Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 3 · Variant 1

1 Solve the inequality [4] 2|x −3| > |3x + 1|.

Mark scheme: 1 EITHER: State or imply non-modular inequality (2(x – 3))2 > (3x + 1)2, or corresponding quadratic equation, or pair of linear equations 2(x – 3) = ±(3x + 1) B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = –7 and x = 1 A1 State answer –7 < x < 1 A1 OR: Obtain critical value x = –7 or x = 1 from a graphical method, or by inspection, or by solving a linear equation or inequality B1 Obtain critical values x = –7 and x = 1 B2 State answer –7 < x < 1 B1 [4] [Do not condone: < for <.]

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2 Solve the equation 1 2 ln x, ln(1 + x2) = + giving your answer correct to 3 significant figures. [4]

Mark scheme: 2 Use law for the logarithm of a power, a quotient, or a product correctly at least once M1 Use ln e = 1 or e = exp(1) M1 Obtain a correct equation free of logarithms, e.g. 1 + x2 = ex2 A1 Solve and obtain answer x = 0.763 only A1 [4] [For the solution x = 0.763 with no relevant working give B1, and a further B1 if 0.763 is shown to be the only root.] [Treat the use of logarithms to base 10 with answer 0.333 only, as a misread.] [SR: Allow iteration, giving B1 for an appropriate formula, e.g. xn+1 = exp((ln(1 + xn2) – 1)/2), M1 for using it correctly once, A1 for 0.763, and A1 for showing the equation has no other root but 0.763.]

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3 Solve the equation 2 sin θ, cos(θ + 60◦) = giving all solutions in the interval [5] 0◦≤θ ≤360◦.

Mark scheme: 3 Attempt use of cos(A + B) formula to obtain an equation in cos θ and sin θ M1 Use trig formula to obtain an equation in tan θ (or cos θ, sin θ or cot θ) M1 Obtain tan θ = 1/(4 + 3 ) or equivalent (or find cos θ, sin θ or cot θ) A1 Obtain answer θ = 9.9° A1 Obtain θ = 189.9°, and no others in the given interval A1 [5] [Ignore answers outside the given interval. Treat answers in radians as a misread (0.173, 3.31).]  [The other solution methods are via cos θ = ±(4 + 3 )/  1 + (4 + 3 )2   or  2  sin θ = ±1/ 1 + (4 + 3 ) .]  

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4 (i) By sketching suitable graphs, show that the equation 4x2 cotx −1 = has only one root in the interval 0 x 12π. [2] < < (ii) Verify by calculation that this root lies between 0.6 and 1. [2] (iii) Use the iterative formula 1 2 xn+1 = √(1 + cotxn) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 4 (i) Make recognisable sketch of a relevant graph over the given range B1 Sketch the other relevant graph on the same diagram and justify the given statement B1 [2] (ii) Consider sign of 4x2 – 1 – cot x at x = 0.6 and x = 1, or equivalent M1 Complete the argument correctly with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.73 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 (0.725, 0.735) A1 [3] GCE A/AS LEVEL – October/November 2010 9709 31 dx

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1 x2 5 Let I dx. = ä 0 √(4 −x2) (i) Using the substitution x 2 sin θ, show that = 16π I 4 sin2θ dθ. = ã 0 [3] (ii) Hence find the exact value of I. [4]

Mark scheme: dx 5 (i) State or imply dx = 2 cos θ dθ, or = 2 cos θ, or equivalent B1 dθ Substitute for x and dx throughout the integral M1 Obtain the given answer correctly, having changed limits and shown sufficient working A1 [3] (ii) Replace integrand by 2 – 2 cos 2θ, or equivalent B1 Obtain integral 2θ – sin 2θ, or equivalent B1√ Substitute limits correctly in an integral of the form aθ ± b sin 2θ, where ab Þ 0 M1 1 3 Obtain answer π − or exact equivalent A1 [4] 3 2 [The f.t. is on integrands of the form a + c cos 2θ, where ac Þ 0.]

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6 The complex number is given by ß i. ß = (√3) + (i) Find the modulus and argument of [2] ß. (ii) The complex conjugate of is denoted by Showing your working, express in the form x iy, where x and y are real, ß ß*. + (a) 2ß + ß*, (b) . iß* [4] ß (iii) On a sketch of an Argand diagram with origin O, show the points A and B representing the complex numbers and respectively. Prove that angle AOB 16π. [3] ß iß* =

Mark scheme: 6 (i) State modulus is 2 B1 State argument is 16 π , or 30°, or 0.524 radians B1 [2] (ii) (a) State answer 3 3 + i B1 (b) EITHER: Multiply numerator and denominator by 3 −,i or equivalent M1 Simplify denominator to 4 or numerator to 2 3 + i2 A1 Obtain final answer 12 3 + 12 i , or equivalent A1 OR 1: Obtain two equations in x and y and solve for x or for y M1 Obtain x = 12 3 or y = 12 A1 Obtain final answer 12 3 + 12 i , or equivalent A1 OR 2: Using the correct processes express iz*/z in polar form M1 Obtain x = 12 3 or y = 12 A1 Obtain final answer 12 3 + 12 i , or equivalent A1 [4] (iii) Plot A and B in relatively correct positions B1 EITHER: Use fact that angle AOB = arg(iz*) – arg z M1 Obtain the given answer A1 OR 1: Obtain tan AOˆ B from gradients of OA and OB and the correct tan(A – B) formula M1 Obtain the given answer A1 OR 2: Obtain cos AOˆ B by using correct cosine formula or scalar product M1 Obtain the given answer A1 [3] GCE A/AS LEVEL – October/November 2010 9709 31

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7 With respect to the origin O, the points A and B have position vectors given by OA i 2j 2k and −−→ = + + OB 3i 4j. The point P lies on the line AB and OP is perpendicular to AB. −−→ = + (i) Find a vector equation for the line AB. [1] (ii) Find the position vector of P. [4] (iii) Find the equation of the plane which contains AB and which is perpendicular to the plane OAB, giving your answer in the form ax by d. [4] + + cß =

Mark scheme: 7 (i) State correct equation in any form, e.g. r = i + 2j + 2k + λ(2i + 2j – 2k) B1 [1] (ii) EITHER: Equate a relevant scalar product to zero and form an equation in λ M1 OR 1: Equate derivative of OP2 (or OP) to zero and form an equation in λ M1 OR 2: Use Pythagoras in OAP or OBP and form an equation in λ M1 State a correct equation in any form A1 Solve and obtain λ = − 16 or equivalent A1 Obtain final answer OP = 23 i + 53 j + 73 k , or equivalent A1 [4] (iii) EITHER: State or imply OP is a normal to the required plane M1 State normal vector 2i + 5j + 7k, or equivalent A1√ Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1 Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 1: Find a vector normal to plane AOB and calculate its vector product with a direction vector for the line AB M1* Obtain answer 2i + 5j + 7k, or equivalent A1 Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 2: Set up and solve simultaneous equations in a, b, c derived from zero scalar products of ai + bj + ck with (i) a direction vector for line AB, (ii) a normal to plane OAB M1* Obtain a : b : c = 2 : 5 : 7, or equivalent A1 Substitute coordinates of a relevant point in 2x + 5y + 7z = d and evaluate d M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 3: With Q (x, y, z) on plane, use Pythagoras in OPQ to form an equation in x, y and z M1* Form a correct equation A1√ Reduce to linear form M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 OR 4: Find a vector normal to plane AOB and form a 2-parameter equation with relevant vectors, e.g., r = i + 2j + 2k + λ(2i – 2j + 2k) + µ(8i – 6j + 2k) M1* State three correct equations in x, y, z, λ and µ A1 Eliminate λ and µ M1(dep*) Obtain answer 2x + 5y + 7z = 26, or equivalent A1 [4] GCE A/AS LEVEL – October/November 2010 9709 31 A Bx + C

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3x 8 Let f(x) = (1 + x)(1 + 2x2). (i) Express in partial fractions. [5] f(x) (ii) Hence obtain the expansion of in ascending powers of x, up to and including the term in x3. f(x) [5] [Questions 9 and 10 are printed on the next page.]

Mark scheme: A Bx + C 8 (i) State or imply the form + 2 B1 1 + x 1 + 2 x Use any relevant method to evaluate a constant M1 Obtain one of A = –1, B = 2, C = 1 A1 Obtain a second value A1 Obtain the third value A1 [5] (ii) Use correct method to obtain the first two terms of the expansion of (1 + x )−1 or (1 + 2 x 2 )−1 M1 Obtain correct expansion of each partial fraction as far as necessary A1√ + A1√ Multiply out fully by Bx + C, where BC Þ 0 M1 Obtain answer 3x – 3x2 – 3x3 A1 [5] − 1  [Symbolic binomial coefficients, e.g.,   are not sufficient for the first M1. The f.t.  1  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.] [If a constant D is added to the correct form, give M1A1A1A1 and B1 if and only if D = 0 is stated.] [If an extra term D/(1 + 2x2) is added, give B1M1A1A1, and A1 if C + D = 1 is resolved to 1/(1 + 2x2).] [In the case of an attempt to expand 3x(1 + x)–1(1 + 2x2)–1, give M1A1A1 for the expansions up to the term in x2, M1 for multiplying out fully, and A1 for the final answer.] [For the identity 3x ≡ (1 + x + 2x2 + 2x3)(a + bx + cx2 + dx3) give M1A1; then M1A1 for using a relevant method to find two of a = 0, b = 3, c = –3 and d = –3; and then A1 for the final answer in series form.]

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9 y x O 2 M The diagram shows the curve y x3 ln x and its minimum point M. = (i) Find the exact coordinates of M. [5] (ii) Find the exact area of the shaded region bounded by the curve, the x-axis and the line x 2. [5] =

Mark scheme: 9 (i) Use correct product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and find non-zero x M1 1 Obtain x = exp (− 3 ) , or equivalent A1 Obtain y = –l/(3e), or any ln-free equivalent A1 [5] 1 (ii) Integrate and reach kx 4 ln x + l ∫ x 4 . x dx M1 Obtain 14 x 4 ln x − 14 ∫ x 3 dx A1 Obtain integral 14 x 4 ln x − 161 x 4 , or equivalent A1 Use limits x = 1 and x = 2 correctly, having integrated twice M1 15 Obtain answer 4 ln 2 − , or exact equivalent A1 [5] 16 GCE A/AS LEVEL – October/November 2010 9709 31 dx ( )

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10 A certain substance is formed in a chemical reaction. The mass of substance formed t seconds after the start of the reaction is x grams. At any time the rate of formation of the substance is proportional dx to When t 0, x 0 and 1. dt (20 −x). = = = (i) Show that x and t satisfy the differential equation dx dt = 0.05(20 −x). [2] (ii) Find, in any form, the solution of this differential equation. [5] (iii) Find x when t 10, giving your answer correct to 1 decimal place. [2] = (iv) State what happens to the value of x as t becomes very large. [1]

Mark scheme: dx 10 (i) State or imply = k (20 − x ) B1 dt Show that k = 0.05 B1 [2] (ii) Separate variables correctly and integrate both sides B1 Obtain term –ln(20 – x), or equivalent B1 Obtain term 201 t , or equivalent B1 Evaluate a constant or use limits t = 0, x = 0 in a solution containing terms a ln(20 – x) and bt M1* Obtain correct answer in any form, e.g. ln 20 – ln(20 – x) = 201 t A1 [5] (iii) Substitute t = 10 and calculate x M1(dep*) Obtain answer x = 7.9 A1 [2] (iv) State that x approaches 20 B1 [1]

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