Cambridge A Level Mathematics 9709 — 2014 May/June Paper 3 · Variant 1

1 (i) Simplify sin sec [2] 2! !. (ii) Given that 3 cos 7 cos 0, find the exact value of cos [3] 2" + " = ".

Mark scheme: 1 (i) State sin2α = 2sinα cosα and secα = 1/cosα B1 Obtain 2sinα B1 [2] (ii) Use cos2β = 2cos2β –1 or equivalent to produce correct equation in cosβ B1 Solve three-term quadratic equation for cosβ M1 1 Obtain cos β = only A1 [3] 3 d u 2

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4 The variables x and y are related by the differential equation dy 6ye3x . dx = 2 e3x + Given that y 36 when x 0, find an expression for y in terms of x. [6] = =

Mark scheme: 4 Separate variables correctly and recognisable attempt at integration of at least one side M1 Obtain ln y, or equivalent B1 3 x Obtain k l n (2 + e ) B1 x k 3 x Use y(0) = 36 to find constant in y = A(2 + 3e ) or l ny = k l n (2 + e ) + c or equivalent M1*  k  3 x *M1 Obtain equation correctly without logarithms from l ny = l n  A( 2 + e )    3 x 2 Obtain y = 4 (2 + e ) A1 [6] GCE A LEVEL – May/June 2014 9709 31 2

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9i 5 The complex number is defined by . Find, showing all your working, 9ï3 + Ï Ï = ï3 −i (i) an expression for 0 and [5] Ï in the form rei1, where r > −0 < 1 ≤0, (ii) the two square roots of 0 and [3] Ï, giving your answers in the form rei1, where r > −0 < 1 ≤0.

Mark scheme: 5 (i) Either Multiply numerator and denominator by 3 + i and use i2 = –1 M1 Obtain correct numerator 18 + 18 i3 or correct denominator 4 B1 9 9   Obtain + i3 or  18 + 18 i3  / 4 A1 2 2   Obtain modulus or argument M1 1 π i Obtain 9e 3 A1 [5] OR Obtain modulus and argument of numerator or denominator, or both moduli or both arguments M1 1 1 Obtain moduli and argument 18 and π or 2 and − π 6 6 1 1 or moduli 18 and 2 or arguments π and − π (allow degrees) B1 6 6 1 1 π i − π i Obtain 18e 6 ÷ 2e 6 or equivalent A1 Divide moduli and subtract arguments M1 1 π i Obtain 9e 3 A1 [5] 1 π i (ii) State 3e 6 , following through their answer to part (i) B1 1 1 π i ± π i State 3e 6 2 , following through their answer to part (i) B1 5 − π i Obtain 3e 6 B1 [3]

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7 The straight line l has equation r 4i 2k 2i 6k . The plane p passes through the point 4, 2 and is perpendicular to=l. −j + + , −3j + −1, (i) Find the equation of p, giving your answer in the form ax by d. [2] + + cÏ = (ii) Find the perpendicular distance from the origin to p. [3] (iii) A second plane q is parallel to p and the perpendicular distance between p and q is 14 units. Find the possible equations of q. [3]

Mark scheme: 7 (i) Obtain 2x – 3y + 6z for LHS of equation B1 Obtain 2x – 3y + 6z = 23 B1 [2] (ii) Either Use correct formula to find perpendicular distance M1 ± 23 Obtain unsimplified value , following answer to (i) A1 2 2 + (− 3)2 + 6 2 23 Obtain or equivalent A1 [3] 7 GCE A LEVEL – May/June 2014 9709 31 OR 1 Use scalar product of (4, –1, 2) and a vector normal to the plane M1 (8 + 3 + 12 ) Use unit normal to plane to obtain ± A1 49 23 Obtain or equivalent A1 [3] 7 OR 2 Find parameter intersection of p and r = µ (2i – 3j + 6k) M1 23  46 69 138  Obtain µ = [and  , − ,  as foot of perpendicular] A1 49  49 49 49  23 Obtain distance or equivalent A1 [3] 7 (iii) Either Recognise that plane is 2x – 3y + 6z = k and attempt use of formula for perpendicular distance to plane at least once M1 23 −k Obtain = 14 or equivalent A1 7 Obtain 2x – 3y + 6z = 121 and 2x – 3y + 6z = –75 A1 [3] OR Recognise that plane is 2x – 3y + 6z = k and attempt to find at least one point on q using l with λ = ±2 M1 Obtain 2x – 3y + 6z = 121 A1 Obtain 2x – 3y + 6z = –75 A1 [3]

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10 y T1 T3 x O T4 T2 The diagram shows the curve y 2x sin 4x for x The stationary points are labelled T1, T2, = 10e−1 ≥0. … as shown. T3, (i) Find the x-coordinates of T1 and T2, giving each x-coordinate correct to 3 decimal places. [6] (ii) It is given that the x-coordinate of Tn is greater than 25. Find the least possible value of n. [4]

Mark scheme: 10 (i) Use of product or quotient rule M1 1 1 − x − x Obtain − 5e 2 sin 4 x + 40e 2 cos 4 x A1 dy Equate to zero and obtain tan4z = k or R cos(4x ± α) M1 dx  −1 1  Obtain tan 4x = 8 or 65 cos 4 x ± tan  A1  8  Obtain 0.362 or 20.7° A1 Obtain 1.147 or 65.7° A1 [6] 1 (ii) State or imply that x-coordinates of Tn are increasing by π or 45° B1 4 Attempt solution of inequality (or equation) of form x1 + (n – 1)kπ . 25 M1 4 Obtain n > (25 − .0362 ) + 1 , following through on their value of x1 A1 π n = 33 A1 [4]

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