Cambridge A Level Mathematics 9709 — 2012 May/June Paper 3 · Variant 1

1 Solve the equation 10, giving your answer correct to 3 significant figures. [3] |4 −2x| =

Mark scheme: 1 State or imply 4 − 2 x = −10 and 10 B1 Use correct method for solving equation of form 2 x = a M1 Obtain 3.81 A1 [3] 2 1

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1 2 (i) Expand in ascending powers of x, up to and including the term in x2, simplifying the coefficients.√(1 −4x) [3] 1 2x (ii) Hence find the coefficient of x2 in the expansion of [2] + √(4 −16x).

Mark scheme: 2 M12 (i) Either Obtain correct (unsimplified) version of x or x2 term from 1( − 4 x 1) Obtain 1 + 2x A1 Obtain + 6x2 A1 − 32 Or Differentiate and evaluate f(0) and f′(0) where f′(x) = k 1( −x4 ) M1 Obtain 1 + 2x A1 Obtain + 6x2 A1 [3] (ii) Combine both x2 terms from product of 1 + 2x and answer from part (i) M1 Obtain 5 A1 [2]

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3 The polynomial is defined by p(x) x3 4a, p(x) = −3ax + where a is a constant. (i) Given that is a factor of find the value of a. [2] (x −2) p(x), (ii) When a has this value, (a) factorise completely, [3] p(x) (b) find all the roots of the equation 0. [2] p(x2) =

Mark scheme: 3 (i) Substitute x = 2 and equate to zero, or divide by x – 2 and equate constant remainder to zero, or equivalent M1 Obtain a = 4 A1 [2] (ii) (a) Find further (quadratic or linear) factor by division, inspection or factor theorem or equivalent M1 Obtain x2 + 2x – 8 or x + 4 A1 State (x – 2)2(x + 4) or equivalent A1 [3] (b) State any two of the four (or six) roots B1 State all roots ( ± 2 , ± i2 ), provided two are purely imaginary B1 [2] 2

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4 The complex number u is defined by u . (1 + 2i)2 2 i = + (i) Without using a calculator and showing your working, express u in the form x iy, where x and y are real. + [4] (ii) Sketch an Argand diagram showing the locus of the complex number such that ß |ß −u| = |u|.[3]

Mark scheme: 4 (i) Either Expand (1 + 2i)2 to obtain –3 + 4i or unsimplified equivalent B1 Multiply numerator and denominator by 2 – i M1 Obtain correct numerator –2 + 11i or correct denominator 5 A1 2 11 Obtain − + i or equivalent A1 5 5 Or Expand (1 + 2i)2 to obtain –3 + 4i or unsimplified equivalent B1 Obtain two equations in x and y and solve for x or y M1 2 Obtain final answer x = − A1 5 11 Obtain final answer y = A1 [4] 5 (ii) Draw a circle M1 Show centre at relatively correct position, following their u A1 Draw circle passing through the origin A1 [3] GCE AS/A LEVEL – May/June 2012 9709 31 1 1 2 1

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5 y a x O The diagram shows the curve y 8 sin 2x1 2x1 = −tan for 0 π. The x-coordinate of the maximum point is α and the shaded region is enclosed by the curve≤xand<the lines x α and y 0. = = (i) Show that α 23π. [3] = (ii) Find the exact value of the area of the shaded region. [4]

Mark scheme: 1 1 2 1 5 (i) Differentiate to obtain 4 cos x − sec x B1 2 2 2 1 Equate to zero and find value of cos x M1 2 1 1 2 Obtain cos x = and confirm α = π A1 [3] 2 2 3 1 (ii) Integrate to obtain − 16 cos x … B1 2 1 … + 2 ln cos x or equivalent B1 2 2 1 1 Using limits 0 and π in a cos x + b ln cos x M1 3 2 2 1 Obtain 8+ 2 ln or exact equivalent A1 [4] 2 dy 2

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6 The equation of a curve is 3x2 y2 45. −4xy + = (i) Find the gradient of the curve at the point [4] (2, −3). (ii) Show that there are no points on the curve at which the gradient is 1. [3]

Mark scheme: dy 6 (i) Obtain 2 y as derivative of y2 B1 dx d y Obtain − 4 y − 4 x as derivative of –4xy B1 d x dy Substitute x = 2 and y = –3 and find value of dx d ( 45) (dependent on at least one B1 being earned and = 0 ) M1 dx 12 Obtain or equivalent A1 [4] 7 dy dy (ii) Substitute = 1 in an expression involving , x and y and obtain ay = bx M1 dx dx Obtain y = x or equivalent A1 Uses y = x in original equation and demonstrate contradiction A1 [3]

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7 The variables x and y are related by the differential equation dy 6xe3x . dx = y2 It is given that y 2 when x 0. Solve the differential equation and hence find the value of y when x 0.5, giving your= answer correct= to 2 decimal places. [8] =

Mark scheme: 7 Separate variables correctly and attempt integration on at least one side M1 1 3 Obtain y or equivalent on left-hand side A1 3 Use integration by parts on right-hand side (as far as axe 3 x + ∫ b e 3 x d x ) M1 Obtain or imply 2 xe 3 x + ∫ 2 e 3 x d x or equivalent A1 2 Obtain 2 xe 3 x − e 3 x A1 3 Substitute x = 0, y = 2 in an expression containing terms Ay3, Bxe3x, Ce3x, where ABC ≠ 0, and find the value of c M1 1 3 3 x 2 3 x 10 Obtain y = 2 xe − e + or equivalent A1 3 3 3 Substitute x = 0.5 to obtain y = 2.44 A1 [8] GCE AS/A LEVEL – May/June 2012 9709 31 2 

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1 2 8 The point P has coordinates 4, and the line l has equation r 3 λ 1 (−1, 11) = ! + 3 !. −4 (i) Find the perpendicular distance from P to l. [4] (ii) Find the equation of the plane which contains P and l, giving your answer in the form ax by d, where a, b, c and d are integers. [5] + + cß =

Mark scheme:   8 (i) Either Obtain ±  − 1  for vector PA (where A is point on line) or equivalent B1   − 15  Use scalar product to find cosine of angle between PA and line M1 42 Obtain or equivalent A1 14 × 230 Use trigonometry to obtain 104 or 10.2 or equivalent A1  2 n + 2    Or 1 Obtain ±  n − 1  for PN (where N is foot of perpendicular) B1   3n − 15   Equate scalar product of PN and line direction to zero Or equate derivative of PN 2 to zero Or use Pythagoras’ theorem in triangle PNA to form equation in n M1 Solve equation and obtain n = 3 A1 Obtain 104 or 10.2 or equivalent A1  2    Or 2 Obtain ±  − 1  for vector PA (where A is point on line) B1   − 15   Evaluate vector product of PA and line direction M1  12    Obtain ±  − 36  A1   − 4   Divide modulus of this by modulus of line direction and obtain 104 or 10.2 or equivalent A1  2    Or 3 Obtain ±  − 1  for vector PA (where A is point on line) B1   − 15   Evaluate scalar product of PA and line direction to obtain distance AN M1 Obtain 3 14 or equivalent A1 Use Pythagoras’ theorem in triangle PNA and obtain 104 or 10.2 or equivalent A1  2    Or 4 Obtain ±  − 1  for vector PA (where A is point on line) B1   − 15   Use a second point B on line and use cosine rule in triangle ABP to find angle A or angle B or use vector product to find area of triangle M1 Obtain correct answer (angle A = 42.25…) A1 Use trigonometry to obtain 104 or 10.2 or equivalent A1 [4] GCE AS/A LEVEL – May/June 2012 9709 31 (ii) Either Use scalar product to obtain a relevant equation in a, b, c, e.g. 2a + b + 3c = 0 or 2a – b – 15c = 0 M1 State two correct equations in a, b and c A1 Solve simultaneous equations to obtain one ratio M1 Obtain a : b : c = –3 : 9 : –1 or equivalent A1 Obtain equation –3x + 9y – z = 28 or equivalent A1  2   2   8        Or 1 Calculate vector product of two of  1  ,  − 1  and  2  or equiv M1       3 − 15 −6       Obtain two correct components of the product A1  − 3    Obtain correct  9  or equivalent A1   − 1   Substitute in –3x + 9y – z = d to find d or equivalent M1 Obtain equation –3x + 9y – z = 28 or equivalent A1 Or 2 Form a two-parameter equation of the plane M1  1   2   2        Obtain r =  3  + s  1  + t  − 1  or equivalent A1       − 4 3 − 15       State three equations in x, y, z, s, t A1 Eliminate s and t M1 Obtain equation 3x – 9y + z = –28 or equivalent A1 [5] B C

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4x2 5x 3 9 By first expressing in partial fractions, show that + + 2x2 5x 2 + + 4 4x2 5x 3 dx 8 9. + + 2x2 5x 2 = −ln [10] ä 0 + +

Mark scheme: B C 9 State or imply form A + + B1 2 x + 1 x + 2 State or obtain A = 2 B1 Use correct method for finding B or C M1 Obtain B = 1 A1 Obtain C = –3 A1 1 Obtain 2 x + ln( 2 x + )1 − 3 ln( x + 2) [Deduct B1 for each error or omission] B3 2 Substitute limits in expression containing aln(2x + 1) + bln(x + 2) M1 1 Show full and exact working to confirm that 8 + ln 9 − 3 ln 6 + 3 ln 2 , or an equivalent 2 expression, simplifies to given result 8 – ln 9 A1 [10] [SR: If A omitted from the form of fractions, give B0B0M1A0A0 in (i); B0 B1 B1 M1A0 in (ii).] M Nx Px Q [SR: For a solution starting with + or + , give B0B0M1A0A0 in (i); 2 x + 1 x + 2 2 x + 1 x + 2 B1 B1 B1 , if recover correct form, M1A0 in (ii).] B Dx + E [SR: For a solution starting with + , give M1A1 for one of B = 1, D = 2, E = 1 2 x + 1 x + 2 and A1 for the other two constants; then give B1B1 for A = 2, C = −3.] Fx + G C [SR: For a solution starting with + , give M1A1 for one of C = −3, F = 4, G = 3 2 x + 1 x + 2 and A1 for the other constants or constant; then give B1B1 for A = 2, B = 1.] GCE AS/A LEVEL – May/June 2012 9709 31 4 3 2

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10 (i) It is given that 2 tan 2x 5 tan2x 0. Denoting tan x by t, form an equation in t and hence show that either t 0 or t + = [4] = = 3√(t + 0.8). (ii) It is given that there is exactly one real value of t satisfying the equation t Verify by calculation that this value lies between 1.2 and 1.3. = 3√(t + 0.8). [2] to find the value of t correct to 3 decimal places. Give (iii) Use the iterative formula tn+1 = 3√(tn + 0.8) the result of each iteration to 5 decimal places. [3] (iv) Using the values of t found in previous parts of the question, solve the equation 2 tan 2x 5 tan2x 0 + = for [3] −π ≤x ≤π.

Mark scheme: 10 (i) Use correct identity for tan 2 x and obtains at4 + bt3 + ct2 + dt = 0, where b may be zero M1 Obtain correct horizontal equation, e.g. 4t + 5t2 – 5t4 = 0 A1 Obtain kt(t3 + et + f) = 0 or equivalent M1 Confirm given results t = 0 and t = 3 t + 8.0 A1 [4] (ii) Consider sign of t − 3 t + 8.0 at 1.2 and 1.3 or equivalent M1 Justify the given statement with correct calculations (–0.06 and 0.02) A1 [2] (iii) Use the iterative formula correctly at least once with 1.2 < tn < 1.3 M1 Obtain final answer 1.276 A1 Show sufficient iterations to justify answer or show there is a change of sign in interval (1.2755, 1.2765) A1 [3] (iv) Evaluate tan–1 (answer from part (iii)) to obtain at least one value M1 Obtain –2.24 and 0.906 A1 State –π, 0 and π B1 [3] [SR If A0, B0, allow B1 for any 3 roots]

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