Cambridge A Level Mathematics 9709 — 2008 May/June Paper 2 · Variant 1

1 Solve the inequality |3x - 1| < 2. [3]

Mark scheme: 1 EITHER State or imply non-modular inequality (3x – 1)2 < 22, or corresponding equation or pair of linear equations M1 1 Obtain critical values – and 1 A1 3 1 State correct answer – < x < 1 A1 3 OR State one critical value, e.g. x = 1, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 State the other critical value correctly B1 1 State correct answer – < x < 1 B1 [3] 3

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2 Use logarithms to solve the equation 4x = 2(3x), giving your answer correct to 3 significant figures. [4] 16π

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain xln 4 = ln 2 + xln 3, or equivalent A1 Solve for x M1 (dep*) Obtain answer x = 2.41 A1 [4] 1

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3 Find the exact value of (cos 2x + sin x) dx. [5] 0

Mark scheme: 3 Obtain integral 12 sin2x – cos x B1 + B1 Substitute limits correctly in an integral of the form a sin 2x + b cos x M1 1 π ) M1 Use correct exact values, e.g. of cos (6 1 Obtain answer 1 – 3 , or equivalent A1 [5] 4

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4 The polynomial 2x3 + 7x2 + ax + b, where a and b are constants, is denoted by p(x). It is given that (x + 1) is a factor of p(x), and that when p(x) is divided by (x + 2) the remainder is 5. Find the values of a and b. [5]

Mark scheme: 4 Substitute x = –1, equate to zero and obtain a correct equation in any form B1 Substitute x = –2 and equate to 5 M1 Obtain a correct equation in any form A1 Solve a relevant pair of equations for a or for b M1 Obtain a = 2 and b = –3 A1 [5]

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5 (i) Express 5 cos θ −sin θ in the form R cos(θ + α), where R > 0 and 0◦< α < 90◦, giving the exact value of R and the value of α correct to 2 decimal places. [3] (ii) Hence solve the equation 5 cos θ −sin θ = 4, giving all solutions in the interval 0◦≤θ ≤360◦. [4]

Mark scheme: 5 (i) State R = 26 B1 Use trig formula to find a M1 Obtain α = 11.31° with no errors seen A1 [3] 4 (ii) Carry out evaluation of cos–1 ( ) (≈ 38.3288…°) M1 26 Obtain answer 27.0° A1 Carry out correct method for second answer M1 Obtain answer 310.4° and no others in the range A1√ [4] [Ignore answers outside the given range.]

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6 It is given that the curve y = (x −2)ex has one stationary point. (i) Find the exact coordinates of this point. [5] (ii) Determine whether this point is a maximum or a minimum point. [2]

Mark scheme: 6 (i) Use product rule M1* Obtain correct derivative in any form, e.g. (x – 1)ex A1 Equate derivative to zero and solve for x M1* (dep) Obtain x = 1 A1 Obtain y = –e A1 [5] (ii) Carry out a method for determining the nature of a stationary point M1 Show that the point is a minimum point, with no errors seen A1 [2] GCE A/AS LEVEL – May/June 2008 9709 02 dy 2

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7 The equation of a curve is x2 + y2 −4xy + 3 = 0. dy 2y −x (i) Show that = [4] dx y −2x. (ii) Find the coordinates of each of the points on the curve where the tangent is parallel to the x-axis. [5]

Mark scheme: dy 7 (i) State 2y as derivative of y2, or equivalent B1 dx dy State 4y + 4x as derivative of 4xy, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain given answer correctly A1 [4] [The M1 is dependent on at least one of the B marks being obtained.] (ii) State or imply that the coordinates satisfy 2y – x = 0 B1 Obtain an equation in x2 (or y2) M1 Solve and obtain x2 = 4 (or y2 = 1) A1 State answer (2, 1) A1 State answer (–2, –1) A1 [5] 1 2

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1 8 The constant a, where a > 1, is such that x + dx = 6. x 1 (i) Find an equation satisfied by a, and show that it can be written in the form a = √(13 −2 ln a). [5] (ii) Verify, by calculation, that the equation a = √(13 −2 ln a) has a root between 3 and 3.5. [2] (iii) Use the iterative formula = √(13 −2 ln an+1 an), with a1 = 3.2, to calculate the value of a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 8 (i) Obtain terms 12 x2 and ln x B1 + B1 Substitute limits correctly M1 a2 + ln a – 1 Obtain correct equation in any form, e.g. 1 = 6 A1 2 2 Obtain given answer correctly A1 [5] (ii) Consider sign of a – (13 − 2ln a ) at a = 3 and a = 3.5, or equivalent M1 Complete the argument correctly with correct calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.26 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (3.255, 3.265) B1 [3]

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