Cambridge A Level Mathematics 9709 — 2014 May/June Paper 3 · Variant 3

10 4 The equation x has one positive real root, denoted by = e2x !. −1 (i) Show that lies between x 1 and x 2. [2] ! = = (ii) Show that if a sequence of positive values given by the iterative formula P Q 1 10 2 + xn+1 = ln 1 xn converges, then it converges to [2] !. (iii) Use this iterative formula to determine correct to 2 decimal places. Give the result of each ! iteration to 4 decimal places. [3]

Mark scheme: 4 (i) Consider sign of x − 10 /( e 2 x − )1 at x = 1 and x = 2 M1 Complete the argument correctly with correct calculated values A1 2 (ii) State or imply α = 12 ln(1 + 10 / α ) B1 Rearrange this as α = 10 /( e 2α − )1 or work vice versa B1 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.14 A1 Show sufficient iterations to 4 d.p. to justify 1.14 to 2 d.p., or show there is a sign change in the interval (1.135, 1.145) A1 3

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6 y M x O 2 The diagram shows the curve x2 y2 2 x2 and one of its maximum points M. Find the + = −y2 coordinates of M. [7]

Mark scheme: 6 Obtain correct derivative of RHS in any form B1 Obtain correct derivative of LHS in any form B1 d y Set equal to zero and obtain a horizontal equation M1 d x Obtain a correct equation, e.g. x 2 + y 2 = 1 , from correct work A1 By substitution in the curve equation, or otherwise, obtain an equation in x 2 or y 2 M1 Obtain x = 12 3 A1 Obtain y = 12 A1 7 2

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37 (a) The complex number −5i is denoted by u. Showing your working, express u in the form 1 4i + x iy, where x and y are real. [3] + (b) (i) On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequalities and . [4] Ï −2 −i ≤1 Ï −i ≤ Ï −2 (ii) Calculate the maximum value of arg for points lying in the shaded region. [2] Ï

Mark scheme: 7 (a) EITHER: Multiply numerator and denominator by 1 − 4i, or equivalent, and use i2 = −1 M1 Simplify numerator to −17 −17i , or denominator to 17 A1 Obtain final answer −1 −i A1 OR: Using i2 = −1 , obtain two equations in x and y, and solve for x or for y M1 Obtain x = −1 or y = −1, or equivalent A1 Obtain final answer −1 − i A1 3 (b) (i) Show a point representing 2 + i in relatively correct position B1 Show a circle with centre 2 + i and radius 1 B1 Show the perpendicular bisector of the line segment joining i and 2 B1 Shade the correct region B1 4 (ii) State or imply that the angle between the tangents from the origin to the circle is required M1 Obtain answer 0.927 radians (or 53.1°) A1 2

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9 y M x O 120 The diagram shows the curve y e2 sinx cosx for 0 and its maximum point M. = ≤x ≤120, (i) Using the substitution u sin x, find the exact value of the area of the shaded region bounded by = the curve and the axes. [5] (ii) Find the x-coordinate of M, giving your answer correct to 3 decimal places. [6]

Mark scheme: 9 (i) Substitute for x and dx throughout using u = sinx and du = cos x dx, or equivalent M1 Obtain integrand e 2 u A1 Obtain indefinite integral 12 e 2 u A1 Use limits u = 0, u = 1 correctly, or equivalent M1 Obtain answer 12 e( 2 − )1 , or exact equivalent A1 5 GCE A LEVEL – May/June 2014 9709 33 (ii) Use chain rule or product rule M1 Obtain correct terms of the derivative in any form, e.g. 2cosx e 2 sin x cos x − e 2 sin x sin x A1 + A1 Equate derivative to zero and obtain a quadratic equation in sin x M1 Solve a 3-term quadratic and obtain a value of x M1 Obtain answer 0.896 A1 6

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10 The line l has equation r i 2j 3i 2k and the plane p has equation 2x 3y 18. = + −k + , −2j + + −5Ï = (i) Find the position vector of the point of intersection of l and p. [3] (ii) Find the acute angle between l and p. [4] (iii) A second plane q is perpendicular to the plane p and contains the line l. Find the equation of q, giving your answer in the form ax by d. [5] + + cÏ =

Mark scheme: 10 (i) Express general point of l in component form, e.g. 1( + 3λ , 2 − 2 λ , − 1 + 2 λ ) B1 Substitute in given equation of p and solve for λ M1 Obtain final answer − 12 i + 3 j − 2k , or equivalent, from λ = − 12 A1 3 (ii) State or imply a vector normal to the plane, e.g. 2 i + 3 j − 5k B1 Using the correct process, evaluate the scalar product of a direction vector for l and a normal for p M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse sine or cosine of the result M1 Obtain answer 23.2° (or 0.404 radians) A1 4 (iii) EITHER: State 2a + 3b − 5c = 0 or 3a − 2b + 2c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Obtain a : b : c = 4 : 19 : 13, or equivalent A1 Substitute coordinates of a relevant point in 4x + 19y + 13z = d, and evaluate d M1 Obtain answer 4x + 19y + 13z = 29, or equivalent A1 OR1: Attempt to calculate vector product of relevant vectors, e.g. (2i + 3j − 5k)×(3i − 2j + 2k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. −4i −19j −13k A1 Substitute coordinates of a relevant point in 4x + 19y + 13z = d M1 Obtain answer 4x + 19y + 13z = 29, or equivalent A1 OR2: Attempt to form a 2-parameter equation with relevant vectors M1 State a correct equation, e.g. r = i + 2 j − k + λ ( 2 i + 3 j − 5k ) + µ (3i − 2 j + 2k ) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain answer 4x + 19y + 13z = 29, or equivalent A1 OR3: Using a relevant point and relevant direction vectors, form a determinant equation for the plane M1 x − 1 y − 2 z + 1 State a correct equation, e.g. 2 3 − 5 = 0 A1 3 − 2 2 Attempt to expand the determinant M1 Obtain correct values of two cofactors A1 Obtain answer 4 x + 19 y + 13 z = 29 , or equivalent A1 5

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