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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w25_qp_259709_w25_ms_259709/25/O/N/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
Find y 6 sin x dx . 2 [3] * 0000800000004 * DFD 4
Mark scheme
1 Use identity to express integrand in the form k1 + k2 cos 2 x M1 Obtain correct 3 − 3cos2x A1 Integrate to obtain 3 x − 32 sin 2 x A1 Condone omission of + c. 3
Question
Solve the equation e 2x `e 2x - 8j = 48 . [3] * 0000800000005 * DFD 5
Mark scheme
2 Attempt solution of quadratic equation in e2 x to obtain at least one value of e2 x M1 Obtain (e 2 x − 12)(e 2 x + 4) = 0 OE, and hence at least e2 x = 12 A1 Obtain x = 12 ln12 or 1.24 and no other solution A1 3
Question
(a) Solve the equation 2x - 3 = 5x + 2 . [3] (b) Hence solve the equation 2 sec i - 3 = 5 sec i + 2 for r 1 i 1 2r . Give your answer correct to 3 significant figures. [3] * 0000800000006 * DFD 6
Mark scheme
3(a) Solve 2x − 3 = 5x + 2 to obtain − 53 B1 Or exact equivalent. Attempt solution of linear equation where 2x and 5x have different signs M1 Obtain 17 A1 OE Alternative Method for Question 3(a) State or imply non-modulus equation (2 x − 3) 2 = (5 x + 2) 2 B1 Attempt complete solution of three-term quadratic equation M1 Obtain − 53 and 17 or equivalents A1 OE 3 3(b) State or imply cosθ = − 53 B1 FT Following an appropriate answer from part (a). Attempt correct process for finding third quadrant angle M1 Condone working in degrees for this mark. Obtain 4.07 A1 Or greater accuracy 4.0688… 3
Question
Solve the equation cot i tan (i + 45°) = 7 for 0° 1 i 1 90° . [5] * 0000800000007 * DFD 7
Mark scheme
4 Attempt to express equation in terms of tanθ only M1* 1 tan θ + tan 45 A1 OE Obtain × =7 tan θ 1 − tan θ tan 45 Simplify to obtain equation 7 tan 2 θ − 6 tan θ + 1 = 0 A1 Attempt solution of three-term quadratic equation in tanθ to obtain at least one DM1 value of θ Obtain tanθ = 17 (3 ± 2) or equivalent and hence 12.8 and 32.2, and no others in the A1 Or greater accuracy 12.764…, 32.235… given range 5
Question
y A B O x 1 The diagram shows the curve with equation y = 8e - 2 x - 1. The curve meets the axes at the points A and B. The shaded region is bounded by the curve and the line segment AB. (a) Show that the x-coordinate of B is 6 ln 2. [2] (b) Find the area of the shaded region. Give your answer in the form p ln 2 - q , where p and q are positive integers. [5] * 0000800000008 * DFD 8
Mark scheme
5(a) −1 x Attempt solution of 8e 2 − 1 = 0 with use of a relevant logarithm property M1 1 E.g. ln x = ln8 2 Confirm x = 6ln 2 A1 AG – necessary detail needed. Need to see ln8 = 3ln 2 or 64 = 26 OE used. Alternative Method for Question 5(a) −1 x Substitute x = 6ln 2 in 8e 2 − 1 and show use of a relevant logarithm property M1 ln 1 Obtain 8e−3ln 2 − 1 and hence 8e 8 − 1 or equivalent and verify answer 0 A1 AG – necessary detail needed. 2 5(b) −1 x Integrate 8e 2 − 1 to obtain the form k e− 2 x − x 1 M1 −1 x A1 Obtain correct −16e 2 − x Apply limits 0 and 6ln 2 to obtain −16e 2 − 1 ×6ln 2 − 6ln 2 + 16 A1 OE 14 − 6ln 2 Attempt area of triangle minus area under curve M1 Allow decimals for this mark (14.56). Obtain 12 × 6ln 2 × 7 − (14 − 6ln 2) or equivalent and hence 27ln 2 − 14 A1 5
Question
A curve has parametric equations x = tan i, y = sin i - 2 sin 3 i , for 0 1 i 1 12 r . dy (a) Show that = 6 cos 5 i - 5 cos 3 i . [4] dx * 0000800000009 * DFD 9 (b) Find the equation of the normal to the curve at the point where it crosses the x-axis. Give your answer in the form y = mx + c , where m and c are exact constants. [5] * 0000800000010 * DFD 10
Mark scheme
6(a) Attempt use of chain rule for differentiation of y or simplification of y and use of the M1 product rule. dx dy A1 Obtain = sec2 θ and = cosθ − 6sin 2 θ cosθ dθ dθ dy dx dy M1 Attempt expression for in terms of cosθ only using their and dx dθ dθ Obtain 6cos5 θ − 5cos3 θ A1 AG – necessary detail needed. 4 6(b) Attempt solution of y = 0 to find value of θ M1 Allow if degrees used Obtain sin 2 θ = 12 and hence θ = 14 π A1 Must be in radians here 1 A1 Obtain gradient of curve at the point is − or exact equivalent 2 Attempt equation of normal using negative reciprocal for their gradient M1 Obtain y = 2 x − 2 or exact equivalent A1 5
Question
The polynomial p (x) is defined by p (x) = 2x 4 + kx 3 + kx 2 + 17x + 18 , where k is a constant. It is given that (x + 2) is a factor of p (x) . (a) Find the value of k. [2] It is given that the equation p (x) = 0 has exactly two real roots, denoted by a and b , where a is an integer and b is not an integer. (b) State the value of a and show that b satisfies the equation x = 3 - 2x - 4.5 . [4] * 0000800000011 * DFD 11 (c) Show by calculation that -1.4 1 b 1 -1.0 . [2] (d) Use an iterative formula, based on the equation in part (b), to find the value of b correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] * 0000800000012 * DFD 12
Mark scheme
7(a) Substitute x = −2 , equate to zero and attempt solution M1 Obtain 32 − 8k + 4k − 34 + 18 = 0 or equivalent and hence k = 4 A1 2 7(b) State α = −2 B1 Divide by x + 2 at least as far as 2 x3 + mx M1 Or equivalent method, e.g. inspection. Obtain 2 x3 + 4 x + 9 A1 Rearrange 2 x3 + 4 x + 9 = 0 to confirm x = 3 −2 x − 4.5 A1 AG – necessary detail needed. 4 7(c) Consider sign of x − 3 −2 x − 4.5 or equivalent for −1.4 and −1.0 M1 Or Example 1: f ( x ) = 2x3 + 4x + 9, finding f ( −1.4 ) and f ( −1) . Example 2: f ( x ) = 3 −2x − 4.5, finding f ( −1.4 ) and f ( −1) . Obtain −0.2... and 0.3… or equivalents and justify conclusion A1 Example 1: f ( −1) = 3, f ( −1.4 ) = −1.2... with a correct conclusion. Example 2: f ( −1) = −1.357.... so −1 > −1.357... f ( −1.4 ) = −1.193.... so −1.4 < −1.19... with a correct conclusion. AG – necessary detail needed. 2 7(d) Use iterative process correctly at least once M1 Obtain final answer −1.26 A1 Answer required to exactly 3 significant figures. Show sufficient iterations to 5 sf to justify answer or show sign change in interval A1 [−1.265, − 1.255] 3
Question
The equation of a curve is y = 4e 1 - 2x 3x - 1 . Find the exact coordinates of the stationary point of the curve. [6] * 0000800000013 * DFD 13 * 0000800000014 * DFD 14 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. DFD 15 * 0000800000016 * DFD 16
Mark scheme
8 Differentiate to obtain the form k1e1−2 x (3x − 1) 2 + k2e1−2 x (3x −1)− 2 1 1 *M1 Where k1k2 ≠ 0. Obtain correct first term −8e1−2 x (3x − 1) 2 1 A1 OE. Allow unsimplified Obtain correct second term + 6e1−2 x (3x −1)− 2 1 A1 OE. Allow unsimplified Equate first derivative to zero and attempt solution as far as 3x − 1 = k DM1 OE. Must be in the form 1 −1 k1e1−2 x (3x −1) 2 + k2e1−2 x (3x −1) 2 . Obtain 3 x − 1 = 34 OE, and hence x = 127 A1 Or exact equivalent. Obtain y = 2 3e− 6 1 A1 Or exact equivalent. 6
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Mathematics 9709/25 for October/November 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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