Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 2 · Variant 3

1 Solve the inequality 8. [3] |3x + 1| >

Mark scheme: 1 EITHER State or imply non-modular inequality (3x + 1)2 > 82, or corresponding equation or pair of linear equations M1 7 Obtain critical values or –3 A1 3 7 State correct answer x < –3 or x > Al 3 OR State one critical value, e.g. x = –3, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 State the other critical value correctly B1 7 State correct answer x < –3 or x > B1 [3] 3

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2 The sequence of values given by the iterative formula 7xn 5 8 xn+1 = , + 2x4n with initial value x1 1.7, converges to α. = (i) Use this iterative formula to determine α correct to 2 decimal places, giving the result of each iteration to 4 decimal places. [3] (ii) State an equation that is satisfied by α and hence show that α [2] 5√20. =

Mark scheme: 2 (i) Use the iterative formula correctly at least once M1 Obtain final answer 1.82 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.815, 1.825) B1 [3] 7 x 5 (ii) State equation x = + 4 , or equivalent B1 8 2 x Derive the exact answer α (or x) = 5 20 B1 [2]

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3 The polynomial x3 4x2 ax 2, where a is a constant, is denoted by It is given that the remainder when + is divided+ +by is equal to the remainder when p(x).is divided by p(x) (x + 1) p(x) (x −2). (i) Find the value of a. [3] (ii) When a has this value, show that is a factor of and find the quotient when is divided by (x −1) p(x) p(x)[3] (x −1).

Mark scheme: 3 (i) Substitute x = –l OR x = 2 correctly M1 Equate remainders to obtain correct equation 5 – a = 26 + 2a or equivalent Al Obtain a = –7 A1 [3] (ii) Attempt division by x – 1 and reach a partial quotient of x2 + kx M1 Obtain quotient x2 + 5x – 2 A1 EITHER Show remainder is zero OR substitute x = 1 to obtain zero B1 [3] 1 2

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4 (a) Find dx. [2] ã e1−2x (b) Express sin23x in terms of cos 6x and hence find sin23x dx. [4] ã

Mark scheme: 4 (a) Obtain integral of the form ke1 – 2x with any non-zero k M1 Correct integral A1 [2] (b) Attempt to use double angle formula to expand cos (3x + 3x) M1 1 1 State correct expression − cos6x or equivalent A1 2 2 Integrate an expression of the form a + b cos6x, where ab ≠ 0, correctly M1 1 1 State correct integral x – sin6x, or equivalent A1 [4] 2 12

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5 ln y (2.2, 1.2) (1.4, 0.8) x O The variables x and y satisfy the equation y where A and b are constants. The graph of ln y against x is a straight line passing through= theA(bx),points and as shown in the diagram. Find the values of A and b, correct to 2 decimal places.(1.4, 0.8) (2.2, 1.2), [6]

Mark scheme: 5 State or imply ln y = ln A + x ln b B1 Form a numerical expression for the gradient of the line M1 Obtain b = 1.65 A1 Use gradient and one point correctly to find In A M1 Obtain ln A = 0.1 A1 Obtain A = 1.11 A1 [6] GCE A/AS LEVEL – October/November 2010 9709 23

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6 (i) Express 2 sin θ θ in the form R where R 0 and α 90◦, giving the exact value of R and the−cosvalue of α correct sin(θto 2 decimal−α), places. > 0◦< < [3] (ii) Hence solve the equation 2 sin θ θ −cos = −0.4, giving all solutions in the interval [4] 0◦≤θ ≤360◦.

Mark scheme: 6 (i) State R = 5 B1 Use trig formula to find α M1 Obtain α = 26.57° with no errors seen A1 [3] ± 4.0  ≈ ±10.3048)° M1 (ii) Carry out evaluation of sin–1   (  5  Obtain answer 16.3° A1 Carry out correct method for second answer M1 Obtain answer 216.9° and no others in the range A1 [4]

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7 y M x O 1 ln x The diagram shows the curve y and its maximum point M. = x2 (i) Find the exact coordinates of M. [5] (ii) Use the trapezium rule with three intervals to estimate the value of 4 ln x dx, x2 ä 1 giving your answer correct to 2 decimal places. [3]

Mark scheme: 7 (i) Use product or quotient rule M1* Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1*(dep) Obtain x = e0.5 or e A1 1 Obtain , or equivalent A1 [5] 2e (ii) State or imply correct ordinates 0, 0.17328..., 0.12206..., 0.08664... B1 Use correct formula, or equivalent, correctly with h = 1 and four ordinates M1 Obtain answer 0.34 with no errors seen A1 [3] dy 2

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8 The equation of a curve is x2 2xy 8 0. + −y2 + = (i) Show that the tangent to the curve at the point is parallel to the x-axis. [4] (−2, 2) (ii) Find the equation of the tangent to the curve at the other point on the curve for which x giving your answer in the form y mx c. = −2,[5] = +

Mark scheme: dy 8 (i) State 2y as derivative of y2, or equivalent B1 dx dy State 2y + 2x as derivative of 2xy, or equivalent B1 dx dy Substitute x = –2 and y = 2 and evaluate M1 dx Obtain zero correctly and make correct conclusion A1 [4] (ii) Substitute x = –2 into given equation and solve M1 Obtain y = –6 correctly A1 dy Obtain = 2 correctly B1 d x Form the equation of the tangent at (–2, –6) M1 Obtain answer y = 2x – 2 Al [5]

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