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2 Formulae FSMQ Additional Maths (6993) Binomial series n n n n-1 n n- n n-r r n 1 r ( a+ b ) = a + C a b+ C a b + + C a b + + b, for positive integers, n, where n n n! C r = nc r = =, r n r r!( n - r )! The binomial distribution n x If X ~B( n, p ) then P( X = x ) = p ( 1- p) x Numerical methods Trapezium rule: b n-x y x 1 d h {( y + y a 0 n )+ ( y 1 + y + + y n -1 )}, where b - a h= n OCR

3 3 Answer all questions 1 A sequence is defined by the rule u = u. 1 n n Determine the value of u 6 given that u =1. [] 3 Find the coefficient of x 3 in the expansion of (+3 x) 5, giving your answer as simply as possible. [4] OCR Turn over

4 4 3 3 You are given that y = x +x - 7. (a) Find d y dx. [] (b) Use your result to part (a) to show that the graph of points. 3 y = x +x - 7 has no turning [] 3(a) 3(b) 4 In this question you must show detailed reasoning. 4 Find the value of ( x +3) d x. [4] 1 OCR

5 5 5 y O The London Eye can be considered to be a circular frame of radius 67.5 m, on the circumference of which are capsules carrying a number of people round the circle. Take a coordinate system where O is the base of the circle and Oy is a diameter. At any time after starting off round the frame, the capsule will be at height h metres when it has rotated. (a) Sketch a graph of h against [] (b) Give an expression for h in terms of. [] (c) Find values of when h = 100. [3] h x 5(a) 5(b) 5(c) OCR Turn over

6 6 6 (a) Simplify x 6 - x+ x-1. [3] In this question you must show detailed reasoning. x 6 (b) Solve - = 4 giving your answer in exact form. [4] x+ x-1 6(a) 6(b) OCR

7 7 7 In this question you must show detailed reasoning. (a) Express x +8x -1 in the form a( x + p) + q. [4] (b) Hence find the minimum value of x +8x -1. [1] 7(a) 7(b) 8 A triangle ABC is such that AB = 5 cm, BC = 8 cm and CA = 7 cm. 8 Show that one angle is 60. [4] OCR Turn over

8 8 9 In this question you must show detailed reasoning. 3 (a) Show that ( x - 3) is a factor of x - 5 x + x +15. [1] (b) Hence solve the equation 3 x - 5 x + x +15 =0. [4] 9(a) 9(b) OCR

9 9 10 A security keypad uses three letters A, B and C and four digits 1-4. A passcode is created using four inputs. (a) If there are no restrictions, how many different passcodes are possible? [1] (b) If there must be exactly two letters and two digits, with no repeats, how many different passcodes are possible? [3] 10(a) 10(b) OCR Turn over

10 10 11 y x The graph shows the curve y = x. (a) Use the trapezium rule with two strips to estimate the area enclosed by the curve, the x-axis, x = and x =4, stating if this is an over or under estimation. [4] (b) (i) Use the chord from (,) to (4,4) to estimate the gradient of y = x x =3. at [1] (ii) Determine an estimate for the gradient of the curve y = x at x =3 which is an improvement of the estimate found in part (b)(i). [] 11(a) OCR

11 11 11(b)(i) 11(b)(ii) OCR Turn over

12 1 1 China cups are packed in boxes of 10. It is known that 1 in 8 are cracked. Find the probability that in a box of 10, chosen at random, (a) exactly 1 cup is cracked, [3] (b) at least cups are cracked. [4] 1(a) 1(b) OCR

13 13 13 E F D 0 0 C A X B The diagram shows a triangular prism. The rectangle ABCD is horizontal and ABFE is a square inclined at 0 to the horizontal such that E is vertically above D and F is vertically above C. The area of the square ABFE is 1600 m. X is a point on AB such that AX = 1 m. Calculate (a) the area of ABCD, [3] (b) the angle between the lines XE and XF. [5] 13(a) 13(b) OCR Turn over

14 14 14 An object is falling through a liquid. The distance fallen is modelled by the formula 3 s = 48t - t until it comes to rest, where s is the distance fallen in centimetres and t is the time in seconds measured from the point when the object entered the liquid. (a) Find (i) the acceleration when t =1, [5] (ii) the time when the object comes to rest, [] (iii) the distance fallen when the object comes to rest. [] (b) 14(a)(i) Sketch the velocity/time graph for the period of time until the object comes to rest. [1] 14(a)(ii) OCR

15 15 14(a)(iii) 14(b) OCR Turn over

16 16 15 John bought a car in January 015 for He investigated how the value of his car might depreciate over the years. He searched the internet and found the following data. Number of years after buying car (t years) Value ( v) He believes that the relationship between the age of the car and its value can be t modelled by the equation v = ka where v is the value in pounds and t is the age in years. (a) Write down the value of k. [1] (b) Show that the equation can be rewritten in the form log v =log k + t loga. [] John plotted log10v against t and obtained the following graph. 5 log10v x x x x x t (c) Use the graph above to estimate a value for a. [3] (d) Use this model to estimate the age of John s car when its value drops below [3] 15(a) OCR

17 17 15(b) 15(c) 15(d) OCR Turn over

18 18 16 A bicycle factory produces two models of bicycle, A and B. Model A requires 0 hours of unskilled and 10 hours of skilled labour. Model B requires 15 hours of unskilled and 5 hours of skilled labour. The factory employs 10 unskilled and 8 skilled labourers, each of whom work a 40 hour week. (a) Suppose the factory makes x model A and y model B per week. (i) Show that the restriction of unskilled labour results in the inequality 4 x+3y 80. [3] (ii) Find a similar inequality from the restriction on skilled labour. [] (b) Draw graphs for the two inequalities and shade the feasible region. [3] (c) Show that making 15 model A bicycles and 5 model B bicycles is possible. [1] (d) The factory makes a profit of 40 on model A and 60 on model B. (i) Write down the objective function. [1] (ii) Find the number of each that should be made to maximise the profit. [3] 16(a)(i) 16(a)(ii) OCR

19 19 16(b) 16(c) There is a spare copy of this graph on page 0. If you wish to offer a second attempt, then you must cross through the attempt on this page. 16(d)(i) 16(d)(ii) OCR Turn over

21 day June 0XX Morning/Afternoon Level 3 Free Standing Mathematics Qualification: Additional Maths 6993 Paper 1 SAMPLE MARK SCHEME Duration: hours MAXIMUM MARK 100 This document consists of 16 pages OCR

22 6993 Mark Scheme June 0XX 1. Annotations and abbreviations Annotation in RM Meaning Assessor and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Text Instructions Other abbreviations Meaning in mark scheme E1 Mark for explaining a result or establishing a given result dep* Mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR The statement In this question you must show detailed reasoning applies to this question.

25 6993 Mark Scheme June 0XX 1 Question Answer Marks AOs Guidance u = 3, u = 45, M1 AO 4 5 u 6 = 89 3 C 3 x [] M1 AO3 Choose correct term 4 and 7 soi 10 soi = (a) d y = 3 x + dx 3 (b) 3 x + 0 for any x, so no turning points 4 DR 3 x +3x 3 1 [4] M1 [] B1 B1 [] M1 AO AO AO Integrate powers increased by 1 ignore limits = M1 Subtract in correct order and substitute limits 1 =5 3 [4] 5

26 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 5 (a) B1 AO3 Max point at 180 degrees and 0 height at 0 and 360 degrees B1 AO correct shaped curve [] 5 (b) h = 67.5(1- cos θ ) B1 B1 AO3 attempt to use cos ratio correct expression 5 (c) h cos θ = [] M1 100 cos θ =1- cos θ = θ =119 and 41 [3] 6

27 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 6 (a) x( x -1) - 6( x + ) ( x+ )( x-1) M1 x - x - 6x -1 ( x+ )( x-1) x - 7x-1 ( x+ )( x-1) [3] 6 (b) DR x - 7x -1 = 4( x +)( x -1) M1 AO3 0 =3 x +11 x ± M1 AO oe completing the square x = method For both x = or x = 6 6 [4] 7 (a) DR = x + 4x- 6 x x = = x = ( x + ) - 0 M1 M1 [4] 7 (b) DR (When x = ) = 0 B1 AO [1] Taking out factor of +4 and -10 soi 7

28 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 8 Triangle is not equilateral and so the angle required is middle angle B1 AO3 Award if cosine rule applied more than once cos B = = = M1 AO B = 60 [4] 9 (a) DR 3 f(3) = =0 B1 [1] 9 (b) DR ( x - 3)( x - x - 5) M1 AO3 ± 4 M1 = x - x - 5 = 0 x = x = 3, 3.45, [4] 10 (a) 7 4 ( = 401) B1 [1] 10 (b) letters = 3 B1 AO digits = 6 Total selection = 18 These can be arranged in 4! Ways B1 No of possibilities = 4! 18 = 43 B1 [3] attempt to divide by ( x - 3) Solve quadratic Must have all three roots For 3 and 6 soi For 4! soi Dep on both previous B 8

29 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 11 (a) x y B1 All values in table correct Area = 1 + their y Area = 5.83 M1 AO Overestimate 11 (b) (i) 4 - Grad chord = = (b) (ii) Choose any pair of values of x either side of x =3in range [,4] their y1-their y Grad chord = their x -their x (Actual answer is 1 1.ln() = [4] B1 [1] M1 [] AO3 AO3 Any answer from correct method nearer to value than in (i) 9

30 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 1 (a) M1 AO3 Binomial term with correct 10 powers 8 8 Coefficient soi [3] 1 (b) P( X ) =1- P(0) - P(1) M1 AO Subtract two terms from 1 oe P( X ) =1- ( ) M1 AO For attempt at P(0) both soi [4] 13 (a) AE = 40 AD = 40cos0 = Area ABCD = their AD 40 = 1500 m (3sf) B1 M1 AO [3] 13 (b) By Pythagoras theorem EX = M1 AO3 EX = 1744 m Similarly FX = = 384 Cosine Rule M1 AO3 cos EXF = cos EXF = EXF =51.7 [5] 10

31 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 14 (a) (i) ds M1 AO3 v= d t v = 48-3t dv M1 a= d t a= -6t a = -6 cms - [5] 14 (a) (ii) v = 48-3 t ; v =0 M1 AO3 FT their expression for v 3 t = 48 t = 4 seconds [] 14 (a) (iii) 3 s = 48t - t ; t = 4 M1 AO3 s 3 4 = 48(4)-(4) =18 Must include units [] 14 (b) B1 AO Shape and intercepts and nothing outside range [1]

32 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 15 (a) k = 8000 B1 [1] 15 (b) Take logs M1 AO One of the log laws must t t log v = log( ka ) = log k +loga be seen AO = log k + t loga AG 15 (c) Attempt to find gradient Loga in range [-0.1, -0.09] a in range [0.79, 0.81] [] M1 AO3 15 (d) Using their equation When v = 3000 t 3000 = 8000(0.8) t (0.8) = log0.107 t = =10 log0.8 So 10 years [3] B1 M1 [3] AO3 AO 1

33 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 16 (a) (i) Number of unskilled labour hours for Model A B1 AO3 0x Number of unskilled labour hours for Model B 15y Total number of available unskilled labour hours is = 400 B1 0 x +15y x + 3y 80 AO AG [3] 16 (a) (ii) Number of skilled labour hours for Model A M1 AO3 10x Number of skilled labour hours for Model B 5y Total number of available unskilled labour hours is 8 40 =30 10 x +5y 30 x +5y 64 AO [] 16 (b) B1 4 x+3y 80 B1 x+5y 64 B1 AO3 shading [3] 16 (c) (15,5) is within feasible region B1 AO3 [1] 13

34 6993 Mark Scheme June 0XX Question Answer Marks AOs Guidance 16 (d) (i) P = 40 x +60y B1 16 (d) (ii) x y 4x+3y x+5y P [1] M1 M1 AO3 AO 14 model A and 7 Model B AO3 [3] Attempt to consider vertices of feasible region Attempt at considering integer points close to vertices 14

35 6993 Mark Scheme June 0XX Question AO AO a 3 b a b c 1 6 a 3 6 b a 4 7 b a 1 9 b a 1 10 b 1 11 a b i 1 11 b ii a 1 1 b 13 a 1 13 b 3 14 a i a ii a iii b 1 15 a 1 15 b 15 c 1 15 d a i a ii b 1 16 c 1 16 d i 1 16 d ii 1 Total

36 6993 Mark Scheme June 0XX BLANK PAGE 16

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