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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s25_qp_219709_s25_ms_219709/21/M/J/25
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
Given that y = 6x cos `x 2 + 1j, find an expression for . [2] dx
Mark scheme
1 Attempt use of product rule M1 Allow for ( ) 6cos x 2 + 1 − 12 x sin x 2 + 1 ( ) 6cos ( x + 1) − 12sin ( x + 1) 2 2 6cos ( x + 1) − 6 x sin ( x + 1) 2 2 6cos ( x + 1) − 6 x sin ( x + 1) 2 2 2 Obtain 6cos( x 2 + 1) − 12 x 2 sin( x 2 + 1) A1 Allow unsimplified. 2
Question
(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 < log4 0.05. ln 0.05 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 (3x + 8) 2 = 92 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
Question
(a) Sketch, on a single diagram, the graphs of y = 3e -2x and y = sec x for values of x such that 0 G x 1 12 r . [2] (b) Show that the x-coordinate of the point of intersection of the two graphs satisfies the equation x = 12 ln (3 cos x) . [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) Equate and arrange at least as far as e−2 x = ... or e2 x = ... with cos x present M1 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
Question
y O x The diagram shows the curve with equation y = 6e 2x - e 3x . The shaded region is bounded by the axes and the curve. (a) Find the exact x-coordinate of the maximum point. [3] p (b) Find the area of the shaded region. Give your answer in the form , where p and q are integers. q [4]
Mark scheme
4(a) Differentiate to obtain 12e2 x − 3e3x B1 Attempt solution for x of equation of form k1e2 x + k2 e3 x = 0 (but not for y = 0 ), to M1 SOI obtain e± x = k Obtain x = ln 4 or x = 2ln 2 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 k3e 2 x + k4 e3 x M1 Where k3 ≠ 6 and k4 ≠ −1 Obtain 3e 2 x − 13 e3 x A1 May see inclusion of + c. Apply limits to obtain answer 100 3 only A1 4
Question
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. [4] (b) Hence factorise p(x) completely. [3] (c) Hence solve the equation p(3 cosec i) = 0 for 90° 1 i 1 270° . [2]
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 2x2 + 12x + 16 Allow one sign error. Obtain quotient x2 + 6 x + 8 A1 Allow 2x2 + 12x + 16 from synthetic division. Conclude (2 x − 3)( x + 2)( x + 4) A1 3 5(c) Use factorised form to determine value of sin θ , with −1 < sinθ < 1 M1 Obtain finally sin θ = − 34 only and hence angle 228.6 A1 Or greater accuracy 228.590… Ignore solutions outside the range. 2
Question
The parametric equations of a curve are 2t + 1 x= , y = 2 ln (3t + 4) , 3t + 4 4 where t 2- . 3 dy (a) Show that can be expressed in the form c (3t + 4) and state the value of the constant c. [5] dx (b) It is given that the gradient of the curve at the point (a, ln 100) is m. Find the values of a and m. [4] (c) State whether the curve represents a decreasing function or an increasing function or neither. Give a reason for your answer. [1]
Mark scheme
6(a) Attempt use of quotient rule (or equivalent) for derivative of x M1 5 A1 Obtain or unsimplified equivalent (3t + 4) 2 6 B1 Allow unsimplified. Obtain for derivative of y 3t + 4 dy M1 Attempt to express in terms of t dx Obtain 65 (3t + 4) stating or implying that c = 65 A1 Form of answer given. Necessary detail needed. 5 6(b) Attempt solution of 2ln(3t + 4) = ln100 M1 Must be using logarithms correctly. Obtain 3t + 4 = 10 (or equivalent) and hence t = 2 (only) A1 May be implied by only one (correct) value for a. Substitute to obtain a = 12 A1 Substitute to obtain m = 12 A1 4 6(c) State increasing, with reference to positive gradient B1 1
Question
(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 (c) Find the exact value of y 3r - 31 r 1 sin 2 t + 4 cos 2 b2 tl cos t dt . [4] 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) 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 cos2 x + 4cos 2 x(cos 2 x − sin 2 x) A1 OE Confirm 4cos4 x A1 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) t M1* State or imply integrand is 2 cos 2 12 t or 4cos 4 2 Use double-angle identity to express in terms of cos t 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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Mathematics 9709/21 for May/June 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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