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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w23_qp_229709_w23_ms_229709/22/O/N/23
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
When the polynomial ax3 + 4ax2 − 7x − 5 is divided by x + 2, the remainder is 33. Find the value of the constant a. [2]
Mark scheme
1 Substitute x = −2 and equate to 33 M1 OE (long or synthetic division). Note: Long division and synthetic division give a remainder of 8a + 14 − 5 . Allow one sign error for M1. Obtain −8a + 16a + 14 − 5 = 33 and hence a = 3 A1 2
Question
Solve the equation sec 1 cos 1 − 60Å = 4 for −180Å < 1 < 180Å. [5]
Mark scheme
2 Attempt to express equation in terms of cosθ and sinθ only, using a correct M1 1 identity and secθ = cosθ Obtain cosθ cos60 + sinθ sin60 = 4cosθ A1 OE Either divide by cosθ to find value of tanθ or use another correct method M1 May still involve cos60 and sin 60 . to find sinθ or cosθ 7 A1 Obtain tan θ = 73 or tanθ = 4.04... or sin θ = or sinθ = 0.970... 52 3 or cos θ = or cosθ = 0.240... 52 Obtain −103.9 and 76.1 A1 Or greater accuracy; and no others between –180 and 180. 5
Question
y A B x O 2 − 1x The diagram shows the curve with equation y = 6e 2 . The points on the curve with x-coordinates 0 and 2 are denoted by A and B respectively. The shaded region is enclosed by the curve, the line through A parallel to the x-axis and the line through B parallel to the y-axis. (a) Find the exact gradient of the curve at B. [2] (b) Find the exact area of the shaded region. [3]
Mark scheme
3(a) −1 x M1 For any non-zero k except 6. Differentiate to obtain form ke 2 3 A1 Substitute x = 2 to obtain −3e −1 or − e 2 3(b) −1 x B1 OE Integrate to obtain −12e 2 − x M1 For any non-zero k except 6 Use limits 0 and 2 correctly to an integral of the form ke 2 , retaining −1 x or equivalent perhaps involving integration of 6 − 6e 2 . exactness 12 A1 Subtract from 12 to obtain final answer 12e−1 or e 3
Question
(a) Sketch, on the same diagram, the graphs of y = 3 − x and y = 9 − 2x. [2] (b) Solve the inequality 3 − x > 9 − 2x. [3] (c) Use logarithms to solve the inequality 23x−10 < 500. Give your answer in the form x < a, where the value of a is given correct to 3 significant figures. [3] (d) List the integers that satisfy both of the inequalities 3 − x > 9 − 2x and 23x−10 < 500. [1]
Mark scheme
4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 9 − 2 x with steeper negative B1 Dependent on first B mark, appropriately positioned gradient with respect to first graph. 2 4(b) Solve linear equation or inequality with signs of x and 2x different M1 Obtain critical value 4 A1 Conclude x > 4 only A1 6 must be discounted. Alternative Method for Question 4(b) State or imply non-modulus equation (or inequality) (3 − x) 2 = (9 − 2 x) 2 B1 Attempt solution of three-term quadratic equation (or inequality) M1 Dependent on previous B1. Conclude x > 4 only A1 6 must be discounted. 3 4(c) State or imply (3x − 10)ln 2 < ln 500 B1 Or equivalent perhaps involving different logarithm base. Obtain critical value 6.32 B1 Obtain x < 6.32 B1 Or greater accuracy. 3 4(d) State 5 and 6 only B1 1
Question
(a) Find the quotient when 6x3 − 5x2 − 24x − 4 is divided by 2x + 1, and show that the remainder is 6. [3] (b) Hence find 6x3 − 5x2 − 24x − 4 7 Ô 2x + 1 dx, 2 giving your answer in the form a + ln b, where a and b are integers. [5]
Mark scheme
5(a) Carry out division at least as far as 3x 2 + k1 x M1 OE (e.g. by inspection). Obtain quotient 3x 2 − 4 x − 10 A1 Confirm given result of remainder is 6 with sufficient detail A1 AG SC If remainder = 6 shown using remainder theorem allow B1. 3 5(b) Integrate to obtain at least x 3 and term of form k2 ln(2 x + 1) *M1 ln term must be added. Obtain x3 − 2 x 2 − 10 x + 3ln(2 x + 1) A1 Apply limits correctly to expression with four terms DM1 Apply appropriate logarithm properties correctly to obtain the form k3 ln a DM1 Obtain 195 + ln 27 A1 5
Question
y B A x O The diagram shows the curve with parametric equations x = 3 ln 2t − 3, y = 4t ln t. The curve crosses the y-axis at the point A. At the point B, the gradient of the curve is 12. (a) Find the exact gradient of the curve at A. [5] (b) Show that the value of the parameter t at B satisfies the equation t= + . 9 3 1 + ln t 2 [2] (c) Use an iterative formula, based on the equation in (b), to find the value of t at B, giving your answer correct to 3 significant figures. Use an initial value of 5 and give the result of each iteration to 5 significant figures. [3]
Mark scheme
6(a) dx 6 B1 Obtain = dt 2t − 3 dy M1 Allow unsimplified. Use product rule to find dt dy (4ln t + 4)(2t − 3) A1 OE Obtain = dx 6 Attempt to find t corresponding to point A using a complete and correct M1 method Obtain t = 2 and hence gradient is 23 ln 2 + 23 A1 Or exact equivalent. 5 6(b) dy k M1 Equate to 12 and attempt rearrangement to 2t − 3 = dx 4ln t + 4 9 3 A1 AG Confirm given result t = + with sufficient detail 1 + ln t 2 2 6(c) Use iteration process correctly at least once M1 Need to see 4.9626 . Obtain final answer 4.96 A1 Answer required to exactly 3 s.f. Show sufficient iterations to 5 sf to justify answer or show sign change in A1 interval [4.955, 4.965] 3
Question
(a) Prove that sin 2x cot x + 3 tan x 4 − 2 cos 2x. [4] 1 π + 3 tan 1 π. (b) Hence find the exact value of cot 12 [2] 12 (c) y A B x O a b The diagram shows the curve with equation y = 4 − 2 cos 2x, for 0 < x < 2π. At the point A, the gradient of the curve is 4. The point B is a minimum point. The x-coordinates of A and B are a and b respectively. Show that Ó 4 − 2 cos 2x dx = 3π + 1. b [5] a Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) Attempt to express LHS in terms of sin x and cos x only M1* Obtain 2cos2 x + 6sin 2 x A1 OE Use relevant identities to express in terms of cos2x only DM1 Confirm given result 4 − 2cos2x with sufficient detail A1 AG 4 7(b) π M1 4 − 2cos 2 Substitute x = 121 π and attempt exact value of 12 π sin 2 12 Obtain 2(4 − 2cos 16 π) or equivalent and hence 8 − 2 3 A1 2 7(c) Differentiate to obtain 4sin 2x B1 B2 if both limits obtained by symmetry or observation. Equate to 4 and obtain limits 14 π and π B1 Integrate to obtain 4x − sin 2x B1 Use limits their a and b to find exact value for integral of form k1 x + k2 sin 2 x M1 Angles must be in radians in terms of π . Obtain 4π − sin 2π − (π − sin 12 π) and confirm given result 3π + 1 with A1 AG sufficient detail 5
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Mathematics 9709/22 for October/November 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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