Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 3 · Variant 3

1 Solve the equation 1 ln x, ln(x + 5) = + giving your answer in terms of e. [3]

Mark scheme: 1 State or imply 1n e = 1 B1 Apply at least one logarithm law for product or quotient correctly M1 (or exponential equivalent) 5 Obtain x + 5= ex or equivalent and hence A1 [3] e − 1

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α 0 and Give the value2 (i) Express 24 sin θ cos θ in the form R where R 0◦< −7 sin(θ −α), > < 90◦. of α correct to 2 decimal places. [3] (ii) Hence find the smallest positive value of θ satisfying the equation 24 sin θ cos θ 17. −7 = [2]

Mark scheme: 2 (i) State or imply R = 25 B1 Use correct trigonometric formula to find ~ M1 Obtain 16.26 ° with no errors seen A1 [3] 17 (ii) Evaluate of sin − 1 ( = 42.84…°) M1 R Obtain answer 59.1 ° A1 [2]

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3 The parametric equations of a curve are 4t x y 2 = 2t 3, = ln(2t + 3). + dy (i) Express in terms of t, simplifying your answer. [4] dx (ii) Find the gradient of the curve at the point for which x 1. [2] =

Mark scheme: 3 (i) Either Use correct quotient rule or equivalent to obtain dx 4( 2t + )3 − 8t = or equivalent B1 dt ( 2t + 2)3 dy 4 Obtain = or equivalent B1 dt 2t + 3 dy dy dt = or equivalent M1 Use dx dx dt 1 Obtain (2t + 3 ) or similarly simplified equivalent A1 3 3 x Or Express t in terms of x or y e.g. t = B1 4 − 2 x  6  Obtain Cartesian equation e.g. y = 21n   B1  2 − x  dy 2 Differentiate and obtain = M1 dx 2 − x 1 Obtain (2t + 3 ) or similarly simplified equivalent A1 [4] 3 3 (ii) Obtain 2t = 3 or t = B1 2 dy Substitute in expression for and obtain 2 B1 [2] dx GCE A LEVEL – October/November 2012 9709 33

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4 The variables x and y are related by the differential equation 6xy. (x2 + 4)dydx = It is given that y 32 when x 0. Find an expression for y in terms of x. [6] = =

Mark scheme: 4 Separate variables correctly and integrate one side M1 Obtain ln y = ... or equivalent A1 Obtain = 31n ( x 2 + 4) or equivalent A1 Evaluate a constant or use x = 0, y = 32 as limits in a solution M1 containing terms a ln y and b ln ( x 2 + 4 ) Obtain ln y = 31n ( x 2 + 4) + ln 32 − 31n 4 or equivalent A1 1 2 Obtain y = (x + 4 ) or equivalent A1 [6] 2

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6 y x a O b The diagram shows the curve y x4 2x3 2x2 which crosses the x-axis at the points = + + −4x −16, (α, 0) and where α β. It is given that α is an integer. (β, 0) < (i) Find the value of α. [2] (ii) Show that β satisfies the equation x [3] = 3√(8 −2x). (iii) Use an iteration process based on the equation in part (ii) to find the value of β correct to 2 decimal places. Show the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Find y for x = –2 M1 Obtain 0 and conclude that ~== –2 A1 [2] (ii) Either Find cubic factor by division or inspection or equivalent M1 Obtain x 3 + 2 x − 8 A1 Rearrange to confirm given equation x = 3 8 − 2 x A1 Or Derive cubic factor from given equation and form product with (x – ~) M1 ( x + 2 )(x 3 + 2 x − 8 ) A1 Obtain quartic x 4 + 2 x 3 + 2 x 2 − 4 x − 16 ( = 0) A1 Or Derive cubic factor from given equation and divide the quartic by the cubic M1 (x 4 + 2 x 3 + 2 x 2 − 4 x − 16 ) ÷ (x 3 + 2 x − 8 ) A1 Obtain correct quotient and zero remainder A1 [3] (iii) Use the given iterative formula correctly at least once M1 Obtain final answer 1.67 A1 Show sufficient iterations to at least 4 d.p. to justify answer 1.67 to 2 d.p. or show there is a change of sign in interval (1.665, 1.675) A1 [3] GCE A LEVEL – October/November 2012 9709 33

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7 y x O The diagram shows part of the curve y sin32x cos32x. The shaded region shown is bounded by the = curve and the x-axis and its exact area is denoted by A. (i) Use the substitution u sin 2x in a suitable integral to find the value of A. [6] = kπ (ii) Given that dx 40A, find the value of the constant k. [2] ã 0 |sin32x cos32x| = [Questions 8, 9 and 10 are printed on the next page.]

Mark scheme: 7 (i) State or imply du = 2cos2x dx or equivalent B1 Express integrand in terms of u and du M1 1 3 2 Obtain u (1 − u ) du or equivalent A1 ∫ 2 Integration to obtain an integral of the form k 1 u 4 + k 2 u 6 , k 1 , k 2 ≠ 0 M1 1 Use limits 0 and 1 or (if reverting to x) 0 and π correctly DM1 4 1 Obtain , or equivalent A1 [6] 24 (ii) Use 40 and upper limit from part (i) in appropriate calculation M1 Obtain k = 10 with no errors seen A1 [2]

More questions on Integration

8 Two lines have equations 5 1 p 2 r 1 s and r 4 t 5 = + −1 = + ! 3 ! ! !, −4 −2 −4 where p is a constant. It is given that the lines intersect. (i) Find the value of p and determine the coordinates of the point of intersection. [5] (ii) Find the equation of the plane containing the two lines, giving your answer in the form ax by d, where a, b, c and d are integers. [5] + + cß =

Mark scheme: 8 (i) State or imply general point of either line has coordinates (5 + s, 1 – s, – 4 + 3s) or B1 (p + 2t, 4 + 5t, – 2 – 4t) Solve simultaneous equations and find s and t M1 Obtain s = 2 and t = – 1 or equivalent in terms of p A1 Substitute in third equation to find p = 9 A1 State point of intersection is (7, – 1, 2) A1 [5] (ii) Either Use scalar product to obtain a relevant equation in a, b, c e.g. a – b + 3c = 0 or 2a + 5b – 4c = 0 M1 State two correct equations in a, b, c A1 Solve simultaneous equations to obtain at least one ratio DM1 Obtain a : b : c = – 11 : 10 : 7 or equivalent A1 Obtain equation –11x + 10y + 7z = –73 or equivalent with integer coefficients A1  1   2      Or 1 Calculate vector product of − 1 and 5 M1          3   −4  Obtain two correct components of the product A1  −11    Obtain correct 10 or equivalent A1      7  Substitute coordinates of a relevant point in r.n = d to find d DM1 Obtain equation –11x + 10y + 7z = –73 or equivalent with integer coefficients A1 Or 2 Using relevant vectors, form correctly a two-parameter equation for the plane M1  5   1   2        Obtain r = 1 + λ −1 + µ 5 or equivalent A1              −4   3   −4  State three equations in x, y, z, λ , µ A1 Eliminate λ and µ DM1 Obtain 11x – 10y – 7z = 73 or equivalent with integer coefficients A1 [5] GCE A LEVEL – October/November 2012 9709 33 A Bx + C

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9 8x29 (i) Express −7x + in partial fractions. [5] (3 −x)(1 + x2) 9 8x2 (ii) Hence obtain the expansion of −7x + in ascending powers of x, up to and including the (3 −x)(1 + x2) term in x3. [5]

Mark scheme: A Bx + C 9 (i) State or imply form + B1 3 − x 1 + x 2 Use relevant method to determine a constant M1 Obtain A = 6 A1 Obtain B = –2 A1 Obtain C = 1 A1 [5] (ii) Either Use correct method to obtain first two terms of expansion − 1 −1 1  2 −1 of (3 −x ) or − 1 x  or ( 1 + x ) M1  3  A  1 1 2 1 3  Obtain  1 + x + x + x  A1 3  3 9 27  Obtain (Bx + C)(1 – x2) A1 Obtain sufficient terms of the product (Bx + C)(1 – x2), B , C ≠ 0 and add the two expansions M1 4 7 2 56 3 Obtain final answer 3 − x − x + x A1 3 9 27 Or Use correct method to obtain first two terms of expansion − 1 −1 1  2 −1 of (3 −x ) or − 1 x  or ( 1 + x ) M1  3  1  1 1 2 1 3  Obtain  1 + x + x + x  A1 3  3 9 27  Obtain (1 – x2) A1 Obtain sufficient terms of the product of the three factors M1 4 7 2 56 3 Obtain final answer 3 − x − x + x A1 [5] 3 9 27 2

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10 (a) Without using a calculator, solve the equation iw2 [3] = (2 −2i)2. (b) (i) Sketch an Argand diagram showing the region R consisting of points representing the complex numbers where ß |ß −4 −4i| ≤2. [2] (ii) For the complex numbers represented by points in the region R, it is given that p and α ≤|ß| ≤q ≤arg ß ≤β. Find the values of p, q, α and β, giving your answers correct to 3 significant figures. [6]

Mark scheme: 10 (a) Expand and simplify as far as i w 2 = − i8 or equivalent B1 Obtain first answer i 8 , or equivalent B1 Obtain second answer − i 8 , or equivalent and no others B1 [3] (b) (i) Draw circle with centre in first quadrant M1 Draw correct circle with interior shaded or indicated A1 [2] (ii) Identify ends of diameter corresponding to line through origin and centre M1 Obtain p = 3.66 and q = 7.66 A1 Show tangents from origin to circle M1 −1  1  Evaluate sin  2  M1  4  1 −1  1  Obtain α = π − sin  2  or equivalent and hence 0.424 A1 4  4  1 −1  1  Obtain β = π + sin  2  or equivalent and hence 1.15 A1 [6] 4  4 

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