Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 2 · Variant 2

1 Use logarithms to show that the equation 5 8 y = 6 7 x can be expressed in the form y = kx . Give the value of the constant k correct to 3 significant figures. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Apply logarithms correctly and apply power law at least once *M1 7ln6 DM1 May be implied by 0.97, 0.973 seen. Obtain y = x or equivalent perhaps involving decimals 8ln5 May use a different base. Obtain k = 0.974 or state y = 0.974 x A1 Or greater accuracy. 3

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3 A curve has equation 6e -x y 2 + e 2 x - 12y + 7 = 0 . Find the gradient of the curve at the point ( ln3, 2) . [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 Use product rule to differentiate 6e−x y 2 M1 Allow correct use of quotient rule. d y Must have from implicit differentiation. d x − x 2 − x dy A1 SOI Obtain −6e y + 12e y dx 2 x dy B1 May see separately. Obtain +2e − 12 dy dx 2 x +2e − 12 + 7 scores B0. dx dy M1 (Correct equation: Rearrange correctly to obtain = ... dy 2 x dy dx − x 2 − x −6e y + 12e y + 2e − 12 = 0) dx dx d y Must be at least one from implicit differentiation d x present. May already have used substitution dy 6 y 2 e − x − 2e 2 x = . dx 12e − x y − 12 d y M0 for inclusion of an incorrect . d x Substitute for x and y to obtain an exact final answer M1 d y Must be at least one from implicit differentiation d x present. Obtain 52 or 2.5 A1 6

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4 (a) Sketch the graphs of y = 1 + e 2 x 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 = 1 ln ( 3 - x) . 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(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. 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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 + e 2 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

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5 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. 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(b) Factorise p ( )x and hence show that the equation p ( )x = 0 has exactly one real root. 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(c) Solve the equation p b 1 cosec il = 0 for - 90° 1 i 1 90° . 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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 − 3 x + 4) A1 SOI Conclude with reference to root –2, discriminant is –23 and no further root A1 Or complete equivalent. 3 5(c) State cosec= −4 B1 1 May be implied by sin= − 4 1 M1 Allow for 14.5, 14.4. Attempt to find at least one value of  from sin=  4 Obtain –14.5 only and no others in the range A1 Or greater accuracy (14.4775…). 3

More questions on Quadratics

6 y 3 2 y = 5x + 7 B 27 A y = 2 x + 5 x O 2 3 2 27 The diagram shows the curves with equations y = 5x + 7 and y = for x H 0 . 2x + 5 The curves meet at the point ( 2, 3) . 3 2 Region A is bounded by the curve y = 5x + 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. 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(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. 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(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. 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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

More questions on Integration

7 (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° . 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(b) Hence find the smallest positive value of i satisfying the equation 1 + 4 sin i sin ( i + 60°) = 0 . 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Mark scheme: 7(a) Expand to obtain 2sin 2 + 2 3sincos B1 Needs to be simplified but may be implied by later work. Attempt to express in terms of sin2 and cos2 *M1 Needs to be in the form k + p cos2+ q sin2. Obtain 1 − cos2+ 3sin 2 A1 Obtain R = 2 A1 FT Following their p sin2+ q cos2 form. Use appropriate trigonometry to find  DM1 For their p sin2+ q cos2 form. Allow for 60 o . Obtain 1 + 2sin(2− 30) A1 Alternative Method for Question 7(a): Expand to obtain 2sin 2 + 2 3sincos B1 Needs to be simplified. For expansion of a + R sin ( 2− ) and comparison with *M1 Need comparison of at least 2 like terms: 2 R sin= 1 , R cos= 3 , a − R sin= 0. 2sin + 2 3sincos R sin= 1 and R cos= 3 leading to values of R and  DM1 Obtain R = 2 A1 FT Obtain = 30 o and a = 1 A1 Obtain 1 + 2sin(2− 30) A1 6 7(b) State sin(2− 30) = − 106 *B1 FT Following their answer to part (a). 1 Must be using sin(2− ) = − − 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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