Cambridge A Level Mathematics 9709 — 2025 May/June Paper 2 · Variant 1

2 dy1 Given that y = 6x cos `x + 1j, find an expression for . [2] dx .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Attempt use of product rule M1 Allow for 6cos x 2 + 1 − 12 x sin x 2 + 1 ( ) ( ) 6cos x 2 + 1 − 12sin x 2 + 1 ( ) ( ) 6cos x 2 + 1 − 6 x sin x 2 + 1 ( ) ( ) 6cos x 2 + 1 − 6 x 2 sin x 2 + 1 ( ) ( ) Obtain 6cos( x 2 + 1) − 12 x 2 sin( x 2 + 1) A1 Allow unsimplified. 2

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2 (a) Use logarithms to solve the inequality 4 x 1 0.05 . Give your answer in the form x 1 a , where the value of a is correct to 3 significant figures. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Solve the inequality 3x + 8 1 9 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Hence state the integers that satisfy both of the inequalities in parts (a) and (b). [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) Apply logarithms to both sides and use relevant logarithm property M1 Allow for x  log 4 0.05. ln0.05 A1 Allow greater accuracy. Obtain x  or equivalent and hence x −2.16 ln 4 2 2(b) Attempt solution of equations 3x + 8 = 9 or of quadratic equation (3 x + 8) 2 = 9 2 M1 Need a complete method to obtain 2 values. Obtain values − 173 and 13 A1 Conclude − 173  x  13 A1 OE 3 2(c) State −5, − 4, − 3 only B1 1

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3 (a) Sketch, on a single diagram, the graphs of y = 3e -2 x and y = sec x for values of x such that 0 G x 1 1 r . [2] 2 (b) Show that the x-coordinate of the point of intersection of the two graphs satisfies the equation x = 1 ln ( 3 cos x) . [2] 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Use an iterative formula, based on the equation in part (b), to find the x-coordinate of the point of intersection correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Sketch decreasing positive curve for y = 3e −2 x B1 Sketch curve for y = sec x B1 Correctly placed with reference to first sketch or correct with ‘1’ marked on y-axis. 2 3(b) −2 x x M1 Equate and arrange at least as far as e = ... or 2e = ... with cos x present Confirm x = 12 ln(3cos x ) A1 AG – necessary detail needed. 2 3(c) Use iterative process correctly at least once M1 Calculator must be in radian mode. Obtain final answer 0.487 A1 Required to precisely 3 decimal places. Show sufficient iterations to justify answer, or show a sign change in the interval A1 Allow iterations to greater accuracy. [0.4865, 0.4875] 3

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4 y x O The diagram shows the curve with equation y = 6e 2 x - e 3 x . The shaded region is bounded by the axes and the curve. (a) Find the exact x-coordinate of the maximum point. 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(b) Find the area of the shaded region. Give your answer in the form p, where p and q are integers. q [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 4(a) 2 x 3 x B1 Differentiate to obtain 12e − 3e Attempt solution for x of equation of form k1e 2 x + k 2 e 3 x = 0 (but not for y = 0 ), to M1 SOI obtain e =x k Obtain x = ln4 or x = 2ln2 A1 x = 1.386 scores A0. 3 4(b) Identify x = ln6 as point where curve meets x-axis B1 May see in part (a) but must be used here. Integrate to obtain the form k 3 e 2 x + k 4 e 3 x M1 Where k3  6 and k4 −1 Obtain 3e 2 x − 13 e 3 x A1 May see inclusion of + c. Apply limits to obtain answer 1003 only A1 4

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5 The polynomial p( )x is defined by p( )x = ax 3 + bx 2 - ax - 24 , where a and b are constants. It is given that ( 2x - 3 ) is a factor of p( )x and that the remainder is -15 when p( )x is divided by ( x + 1 ) . (a) Find the values of a and b. 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(b) Hence factorise p( )x completely. 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(c) Hence solve the equation p(3 cosec i) = 0 for 90° 1 i 1 270° . 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Mark scheme: 5(a) Substitute x = 32 equating to 0, and x = −1 equating to −15 M1 Allow algebraic long division, but must result in two expressions correctly equated to 0 and −15. SC B1 for b = 9 if M0 scored. Obtain 278 a + 94 b − 32 a − 24 = 0 A1 OE Powers of 32 must be evaluated. Obtain −+a b + a − 24 = −15 A1 Solve to obtain a = 2 and b = 9 A1 4 5(b) Divide p( x ) by 2 x − 3 M1 OE method, such as inspection. For algebraic long division, must go as far as the term in x. Allow attempt at synthetic division. 3 2 9 –2 –24 2 3 18 24 2 x 2 + 12 x + 16 Allow one sign error. 2 A1 2 Obtain quotient x + 6 x + 8 Allow 2 x + 12 x + 16 from synthetic division. Conclude (2 x − 3)( x + 2)( x + 4) A1 3 5(c) Use factorised form to determine value of sin, with −1 sin 1 M1 Obtain finally sin= − 34 only and hence angle 228.6 A1 Or greater accuracy 228.590… Ignore solutions outside the range. 2

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7 (a) Prove that sin 2 2x + 4 cos 2 x cos 2x / 4 cos 4 x . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the set of possible values of the constant k for which the equation sin 2 2x + 4 cos 2 x cos 2x + 5 = k has no real solutions. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 r 3 (c) Find the exact value of y 1 sin 2 t + 4 cos 2 b 12 tl cos t td . 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Mark scheme: 7(a) Use correct identities to express in terms of sin x and cos x M1 Allow for 2sin 2 x cos 2 x + 4cos 2 x cos 2 x − sin 2 x , OE. ( ) Obtain 4sin 2 x cos 2 x + 4cos 2 x (cos 2 x − sin 2 x ) A1 OE 4 A1 Confirm 4cos x 3 7(b) Use identity from part (a) to obtain k  5 or k  9 B1 Obtain k ,5 k  9 B1 2 7(c) 2 1 4 t M1* State or imply integrand is 2cos 2 t or 4cos 2 Use double-angle identity to express in terms of cost M1 Obtain 1 + cost and integrate to obtain t + sin t A1 Use limits to obtain 23 π + 3 A1 4

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