Cambridge A Level Mathematics 9709 — 2018 Feb/March Paper 1 · Variant 2

3 On a certain day, the height of a young bamboo plant was found to be 40 cm. After exactly one day its height was found to be 41.2 cm. Two different models are used to predict its height exactly 60 days after it was first measured. Model A assumes that the daily amount of growth continues to be constant at the amount found ³ for the first day. Model B assumes that the daily percentage rate of growth continues to be constant at the ³ percentage rate of growth found for the first day. (i) Using model A, find the predicted height in cm of the bamboo plant exactly 60 days after it was first measured. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Using model B, find the predicted height in cm of the bamboo plant exactly 60 days after it was first measured. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 2 Question Answer Marks Guidance 3(ii) Find rate of growth e.g. 41.2/40 or 1.2/40 *M1 SOI, Also implied by 3% , 0.03 or 1.03 seen ( ) 60 59 40 1 0.03 or their × + DM1 236 A1 Allow 2.36 m 3

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4 A straight line cuts the positive x-axis at A and the positive y-axis at B 0, 2 . Angle BAO 1 radians, = 60 where O is the origin. (i) Find the exact value of the x-coordinate of A. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the equation of the perpendicular bisector of AB, giving your answer in the form y mx c, = + where m is given exactly and c is an integer. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) 1 2 3 x = or 1 2 3 y x − = M1 OE, Allow 1 2 3 y x + − = . Attempt to express tan tan 6 3 or is required or the use of 1/ 3 3 or √ √ ( )2 3 x = A1 OE 2 4(ii) Mid-point (a, b) = (½ their (i), 1) B1FT Expect (√3, 1) Gradient of AB leading to gradient of bisector, m M1 Expect 1/ 3 − √ leading to 3 m = √ Equation is ( ) y theirb m x their a − = − OE DM1 Expect ( ) 1 3 3 y x −= − 3 2 = − y x OE A1 4

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6 P 10 cm O R 2.2 rad Q The diagram shows a sector POQ of a circle of radius 10 cm and centre O. Angle POQ is 2.2 radians. QR is an arc of a circle with centre P and POR is a straight line. (i) Show that the length of PQ is 17.8 cm, correct to 3 significant figures. 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(ii) Find the perimeter of the shaded region. 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Mark scheme: 6(i) 10 sin1.1 2 × (PQ =) 17.8 (17.82…implies M1, A1) AG A1 OR 10sin 2.2 10sin 2.2 sin0.4708 sin 1.1 2 PQ or π =   −     or 200 200cos2.2 − = 17.8 2 6(ii) Angle OPQ = (π/2 ‒ 1.1) [accept 27° ] B1 OE Expect 0.4708 or 0.471. Can be scored in part (i) Arc QR =17.8 × their (π/2 ‒ 1.1) M1 Expect 8.39. (or 8.38). Perimeter 17.8 10 10 arc their QR = − + + M1 26.2 A1 For both parts allow correct methods in degrees 4

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7 E 8 4 C 3 B 1 D C B 3 D k 2 j O 7 A O i A Fig. 1 Fig. 2 Fig. 1 shows a rectangle with sides of 7 units and 3 units from which a triangular corner has been removed, leaving a 5-sided polygon OABCD. The sides OA, AB, BC and DO have lengths of 7 units, 3 units, 3 units and 2 units respectively. Fig. 2 shows the polygon OABCD forming the horizontal base of a pyramid in which the point E is 8 units vertically above D. Unit vectors i, j and k are parallel to OA, OD and DE respectively. (i) Find −−→CE and the length of CE. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A m(ii) Use a scalar product to find angle ECA, giving your answer in the form cos−1 , where m n and n are integers. 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Mark scheme: 7(i) CE = ‒4i ‒ j + 8k | JJJG CE | = ( ) ( ) ( ) 2 2 2 ( 4 1 8 9 their their their − + − + = M1A1 Could use Pythagoras’ theorem on triangle CDE 3 7(ii) JJJG CA = 3i ‒ 3j or JJJG AC = –3i + 3j B1 JJJG CE . JJJG CA = (‒4i ‒ j + 8k).(3i ‒ 3j) = ‒12 + 3 (Both vectors reversed ok) M1 Scalar product of their JJJG CE , JJJG CA . One vector reversed ok for all M marks | JJJG CE |×| JJJG CA | = 16 1 64 + + × 9 9 + M1 Product of moduli of their JJJG CE , JJJG CA 1 1 12 3 1 cos cos 9 18 18 − − − + −     =         1 1 3 9 or e.g. cos , cos 162 1458 − −   − −                 etc. A1A1 A1 for any correct expression, A1 for required form Equivalent answers must be in required form m/√n (m, n integers) 5

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3 28 A curve has equation y 8x. = 12x2 −4x + (i) Find the x-coordinates of the stationary points. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ d2y (ii) Find . 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(iii) Find, showing all necessary working, the nature of each stationary point. 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Mark scheme: 8(i) ½ d / d 6 8 y x x x = − + Set to zero and attempt to solve a quadratic for ½ x M1 Could use a substitution for ½ x or rearrange and square correctly* ½ ½ 4 or 2 x x = = [ 2 4 x and x = = gets M1 A0] A1 Implies M1. ‘Correct’ roots for their d / d y x also implies M1 16or 4 = x A1FT Squares of their solutions *Then A1,A1 for each answer 5 Question Answer Marks Guidance 8(ii) 2 2 d / d 1 3 − = − y x x ½ B1FT FT on their dy/dx, providing a fractional power of x is present 1 8(iii) (When x = 16) 2 2 d / d y x = 1/4 > 0 hence MIN M1 Checking both of their values in their 2 2 d / d y x (When x = 4) 2 2 d / d y x = ‒1/2 < 0 hence MAX A1 All correct Alternative methods ok but must be explicit about values of x being considered 2

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1 9 A curve has equation y c and a line has equation y cx where c is a constant. = x + = −3, (i) Find the set of values of c for which the curve and the line meet. 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(ii) The line is a tangent to the curve for two particular values of c. For each of these values find the x-coordinate of the point at which the tangent touches the curve. 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Mark scheme: 9(i) 2 2 1 3 3 1 0 cx cx x cx x c + = − → − + − = equality Use ( ) ( ) 2 2 2 2 4 3 4 10 9 or 5 16 b ac c c c c c   − = + + = + + + −   M1 Select their correct coefficients which must contain ‘c’ twice Ignore = 0, < 0, >0 etc. at this stage (Critical values) ‒1, ‒9 A1 SOI 9, 1 − − c c - . A1 4 Question Answer Marks Guidance 9(ii) Sub their c to obtain a quadratic ( ) 2 1 2 1 0 c x x   = −→− − − =   M1 1 = − x A1 Sub their c to obtain a quadratic ( ) ( 2 [ 9 9 6 1 0 c x x  = − →− + − =  M1 1/ 3 = x A1 [Alt 1: 2 / 1/ dy dx x c = − = , when 1 1, 1, 9, 3 c x c x = − = ± = − = ± Give M1 for equating the gradients, A1 for all four answers and M1A1 for checking and eliminating] [Alt 2: 2 / 1/ dy dx x c = − = leading to ( ) 2 2 1/ 1/ x ( 1/ x ) x 3 x − = − − Give M1 A1 at this stage and M1A1 for solving] 4

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