Cambridge A Level Mathematics 9709 — 2009 Oct/Nov Paper 2 · Variant 1

1 Solve the inequality [4] |2x + 3| < |x −3|.

Mark scheme: 1 EITHER: Obtain a non-modular inequality from (2x + 3)2 < (x – 3)2, or corresponding quadratic equation, or pair of linear equations 2x + 3 = ±(x – 3) M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = –6 and x = 0 A1 State answer –6 I x I 0 A1 OR: obtain the critical value x = –6 from a graphical method or by inspection, or by solving a linear equation or inequality B1 Obtain the critical value x = 0 similarly B2 State answer –6 I x I 0 B1 [4] 2

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2 Solve the equation 2 ln x, giving your answer correct to 3 significant figures. [4] ln(3 −x2) =

Mark scheme: 2 Use lnx2 = 2lnx B1 Obtain 3 – x2 = x2, or equivalent B1 Solve for x M1 Obtain answer x – 1.22, having rejected x = –1.22 A1 [4] 1

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3 The polynomial 4x3 ax where a is a constant, is denoted by It is given that −8x2 + −3, p(x). (2x + 1) is a factor of p(x). (i) Find the value of a. [2] (ii) When a has this value, factorise completely. [4] p(x)

Mark scheme: 3 (i) Substitute x = − 1 and equate to zero M1 2 Obtain a = –11 A1 [2] (ii) EITHER: Attempt division by 2x + 1 reaching a partial quotient 2x2 – 5x M1 Obtain quadratic factor 2x2 – 5x – 3 A1 Obtain complete factorisation (2x + 1)2(x – 3) A1 + A1 OR: Obtain factor (x – 3) by inspection or factor theorem B2 Attempt division by (x – 3) reaching a partial quotient 4x2 + 4x M1 Obtain complete factorisation (2x + 1)2(x – 3) A1 [4]

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4 (i) Show that the equation 2 sin x can be written in the form tan x k, where k is a sin(60◦−x) = = constant. [4] (ii) Hence solve the equation [2] 2 sin x, for x 0◦< sin(60◦−x) = < 360◦.

Mark scheme: 4 (i) Use trig formulae to express equation in terms of sin x and cos x M1 Use cos 60° = 1 and sin 60° = 3 , or equivalent M1 2 2 Obtain equation in sin x and cos x in any correct form A1 Obtain tan x = 3 / 5 , or 0.3464…, or equivalent A1 [4] (ii) Obtain answer x = 19.1° B1 Obtain answer x = 199.1° and no others in the range B1√ [2] [ignore answers outside the given range.]

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5 (i) Express cos2 2x in terms of cos 4x. [2] 18π (ii) Hence find the exact value of cos2 2x dx. [4] ã 0

Mark scheme: 5 (i) Use double angle formulae and obtain a + bcos 4x M1 Obtain answer 1 + 1 cos 4x, or equivalent A1 [2] 2 2 (ii) Integrate and obtain 1 x + 1 sin 4x A1√ + A1√ 2 8 Substitute limits correctly M1 Obtain answer 1 π + 1 , or exact equivalent A1 [4] 16 8 GCE A/AS LEVEL – October/November 2009 9709 21

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6 The curve with equation y x ln x has one stationary point. = (i) Find the exact coordinates of this point, giving your answers in terms of e. [5] (ii) Determine whether this point is a maximum or a minimum point. [2]

Mark scheme: 6 (i) Use product rule M1* Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1(dep*) Obtain x = 1/e, or exact equivalent A1 Obtain y = –1/e, or exact equivalent A1 [5] (ii) Carry out complete method for determining the nature of a stationary point M1 Show that at x = 1/e there is a minimum point, with no errors seen A1 [2]

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7 y 1 R x O p 1 The diagram shows the curve y The shaded region R is bounded by the curve and the lines y = e−x. = and x p, where p is a constant. = (i) Find the area of R in terms of p. [4] (ii) Show that if the area of R is equal to 1 then p 2 = −e−p. [1] (iii) Use the iterative formula pn+1 = 2 −e−pn, 2, to calculate the value of p correct to 2 decimal places. Give the result with initial value p1 = of each iteration to 4 decimal places. [3]

Mark scheme: 7 (i) EITHER: Integrate 1 – e–x obtaining x ± e–x M1 Obtain indefinite integral x – e–x A1 Substitute limits x = 0, x = p correctly M1 Obtain answer p + e–p – 1, or equivalent A1 OR: Integrate e–x obtaining ± e–x M1 Substitute limits x = 0, x = p correctly M1 Obtain area below curve is 1 – e–p A1 Obtain answer p + e–p – 1, or equivalent A1 [4] (ii) Show that p + e–p – 1 = 1 is equivalent to p = 2 – e–p or vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.84 A1 Show sufficient iterations to justify its accuracy to 2 d.p. A1 [3]

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8 The equation of a curve is y2 2xy 2. + −x2 = (i) Find the coordinates of the two points on the curve where x 1. [2] = (ii) Show by differentiation that at one of these points the tangent to the curve is parallel to the x-axis. Find the equation of the tangent to the curve at the other point, giving your answer in the form ax by c 0. [7] + + =

Mark scheme: 8 (i) EITHER: Substitute x = 1 and attempt to solve 3-term quadratic in y M1 Obtain answers (1, 1) and (1, –3) A1 OR: State answers (1, 1) and (1, –3) B1 + B1 [2] dy (ii) State 2y as derivative of y2 B1 dx dy State 2y + 2x as derivative of 2xy B1 dx dy Substitute for x and y, and solve for M1 dx d y Obtain = 0 when x = 1 and y = 1 A1 dx d y Obtain = –2 when x = 1 and y = –3 A1√ dx Form the equation of the tangent at (1, –3) M1 Obtain answer 2x + y + 1 = 0 A1 [7]

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