Cambridge A Level Mathematics 9709 — 2012 May/June Paper 3 · Variant 2

1 Solve the equation 2 ln(3x + 4) = ln(x + 1), giving your answer correct to 3 significant figures. [4]

Mark scheme: 1 EITHER: Use law of the logarithm of a power or quotient and remove logarithms M1 Obtain a 3-term quadratic equation x 2 −x − 3 = 0 , or equivalent A1 Solve 3-term quadratic obtaining 1 or 2 roots M1 Obtain answer 2.30 only A1  1  OR1: Use an appropriate iterative formula, e.g. x n +1 = exp ln (3 x n + 4 )  − 1 correctly at  2  least once M1 Obtain answer 2.30 A1 Show sufficient iterations to at least 3 d.p. to justify 2.30 to 2 d.p., or show there is a sign change in the interval (2.295, 2.305) A1 Show there is no other root A1 OR2: Use calculated values to obtain at least one interval containing the root M1 Obtain answer 2.30 A1 Show sufficient calculations to justify 2.30 to 3 s.f., e.g. show it lies in (2.295, 2.305) A1 Show there is no other root A1 [4] 1 2 1

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2 C q M a A B In the diagram, ABC is a triangle in which angle ABC is a right angle and BC a. A circular arc, with centre C and radius a, joins B and the point M on AC. The angle ACB is θ radians.= The area of the sector CMB is equal to one third of the area of the triangle ABC. (i) Show that θ satisfies the equation tan θ 3θ. = [2] (ii) This equation has one root in the interval 0 θ 12π. Use the iterative formula < < θn+1 = tan−1(3θn) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 1 2 1 2 (i) Using the formulae r θ and bh , form an equation an a and θ M1 2 2 Obtain given answer A1 [2] (ii) Use the iterative formula correctly at least once M1 Obtain answer θ = 1.32 A1 Show sufficient iterations to 4 d.p. to justify 1.32 to 2 d.p., or show there is a sign change in the interval (1.315, 1.325) A1 [3] GCE AS/A LEVEL – May/June 2012 9709 32 1 1 2

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r 1 3 Expand in ascending powers of x, up to and including the term in x2, simplifying the −x 1 x coefficients. + [5]

Mark scheme: 2 2 − 23 EITHER: State a correct unsimplified term in x or x of 1( − x ) or 1( + x ) B1 1 State correct unsimplified expansion of 1( − x ) 2 up to the term in x 2 B1 − 12 2 State correct unsimplified expansion of 1( + x ) up to the term in x B1 1 2 − 12 Obtain sufficient terms of the product of the expansions of 1( − x ) and 1( + x ) M1 1 2 Obtain final answer 1 − x + x A1 2 2 − 12 OR1: State that the given expression equals 1( − x )(1 − x ) and state that the first term of 2 − 12 the expansion of 1( −x ) is 1 B1 2 2 − 12 State correct unsimplified term in x of 1( −x ) B1 2 − 12 2 State correct unsimplified expansion of 1( −x ) up to the term in x B1 Obtain sufficient terms of the product of (1 – x) and the expansion M1 1 2 Obtain final answer 1 − x + x A1 2 1 OR2: State correct unsimplified expansion of 1( + x ) 2 up to the term in x 2 B1 Multiply expansion by (1 – x) and obtain 1 – 2x + 2x2 B1 Carry out correct method to obtain one non-constant term of the expansion of 2 2 M1 (1 − 2 x + 2 x 1 ) Obtain a correct unsimplified expansion with sufficient terms A1 1 2 Obtain final answer 1 − x + x A1 [5] 2 1 2 by the EITHER scheme.] [Treat 1( + x ) −1 1( − x 2 )  1  [Symbolic coefficients, e.g.  2  , are not sufficient for the B marks.]  2 

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4 Solve the equation cosec 2θ sec θ cot θ, = + θ giving all solutions in the interval [6] 0◦< < 360◦.

Mark scheme: 4 Use trig formulae to express equation in terms of cos θ and sin θ M1 Use Pythagoras to obtain an equation in sin θ M1 Obtain 3-term quadratic 2 sin 2 θ − 2 sin θ − 1 = 0 , or equivalent A1 Solve a 3-term quadratic and obtain a value of θ M1 Obtain answer, e.g. 201.5° A1 Obtain second answer, e.g. 338.5°, and no others in the given interval A1 [6] [Ignore answers outside the given interval. Treat answers in radians (3.52, 5.91) as a misread and deduct A1 from the marks for the angles.]

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5 The variables x and y satisfy the differential equation dy e2x+y, dx = and y 0 when x 0. Solve the differential equation, obtaining an expression for y in terms of x. [6] = =

Mark scheme: 5 Separate variables correctly and attempt integration of both sides B1 Obtain term −e − y , or equivalent B1 1 2 x Obtain term e , or equivalent B1 2 Evaluate a constant, or use limits x = 0, y = 0 in a solution containing terms a e − y and b e 2 x M1 − y 1 2 x 3 Obtain correct solution in any form, e.g. − e = e − A1 2 2 Rearrange and obtain y = ln( 2 /(3 − e 2x )) , or equivalent A1 [6] GCE AS/A LEVEL – May/June 2012 9709 32 2

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6 The equation of a curve is y 3 sin x 4 cos3x. = + (i) Find the x-coordinates of the stationary points of the curve in the interval 0 x π. [6] < < (ii) Determine the nature of the stationary point in this interval for which x is least. [2]

Mark scheme: 6 (i) State derivative in any correct form, e.g. 3 cos x − 12 cos 2 x sin x B1 + B1 Equate derivative to zero and solve for sin 2x, or sin x or cos x M1 1 Obtain answer x = π A1 12 5 Obtain answer x = π A1 12 1 Obtain answer x = π and no others in the given interval A1 [6] 2 (ii) Carry out a method for determining the nature of the relevant stationary point M1 1 Obtain a maximum at π correctly A1 [2] 12 [Treat answers in degrees as a misread and deduct A1 from the marks for the angles.]

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7 Throughout this question the use of a calculator is not permitted. The complex number u is defined by 1 2i u + 1 = −3i. (i) Express u in the form x iy, where x and y are real. [3] + (ii) Show on a sketch of an Argand diagram the points A, B and C representing the complex numbers u, 1 2i and 1 respectively. [2] + −3i (iii) By considering the arguments of 1 2i and 1 show that + −3i, 3 4π. tan−12 + tan−13 = [3]

Mark scheme: 7 (i) EITHER: Multiply numerator and denominator by 1 + 3i, or equivalent M1 Simplify numerator to –5 + 5i, or denominator to 10, or equivalent A1 1 1 Obtain final answer − + i , or equivalent A1 2 2 OR: Obtain two equations in x and y, and solve for x or for y M1 1 1 Obtain x = − or y = , or equivalent A1 2 2 1 1 Obtain final answer − + i , or equivalent A1 [3] 2 2 (ii) Show B and C in relatively correct positions in an Argand diagram B1 Show u in a relatively correct position B1 [2] (iii) Substitute exact arguments in the LHS arg(1 + 2i) − arg(1 − 3i) = arg u, or equivalent M1 3 Obtain and use arg u = π A1 4 Obtain the given result correctly A1 [3] GCE AS/A LEVEL – May/June 2012 9709 32

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9 y R e x O 1 The diagram shows the curve y x 2 ln x. The shaded region between the curve, the x-axis and the line x e is denoted by R. = = (i) Find the equation of the tangent to the curve at the point where x 1, giving your answer in the form y mx c. = [4] = + (ii) Find by integration the volume of the solid obtained when the region R is rotated completely about the x-axis. Give your answer in terms of π and e. [7] [Question 10 is printed on the next page.]

Mark scheme: 9 (i) Use correct product rule M1 ln x x Obtain derivative in any correct form, e.g. + A1 2 x x Carry out a complete method to form an equation of the tangent at x = 1 M1 Obtain answer y = x – 1 A1 [4] (ii) State or imply that the indefinite integral for the volume is π ∫ x (ln x ) 2 d x B1 ln x 2 2 2 Integrate by parts and reach ax (ln x ) + b ∫ x . x dx M1* 1 Obtain x 2 (ln x ) 2 − ∫ x ln x dx , or unsimplified equivalent A1 2 1 Attempt second integration by parts reaching cx 2 ln x + d ∫ x 2 . x d x M1(dep*) 1 2 2 1 2 1 2 Complete the integration correctly, obtaining x (ln x ) − x ln x + x A1 2 2 4 Substitute limits x = 1 and x = e, having integrated twice M1(dep*) 1 2 Obtain answer π e( − )1 , or exact equivalent A1 [7] 4 [If π omitted, or 2π or π/2 used, give B0 and then follow through.] [Integration using parts x ln x and ln x is also viable.] GCE AS/A LEVEL – May/June 2012 9709 32

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10 Two planes, m and n, have equations x 2y 1 and 2x 7 respectively. The line l has equation r i j j + −2ß = −2y + ß = = + −k + λ(2i + + 2k). (i) Show that l is parallel to m. [3] (ii) Find the position vector of the point of intersection of l and n. [3] (iii) A point P lying on l is such that its perpendicular distances from m and n are equal. Find the position vectors of the two possible positions for P and calculate the distance between them. [6] [The perpendicular distance of a point with position vector x1i y1 j from the plane + + ß1k by1 ax by d is |ax1 + + cß1 −d| .] b2 + + cß = √(a2 + + c2)

Mark scheme: 10 (i) EITHER: Substitute coordinates of a general point of l in given equation of plane m M1 Obtain equation in λ in any correct form A1 Verify that the equation is not satisfied for any value of λ A1 OR1: Substitute for r in the vector equation of plane m and expand scalar product M1 Obtain equation in λ in any correct form A1 Verify that the equation is not satisfied for any value of λ A1 OR2: Expand scalar product of a normal to m and a direction vector of l M1 Verify scalar product is zero A1 Verify that one point of l does not lie in the plane A1 OR3: Use correct method to find perpendicular distance of a general point of l from m M1 Obtain a correct unsimplified expression in terms of λ A1 Show that the perpendicular distance is 4/3, or equivalent, for all λ A1 OR4: Use correct method to find the perpendicular distance of a particular point of l from m M1 Obtain answer 4/3, or equivalent A1 Show that the perpendicular distance of a second point is also 4/3, or equivalent A1 [3] (ii) EITHER: Express general point of l in component form, e.g. (1 + 2λ, 1 + λ, −1 + 2λ) B1 Substitute in given equation of n and solve for λ M1 Obtain position vector 5i + 3j + 3k from λ = 2 A1 OR: State or imply plane n has vector equation r.(2i – 2j + k) = 7, or equivalent B1 Substitute for r, expand scalar product and solve for λ M1 Obtain position vector 5i + 3j + 3k from λ = 2 A1 [3] (iii) Form an equation in λ by equating perpendicular distances of a general point of l from m and n M1* Obtain a correct modular or non-modular equation in λ in any form A1 Solve for λ and obtain a point, e.g. 7i + 4j + 5k from λ = 3 A1 Obtain a second point, e.g. 3i + 2j + k from λ = 1 A1 Use a correct method to find the distance between the two points M1(dep*) Obtain answer 6 A1 [6] [The f.t. is on the components of l.]

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