Cambridge A Level Mathematics 9709 — 2022 May/June Paper 2 · Variant 1

1 The variables x and y satisfy the equation y where a is an integer. As shown in the diagram, the graph of ln y against x is a straight line =passing42x−a, through the point 0, , where the second coordinate is given correct to 3 significant figures. −20.8 (a) Show that the gradient of the straight line is ln 16. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Determine the value of a. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) State or imply equation is ln (2 )ln 4 y x a State gradient is 2ln4 and confirm ln16 B1 AG – necessary detail needed 2 1(b) Substitute for ln y and attempt value of a M1 Allow if ln 2 ln4   y x a Obtain 15  a A1 Integer answer required, but condone 15.0 2

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2 (a) Express the equation 7 tan 4 cot sec 0 in terms of sin only. [3] 1 + 1 −13 1 = 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 7 tan 4 cot sec 0 for [3] 1 + 1 −13 1 = 0Å < 1 < 360Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) M1 with at least two of the terms correct and no missing s Condone use of x instead of  Obtain   2 2 7sin 4cos 13sin 0       A1 SOI, OE Obtain 2 3sin 13sin 4 0      A1 Allow if missing s are recovered SC Allow full marks for 4 3sin 13 0 sin      Must be in terms of  for final A mark 3 Question Answer Marks Guidance 2(b) Attempt solution of 3-term quadratic equation for sin M1 Obtain 1 3 sin and hence 19.5 A1 or greater accuracy Obtain second value 160.5 A1 or greater accuracy; and no other values within the given range FT on o o 180 19.5 their 3

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3 The diagram shows the curve with equation y 3 sin x sin 2x for 0 The curve meets the x-axis at the origin and at the points with x-coordinates= −3a and ≤x ≤π. π. (a) Find the exact value of a. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the area of the shaded region. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Attempt to find x-value from 3sin 3sin 2 0 x x using identity for sin 2x M1 Obtain at least 1 2 cos  x A1 Obtain 1 3  A1 SC B3 can be spotted from sin sin 2  x x 3 3(b) Integrate to obtain form 1 2 cos cos2  k x k x *M1 non-zero constants 1 2 , k k M0 for 3cos 6cos2  x x Obtain correct 3 2 3cos cos2   x x A1 Attempt value of integral using their lower limit (in radians) and  correctly DM1 Allow one sign error Obtain 27 4 A1 OE 4

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5 (a) By sketching the graphs of y 5 and y 3 ln x = −2x = on the same diagram, show that the equation 5 3 ln x has exactly two roots. [3] −2x = (b) Show that the value of the larger root satisfies the equation x 2.5 1.5 ln x. [1] = + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Show by calculation that the value of the larger root lies between 4.5 and 5.0. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Use an iterative formula, based on the equation in part (b), to find the value of the larger root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Draw correct sketch of 5 2   y x *B1 with vertex on positive x-axis Draw correct sketch of 3ln  y x *B1 Indicate the two roots either on the diagram or by a statement DB1 3 Question Answer Marks Guidance 5(b) State 2 5 3ln   x x and rearrange to confirm 2.5 1.5ln   x x B1 AG – necessary detail needed 1 5(c) Consider sign of 2.5 1.5ln   x x , or equivalent, for 4.5 and 5.0 M1 Obtain 0.25...  and 0.08... or equivalents and justify conclusion A1 AG – necessary detail needed Alternative method for question 5(c) Consider sign of 5 2 3ln   x x , or equivalent, for 4.5 and 5.0 M1 Obtain 0.51...  and 0.17... or equivalents and justify conclusion A1 AG – necessary detail needed 2 5(d) Use iteration process correctly at least once M1 Obtain final answer 4.88 A1 Answer required to exactly 3 s.f. Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval [4.875, 4.885] A1 3

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