Cambridge A Level Mathematics 9709 — 2013 May/June Paper 2 · Variant 1

1 Solve the equation 1, giving answers correct to 2 decimal places where appropriate. [5] |2x −7| =

Mark scheme: x 21 Either State or imply non-modular equation (2 − 7 ) = 12, or corresponding pair of equations M1 Obtain 2x = 8 and 2x = 6 A1 State answer 3 B1 Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1 State answer 2.58 A1 Or State or imply one value for 2x, e.g. 8, by solving an equation or by inspection B1 State answer 3 B1 State second value for 2x B1 Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1 State answer 2.58 A1 [5] 2

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2 Solve the equation ln x ln 5. [5] ln(3 −2x) −2 =

Mark scheme: 2 Use 2 ln x = ln(x2) M1 Use law for addition or subtraction of logarithms M1 Obtain correct quadratic equation in x A1 Make reasonable solution attempt at a 3-term quadratic DM1 (dependent on previous M marks) 3 State x = and no other solutions A1 [5] 5

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3 (i) Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x). (ii) Hence show that 13π π 3√3 12 sin2x cos2x dx . ã 1 = 8 + 16 [3] 4π

Mark scheme: 3 (i) Either Use sin 2x = 2sin x cos x to convert integrand to k sin2 2x M1 Use cos 4x = 1 – 2 sin2 2x M1 1 1 State correct expression − cos 4 x or equivalent A1 2 2 Or 1 − cos 2 x 1 − cos 2 x Use cos2 x = and/or x = to obtain an equation in cos 2 x only M1 2 2 1 + cos 4 x Use cos2 2 x = M1 2 1 1 State correct expression − cos 4 x or equivalent A1 [3] 2 2 3 3 (ii) State correct integral x − sin 4 x , or equivalent B1 2 8 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [3] 3

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4 The polynomial ax3 bx 9, where a and b are constants, is denoted by It is given that −5x2 + + p(x). is a factor of and that when is divided by the remainder is 8. (2x + 3) 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: 3 4 (i) Substitute x = −2 , equate to zero M1 Substitute x = −1 and equate to 8 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 = −6 A1 [5] GCE AS LEVEL – May/June 2013 9709 21 (ii) Attempt either division by 2x + 3 and reach a partial quotient of x 2 + kx , use of an identity or observation M1 Obtain quotient x 2 −x4 + 3 Obtain linear factors x – 1 and x – 3 A1 [Condone omission of repetition that 2x + 3 is a factor.] A1 [If linear factors x – 1, x − 3 obtained by remainder theorem or inspection, award B2 + B1.] [3]

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5 The parametric equations of a curve are x e2t, y 4tet. = = dy (i) Show that 2(t + 1) . [4] dx = et (ii) Find the equation of the normal to the curve at the point where t 0. [4] =

Mark scheme: 5 (i) Use product rule to differentiate y M1 Obtain correct derivative in any form A1 d y d y d x Use = ÷ M1 d x d t d t Obtain given answer correctly A1 [4] d y (ii) Substitute t = 0 in and both parametric equations B1 d x dy Obtain = 2 and coordinates (1, 0) B1 dx dy Form equation of the normal at their point, using negative reciprocal of their M1 dx 1 1 State correct equation of normal y = − x + or equivalent A1 [4] 2 2

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6 (i) By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is in radians, has only one root for 0 2π. [2] ≤x ≤1 (ii) Verify by calculation that this root lies between x 0.7 and x 0.9. [2] = = (iii) Show that this root also satisfies the equation 1 2 tan x x + . = 4 tan x [1] 1 2 tan xn (iv) Use the iterative formula + to determine this root correct to 2 decimal places. xn+1 = 4 tan xn Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = cot x or y = 4x – 2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of 4x – 2 – cot x at x = 0.7 and x = 0.9, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] 1+ 2 tan x (iii) Show that given equation is equivalent to x = , or vice versa B1 [1] 4 tan x (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.76 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.755, 0.765) B1 [3] GCE AS LEVEL – May/June 2013 9709 21

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α7 (i) Express 5 sin 2θ 0 and giving the 2 cos 2θ in the form R where R 0◦< + sin(2θ + α), > < 90◦, exact value of R and the value of α correct to 2 decimal places. [3] Hence (ii) solve the equation 5 sin 2θ 2 cos 2θ 4, + = giving all solutions in the interval [5] 0◦≤θ ≤360◦, 1 (iii) determine the least value of as θ varies. [2] sin 2θ 4 cos (10 + 2θ)2

Mark scheme: 7 (i) State R = 29 B1 Use trig formula to find α M1 Obtain α = 21.80 0 with no errors seen A1 [3] −1  4  0 (ii) Carry out evaluation of sin   (≈ 47.97 ) M1  R  Carry out correct method for one correct answer M1 Obtain one correct answer e.g. 13.1o A1 Carry out correct method for a further answer M1 Obtain remaining 3 answers 55.10, 193.10, 235.10 and no others in the range A1 [5] (iii) Greatest value of 10 sin 2θ + 4 cos 2θ = 2√29 M1 1 A1 [2] 116

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