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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_259709_s25_ms_259709/25/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
Show that [3] 2
Mark scheme
1 Integrate to obtain 2ln(4 x + 1) B1 Apply limits correctly to k ln(4 x + 1) and use at least one relevant logarithm property M1 Obtain 2ln 45 − 2ln9 = 2ln5 or equivalent, and conclude ln 25 A1 3
Question
(a) Sketch on the same diagram the graphs of y = 2x - 9 and y = 4x - 5. [2] (b) Solve the inequality 2x - 9 1 4x - 5. [3] 8x
Mark scheme
2(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of y = 4 x − 5 B1 Correctly placed with reference to modulus graph and with steeper gradient. 2 2(b) Solve linear equation or inequality with signs of 2x and 4x different M1 Obtain −2x + 9 = 4x − 5 and hence x = 73 A1 OE Conclude x > 73 A1 7 7 OE, e.g. , ∞ , or , ∞ . 3 3 Alternative Method for Question 2(b) State or imply non-modulus equation (or inequality) (2 x − 9)2 = (4 x − 5) 2 B1* Attempt solution of three-term quadratic equation (or inequality) DM1 12x2 − 4x − 56 = 0 leading to 73 and −2. Obtain at least 73 and conclude x > 73 A1 Must be from correct work. 7 7 OE, e.g. , ∞ or , ∞ 3 3 3
Question
Find the coordinates of the stationary points of the curve with equation y = - 6x + 5. [5] 2x + 3
Mark scheme
3 8x −12 x + 15 2 *M1 Attempt use of quotient rule (or equivalent) to differentiate or 2x + 3 ( 2 x + 3) dy 24 A1 OE Obtain = −6 Allow unsimplified. dx (2 x + 3) 2 Equate first derivative to zero, and attempt solution of a quadratic equation to find DM1 4x2 + 12x + 5 = 0 two values of x Obtain at least one of the stationary points ( − 52 , 30) and ( − 12 , 6), A1 or both x-values, − 52 and − 12 Obtain both stationary points A1 5
Question
y P O x The diagram shows parts of the curves with equations y = 4e -2x and y = 1 + 0.5 sin 3x . Point P is a point of intersection of the curves, and the shaded region is bounded by the two curves and the y-axis. (a) Show that the x-coordinate of P satisfies the equation x =-0.5 ln (0.25 + 0.125 sin 3x) . [1] (b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Use an initial value of 0.5 and give the result of each iteration to 6 significant figures. [3] (c) Hence find the area of the shaded region. Give your answer correct to 2 significant figures. [4]
Mark scheme
4(a) State 4e−2 x = 1 + 0.5sin3x and confirm x = −0.5ln(0.25 + 0.125sin 3 x) B1 AG Necessary detail needed. 1 4(b) Use iterative process correctly at least once M1 Obtain final answer 0.4912 A1 Answer required to exactly 4 significant figures. Show sufficient iterations to 6 significant figures to justify answer A1 or show a sign change in the interval [0.49115, 0.49125] 3 4(c) Integrate to obtain the form k1e−2 x + k2 x + k3 cos3x *M1 k1, k2 , k3 ≠ 0 OE, with separate integrals or ‘reversed’. Obtain correct −2e−2 x − x + 16 cos3x A1 OE Allow for 2e−2 x + x − 16 cos3x Apply limits 0 and their part (b) answer correctly and attempt evaluation DM1 Obtain 0.61 A1 4
Question
The polynomial p (x) is defined by p (x) = ax 4 + bx 3 + 13x 2 - 35x + 15 , where a and b are constants. It is given that (2x - 1) and (x - 3) are factors of p (x) . (a) Find the values of a and b. [4] (b) Hence factorise p (x) . [3] (c) Find the least positive value of i in radians such that p (cot 2i) = 0 . [2]
Mark scheme
5(a) Substitute x = 12 and x = 3, and equate each to zero to produce two equations M1 SC B1 for a correct equation, if M0 otherwise. Obtain 161 a + 81 b = − 43 A1 OE a + 2b + 12 = 0 Obtain 81a + 27b = −27 A1 OE 3a + b + 1 = 0 Solve simultaneous equations to obtain a = 2 and b = −7 A1 4 5(b) Divide by 2x2 − 7 x + 3 or successively by 2x − 1 and x − 3 M1 OE method, such as inspection. Obtain quotient x 2 + 5 ( ) A1 Condone 2 x 2 + 5 from synthetic division. State fully factorised form (2 x − 1)( x − 3)( x 2 + 5) A1 3 5(c) Attempt solution of at least cot 2θ = 3 M1 Obtain tan 2θ = 13 and hence θ = 0.161 A1 Or greater accuracy 0.16087… 2
Question
A curve has equation (x 2 - 3) ln y + 6x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . [3] (b) Find the equation of the tangent to the curve at the point (2, e 2) . Give your answer in the form y = mx + c , where m and c are exact constants. [6]
Mark scheme
6(a) Substitute y = e −1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + +6=0 Following their derivative of ( x 2 − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e2 A1 Obtain equation y = −14e2 x + 29e 2 or equivalent of required form A1 6
Question
(a) Express 4 cos i sin (i + 30°) in the form R cos (2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. [6] (b) Hence solve the equation 12 cos 2z sin (2z + 30°) = 5 for 0° 1 z 1 90° . [5] 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) Expand to obtain the form k1 cosθ sin θ + k2 cos 2 θ M1 Obtain 2 3 cosθ sin θ + 2cos 2 θ with exact coefficients A1 Obtain 3sin 2θ + cos2θ + 1 A1 FT Following their expression in sin θ and cosθ . State R = 2 B1 FT Following their expression in sin 2θ and cos2θ . Use appropriate trigonometry to find value of α M1 Must be of the form a sin 2θ + b cos 2θ . Conclude 2cos(2θ − 60) + 1 A1 6 7(b) Attempt to express equation in form cos(4φ − α ) = k *M1 Use correct process to find one value of φ DM1 Obtain cos(4φ − 60) = 13 and hence 32.6 A1 Or greater accuracy 32.6321… Use correct process to find second value of φ DM1 Obtain 87.4 and no others between 0 and 90 A1 Or greater accuracy 87.3678… 5
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Mathematics 9709/25 for May/June 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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