Cambridge A Level Mathematics 9709 — 2009 Oct/Nov Paper 3 · Variant 2

1 Solve the equation ln 5 x, ln(5 −x) = −ln giving your answers correct to 3 significant figures. [4]

Mark scheme: 1 Use law of the logarithm of a product or quotient and remove logarithms M1 Obtain quadratic equation x2 – 5x + 5 = 0, or equivalent A1 Solve 3-term quadratic obtaining 1 or 2 roots A1 Obtain answers 1.38 and 3.62 A1 [4] 3

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2 The equation x3 0 has one real root. −8x −13 = (i) Find the two consecutive integers between which this root lies. [2] (ii) Use the iterative formula 1 xn+1 = (8xn + 13) 3 to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 2 (i) Evaluate, or consider the sign of, x3 – 8x – 13 for two integer values of x, or equivalent M1 Conclude x = 3 and x = 4 with no errors seen A1 [2] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 3.43 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (3.425, 3.435) A1 [3] 2 dy 2

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3 The equation of a curve is x3 3. −x2y −y3 = dy (i) Find in terms of x and y. [4] dx (ii) Find the equation of the tangent to the curve at the point giving your answer in the form ax by c 0. (2, 1), [2] + + =

Mark scheme: dy 3 (i) State 2xy + x2 as derivative of x2y B1 dx dy State 3y2 as derivative of y3 B1 dx dy Equate derivative of LHS to zero and solve for M1 dx 3 x 2 − 2 xy Obtain answer 2 2 , or equivalent A1 [4] x + 3 y (ii) Find gradient of tangent at (2, 1) and form equation of tangent M1 Obtain answer 8x – 7y – 9 = 0, or equivalent A1√ [2]

More questions on Differentiation

4 The angles α and β lie in the interval and are such that x 0◦< < 180◦, tan α 2 tan β and 3. = tan(α + β) = Find the possible values of α and β. [6]

Mark scheme: 4 Use tan(A ± B) formula and obtain an equation in tan α and tan β M1* Substitute throughout for tan α or for tan β M1(dep*) Obtain 2 tan2 β + tan β – 1 = 0 or tan2 α + tan α – 2 = 0, or equivalent A1 Solve a 3-term quadratic and find an angle M1 Obtain answer α = 45°, β = 26.6° A1 Obtain answer α = 116.6°, β = 135° A1 [6] [Treat answers given in radians as a misread. Ignore answers outside the given range.] [SR: Two correct values of α (or β) score A1; then A1 for both correct α, β pairs]

More questions on Trigonometry

5 The polynomial 2x3 ax2 bx where a and b are constants, is denoted by The result of differentiating with+ respect+ −4,to x is denoted by It is given that is p(x).a factor of and of p(x) p′(x). (x + 2) p(x) p′(x). (i) Find the values of a and b. [5] (ii) When a and b have these values, factorise completely. [3] p(x)

Mark scheme: 5 (i) Substitute x = –2, equate to zero and state a correct equation, e.g. –16 + 4a – 2b – 4 = 0 B1 Differentiate p(x), substitute x = –2 and equate to zero M1 Obtain a correct equation, e.g. 24 – 4a + b = 0 A1 Solve for a or for b M1 Obtian a = 7 and b = 4 A1 [5] (ii) EITHER: State or imply (x + 2)2 is a factor B1 Attempt division by (x + 2)2 reaching a quotient 2x + k or use inspection with unknown factor cx + d reaching c = 2 or d = –1 M1 Obtain factorisation (x + 2)2 (2x – 1) A1 OR: Attempt division by (x + 2) M1 Obtain quadratic factor 2x2 + 3x – 2 A1 Obtain factorisation (x + 2)(x + 2)(2x – 1) A1 [3] [The M1 is earned if division reaches a partial quotient of 2x2 + kx, or if inspection has an unknown factor of 2x2 + ex + f and an equation in e and/or f, or if two coefficients with the correct moduli are stated without working.] GCE A/AS LEVEL – October/November 2009 9709 32 dx 2 2

More questions on Quadratics

6 (i) Use the substitution x 2 tan θ to show that = 2 1 8 4π dx cos2 θ dθ. ä 0 [4] 0 = ã (4 + x2)2 (ii) Hence find the exact value of 2 8 dx. ä 0 [4] (4 + x2)2

Mark scheme: dx 6 (i) State or imply = 2sec2 θ or dx = 2 sec2 θ dθ B1 dθ Substitute for x and dx throughout M1 Obtain any correct form in terms of θ A1 Obtain the given form correctly (including the limits) A1 [4] (ii) Use cos 2A formula, replacing integrand by a + b cos 2θ, where ab ≠ 0 M1* Integrate and obtain 1 θ + 1 sin 2θ A1 2 4 Use limits θ = 0 and θ = 1 π M1(dep*) 4 Obtain answer 1 (π + 2), or exact quivalent A1 [4] 8

More questions on Integration

7 The complex numbers i and 3 i are denoted by u and v respectively. −2 + + (i) Find, in the form x iy, the complex numbers + (a) u v, [1] + u (b) v, showing all your working. [3] u (ii) State the argument of v. [1] In an Argand diagram with origin O, the points A, B and C represent the complex numbers u, v and u v respectively. + (iii) Prove that angle AOB 34π. [2] = (iv) State fully the geometrical relationship between the line segments OA and BC. [2]

Mark scheme: 7 (i) (a) State that u + v is equal to 1 + 2i B1 [1] (b) EITHER: Multiply numerator and denominator of u/v by 3 – i, or equivalent M1 Simplify numerator to –5 + 5i, or denominator to 10 A1 Obtain answer – 1 + 1 i, or equivalent A1 2 2 OR1: Obtain two equations in x and y and solve for x or for y M1 Obtain x = – 1 or y = 1 A1 2 2 Obtain answer – 1 + 1 i, or equivalent A1 2 2 OR2: Using the correct processes express u/v in polar form M1 Obtain x = – 1 or y = 1 correctly A1 2 2 Obtain answer – 1 + 1 i, or equivalent A1 [3] 2 2 (ii) State that the argument of u/v is 3 π (2.36 radians or 135°) B1√ [1] 4 (iii) EITHER: Use facts that angle AOB = arg u – arg v and arg u – arg v = arg(u/v) M1 Obtain given answer A1 OR1: Obtain tan AOˆ B from gradients of OA and OB and the tan (A ± B) formula M1 Obtain given answer A1 OR2: Obtain cos AOˆ B by using the cosine formula or scalar product M1 Obtain given answer A1 [2] (iv) State OA = BC B1 State OA is parallel to BC B1 [2] A Bx + C

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1 x 8 (i) Express in partial fractions. [5] + (1 −x)(2 + x2) 1 x (ii) Hence obtain the expansion of in ascending powers of x, up to and including the + term in x2. (1 −x)(2 + x2) [5]

Mark scheme: A Bx + C 8 (i) State or imply partial fractions are of the form + 2 B1 1 − x 2 + x Use a relevant method to determine a constant M1 Obtain A = 2 , B = 2 and C = 1 A1 + A1 + A1 [5] 3 3 3 GCE A/AS LEVEL – October/November 2009 9709 32 (ii) Use correct method to find first two terms of the expansion of (1 – x)–1, (2 + x2)–1 or (1 + 1 x2)–1 M1 2 Obtain complete unsimplified expansions up to x2 of each partial fraction e.g. 2 (1 + x + x2) 3 and 1 ( 2 x – 1 )(1 – 1 x2) A1√ + A1√ 2 3 3 2 Carry out multiplication of (2 + x2)–1 by ( 2 x – 1 ), or equivalent, provided BC ≠ 0 M1 3 3 Obtain answer 1 + x + 3 x 2 A1 [5] 2 4 [Symbolic binomial coefficients are not sufficient for the first M1. The f.t. is on A, B, C.] [If B or C omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii), max 4/10] [In the case of an attempt to expand (1 + x)(1 – x)–1 (2 + x2)–1, give M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer.] [Allow Maclaurin, giving M1A1√A1√ for differentiating and obtaining f(0) = 1 and f ′(0) = 1, A1√ 2 for f ″(0) = 3 , and A1 for the final answer (the f.t. is on A, B, C if used).] 2

More questions on Algebra

9 The temperature of a quantity of liquid at time t is θ. The liquid is cooling in an atmosphere whose temperature is constant and equal to A. The rate of decrease of θ is proportional to the temperature difference Thus θ and t satisfy the differential equation (θ −A). dθ dt = −k(θ −A), where k is a positive constant. (i) Find, in any form, the solution of this differential equation, given that θ 4A when t 0. [5] = = (ii) Given also that θ 3A when t 1, show that k ln 32. [1] = = = (iii) Find θ in terms of A when t 2, expressing your answer in its simplest form. [3] =

Mark scheme: 9 (i) Separate variables correctly B1 Integrate and obtain term ln(θ – A), or equivalent B1 Integrate and obtain term –kt, or equivalent B1 Use θ = 4A, t = 0 to determine a constant, or as limits M1 Obtain correct answer in any form, e.g. ln(θ – A) = –kt + ln 3A, with no errors seen A1 [5] (ii) Substitute θ = 3A, t = 1 and justify the given statement B1 [1] (iii) Substitute t = 2 and solve for θ in terms of A M1 Remove logarithms M1 Obtain answer θ = 7 A, or equivalent, with no errors seen A1 [3] 3 [The M marks are only available if the solution to part (i) contains terms aln(θ – A) and bt.]

More questions on Differential equations

10 The plane p has equation 2x 16. The plane q is parallel to p and contains the point with position vector i 4j 2k. −3y + 6ß = + + (i) Find the equation of q, giving your answer in the form ax by d. [2] + + cß = (ii) Calculate the perpendicular distance between p and q. [3] (iii) The line l is parallel to the plane p and also parallel to the plane with equation x 5. Given that l passes through the origin, find a vector equation for l. −2y + 2ß =[5]

Mark scheme: 10 (i) Substitute coordinates (1, 4, 2) in 2x – 3y + 6z = d M1 Obtain plane equation 2x – 3y + 6z = 2, or equivalent A1 [2] (ii) EITHER: Attempt to use plane perpendicular formula to find perpendicular from (1, 4, 2) to p M1 2 − 3( 4) + 6( 2) − 16 Obtain a correct unsimplified expression, e.g. A1 ( 2 2 + ( −3) 2 + 6 2 ) Obtain answer 2 A1 OR1: State or imply perpendicular from O to p is 16 , or from O to q is 2 , or 7 7 equivalent B1 Find difference in perpendiculars M1 Obtain answer 2 A1 OR2: Obtain correct parameter value, or position vector or coordinates of foot of perpendicular from (1, 4, 2) to p (µ = ± 2 ; ( 11 , 22 , 26 )) B1 7 7 7 7 Calculate the length of the perpendicular M1 Obtain answer 2 A1 GCE A/AS LEVEL – October/November 2009 9709 32 OR3: Carry out correct method for finding the projection onto a normal vector of a line segment joining a point on p, e.g. (8, 0, 0) and a point on q, e.g. (1, 4, 2) M1 2(8 − )1 − (3 −4) + 6( −2) Obtain a correct unsimplified expression, e.g. A1 ( 2 2 + ( −3) 2 + 6 2 ) Obtain answer 2 A1 [3] (iii) EITHER: Calling the direction vector ai + bj + ck, use scalar product to obtain a relevant equation in a, b and c M1* Obtain two correct equations, e.g. 2a – 3b + 6c = 0, a – 2b + 2c = 0 A1 Solve for one ratio, e.g. a : b M1(dep*) Obtain a : b : c = 6 : 2 : –1, or equivalent A1 State answer r = λ(6i + 2j – k) or equivalent A1√ OR: Attempt to calculate vector product of two normals, e.g. (i – 2j + 2k) × (2i – 3j + 6k) M2 Obtain two correct components A1 Obtain –6i –2j + k, or equivalent A1 State answer r = λ(–6i – 2j + k), or equivalent A1√ [5]

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