Cambridge A Level Mathematics 9709 — 2013 May/June Paper 3 · Variant 2

1 Solve the equation x 13x . [3] −2 =

Mark scheme: 2  1  1 EITHER: State or imply non-modular equation ( x − 2 ) =  x  ,  3  1 or pair of equations x − 2 = ± x M1 3 Obtain answer x = 3 A1 3 Obtain answer x = , or equivalent A1 2 OR: Obtain answer x = 3 by solving an equation or by inspection B1 1 State or imply the equation x – 2 = − , or equivalent M1 3 3 Obtain answer x = , or equivalent A1 [3] 2

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2 The sequence of values given by the iterative formula xn x3n 100 , + xn+1 = 2 x3n 25 + with initial value x1 3.5, converges to = !. (i) Use this formula to calculate correct to 4 decimal places, showing the result of each iteration to 6 decimal places. ! [3] (ii) State an equation satisfied by and hence find the exact value of [2] ! !.

Mark scheme: 2 (i) Use the iterative formula correctly at least once M1 Obtain final answer 3.6840 A1 Show sufficient iterations to at least 6 d.p. to justify 3.6840, or show there is a sign change in the interval (3.68395, 3.68405) A1 [3] x ( x 3 + 100) (ii) State a suitable equation, e.g. x = B1 2( x 3 + 25) State that the value of α is 3 50 , or exact equivalent B1 [2] 2

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5 y M x –a 3a The diagram shows the curve with equation x3 xy2 ay2 0, + + −3ax2 = where a is a positive constant. The maximum point on the curve is M. Find the x-coordinate of M in terms of a. [6]

Mark scheme: dy 5 EITHER: State 2 ay as derivative of ay2 B1 dx dy 2 State y + 2 xy as derivative of xy2 B1 dx d y Equate derivative of LHS to zero and set equal to zero M1 d x Obtain 3 x 2 + y 2 − 6 ax = 0 , or horizontal equivalent A1 Eliminate y and obtain an equation in x M1 Solve for x and obtain answer x = 3a A1 2 3ax 2 − x 3 OR1: Rearrange equation in the form y = and attempt differentiation of one x + a side B1 Use correct quotient or product rule to differentiate RHS M1 Obtain correct derivative of RHS in any form A1 d y Set equal to zero and obtain an equation in x M1 d x Obtain a correct horizontal equation free of surds A1 Solve for x and obtain answer x = 3a A1 1  3ax 2 − x 3  2 OR2: Rearrange equation in the form y =   and differentiation of RHS B1 x + a   Use correct quotient or product rule and chain rule M1 Obtain correct derivative in any form A1 GCE AS/A LEVEL – May/June 2013 9709 32 Equate derivative to zero and obtain an equation in x M1 Obtain a correct horizontal equation free of surds A1 Solve for x and obtain answer x = 3a A1 [6]

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10 The points A and B have position vectors 2i 2k and 5i k respectively. The plane p has equation x y 5. −3j + −2j + + = (i) Find the position vector of the point of intersection of the line through A and B and the plane p. [4] (ii) A second plane q has an equation of the form x by d, where b, c and d are constants. The plane q contains the line AB, and the acute angle+ +betweencÏ = the planes p and q is Find the equation of q. 60Å. [7]

Mark scheme: 10 (i) Carry out a correct method for finding a vector equation for AB M1 Obtain r = 2i – 3j + 2k + λ (3i + j – k) or r = µ (2i + 3j + 2k) + (1 – µ )(5i – 2j + k), or equivalent A1 Substitute components in equation of p and solve for λ or for µ M1 3 1 13 3 1 Obtain λ = or µ = − and final answer i − j + k, or equivalent A1 [4] 2 2 2 2 2 (ii) Either equate scalar product of direction vector of AB and normal to q to zero or substitute for A and B in the equation of q and subtract expressions M1* Obtain 3 + b – c = 0, or equivalent A1 Using the correct method for the moduli, divide the scalar product of the normals to 1 p and q by the product of their moduli and equate to ± , or form horizontal 2 equivalent M1* 1 + b 1 Obtain correct equation in any form, e.g. = ± A1 2 (1 + b 2 + c 2 ) 1( + )1 Solve simultaneous equations for b or for c M1 (dep*) Obtain b = –4 and c = –1 A1 Use a relevant point and obtain final answer x – 4y – z = 12, or equivalent A1 [7] (The f.t. is on b and c.)

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