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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w24_qp_229709_w24_ms_229709/22/O/N/24
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
Use logarithms to show that the equation 5 8y = 6 7x can be expressed in the form y = kx . Give the value of the constant k correct to 3 significant figures. [3]
Mark scheme
1 Apply logarithms correctly and apply power law at least once *M1 7ln 6 DM1 May be implied by 0.97, 0.973 seen. Obtain y = x or equivalent perhaps involving decimals 8ln 5 May use a different base. Obtain k = 0.974 or state y = 0.974 x A1 Or greater accuracy. 3
Question
Let f (x) = 4 sin 2 3x . (a) Find the value of f lb14 rl. [3] (b) Find y f (x) dx . [3]
Mark scheme
2(a) Differentiate to obtain form k sin3x cos3x or k sin 6x M1 Obtain correct 24sin3x cos3x or 12sin6x A1 Substitute x = 14 π to obtain –12 A1 3 2(b) Express integrand in form k1 + k2 cos6 x M1 Where k1k2 ≠ 0. Obtain 2 − 2cos6x A1 Integrate to obtain 2 x − 13 sin 6 x A1 OE 3
Question
A curve has equation 6e -x y 2 + e 2x - 12y + 7 = 0 . Find the gradient of the curve at the point (ln 3, 2) . [6]
Mark scheme
3 Use product rule to differentiate 6e − x y 2 M1 Allow correct use of quotient rule. dy Must have from implicit differentiation. dx dy A1 SOI Obtain −6e− x y 2 + 12e− x y dx dy B1 May see separately. Obtain +2e2 x − 12 dx dy +2e2 x − 12 + 7 scores B0. dx dy M1 (Correct equation: Rearrange correctly to obtain = ... dy dy dx −6e− x y 2 + 12e− x y + 2e2 x − 12 = 0) dx dx dy Must be at least one from implicit differentiation dx present. May already have used substitution dy 6 y 2e− x − 2e2 x = . dx 12e− x y − 12 dy M0 for inclusion of an incorrect . dx Substitute for x and y to obtain an exact final answer M1 dy Must be at least one from implicit differentiation dx present. Obtain 52 or 2.5 A1 6
Question
(a) Sketch the graphs of y = 1 + e 2x and y = x - 4 on the same diagram. [2] (b) The two graphs meet at the point P. Show that the x-coordinate of P satisfies the equation x = 12 ln (3 - x) . [2] (c) Use an iterative formula, based on the equation in part (b), to find the x-coordinate of P correct to 3 significant figures. Use an initial value of 0.45 and give the result of each iteration to 5 significant figures. [3]
Mark scheme
4(a) Show increasing curve above x-axis for y = 1 + e 2 x *B1 And appearing in first and second quadrants. Show V-shaped graph with vertex on positive x-axis and only one point of DB1 With modulus graph crossing y-axis above first graph. intersection with first curve 2 4(b) State or clearly imply 1 + e2 x = 4 − x B1 Arrange to confirm x = 12 ln(3 − x) B1 AG – necessary detail needed. Do not condone incorrect use of logs. 2 4(c) Use iterative process correctly at least once M1 Obtain final answer 0.465 A1 Answer required to exactly 3 sf. Show sufficient iterations to justify answer or show a sign change in the A1 interval [0.4645, 0.4655] 3
Question
The polynomial p (x) is defined by p (x) = ax 3 + bx 2 - ax + 8 , where a and b are constants. It is given that (x + 2) is a factor of p (x) , and that the remainder is 24 when p (x) is divided by (x - 2) . (a) Find the values of a and b. [4] (b) Factorise p (x) and hence show that the equation p (x) = 0 has exactly one real root. [3] (c) Solve the equation p b12 cosec il = 0 for - 90° 1 i 1 90° . [3]
Mark scheme
5(a) Substitute x = −2 and equate to zero M1 −8a + 4b + 2a + 8 = 0 Substitute x = 2 and equate to 24 M1 8a + 4b − 2a + 8 = 24 Obtain −6a + 4b + 8 = 0 and 6a + 4b − 16 = 0 A1 OE Obtain a = 2 and b = 1 A1 4 5(b) Divide by x + 2 at least as far as the x term M1 OE Obtain ( x + 2)(2 x 2 − 3x + 4) A1 SOI Conclude with reference to root –2, discriminant is –23 and no further root A1 Or complete equivalent. 3 5(c) State cosecθ = −4 B1 1 May be implied by sin θ = − 4 1 M1 Allow for 14.5, 14.4. Attempt to find at least one value of θ from sin θ = ± 4 Obtain –14.5 only and no others in the range A1 Or greater accuracy (14.4775…). 3
Question
y 3 y = 5x 2 + 7 B 27 A y= 2x + 5 O 2 x 3 27 The diagram shows the curves with equations y = 5x 2 + 7 and y = for x H 0 . 2x + 5 The curves meet at the point (2, 3) . 3 Region A is bounded by the curve y = 5x 2 + 7 and the straight lines x = 0 , x = 2 and y = 0 . Region B is bounded by the two curves and the straight line x = 0 . (a) Use the trapezium rule with two intervals to find an approximation to the area of region A. Give your answer correct to 3 significant figures. [3] (b) Find the exact total area of regions A and B. Give your answer in the form k ln m, where k and m are constants. [3] (c) Deduce an approximation to the area of region B. Give your answer correct to 3 significant figures. [1] (d) State, with a reason, whether your answer to part (c) is an over-estimate or an under-estimate of the area of region B. [2]
Mark scheme
6(a) Use y-values 3 7, 3 12 and 3 27, or decimal equivalents B1 1.913, 2.289, 3 Use correct formula, or equivalent, with h = 1 M1 May see 2 separate trapezia (2.1….+ 2.65…). Obtain 4.75 A1 3 6(b) Integrate to obtain form k ln(2 x + 5) M1 Obtain correct 272 ln(2 x + 5) A1 Condone inclusion of working for area in part (a). Apply limits 0 and 2 to obtain 272 ln 95 or exact equivalent of required form and A1 no extra terms 3 6(c) Obtain 3.19 B1 FT Following answers to (b) and (a). If incorrect answers for (a) and/or (b), then need to see their (b) (with no extra terms) – their (a). 1 6(d) State under-estimate … *B1 … because (a) is over-estimate due to tops of trapezia being above curve DB1 Or similarly detailed comment. 2
Question
(a) Express 4 sin i sin (i + 60°) in the form a + R sin (2i - a) , where a and R are positive integers and 0° 1 a 1 90° . [6] (b) Hence find the smallest positive value of i satisfying the equation 1 5 + 4 sin i sin (i + 60°) = 0 . [3] 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) Expand to obtain 2sin 2 θ + 2 3 sin θ cos θ B1 Needs to be simplified but may be implied by later work. Attempt to express in terms of sin 2θ and cos2θ *M1 Needs to be in the form k + p cos 2θ + q sin 2θ . Obtain 1 − cos 2θ + 3 sin 2θ A1 Obtain R = 2 A1 FT Following their p sin 2θ + q cos 2θ form. Use appropriate trigonometry to find α DM1 For their p sin 2θ + q cos 2θ form. Allow for 60o. Obtain 1 + 2sin(2θ − 30) A1 Alternative Method for Question 7(a): Expand to obtain 2sin 2 θ + 2 3 sin θ cos θ B1 Needs to be simplified. For expansion of a + R sin ( 2θ − α ) and comparison with *M1 Need comparison of at least 2 like terms: R sinα = 1 , R cos α = 3 , a − R sinα = 0. 2sin 2 θ + 2 3 sin θ cos θ R sinα = 1 and R cos α = 3 leading to values of R and α DM1 Obtain R = 2 A1 FT Obtain α = 30o and a = 1 A1 Obtain 1 + 2sin(2θ − 30) A1 6 7(b) State sin(2θ − 30) = − 106 *B1 FT Following their answer to part (a). 1 Must be using sin(2θ − α ) = − − their (a). Must have a 5 value for a. Attempt complete method to find smallest positive value DM1 Obtain 123.4 only A1 Or greater accuracy. 3
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Mathematics 9709/22 for October/November 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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