Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 3 · Variant 1

1 Find the set of values of x satisfying the inequality [4] 3|x −1| < |2x + 1|.

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

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2 Solve the equation 5x 5x−1 = −5, giving your answer correct to 3 significant figures. [4]

Mark scheme: 2 EITHER Use laws of indices correctly and solve for 5x or for 5–x or for 5x–1 M1 5 Obtain 5x or for 5–x or for 5x–1 in any correct form, e.g. 5x = A1 1 – 1 5 Use correct method for solving 5x = a, or 5–x = a, or 5x–1 = a, where a 0 M1 Obtain answer x = 1.14 A1 ln 5x–1+5 OR Use an appropriate iterative formula, e.g. xn+1 = ln 5 , correctly, at least once M1 Obtain answer 1.14 A1 Show sufficient iterations to at least 3 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 Show there is no other root A1 [4] [For the solution x = 1.14 with no relevant working give B1, and a further B1 if 1.14 is shown to be the only solution.]

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3 Solve the equation 2 sin(θ + 45◦) = cos(θ −30◦), θ [5] giving all solutions in the interval < 180◦. 0◦<

Mark scheme: 3 Attempt use of sin (A + B) and cos (A – B) formulate to obtain an equation in cos θ and sin θ M1 Obtain a correct equation in any form A1 Use trig. formula to obtain an equation in tan θ (or cos θ, sin θ or cot θ) M1 √6 –1 Obtain tan θ = 1 – √2 , or equivalent (or find cost θ, sin θ or cot θ) A1 Obtain answer θ = 105.9°, and no others in the given interval A1 [5] [Ignore answers outside the given material] 3

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4 When where a is a positive constant, is expanded in ascending powers of x, the coefficients of x and(1 x3+ ax)−2,are equal. (i) Find the exact value of a. [4] (ii) When a has this value, obtain the expansion up to and including the term in x2, simplifying the coefficients. [3]

Mark scheme: 4 (i) Obtain correct unsimplified terms in x and x3 B1 + B1 Equate coefficients and solve for a M1 1 Obtain final answer a = , or exact equivalent A1 [4] √2 (ii) Use correct method and value of a to find the first two terms of the expansion (1 + ax)–2 M1 Obtain 1 – √2x, or equivalent A1 3 Obtain term 2 x A1 [3] –2 a, are not sufficient for the first B marks] [Symbolic coefficients, e.g. 1 [The f.t. is solely on the value of a.] GCE AS/A LEVEL – October/November 2012 9709 31

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1 dy 5 (i) By differentiating show that if y sec x then sec x tan x. [2] cos x, dx = = 1 (ii) Show that x tan x. [1] secx x ≡sec + −tan 1 (iii) Deduce that sec2x 2 sec x tan x. [2] x ≡2 −1 + (sec −tan x)2 14π 1 1 (iv) Hence show that dx [3] ä 0 = 4(8√2 −π). (secx −tan x)2

Mark scheme: 5 (i) Use correct quotient or chain rule M1 Obtain the given answer correctly having shown sufficient working A1 [2] (ii) Use a valid method, e.g. multiply numerator and denominator by sec x + tan x, and a version of Pythagoras to justify the given identity B1 [1] (iii) Substitute, expand (sec x + tan x)2 and use Pythagoras once M1 Obtain given identity A1 [2] (iv) Obtain integral 2 tan x – x + 2 sec x B1 Use correct limits correctly in an expression of the form a tan x + bx + c sec x, or equivalent, where abc 0 M1 Obtain the given answer correctly A1 [3]

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6 The variables x and y are related by the differential equation dy x 1 dx = −y2. When x 2, y 0. Solve the differential equation, obtaining an expression for y in terms of x. [8] = =

Mark scheme: 6 Separate variables correctly and attempt integration of one side B1 Obtain term ln x B1 1 B State or imply 1 – ≡ 1 – 1 and use a relevant method to find A or B M1 1 1 Obtain A = 2 , B = 2 1 1 Integrate and obtain – 2 ln (1 – y) + 2 ln (1 + y), or equivalent A1 1 + y 1 – y [If the integral is directly stated as k1 ln 1 – y or k2 ln 1 + y give M1, and then A2 for 1 1 k1 = 2 or k2 = – 2] Evaluate a constant, or use limits x = 2, y = 0 in a solution containing terms a ln x, b ln (1 – y) and c ln (1 + y), where abc 0 M1 [This M mark is not available if the integral of 1/(1 – y2) is initially taken to be of the form k ln (1 – y2)] 1 1 + y Obtain solution in any correct form, e.g. 2 ln 1 – y = ln x – ln 2 A1 x2 – 4 Rearrange and obtain y = x2 4, or equivalent, free of logarithms A1 [8] 1 1 dy

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7 The equation of a curve is 1. ln(xy) −y3 = dy y (i) Show that [4] dx = x(3y3 −1). (ii) Find the coordinates of the point where the tangent to the curve is parallel to the y-axis, giving each coordinate correct to 3 significant figures. [4]

Mark scheme: 1 1 dy 7 (i) EITHER: State or imply as derivative of ln xy, or equivalent B1 x dx dy State or imply 3y2 as derivative of y3, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain the given answer A1 dy OR Obtain xy = exp (1 + y3) and state or imply y + x dxas derivative of xy B1 d State or imply 3y2 exp (1 + y3) as derivative of (1 + y3) B1 dx dy Equate derivatives and solve for M1 dx Obtain the given answer A1 [4] [The M1 is dependent on at least one of the B marks being earned] (ii) Equate denominator to zero and solve for y M1* Obtain y = 0.693 only A1 Substitute found value in the equation and solve for x M1(dep*) Obtain x = 5.47 only A1 [4] GCE AS/A LEVEL – October/November 2012 9709 31

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9 The complex number 1 is denoted by u. The polynomial x4 x2 2x 6 is denoted by + (√2)i + + + p(x). (i) Showing your working, verify that u is a root of the equation 0, and write down a second complex root of the equation. p(x) = [4] (ii) Find the other two roots of the equation 0. [6] p(x) =

Mark scheme: 9 (i) EITHER Substitute x = 1 + √2 i and attempt the expansions of the x2 and x4 terms M1 Use i2 = –1 correctly at least once B1 Complete the verification A1 State second root 1 – √2 i B1 OR 1 State second root 1 – √2 i B1 Carry out a complete method for finding a quadratic factor with zeros 1 ± √2 i M1 Obtain x2 – 2x + 3, or equivalent A1 Show that the division of p(x) by x2 – 2x + 3 gives zero remainder and complete the verification A1 OR 2 Substitute x = 1 + √2 i and use correct method to express x2 and x4 in polar form M1 Obtain x2 and x4 in any correct polar form (allow decimals here) B1 Complete an exact verification A1 State second root 1 – √2 i, or its polar equivalent (allow decimals here) B1 [4] (ii) Carry out a complete method for finding a quadratic factor with zeros 1 ± √2 i M1* Obtain x2 – 2x + 3, or equivalent A1 Attempt division of p(x) by x2 – 2x + 3 reaching a partial quotient x2 + kx, or equivalent M1 (dep*) Obtain quadratic factor x2 – 2x + 2 A1 Find the zeros of the second quadratic factor, using i2 = –1 M1 (dep*) Obtain roots –1 + i and –1 –i A1 [6] [The second M1 is earned if inspection reaches an unknown factor x2 + Bx + C and an equation in B and/or C, or an unknown factor Ax2 + Bx + (6/3) and an equation in A and/or B] [If part (i) is attempted by the OR 1 method, then an attempt at part (ii) which uses or quotes relevant working or results obtained in part (i) should be marked using the scheme for part (ii)] GCE AS/A LEVEL – October/November 2012 9709 31

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10 With respect to the origin O, the points A, B and C have position vectors given by 3 2 1 OA OB and OC −−→ −−→ −−→ = −24 !, = −17 ! = −5 !. −3 The plane m is parallel to OC and contains A and B. −−→ (i) Find the equation of m, giving your answer in the form ax by d. [6] + + cß = (ii) Find the length of the perpendicular from C to the line through A and B. [5]

Mark scheme: 10 (i) EITHER Use scalar product of relevant vectors, or subtract point equations to form two equations in a,b,c, e.g. a – 5b – 3c = 0 and a – b – 3c = 0 M1* State two correct equations in a,b,c A1 Solve simultaneous equations and find one ratio, e.g. a : c, or b = 0 M1 (dep*) Obtain a : b : c = 3 : 0 : 1, or equivalent A1 Substitute a relevant point in 3x + z = d and evaluate d M1 (dep*) Obtain equation 3x + z = 13, or equivalent A1 OR 1 Attempt to calculate vector product of relevant vectors, e.g. (i – 5j – 3k) × (i – j – 3k) M2* Obtain 2 correct components of the product A1 Obtain correct product, e.g. 12i + 4k A1 Substitute a relevant point in 12x + 4z = d and evaluate d M1 (dep*) Obtain 3x + z =13, or equivalent A1 OR 2 Attempt to form 2–parameter equation for the plane with relevant vectors M2* State a correct equation e.g. r = 3i – 2j + 4k + λ(i – 5j – 3k) + µ(i – j –3k) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 (dep*) Obtain equation 3x + z =13, or equivalent A1 [6] (ii) EITHER Find for a point P on AB with a parameter t, e.g. 2i + 3j + 7k + t(–i + j + 3k) B1 Either: Equate scalar product , to zero and form an equation in t Or 1: Equate derivative for CP2 (or CP) to zero and form an equation in t Or 2: Use Pythagoras in triangle CPA (or CPB) and form an equation in t M1 Solve and obtain correct value of t, e.g. t = –2 A1 Carry out a complete method for finding the length of CP M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 1 State (or ) and in component form B1 Using a relevant scalar product find the cosine of CAB (or CBA) M1 22 33 Obtain cost CAB = – , or cos CBA = , or equivalent A1 √11.√62 √11.√117 Use trig to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 2 State (or ) and in component form B1 Using a relevant scalar product find the length of the projection AC (or BC) on AB M1 Obtain answer 2√11 (or), 3√11 or equivalent A1 Use Pythagoras to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 3 State (or ) and in component form B1 Calculate their vector product, e.g. (–2i – 3j – 7k ) × (–i + j + 3k) M1 Obtain correct product, e.g. –2i + 13j – 5k A1 Divide modulus of the product by the modulus of M1 Obtain answer 3√2 (4.24), or equivalent A1 OR 4 State two of , ) and in component form B1 Use cosine formula in triangle ABC to find cos A or cos B M1 44 66 Obtain cos A = – , or cos B = A1 2√11.√62 2√11.√117 Use trig to find the length of the perpendicular M1 Obtain answer 3√2 (4.24), or equivalent A1 [5] [The f.t is on ]

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