Cambridge A Level Mathematics 9709 — 2023 May/June Paper 2 · Variant 3

1 Solve the equation 5 9 17 sec sec21 + tan21 = + 1 for [5] 0Å < 1 < 360Å. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 identity, allow if ‘5’ omitted. Obtain   2 6sec 17sec 14 0      A1 or 2 14cos 17 cos 6 0      . Attempt solution of 3-term quadratic equation to find one value of , from cos ...  M1 Obtain 73.4 A1 or greater accuracy. Obtain 286.6 A1 or greater accuracy; and no others between 0 and 360. 5

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3 y 2 x O 6 6 The diagram shows part of the curve y The shaded region is bounded by the curve and the = 2x 3. + lines x 6 and y 2. = = Find the exact area of the shaded region, giving your answer in the form a b, where a and b are −ln integers. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Integrate to obtain the form ln(2 3) k x Obtain correct 3ln(2 3)  x A1 Allow unsimplified. Apply limits 0 and 6 correctly to obtain ln15 ln3  k k *DM1 Allow unsimplified. Apply relevant logarithm properties correctly to obtain form lnb DM1 Obtain 12 ln125  A1 5

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4 (a) y x O 3 2x. The diagram shows the graph of y = −e−1 2x On the diagram, sketch the graph of y 5x , and show that the equation 3 5x = −4 −e−1 = −4 has exactly two real roots. [2] 2x It is given that the two roots of 3 5x are denoted by and where −e−1 = −4 ! ", ! < ". (b) Show by calculation that lies between 0.36 and 0.37. [2] ! ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 7 to find correct to 4 significant figures. Give the 5 −e−12xn! " (c) Use the iterative formula xn+1 = result of each iteration to 6 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Draw (more or less) correct sketch with vertex on positive x-axis *B1 crossing y-axis above given graph, may be implied by extrapolation. Indicate in some way the two roots DB1 2 4(b) Consider sign of 1 2 3 e 5 4     x x or of 1 2 3 e 5 4     x x for 0.36 and 0.37 M1 but not for sign of 1 2 3 e 5 4     x x . May be implied by 0.035...  and 0.018..., or equivalents. Obtain 0.035...  and 0.018..., or equivalents, and justify conclusion A1 AG necessary detail needed. 2 Question Answer Marks Guidance 4(c) Use iteration process correctly at least once M1 Obtain final answer 1.295 A1 answer required to exactly 4 sf. Show sufficient iterations to 6 sf to justify answer or show sign change in interval [1.2945, 1.2955] A1 3

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5 y C A B x O 2x The diagram shows the curve with equation y x2 4 . The curve crosses the x-axis at the = e−1 −5x + points A and B, and has a maximum at the point C. (a) Find the exact gradient of the curve at B. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact coordinates of C. 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Mark scheme: 5(a) Attempt use of product rule to find first derivative *M1 Obtain 1 1 2 2 2 1 2e ( 5 4) e (2 5)        x x x x x A1 OE Obtain 4  x for point B B1 Substitute 4  x to find the value of the derivative DM1 Obtain 2 3e  A1 or exact equivalent. 5 5(b) Equate their first derivative to zero and simplify as far as quadratic equation *M1 allow if it appears in part (a). Obtain at least 2 9 14 0    x x A1 OE Solve to find relevant x value and substitute to find the value of y DM1 Obtain 7  x and 7 2 18e   y A1 or exact equivalent. 4

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6 (a) Show that 4 sin 1 cos 3 2 sin [4] 1 + 3π 1 −13π  + 21. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact value of 4 sin 17 cos 1 [2] 24π 24π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 18π 1(c) Find the exact value of 4 sin 2x cos 2x dx. [4] Ó 0 + 3π −13π ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) Obtain at least either 1 1 2 2 ( sin 3cos )    or 1 1 2 2 ( cos 3sin )    B1 Allow if implied by decimal values. Expand and simplify with correct use of 2 2 sin cos 1     M1 Use 1 2 sin cos sin 2     M1 Confirm given result 3 2 sin 2  A1 AG necessary detail required. 4 6(b) Identify value of  is 3 8 π *B1 OE Obtain 3 4 3 2sin π  and conclude 3 2  DB1 or exact equivalent. 2 6(c) Identify integrand as 3 2 sin 4  x B1 Integrate to obtain form 1 2 cos 4  k x k x M1 where 1 2 0  k k . Obtain correct 1 2 3 cos4  x x A1 Obtain 1 1 8 2 π 3 A1 or exact equivalent. 4

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