Cambridge A Level Mathematics 9709 — 2009 Oct/Nov Paper 3 · Variant 1

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

Mark scheme: 1 EITHER: State or imply non-modular inequality (2 – 3x)2 < (x – 3)2, or corresponding equation, and make a reasonable solution attempt at a 3-term quadratic M1 Obtain critical value x = – 1 A1 2 Obtain x > – 1 A1 2 Fully justify x > – 1 as only answer A1 2 OR1: State the relevant critical linear equation, i.e. 2 – 3x = 3 – x B1 Obtain critical value x = – 1 B1 2 Obtain x > – 1 B1 2 Fully justify x > – 1 as only answer B1 2 OR2: Obtain the critical value x = – 1 by inspection, or by solving a linear inequality B2 2 Obtain x > – 1 B1 2 Fully justify x > – 1 as only answer B1 2 OR3: Make recognisable sketches of y = 2 – 3x and y = |x – 3| on a single diagram B1 Obtain critical value x = – 1 B1 2 Obtain x > – 1 B1 2 Fully justify x > – 1 as only answer B1 [4] 2 [Condone [ for > in the third mark but not the fourth.]

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3x 32, giving your answer correct to 3 significant figures. [4]2 Solve the equation 3x+2 = +

Mark scheme: 2 EITHER: Use laws of indices correctly and solve a linear equation for 3x, or for 3–x M1 3 2 Obtain 3x, or 3–x in any correct form, e.g. 3x = A1 (3 2 − )1 Use correct method for solving 3±x = a for x, where a > 0 M1 Obtain answer x = 0.107 A1 ln(3 x n + 9 ) OR: State an appropriate iterative formula, e.g. xn+1 = − 2 B1 ln 3 Use the formula correctly at least once M1 Obtain answer x = 0.107 A1 Show that the equation has no other root but 0.107 A1 [4] [For the solution 0.107 with no relevant working, award B1 and a further B1 if 0.107 is shown to be the only root.]

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3 The sequence of values given by the iterative formula 3xn 15 xn+1 = 4 + x3 , n with initial value x1 3, converges to α. = (i) Use this iterative formula to find α correct to 2 decimal places, giving the result of each iteration to 4 decimal places. [3] (ii) State an equation satisfied by α and hence find the exact value of α. [2]

Mark scheme: 3 (i) Use the iterative formula correctly at least once M1 State final answer 2.78 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in an appropriate function in (2.775, 2.785) A1 [3] 3 15 (ii) State a suitable equation, e.g. x = x + B1 4 x 3 State that the exact value of α is 4 60 , or equivalent B1 [2] GCE A/AS LEVEL – October/November 2009 9709 31

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4 A curve has equation y tan x. Find the x-coordinates of the stationary points on the curve in the = e−3x 1 interval Give your answers correct to 3 decimal places. [6] x −12π < < 2π.

Mark scheme: 4 Use product or quotient rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and obtain an equation of the form a sin 2x = b, or a quadratic in tan x, sin2 x, or cos2 x M1* Carry out correct method for finding one angle M1(dep*) Obtain answer, e.g. 0.365 A1 Obtain second answer 1.206 and no others in the range (allow 1.21) A1 [6] [Ignore answers outside the given range.] [Treat answers in degrees, 20.9° and 69.1°, as a misread.]

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5 (i) Prove the identity cos 4θ cos 2θ 3 sin4 θ. [4] −4 + ≡8 (ii) Using this result find, in simplified form, the exact value of 13π sin4 θ dθ. ã 1 [4] 6π

Mark scheme: 5 (i) EITHER: Use double angle formulae correctly to express LHS in terms of trig functions of 2θ M1 Use trig formulae correctly to express LHS in terms of sin θ, converting at least two terms M1 Obtain expression in any correct form in terms of sin θ A1 Obtain given answer correctly A1 OR: Use double angle formulae correctly to express RHS in terms of trig functions of 2θ M1 Use trig formulae correctly to express RHS in terms of cos 4θ and cos 2θ M1 Obtain expression in any correct form in terms of cos 4θ and cos 2θ A1 Obtain given answer correctly A1 [4] (ii) State indefinite integral 1 sin 4θ – 4 sin 2θ + 3θ, or equivalent B2 4 2 (award B1 if there is just one incorrect term) Use limits correctly, having attempted to use the identity M1 Obtain answer 1 (2π – 3 ), or any simplified exact equivalent A1 [4] 32

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6 With respect to the origin O, the points A, B and C have position vectors given by −−→OA i −−→OB 3i 2j and −−→OC 4i 2k. = −k, = + −3k = −3j + The mid-point of AB is M. The point N lies on AC between A and C and is such that AN 2NC. = (i) Find a vector equation of the line MN. [4] (ii) It is given that MN intersects BC at the point P. Find the position vector of P. [4]

Mark scheme: 6 (i) EITHER: State that the position vector of M is 2i + j – 2k, or equivalent B1 Carry out a correct method for finding the position vector of N M1 Obtain answer 3i – 2j + k, or equivalent A1 Obtain vector equation of MN in any correct form, e.g. r = 2i + j – 2k + λ(i – 3j + 3k) A1 OR: State that the position vector of M is 2i + j – 2k, or equivalent B1 Carry out a correct method for finding a direction vector for MN M1 Obtain answer, e.g. i – 3j + 3k, or equivalent A1 Obtain vector equation of MN in any correct form, e.g. r = 2i + j – 2k + λ(i – 3j + 3k) A1 [4] [SR: The use of AN = AC/3 can earn M1A0, but AN = AC/2 gets M0A0.] (ii) State equation of BC in any correct form, e.g. r = 3i + 2j – 3k + µ(i – 5j + 5k) B1 Solve for λ or for µ M1 Obtain correct value of λ, or µ, e.g. λ = 3, or µ = 2 A1 Obtain position vector 5i – 8j + 7k A1 [4] 3

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7 The complex number i is denoted by u. −2 + (i) Given that u is a root of the equation x3 0, where k is real, find the value of k. [3] −11x −k = (ii) Write down the other complex root of this equation. [1] (iii) Find the modulus and argument of u. [2] (iv) Sketch an Argand diagram showing the point representing u. Shade the region whose points represent the complex numbers satisfying both the inequalities ß and 0 |ß| < |ß −2| < arg(ß −u) < 14π. [4]

Mark scheme: 7 (i) Substitute x = –2 + i in the equation and attempt expansion of (–2 + i)3 M1 Use i2 = –1 correctly at least once and solve for k M1 Obtain k = 20 A1 [3] (ii) State that the other complex root is –2 – i B1 [1] GCE A/AS LEVEL – October/November 2009 9709 31 (iii) Obtain modulus 5 B1 Obtain argument 153.4° or 2.68 radians B1 [2] (iv) Show point representing u in relatively correct position in an Argand diagram B1 Show vertical line through z = 1 B1 Show the correct half-lines from u of gradient zero and 1 B1 Shade the relevant region B1 [4] [SR: For parts (i) and (ii) allow the following alternative method: State that the other complex root is –2 – i B1 State quadratic factor x2 + 4x + 5 B1 Divide cubic by 3-term quadratic, equate remainder to zero and solve for k, or, using 3-term quadratic, factorise cubic and obtain k M1 Obtain k = 20 A1] A B C

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5x 38 (i) Express + in partial fractions. [5] (x + 1)2(3x + 2) 5x 3 (ii) Hence obtain the expansion of + in ascending powers of x, up to and including the (x + 1)2(3x + 2) term in x2, simplifying the coefficients. [5]

Mark scheme: A B C 8 (i) State or imply partial fractions are of the form + 2 + B1 x + 1 ( x + )1 3 x + 2 Use any relevant method to obtain a constant M1 Obtain one of the values A = 1, B = 2, C = –3 A1 Obtain a second value A1 Obtain the third value A1 [5] (ii) Use correct method to obtain the first two terms of the expansion of (x + 1)–1, (x + 1)–2, (3x + 2)–1 or (1 + 3 x)–1 M1 2 Obtain correct unsimplified expansion up to the term in x2 of each partial fraction A1√ + A1√ + A1√ Obtain answer 3 − 11 x + 29 x 2 , or equivalent A1 [5] 2 4 8 − 1 [Symbolic binomial coefficients, e.g.   , are not sufficient for the first M1. The f.t. is on A, B, C.]  1  Dx + E C [The form 2 + , where D = 1, E = 3, C = –3, is acceptable. In part (i) give ( x + )1 3 x + 2 B1M1A1A1A1. In part (ii) give M1A1√A1√ for the expansions, and, if DE ≠ 0, M1 for multiplying out fully and A1 for the final answer.] [If B or C omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii), max 4/10] [If D or E omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii), max 4/10] [In the case of an attempt to expand (5x + 3)(x + 1)–2 (3x + 2)–1, give M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer.] [Allow use of Maclaurin, giving M1A1√A1√ for differentiating and obtaining f(0) = 3 and 2 f ′(0) = – 11 , A1√ for f ″(0) = 29 , and A1 for the final answer (the f.t. is on A, B, C if used).] 4 4

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9 y M A x O 4 ln x The diagram shows the curve y and its maximum point M. The curve cuts the x-axis at the = √x point A. (i) State the coordinates of A. [1] (ii) Find the exact value of the x-coordinate of M. [4] (iii) Using integration by parts, show that the area of the shaded region bounded by the curve, the x-axis and the line x 4 is equal to 8 ln 2 [5] = −4.

Mark scheme: 9 (i) State coordinates (1, 0) B1 [1] (ii) Use correct quotient or product rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1 Obtain x = e2 correctly A1 [4] GCE A/AS LEVEL – October/November 2009 9709 31 1 (iii) Attempt integration by parts reaching a x ln x ± a ∫ x x dx M1* 1 Obtain 2 x ln x − 2 ∫ dx A1 x Integrate and obtain 2 x ln x − 4 x A1 Use limits x = 1 and x = 4 correctly, having integrated twice M1(dep*) Justify the given answer A1 [5] dA

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10 In a model of the expansion of a sphere of radius r cm, it is assumed that, at time t seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When t 0, = dr r 5 and 2. = dt = (i) Show that r satisfies the differential equation dr 0.08r2. dt = [4] [The surface area A and volume V of a sphere of radius r are given by the formulae A 4πr2, 4 = V = 3πr3.] (ii) Solve this differential equation, obtaining an expression for r in terms of t. [5] (iii) Deduce from your answer to part (ii) the set of values that t can take, according to this model. [1]

Mark scheme: dA 10 (i) State or imply = kV M1* dt dr dr 4 Obtain equation in r and , e.g. 8πr = k πr3 A1 dt dt 3 dr Use = 2, r = 5 to evaluate k M1(dep*) dt Obtain given answer A1 [4] (ii) Separate variables correctly and integrate both sides M1 1 Obtain terms – and 0.08t, or equivalent A1 + A1 r Evaluate a constant or use limits t = 0, r = 5 with a solution containing terms of the form a and bt M1 r 5 Obtain solution r = , or equivalent A1 [5] 1( − 4.0t ) (iii) State the set of values 0 Y t < 2.5, or equivalent B1 [1] [Allow t < 2.5 and 0 < t < 2.5 to earn B1.]

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