Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 2 · Variant 2

1 Solve the inequality . [4] |x + 2| > 12x −2

Mark scheme:  1 1 EITHER State or imply non-modular inequality ( x + 2 )2 >  x − 2  , or corresponding  2  equation or pair of linear equations M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values –8 and 0 A1 State correct answer x < –8 or x > 0 A1 OR Obtain one critical value, e.g. x = –8, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 Obtain the other critical value similarly B2 State correct answer x < –8 or x > 0 B1 [4]

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giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 4x+1 = 52x−3, [4]

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain (x +1)log4 = (2x – 3)log5, or equivalent A1 Solve for x M1(dep*) Obtain answer x = 3.39 A1 [4]

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3 y M x O The diagram shows the curve y x ln x and its minimum point M. = −2 (i) Find the x-coordinate of M. [2] (ii) Use the trapezium rule with three intervals to estimate the value of 5 ln dx, ã 2 (x −2 x) giving your answer correct to 2 decimal places. [3] (iii) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii). [1]

Mark scheme: 3 (i) Obtain correct derivative B1 Obtain x = 2 only B1 [2] (ii) State or imply correct ordinates 0.61370..., 0.80277..., 1.22741..., 1.78112... B1 Use correct formula, or equivalent, correctly with h = 1 and four ordinates Ml Obtain answer 3.23 with no errors seen Al [3] (iii) Justify statement that the trapezium rule gives an over-estimate B1 [1]

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4 Find the exact value of the positive constant k for which k 2k e4x dx ex dx. ã 0 = ã 0 [6]

Mark scheme: 4 State at least one correct integral B1 Use limits correctly to obtain an equation in e2k, e4k M1 Carry out recognizable solution method for quadratic in e2k M1 Obtain e2k = 1 and e2k = 3 A1 Use logarithmic method to solve an equation of the form eλa = b, where b > 0 M1 1 Obtain answer k = ln 3 A1 [6] 2 1

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5 (i) By sketching a suitable pair of graphs, show that the equation 1 sin x, x = [2] where x is in radians, has only one root for 0 x < ≤12π. (ii) Verify by calculation that this root lies between x 1.1 and x 1.2. [2] = = 1 (iii) Use the iterative formula xn+1 = sin xn to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 1 5 (i) Make a recognisable sketch of a relevant graph, e.g. y = sin x or y = B1 x Sketch a second relevant graph and justify the given statement B1 [2] 1 (ii) Consider sign of − sin x at x = 1.1 and x = 1.2, or equivalent M1 x Complete the argument correctly with appropriate calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.11 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.105, 1.115) B1 [3] GCE AS/A LEVEL – October/November 2011 9709 22 dx dy 2

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6 The parametric equations of a curve are x 1 2 sin2θ, y 4 tan θ. = + = dy 1 (i) Show that . [3] dx = sin θ cos3θ (ii) Find the equation of the tangent to the curve at the point where θ 14π, giving your answer in = the form y mx c. [4] = +

Mark scheme: dx dy 6 (i) State = 4 sin θ cos θ or equivalent (nothing for = 4 sec 2 θ ) B1 dt dx dy dy dx Use = ÷ M1 dx dθ dθ Obtain given answer correctly A1 [3] π dy (ii) Substitute θ = in and both parametric equations M1 4 dx dy Obtain = 4 and coordinates (2, 4) A1 dx Form equation of tangent at their point M1 State equation of tangent in correct form y = 4x – 4 A1 [4]

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7 The polynomial ax3 b, where a and b are constants, is denoted by It is given that −3x2 −11x + p(x). is a factor of and that when is divided by the remainder is 12. (x + 2) p(x), p(x) (x + 1) (i) Find the values of a and b. [5] (ii) When a and b have these values, factorise completely. [3] p(x)

Mark scheme: 7 (i) Substitute x = –2, equate to zero and obtain a correct equation in any form B1 Substitute x = –1 and equate to 12 M1 Obtain a correct equation in any form A1 Solve a relevant pair of equations for a or b M1 Obtain a = 2 and b = 6 A1 [5] (ii) Attempt division by x + 2 and reach a partial quotient of 2x2 – 7x M1 Obtain quotient 2x2 – 7x + 3 A1 Obtain linear factors 2x – 1 and x – 3 A1 [Condone omission of repetition that x + 2 is a factor.) [If linear factors 2x – 1, x – 3 obtained by remainder theorem or inspection, award B2 + B1.] S.C. M1A1√ if a, b not both correct [3]

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α8 (i) Express 5 cos θ 0 and giving the exact sin θ in the form R where R −3 cos(θ + α), > < 90◦, 0◦< 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, −3 = giving all solutions in the interval [4] 0◦≤θ ≤360◦. (iii) Write down the least value of 15 cos θ sin θ as θ varies. [1] −9

Mark scheme: 8 (i) State R = 34 B1 Use trig formula to find α M1 Obtain α = 30.96º with no errors seen A1 [3] ± 4    (ii) Carry out evaluation of cos–1 (≈ 46.6861º or 313.3139º) M1  R  Obtain answer 15 .7º A1 Carry out correct method for second answer M1 Obtain answer 282.3º or 282.4º and no others in the range A1 [4] (iii) State − 3 34 (= –3R) B1√ [1]

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