in set notation. (Total 2 marks) Question 2. Solve 2 21 + 4 Display your answer on a sketch of the graph of the solution (Total 2 marks)

Question 3. For which values of x is the expression 2 4 7 less than or equal to the expression 3 3 2 2? Represent the possible values of on a number line. Question 4. Find the set(s) of all values for which 15 +39 +3 > 7 Display your answer on a sketch of the graph of the solution TOTAL /10

Simultaneous equations (non-linear) 2 Grade 7 Objective: Question 1. Solve two simultaneous equations (one linear, one quadratic) algebraically and approximately graphically Solve this pair of simultaneous equations. = 2 +9 +4 =2 8 (Total 2 marks) Question 2. Solve these simultaneous equations. = 6 2 =8 (Total 2 marks)

Question 3. Calculate the solutions to these simultaneous equations. 2 + 2 =29 = 7 Question 4. Use graphical methods to find the approximate solutions to this pair of simultaneous equations 2 5. = 2 3 4 2 = 4 TOTAL /10

Solve quadratic inequalities 2 Grade 7 Objective: Solve quadratic inequalities in one variable Question 1. Solve a) 2 81 > 0 b) 4 2 49 < 0 c) 2 2 11 0 Question 2. Solve the inequality 2 9 + 18 0 (Total 2 marks)

Question 3. Solve the inequality 2 < 24 + 2 (Total 2 marks) Question 4. Solve 3 2 8 > 2 TOTAL /10

Translations and reflections of a function 2 Grade 7 Objective: Sketch translations and reflections of a function Question 1 The graph of = ( ) is shown below. Below each sketch below, write down the equation of the transformed graph y = y = (4) (Total 4 marks)

Question 2 The graph of = ( ) is shown on the grid below. (a) On the same grid draw the graph of = ( ) (2) (b) On grid above draw the graph of =3 ( ) (2)

Trigonometric Graphs 2 Grade 8 Objective: Recognise, sketch, and interpret graphs of trigonometric functions Question 1 Sketch the graph of y = tan x for 0 360 (3) Question 2 Here is the graph of y = cos x for 0 360 On the axes above, sketch the graph =cos(2 ) 2 for 0 360 (3)

Question 3 The graph of y = sin x for 0 360 is shown below. What are the coordinates of the 4 points labelled on the graph? (, ) (, ) (, ) (, ) (4) (Total 4 marks) TOTAL /10

Turning points and completing the square 2 Grade 7 Objective: Deduce turning points by completing the square Question 1 The graph of y = f(x) is shown below. Write down the turning point of the graph. (, ) (2) (Total 2 marks)

Question 2 The expression 2 8 +7 can be written in the form ( ) 2 (a) Find the values of p and q. p = q = (3) The equation of a curve is = ( ) where ( )= 2 8 +7 The diagram shows a sketch of the graph = ( ). B B is the minimum point of the curve. (b) Write down the coordinates of B. (, ) (1)

Question 3 Use completing the square to find the minimum point of the curve = 2 +8 +1. (4) (Total 4 marks) TOTAL /10

General iterative processes 2 Grade 7 Objective: Work with general iterative processes Question 1. The cubic equation x 3 + 2x 5 = 0 has a solution which lies between 1 and 2. Use the decimal search method and the table below to find the solution correct to 1dp. x Value of x 3 + 2x 5 Positive or Negative? Question 2. Xn+1 = 1 + 1 2 with X1 = 1.4 (a) Work out the values of X2 and X3... (b) Work out the solution correct to 2 decimal places.... (2)... (1)

PLC Papers Created For:

Algebra and proof 2 Grade 8 Solutions Objective: Use algebra to construct proofs Question 1 a) If n is a positive integer explain why the expression 2n + 1 is always an odd number. 2n is a multiple of 2 so it must be even so 2n + 1 is the number after an even number so it must be odd. b) Use algebra to prove that the product of two odd numbers is also odd. Question 2 (2n + 1) (2m + 1) = 4mn + 2n + 2m + 1 = 2 ( 2mn + n + m) + 1 2 ( 2mn + n + m) must be even so 2 ( 2mn + n + m) + 1 must be odd a) If x > 3 and prove that F > 1 Expand and simplify brackets Factorise Explain why factorised part is even State result must be odd Factorise numerator Factorise denominator Simplify fraction Explain why F > 1 (1) (4) x + 2 > x so numerator is bigger than denominator hence F > 1 b) Explain what happens if x = 3 If x = 3 then x 3 = 0 If you divide by x 3 you are dividing by 0 so F is undefined (May write you can t divide by 0) (4) (1) Total /10

Composite Functions 2 Grade 7 Solutions Objective: notation Interpret the succession of two functions as a composite function including the correct Question 1. The functions f and g are such that ( ) = 4 ( ) = 2 + 1 Write in its simplest form the function (a) ( ) ( ) = ( + ) (M1) ( ) = + (A1). (2) (b) ( ) ( ) = ( ) + (M1) ( ) = + (A1). (2) (Total 4 marks)

Question 2. The functions p and r are such that ( ) = 2 2 ( ) = 4-3 (a) Write in its simplest form, the function ( ) ( ) = ( ) (M1) = (A1) (b) Calculate the value of. (2) (1) ( ) = 2 (A1) ( ) = = 8-3 = 5 (A1) (c) Calculate the value of. (2) ( 1) ( ) = = = (A1) ( ) = ( ) = = (A1). (2) (Total 6 marks)

Total /10

Expand the product of two or more binomials 2 Grade 7 Solutions Objective: Expand the product of two or more binomials Question 1. (a) Show that ( 2)( + 1)(2 3)= + ( )( + ) = + = ( ) ( ) = + + = +. (3) (b) Show that (3 1)( + 5)(3 1) = 9 3 + 39 2 29 + 5 ( )( + ) = + = + ( + )( ) = + + = + +. (3) (Total 6 marks)

Question 2. Work out (2 1) 2 (3 4 ) 3 3 + 7 ( )( ) = + = + ( + ) ( ) = + + = + + + + + = + +. (Total 4 marks) Total /10

Exponential Graphs 2 Grade 8 Solutions Objective: Recognise, sketch, and interpret graphs of exponential functions Question 1 For each of the 3 statements, indicate whether it is TRUE or FALSE. You explain your answers. (a) The graph =2 passes through the point (2, 2). 2 2 = 4 not 2 (or explanation that leads to this) FALSE (C1) (b) The graph =5 passes through the point (-1, 0.2). 5-1 = 1/5 = 0.2 (or explanation that leads to this) TRUE (C1) (c) The graph =10 passes through the point (3, 100). 10 3 = 1000 not 100 (or explanation that leads to this) FALSE (C1)

Question 2 Here are the equations of six different graphs: 5 +2 8=0 =5 = 5 =( +5) 2 1 =5 3 +12 =5 Match one of the equations to each of the following graphs: = 5 (A1) =( +5) 2 1 (A1) =5 (A1) (3)

Question 3 (a) Complete the table of values for y = 2 x x 2 1 0 1 2 y ¼ o.e. ½ o.e. 1 2 4 (b) On the grid, draw the graph of y = 2 x Any 3 correct All correct (M1) (A1) (2) Points plotted correctly from their table Fully correct graph (smooth) (M1) (G1) (2) (Total 4 marks) TOTAL /10

Factorising difficult quadratic expressions 2 Grade 7 Solutions Objective: Factorise a quadratic expression of the form ax 2 + bx + c Question 1 Factorise x 2-17x+30 (x-15)(x-2) A2 (2 Marks) Question 2 Factorise 5x 2 +14x+8 5x 2 +10x +4x+8 5x(x+2) +2(x+2) M1 (5x+2)(x+2) A2 (3 Marks) Question 3 Factorise 9x 2-25 (3x+5)(3x-5) A2 (2 Marks) Question 4 Factorise 5x 2-8x-4 5x 2-10x +2x-4 5x(x-2) +2(x-2) M1 (5x+2)(x-2) A2 (3 Marks) Total marks / 10

Geometric Sequences 2 Grade 7 Solutions Objective: Recognise and use geometric sequences (r n, where n is an integer and r can be a surd) Question 1. Find the 5 th and 6 th terms of the sequences below. (a) 0.3, 0.9, 2.7, 8.1,,, 24.3, 72.9 (A1) (1) Common ratio is 3 8.1 3 = 24.3 and 24.3 3 = 72.9 (b) -5, 1, -0.2, 0.04,,, -0.008, 0.0016 (A1) (1) Common ratio is -0.2-0.2 0.04 = -0.008 and -0.2-0.008 = 0.0016 Question 2. (Total 2 marks) (a) Write down the first four terms of the geometric sequence with nth term 3 n. 3 1, 3 2, 3 3, 3 4, (M1) 3, 9, 27, 81, (A1)... (2) (b) State the term-to-term rule of the sequence. The term-to-term rule is multiply the previous term by 3.... (1)

Question 3. In this geometric sequence, the first term is 3 and the term-to-term rule is multiply by 3. Continue the sequence for three more terms. 5, 5, 5 5,,,, 25, 25 5, 125 5 5 5 = 5 5 5 = 5 5 =25 (M1) 25 5 = 25 5 (A1) 25 5 5 = 25 5 5 = 25 5 = 125 (A1) Question 4. Work out the missing terms in this geometric sequence. 1 8, 3 8, 1 1 8, 3 3 8,10 1 8 Common ratio is 3. (M1) 3 8 3 = 1 2 and 3 3 8, 3 = 10 1 8 1 8 and 10 1 8 (A1) (Total 2 marks) Total /10

Gradients and area under a graph 2 Grade 8 Solutions Objective: Calculate or estimate the gradient of a graph and the area under a graph Question 1 A straight line has been drawn on a grid. Calculate the gradient of the line. = 4 2 (M1) m = -2 (A1) (2) (Total 2 marks) Question 2 Work out the gradient of the line 5 3 = 20 Correct attempt to make y the subject: = 3 5 + 20 (M1) = 3 5 (A1) (2) (Total 2 marks)

Question 3 The graph of = 3 + 3 2 2 1 is drawn on the grid below. Calculate an estimate to the gradient of the curve at the point Q(-1, 3). Consider points just above and just below, i.e. x = -1.1 and x = -0.9 (M1) (-1.1, 3.499) and (-0.9, 2.501) = 2.501 3.499 0.9+1.1 (M1) = 4.99 m = -4.99 (or -5) (A1) (3)

Question 4 The scatter graph shows the cost of cars in a used car showroom. (a) Draw a line of best fit and calculate the gradient of this line. Using their line, = 8000 8 or use of any other points (M1) m = -1000 (A1) (2) (b) Give an interpretation of this gradient. The value of a car goes down by 1000 every year it gets older (or similar explanation) (C1) (1) TOTAL /10

Quadratic equations (completing the square) 2 Grade 8 SOLUTIONS Objective: Solve quadratic equations by completing the square. Question 1. Rewrite 2 + 6 + 7 in the form ( + ) 2 2 + 6 + 9 9 + 7 ( + 3) 2 2 (A1) (Total 1 mark) Question 2. Solve 2 10 + 9 = 0 by completing the square. 2 10 + 25 25 + 9 = 0 ( 5) 2 16 = 0 (M1) ( 5) 2 = 16 5 = ±4 = 5 ± 4 = 9 = 1 (A1) (Total 2 marks)

Question 3. Solve 2 8 12 = 0 by completing the square. Leave your answers in surd form. 2 8 + 16 16 12 = 0 ( 4) 2 28 = 0 (M1) ( 4) 2 = 28 4 = ± 28 = 4 ± 28 = 4 ± 2 7 (M1) (A1) Question 4. Solve 4 2 + 28 24 = 0 by completing the square. Give your answers to 3 significant figures. 2 + 7 6 = 0 (M1) 2 + 7 + 12.25 12.25 6 = 0 ( + 3.5) 2 18.25 = 0 (M1) ( + 3.5) 2 = 18.25 + 3.5 = ± 18.25 = 3.5 ± 18.25 = 0.772 = 7.77 (M1) (A1) (Total 4 marks) TOTAL /10

Quadratic equations (needing re-arrangement) 2 Grade 7 Solutions Objective: Solve quadratic equations that need rearrangement Question 1. Solve 6(18+4 ) +2 = 12 108 + 24 = 12 ( + 2) 108 + 24 = 12 2 + 24 108 = 12 2 (M1) 9 = 2 = ±3 (A1) (Total 2 marks) Question 2. Show that 10 4 = 1 1 2 can be written as 2 9 + 18 = 0 and hence calculate the two solutions. 10( 2) 4( 1) = 1( 1)( 2) 10 20 4 + 4 = 2 3 + 2 6 16 = 2 3 + 2 0 = 2 9 + 18 (M1) 0 = ( 3)( 6) = 3 = 6 (A1) (Total 2 marks)

Question 3. Solve 21 5 = 4 +2 +1 21( + 1) 5( + 2) = 4( + 2)( + 1) (M1) 21 + 21 5 10 = 4( 2 + 3 + 2) 16 + 11 = 4 2 + 12 + 8 0 = 4 2 4 3 (M1) 0 = (2 + 1)(2 3) = 1 2 = 3 2 (A1) Question 4. Solve 42 +3 8 +1 = 5 42( + 1) 8( + 3) = 5( + 3)( + 1) (M1) 42 + 42 8 24 = 5( 2 + 4 + 3) 34 + 18 = 5 2 + 20 + 15 0 = 5 2 14 3 (M1) 0 = (5 + 1)( 3) = 1 5 = 3 (A1) TOTAL /10

Quadratic equations (quadratic formula) 2 Grade 7 SOLUTION Objective: Solve quadratic equations by using the quadratic equation formula. Question 1. Solve 3 2 5 + 1 = 0. Give your answer to 3 significant figures. = 5± ( 5)2 4 3 1 2 3 (M1) = 0.2324081208 or = 1.434258546 = 0.232 or = 1.43 (A2) = or = Question 2. Solve 1 2 2 + 3 19 = 0. Give your answer to 2 decimal places. = 3± 32 4 1 2 19 2 1 2 (M1) = 3.8556546 or = 9.8556546 = 3.86 or = 9.86 (A2) = or =

Question 3. a) Solve 2 3 2 = 0 Give your answer to 2 decimal places. = 3± ( 3)2 4 1 2 2 1 (M1) = 3.561552813 or = 0.5615528128 = 3.56 or = 0.56 (A2) = or = (3) b) Write down the solutions, correct to 2 decimal places, of 4 2 12 8 = 0 4( 2 3 2) = 0 = 3.56 or = 0.56 (A1) = or = (1) (Total 4 marks) TOTAL /10

Represent quadratic inequalities 2 Grade 7 Solutions Objective: Represent the solution to a quadratic inequality on a number line, using set notation and on a graph Question 1. a) Solve 2 + 11x + 24 < 0 Represent your solution on a number line. ( + 8)( + 3) < 0 8 < < 3 b) Write the integer answers for part a) in set notation. { -7, -6, -5, -4 } (A1) (A1) (Total 2 marks) Question 2. Solve 2 21 + 4 Display your answer on a sketch of the graph of the solution 2 4 21 0 ( 7)( + 3) 0-3 7 (Total 2 marks)

Question 3. For which values of x is the expression 2 4 7 less than or equal to the expression 3 3 2 2? Represent the possible values of on a number line. 2 4 7 3 3 2 2 3 2 10 0 (M1) (3 + 5)( 2) 0 (M1) 5 3 2 (A1) Question 4. Find the set(s) of all values for which 15 +39 +3 > 7 Display your answer on a sketch of the graph of the solution 15 + 39 > (7 )( + 3) 15 + 39 > 21 + 4 2 (M1) 2 + 11 + 18 > 0 ( + 9)( + 2) > 0 (M1) ( 7)( + 3) 0-3 7

TOTAL /10

Simultaneous equations (non-linear) 2 Grade 7 Solutions Objective: Solve two simultaneous equations (one linear, one quadratic) algebraically and approximately graphically Question 1. Solve this pair of simultaneous equations. = 2 + 9 + 4 = 2 8 2 + 9 + 4 = 2 8 2 + 7 + 12 = 0 (M1) ( + 4)( + 3) = 0 = 4 = 16 = 3 = 14 (A1) (Total 2 marks) Question 2. Solve these simultaneous equations. = 6 2 = 8 (2 8) = 6 2 2 8 + 6 = 0 2 4 + 3 = 0 (M1) ( 3)( 1) = 0 = 3 = 2 = 1 = 6 (A1) (Total 2 marks)

Question 3. Calculate the solutions to these simultaneous equations. 2 + 2 = 29 = 7 2 + ( 7) 2 = 29 (M1) 2 + 2 14 + 49 = 29 2 2 14 + 20 = 0 2 7 + 10 = 0 (M1) ( 5)( 2) = 0 = 5 = 2 = 2 = 5 (A1) Question 4. Use graphical methods to find the approximate solutions to this pair of simultaneous equations 2 5. = 2 3 4 2 = 4 TOTAL /10

Solve quadratic inequalities 2 Grade 7 SOLUTIONS Objective: Solve quadratic inequalities in one variable Question 1. Solve a) 2 81 > 0 ( + 9)( 9) > 0 < 9, > 9 (A1) b) 4 2 49 < 0 (2 + 7)(2 7) < 0 7 2 < < 7 2 (A1) c) 2 2 11 0 (2 11) 0 0 11 2 (A1) Question 2. Solve the inequality 2 9 + 18 0 ( 6)( 3) 0 (M1) 3, 6 (A1) (Total 2 marks)

Question 3. Solve the inequality 2 < 24 + 2 2 2 24 < 0 ( 6)( + 4) < 0 (M1) 4 < < 6 (A1) (Total 2 marks) Question 4. Solve 3 2 8 > 2 3 2 2 8 > 0 (M1) (3 + 4)( 2) > 0 (M1) < 4 3, > 2 (A1) TOTAL /10

Translations and reflections of a function 2 Grade 7 Solutions Objective: Sketch translations and reflections of a function Question 1 The graph of = ( ) is shown below. Below each sketch below, write down the equation of the transformed graph y = -f(x) y = f(x+1) - 3 (4) (Total 4 marks)

Question 2 The graph of = ( ) is shown on the grid below. 1.5 1 0.5 0-8 -6-4 -2 0 2 4 6 8 10-0.5-1 -1.5 y = f(x) y = f(-x) (a) On the same grid draw the graph of = ( ) (2) 4 3 2 1 0-8 -6-4 -2 0 2 4 6 8 10-1 -2-3 -4 y = f(x) y = 3f(x) (b) On grid above draw the graph of =3 ( ) (2)

3.5 3 2.5 2 1.5 1 0.5 0-8 -6-4 -2-0.5 0 2 4 6 8 10-1 -1.5 y = f(x) y = -f(x)+2 (c) On grid above draw the graph of = ( ) + 2 (2) (Total 6 marks) TOTAL /10

Trigonometric Graphs 2 Grade 8 Solutions Objective: Recognise, sketch, and interpret graphs of trigonometric functions Question 1 Sketch the graph of y = tan x for 0 360 (3) Question 2 Here is the graph of y = cos x for 0 360 1.5 1 0.5 0-0.5-1 -1.5-2 -2.5-3 -3.5 0 1 2 3 4 5 6 7 y = cos x y = cos(2x) -2 On the axes above, sketch the graph =cos(2 ) 2 for 0 360 (3)

Question 3 The graph of y = sin x for 0 360 is shown below. What are the coordinates of the 4 points labelled on the graph? ( 0, 0 ) ( 90, 1 ) ( 270, -1 ) ( 360, 0 ) (4) (Total 4 marks) TOTAL /10

Turning points and completing the square 2 Grade 7 Solutions Objective: Deduce turning points by completing the square Question 1 The graph of y = f(x) is shown below. Write down the turning point of the graph. ( 2.5, 1.25 ) B1 B1 (2) (Total 2 marks)

Question 2 The expression 2 8 +7 can be written in the form ( ) 2 (a) Find the values of p and q. (x 4) 2-16 + 7 M1 (x 4) 2-9 p = 4 B1 q = -9 B1 SC B1 for -4, 9 (3) The equation of a curve is = ( ) where ( )= 2 8 +7 The diagram shows a sketch of the graph = ( ). B B is the minimum point of the curve. (b) Write down the coordinates of B. ( 4, -9 ) (1) (Total 4 marks)

Question 3 Use completing the square to find the minimum point of the curve = 2 +8 +1. (x + 4) 2 16 + 1 M1 (x + 4) 2 15 M1 Min point at ( -4, -15) B1 B1 (4) (Total 4 marks) TOTAL /10

General iterative processes 2 Grade 7 Solutions Objective: Work with general iterative processes Question 1. The cubic equation x 3 + 2x 5 = 0 has a solution which lies between 1 and 2. Use the decimal search method and the table below to find the solution correct to 1dp. x Value of x 3 + 2x 5 Positive or Negative? 1.1 1.1 3 + 2(1.1) 5 = -1.469 Negative 1.2 1.2 3 + 2(1.2) 5 = -0.872 Negative 1.3 1.3 3 + 2(1.3) 5 = -0.203 Negative 1.4 1.4 3 + 2(1.4) 5 = 0.544 Positive 1.31 1.31 3 + 2(1.31) 5 = -0.131909 Negative 1.32 1.32 3 + 2(1.32) 5 = -0.060032 Negative 1.33 1.33 3 + 2(1.33) 5 = 0.012637 Positive Question 2. Correct iterations that identify change of sign (solution) is between 1.3 and 1.4 (M1) Correct iterations that identify change of sign (solution) is between 1.32 and 1.33 (M1) Xn+1 = 1 + 1 2 with X1 = 1.4 (a) Work out the values of X2 and X3 (b) Work out the solution correct to 2 decimal places. So x = 1.3 to 1dp (A1)... X2 = 1.510204082 (B1) X3 = 1.438458729 (B1 ft from X2)... Using the ANS key to continue to generate terms X9 = 1.466566356 and X10 = 1.465003041 (2) Both round to 1.47 (B1)... (1)

Question 3. This iterative process can be used to find approximate solutions to the equation x 3 3x 1 = 0 to 2dp. Start with a value of x 3 Work out the value of 1 + 3 Is your answer to 2 decimal places the same as your value of x to 2 decimal places? Yes No This is an approximate solution to x 3 3x 1 = 0 Use your answer as the next value of x and start again Use this iterative process to find a solution to 2 decimal places to x 3 3x 1 = 0. Start with x = 2 Substitutes in x = 2 to get x = 1.912931183 (M1) Second pass through the flow diagram to give x = 1.888835126 (M1) Third pass through the flow diagram to give x = 1.88205688 (M1) Fourth pass through the flow diagram to give x = 1.880141328 so x = 1.88 to 2dp (A1)... (Total 4 marks) Total /10