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Questions Assessment Standard Description 1 RC1.1 Factorise cubic a≠1 2 RC1.1 Given nature use discriminant to find coefficient 3 EF1.2 Trig identities 4 EF1.2 Application of addition and double angle formulae 5,6 RC1.2 Solve equations in degrees or radians using trig formulae/identities 7 A1.2 Determine and use the equation of a circle 8 A1.2 Use properties of tangency 9 RC1.4 Integrate algebraic functions expressed in powers of x 10 A1.4 Find the area between curve and x axis 11 A1.4 Find the area between line and curve or two curves Relationships and Calculus Assessment Standard 1.1 1 (x – 2) is a factor of the polynomial f (x) = x3 + 2x2 – kx + 42. (a) (b) 2 Show that k = 29. Fully factorise f (x). (3) (3) The x-axis is a tangent to the graph of y = px2 – 2px + 3. Determine, algebraically, the value of p. #2.1 (3) Expressions and Functions Assessment Standard 1.2 3cos x sin x cos x 3sin x 3 5sin 2x . 3 Show that 4 The diagram shows two right-angled triangles. #2.1 (3) (a) Write down the values of sin x and cos y. (1) 16 (b) Show that the exact value of cos( x y ) is . 65 (3) Relationships and Calculus Assessment Standard 1.2 5 6 1 for 0 x 180 º. 2 Show that 2 cos 2 x 3sin 2 x 1 sin 2 x. Solve the equation cos 2 x (a) (b) 1 Hence solve 2 cos 2 x 3sin 2 x , 0 x 360. 4 (2) (2) (5) Applications Assessment Standard 1.2 7 (a) A circle has radius 4 units and centre (-3,-2). Write down the equation of the circle. 2 2 (b) A circle has equation x y 6 x 8 y 24 0 . Write down the coordinates of its centre and the length of its radius. (2) (2) (#2.2 (c) Determine whether the circles in a) and b) intersect. Justify your ) answer. 8 Determine whether the line y = x is a tangent to the circle ( x – 2)2 + ( y + 1)2 = 1. Justify your answer (3) #2.2 Relationships and Calculus Assessment Standard 1.4 9 Find 5 3 x7 dx, x 0 (3) Applications Assessment Standard 1.4 10 The curve with equation (4) Y = 2x3 -5x2 + x + 2 is shown in the diagram with Q at (1,0). Calculate the shaded area 11 The line with equation y 3 x and the curve with equation y 3 x 2 are shown in the diagram. The line and curve meet at the points where x 0 and x 1 . Calculate the shaded area shown in the diagram. (4) 1 (a) •1 Strategy to start process of evaluating polynomial, eg direct substitution or synthetic division •1 f (2) = … and begins substitution •2 Evaluates polynomial and equates to zero •2 f (2) = 58 2k = 0 or or 2 1 2 k 42 2 8 16 2k 1 4 k 8 2k 58 and –2k + 58 = 0 •3 Completes proof •3 k = 29 Notes: Candidates who substitute k with 29 at •1 and then evaluate to obtain zero can be awarded full credit — for showing that the converse is true — provided a statement such as ‘so (x 2) is a factor when k 29 ’ appears at the end. (b) 2 •1 First linear factor •1 (x – 2)(……….) or (x – 3)(……….) or (x + 7)(……….) •2 Extracts quadratic factor •2 (x2 + 4x – 21) or other consistent with •1 •3 Completes factorisation •3 (x – 2)(x – 3)(x + 7) #2.1 Correct strategy #2.1 Evidence of using b2 – 4ac = 0 •1 Sets up correct equation •1 4p2 – 12p = 0 •2 Factorises •2 4p(p – 3) = 0 •3 Solves equation •3 p not equal to zero so p = 3 Notes: An answer of p = 3 with no working receives no credit. … = 0 must appear somewhere before •3 otherwise #2.1 and •1 are unavailable. •2 is only available for attempting to solve the equation by factorising or by using the quadratic formula. For •2 accept an ‘incomplete’ factorisation, eg 2p(2p – 6) or p(4p – 12) but not 4(p2 – 3p). Candidates must clearly reject the solution p = 0 in order to receive •3. Also, for equation 2 p 2 12 p 0 lose •1 •2 available for 2 p( p 6) 0 or p(2 p 12) 0 but not 2( p 2 6 p) 0 •3 available for p 0 so p 6 3 #2.1 Correct strategy #2.1 Uses a trigonometric identity •1 Expands LHS •1 3cos 2 x 10sin x cos x 3sin 2 x •2 Starts process •2 3 cos 2 x sin 2 x ... leading to 3 ... •3 Completes process •3 10sin x cos x 5 2sin x cos x 5sin 2x Notes: •1 may be awarded for 3cos 2 x 9sin x cos x sin x cos x 3sin 2 x . •2 can only be awarded where there is clear evidence of using cos 2 x sin 2 x 1 •3 can only be awarded for clearly demonstrating the equivalence of 10sin x cos x and 5sin 2x . Candidates who use ‘=’ instead of ‘ ’ should not be penalised. 4 (a) 4 12 sin x and cos y 5 13 •1 Interpret diagram for sin x •1 •2 Interpret diagram for •2 cos y 4(b) sin x 4 5 cos y 12 13 16 65 •3 Use compound angle formula •3 cos(x y) = cos x cos y sin x sin y •4 Substitute and complete •4 3 12 4 5 16 5 13 5 13 65 Notes: In (a) The evidence for •1 and •2 may not be evident until (b). In (b) •3 may be stated or implied by •4. Simply stating the formula for cos(A + B) with no further working gains no marks. Calculating approximate angles using arcsin(sin 1 ) and arccos(cos1 ) gains no credit. 5 x 30, 150 •1 Solve equation for 2x 1 2 2 x 60 and 300 1 •2 Process solutions for x 2 x 30 and 150 Notes: Answers shown in the marking scheme (•1 and •2) can be marked horizontally or vertically. π 5π Accept , or decimal equivalent 0·523598…and 2·617993… 6 6 6(a) •1 Substitutes for cos2x •2 Completes process (b) •3 Rearranges equation •4 Takes square root of both sides •5 Solves equation •6 Knows to consider the negative square root •7 Completes solution •1 2 1 2 sin 2 x 3sin 2 x •2 2 1 2sin 2 x 3sin 2 x •3 sin 2 x 3 4 3 •4 sin x 2 5 • x 60, 120 •6 sin x 3 stated explicitly 2 •7 x 240, 300 Notes: Do not award •1 if brackets are omitted, ie 2 1 2 sin 2 x... If candidates obtain all correct answers to (b) purely from sin x 3 , 2 only 3 points of process may be awarded. Do not penalise solutions outwith 0 x 360 . Do not accept, eg 180 60 for 240 . 7(a) ( x 3)2 ( y 2) 2 16 7(b) 7(c) •1 Interpret centre •1 ( x (3))2 ( y (2))2 .... •2 Interpret radius and complete equation •2 ... 16 Centre 3, 4 radius 1 unit •3 State centre of circle •3 3, 4 4 • Know how to and find radius of circle •4 1 The circles do not intersect, with valid explanation. #2.2 Finds distance #2.2 Distance between centres is 6, sum between centres, finds sum of radii is 5, 6 > 5 so the circles do of radii and communicates not intersect Note: In (a) •2 is not awarded for 42, this must be simplified to 16. 8 •1 set up imtersection •1 (x-2)2 + (x +1)2 = 1 equation 9 •2 simplifies •2 •3 calculates the discriminant #2.2 interprets solution 5 5 3 x7 dx 3x 6 x6 c •3 b2 – 4ac = 12 -4x1x2 = -7 #2.2 -7<0 so no points of contact so line is not a tangent to the circle •1 Express in integrable form •1 5x 7 •2 Integrate the constant •2 3x •3 Integrate term with a negative power •3 5 6 x 6 2x2- 2x + 4 = 0 Note: •1 is the only mark available to candidates who differentiate. 10 •1 know to integrate with correct limits •1 •2 integrates •2 ½ x4 - 5/3 x3 + ½x2 +2x •3 substitutes in the limits •3 ½x14 - 5/3 x13 + ½x12 +2x1-(0) •4 •4 4/3 1 (...)dx 0 Note: •3 and 4 are only available to those who integrate 11 1 or 0.25 square units 4 •1 Know to integrate and state limits •2 Use ‘upper – lower’ •1 1 ...dx 0 •2 'upper lower' stated or implied by •3 •3 Interpret upper – lower •3 x3 x •4Integrate •4 •5 Substitute limits •5 14 12 0 4 0 2 4 4 2 2 •6 Process limits •6 1 or equivalent 4 x4 x2 4 2 Notes: Do not penalise candidates who work with 3 x3 3 x throughout in an unsimplified form. Candidates who use 3 x3 3 x leading to x 3 x lose •3 but all other marks are still available. Differentiation loses •4, •5 and •6. •2 is lost for subtracting the wrong way round and subsequently •6 may be lost for statements such as: 1 1 1 1 1 or sq units ’ or ‘ = ’ or ‘ so ignore negative’ 4 4 4 4 4 9 9 •6 may be gained for statements such as ‘ so area is ’ 2 2 ‘ 0 'lower upper ' is correct and so all six marks are still available 1