Cambridge A Level Mathematics 9709 — 2024 May/June Paper 2 · Variant 1

1 A curve has equation y = 2 tan x - 5 sin x for 0 G x 1 12 r. Find the x-coordinate of the stationary point of the curve. Give your answer correct to 3 significant figures. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 Differentiate to obtain 2 2sec 5cos x x  B1 Equate first derivative to zero and solve correctly at least as far as cos ... x  M1 B1 M1 for o 42.5 x  . Obtain 3 cos 0.4 x  and hence 0.742 x  A1 Or greater accuracy (0.74261…). 3

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2 A curve has equation x 2 ln y + y 2 + 4x = 9 . Find the gradient of the curve at the point (2, 1). [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Attempt use of product rule for differentiating x y Allow if d d y x missing. Obtain 2 d 2 ln d x y x y y x  A1 Obtain d 2 d y y x  B1 Substitute 2, 1 x y   in equation equal to 0 involving at least one d d y x and solve for d d y x M1 Correct equation: 2 d 2 ln d x y x y y x  d 2 d y y x  4 0   Obtain 2 3  A1 Or exact equivalent. Not from wrong working. 5

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3 (a) Sketch on the same diagram the graphs of y = 3x - 8 and y = 5 - x . [2] (b) Solve the inequality 3x - 8 1 5 - x . 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(c) Hence determine the largest integer N satisfying the inequality 3 e 0 .1 N - 8 1 5 - e 0 .1 N . 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Mark scheme: 3(a) Draw V-shaped graph with vertex on positive x-axis in the first quadrant. B1 Draw correct graph of 5 y x   correctly positioned with respect to modulus graph. B1 Two points of intersection. 2 3(b) Solve 3 8 5 x x    to obtain 13 4 B1 Or inequality. Solve linear equation or inequality with signs of 3x and x the same M1 Obtain 3 2 A1 Conclude 3 13 2 4 x   or 3 2 x  and 13 4 x  A1 Allow alternative notation e.g.   3 13 2 4 , . Alternative Method for Question 3(b) State or imply non-modulus equation (or inequality) 2 2 (3 8) (5 ) x x    (B1) Attempt solution of three-term equation (or inequality) (M1) Obtain 3 2 and 13 4 (A1) Conclude 3 13 2 4 x   or 3 2 x  and 13 4 x  (A1) Allow alternative notation e.g.   3 13 2 4 , . 4 3(c) Attempt value of N (maybe non-integer at this stage) for 0.1 13 4 e N their  M1 Allow 0.1 13 4 e N their  (or inequality). Conclude with single integer 11 A1 2

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4 (a) Show that 3 tan 2i + tan ( i + 45 °) / i 2 i . 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(b) Hence solve the equation 3 tan 2i + tan ( i + 45°) = 4 for 0° 1 i 1 180 ° . 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Mark scheme: 4(a) State 2 6tan 3tan 2 1 tan     B1 Allow for 2 2tan 3tan 2 3 1 tan           . State tan 1 tan( 45 ) 1 tan       B1 Attempt to express left-hand side in terms of tan as a single fraction M1 Condone sign errors in identities. Allow if o tan 45 is not evaluated. Allow if 2 separate terms with a common denominator. Confirm 2 2 tan 8tan 1 1 tan       A1 Answer given – necessary detail needed. 4 4(b) Solve quadratic equation 2 5tan 8tan 3 0      to obtain at least one value of  M1 Must be using a correct method. Obtain 17.4 A1 Or greater accuracy. Obtain 117.6 A1 Or greater accuracy; and no others between 0 and 180. 3

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5 A curve has equation y = . The curve has exactly one stationary point P. 1 + 3x dy 1 1 -2 x (a) Find and hence show that the x-coordinate of P satisfies the equation x = 6 + 2 e . 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(b) Show by calculation that the x-coordinate of P lies between 0.35 and 0.45 . 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(c) Use an iterative formula based on the equation in part (a) to find the x-coordinate of P correct to 3 significant figures. Give the result of each iteration to 5 significant figures. 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Mark scheme: 5(a) Differentiate using quotient rule *M1 OE Obtain 2 2 2 (1 3 ) 2e (1 e ) 3 (1 3 ) x x x x     A1 OE Equate first derivative to zero and arrange as far as ... x  DM1 Confirm 2 1 1 6 2 e x x    A1 Answer given – necessary detail needed. Must have exact terms. 4 5(b) Consider sign of 2 1 1 6 2 e x x    M1 OE Obtain 0.06...  (– 0.064959...) and 0.08... (0.0800…) or equivalents and justify conclusion A1 Answer given – necessary detail needed. Alternative Method 1 for Question 5(b) Consider  2 1 1 6 2 f e x x    and obtain   f 0.35 0.42 (0.4149…) and   f 0.45 0.37 (0.36995….) (M1) Conclude   f 0.35 0.45  and   f 0.45 0.35  so root lies in given interval. (A1) Alternative Method 2 for Question 5(b) Consider the sign of their d d y x from part (a) (M1) Obtain 0.19...  (– 0.187...) and 0.21... (0.2139…) or equivalents and justify conclusion (A1) 2 Question Answer Marks Guidance 5(c) Use iterative process correctly at least once M1 Obtain final answer 0.394 A1 Answer required to exactly 3sf. Show sufficient iterations to 5 sf to justify answer or show sign change in interval [0.3935, 0.3945] A1 3

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6 y x O 1 r 6 The diagram shows the curve with equation y = sin 2x + sin 2 2x for 0 G x G 16 r. The shaded region is r and y = 0 . bounded by the curve and the straight lines x = 16 (a) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures. 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(b) The shaded region is rotated completely about the x-axis. Find the exact volume of the solid produced. 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Mark scheme: 6(a) Use y-values 2 2 1 1 1 1 6 6 3 3 (0), sin π sin π, sin π sin π   or decimal equivalents 0 , 0.75 or 0.866, 1.616 or 1.271. Use correct formula, or equivalent, with 1 12 π h  M1 Must be using ‘y’ values. May do as 2 separate trapezia (0.113359 + 0.27976). Obtain 0.39 A1 Allow 0.393 but not greater accuracy. 3 Question Answer Marks Guidance 6(b) Use  2 π (sin2 sin 2 ) d x x x   M1 OE Express integrand in the form 1 2 3 sin 2 cos4 k x k k x   *M1 Where 1 2 3 0 k k k  . Obtain correct 1 1 2 2 sin 2 cos 4 x x   A1 Or π times this. Integrate to obtain 1 1 1 2 2 8 cos 2 sin 4 x x x    A1 FT Following their integrand only if of correct form. Apply limits correctly to obtain exact value DM1 Obtain volume = 2 1 1 1 4 12 16 π π π 3   A1 Or exact equivalent. 6

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7 The polynomial p ( )x is defined by p ( )x = 9 x 3 + 6x 2 + 12x + k , where k is a constant. (a) Find the quotient when p ( )x is divided by ( 3x + 2) and show that the remainder is ( k - 8 ) . 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) Carry out division at least as far as 2 1 3x n  M1 Or equivalent (inspection, …). Obtain quotient 2 3 4 x  A1 Confirm remainder is 8 k  A1 Answer given – necessary detail needed. SC B1 for correct use of factor theorem to show remainder is 8 k  . Alternative Method for Question 7(a) Synthetic division –2/3 9 6 12 k –6 0 –8 9 0 12 8 k  (M1) Allow one sign error. Obtain quotient 2 3 4 x  (A1) Confirm remainder is 8 k  (A1) 3 Question Answer Marks Guidance 7(b) Integrate to obtain at least a term in 3 x and term of form 2 ln(3 2) n x  *M1 Need to be using their answer to part (a). Obtain 3 1 3 4 ( 8)ln(3 2) x x k x     A1 FT on a quotient of 2 9 12 x  . Apply limits correctly to expression with three terms DM1 Obtain 235 a  A1 FT on a quotient of 2 9 12 x  . Equate logarithm term to ln64 and apply appropriate logarithm properties DM1 Obtain 17 k  A1 6 SC 2 marks for use of quotient 2 3 4 x  or 2 9 12 x  to obtain either 235 or 705 if no other marks are available.

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