Cambridge A Level Mathematics 9709 — 2016 May/June Paper 3 · Variant 3

1 Solve the inequality 2 x 3x 1 . [4] −2 > +

Mark scheme: 1 EITHER: State or imply non-modular inequality (2( x − 2)) 2 > (3 x + 1) 2 , or corresponding quadratic equation, or pair of linear equations 2( x − 2) = ± (3 x + 1) B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for x M1 Obtain critical values x = −5 and x = 53 A1 State final answer −<5 x < 53 A1 OR: Obtain critical value x = −5 from a graphical method, or by inspection, or by solving a linear equation or inequality (B1 Obtain critical value x = 53 similarly B2 State final answer −<5 x < 53 B1) [Do not condone ≤ for <.] [4]

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2 The variables x and y satisfy the relation 3y = 42−x. (i) By taking logarithms, show that the graph of y against x is a straight line. State the exact value of the gradient of this line. [3] (ii) Calculate the exact x-coordinate of the point of intersection of this line with the line with equation y 2x, simplifying your answer. [2] =

Mark scheme: 2 (i) State or imply y ln3 = (2 − x )ln 4 B1 State that this is of the form ay = bx + c and thus a straight line, or equivalent B1 ln4 State gradient is − , or exact equivalent B1 ln3 [3] (ii) Substitute y = 2x and solve for x, using a log law correctly at least once M1 Obtain answer x = ln 4 / ln6 , or exact equivalent A1 [2]

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4 The parametric equations of a curve are x t cost, y ln 1 sin t , = + = + where t 1 −120 < < 20. dy (i) Show that sec t. [5] dx = (ii) Hence find the x-coordinates of the points on the curve at which the gradient is equal to 3. Give your answers correct to 3 significant figures. [3]

Mark scheme: dx 4 (i) State = 1 − sin t B1 dt Use chain rule to find the derivative of y M1 dy cos t Obtain = , or equivalent A1 dt 1 + sin t d y dy d x Use = ÷ M1 d x dt dt Obtain the given answer correctly A1 [5] (ii) State or imply t = cos −1 ( 13 ) B1 Obtain answers x = 1.56 and x = − 0.898 B1 + B1 [3]

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5 The variables x and y satisfy the differential equation dy e−2y tan2x, dx = for 0 1 and it is given that y 0 when x 0. Solve the differential equation and calculate the ≤x < 20, 1 = = value of y when x [8] = 40.

Mark scheme: 5 Separate variables and make reasonable attempt at integration of either integral M1 Obtain term 12 e 2 y B1 Use Pythagoras M1 Obtain terms tan x − x A1 Evaluate a constant or use x = 0, y = 0 as limits in a solution containing terms a e ± 2 y and b tan x ,( ab ≠ 0) M1 Obtain correct solution in any form, e.g. 12 e 2 y = tan x − x + 12 A1 Set x = 14 π and use correct method to solve an equation of the form e ± 2 y = a or e ± y = a , where a > 0 M1 Obtain answer y = 0.179 A1 [8]

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6 The curve with equation y x2 cos 2x1 has a stationary point at x p in the interval 0 x = = < < 0. 1 4 (i) Show that p satisfies the equation tan 2p p. [3] = (ii) Verify by calculation that p lies between 2 and 2.5. [2] @ A 4 (iii) Use the iterative formula 2 to determine the value of p correct to 2 decimal tan−1 pn+1 = pn places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Use the product rule M1 Obtain correct derivative in any form A1 Equate 2-term derivative to zero and obtain the given answer correctly A1 [3] (ii) Use calculations to consider the sign of a relevant expression at p = 2 and p = 2.5, or compare values of relevant expressions at p= 2 and p = 2.5 M1 Complete the argument correctly with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 2.15 A1 Show sufficient iterations to 4 d.p. to justify 2.15 to 2 d.p., or show there is a sign change in the interval (2.145,2.155) A1 [3]

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1 x5 7 Let I dx. 3 = Ô0 1 x2 + 2 u 2 (i) Using the substitution u 1 x2, show that I du. [3] −1 2u3 = + = Ô1 (ii) Hence find the exact value of I. [5]

Mark scheme: 7 (i) State or imply du = 2x dx , or equivalent B1 Substitute for x and dx throughout M1 Reduce to the given form and justify the change in limits A1 [3] (ii) Convert integrand to a sum of integrable terms and attempt integration M1 1 1 1 Obtain integral 2 ln u + − 2 , or equivalent A1 + A1 u 4u (deduct A1 for each error or omission) Substitute limits in an integral containing two terms of the form a ln u and bu− 2 M1 Obtain answer 12 ln2 − 165 , exact simplified equivalent A1 [5]

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8 The points A and B have position vectors, relative to the origin O, given by OA i j k and −−→ = + + OB 2i 3k. The line l has vector equation r 2i 2j k . −−→ = + = −2j −k + - −i + + (i) Show that the line passing through A and B does not intersect l. [4] 1 (ii) Show that the length of the perpendicular from A to l is ï2. [5]

Mark scheme: 8 (i) State a correct equation for AB in any form, e.g. r = i + j + k + λ ( i −+j 2k ) , or equivalent B1 Equate at least two pairs of components of AB and l and solve for λ or for µ M1 Obtain correct answer for λ or for µ , e.g. λ = −1 or µ = 2 A1 Show that not all three equations are not satisfied and that the lines do not intersect A1 [4] (ii) EITHER: Find AP (or PA) for a general point P on l, e.g. (1 − µ ) i + ( −+3 2 µ ) j + ( −+2 µ )k B1 Calculate the scalar product of AP and a direction vector for l and equate to zero M1 Solve and obtain µ = 32 A1 Carry out a method to calculate AP when µ = 32 M1 1 Obtain the given answer correctly A1 2 OR 1:Find AP (or PA) for a general point P on l (B1 Use correct method to express AP 2 (or AP) in terms of µ M1 Obtain a correct expression in any form, e.g. (1 − µ ) 2 + ( −+3 2 µ ) 2 + ( −+2 µ ) 2 A1 Carry out a complete method for finding its minimum M1 Obtain the given answer correctly A1) OR 2:Calling (2, −2, −1) C, state AC (or CA) in component form, e.g. i − 3 j − 2k (B1 Use a scalar product to find the projection of AC ( or CA) on l M1 9 Obtain correct answer in any form, e.g. A1 6 Use Pythagoras to find the perpendicular M1 Obtain the given answer correctly A1) OR 3:State AC ( or CA) in component form (B1 Calculate vector product of AC and a direction vector for l, e.g.( i − 3 j − 2k ) × ( −+i 2 j + k ) M1 Obtain correct answer in any form, e.g. i + j − k A1 Divide modulus of the product by that of the direction vector M1 Obtain the given answer correctly A1) [5] u

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9 Throughout this question the use of a calculator is not permitted. The complex numbers 3i and 2 are denoted by u and v respectively. In an Argand diagram with origin O, the points−1A,+ B and C represent−i the numbers u, v and u v respectively. + (i) Sketch this diagram and state fully the geometrical relationship between OB and AC. [4] u (ii) Find, in the form x iy, where x and y are real, the complex number . [3] v + (iii) Prove that angle AOB 3 [2] = 40.

Mark scheme: u 9 (i) EITHER: Multiply numerator and denominator of by 2 + i, or equivalent M1 v Simplify the numerator to −5 +5i or denominator to 5 A1 Obtain final answer −1 + I A1 OR: Obtain two equations in x and y and solve for x or for y (M1 Obtain x = −1 or y = 1 A1 Obtain final answer −1 + I A1) [3] (ii) Obtain u + v = 1 + 2i B1 In an Argand diagram show points A, B, C representing u, v and u + v respectively B1 State that OB and AC are parallel B1 State that OB = AC B1 [4] (iii) Carry out an appropriate method for finding angle AOB, e.g. find arg(u / v ) M1 Show sufficient working to justify the given answer 34π A1 [2] A B C

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