Cambridge A Level Mathematics 9709 — 2010 May/June Paper 3 · Variant 1

1 Solve the inequality where a is a positive constant. [4] |x + 3a| > 2|x −2a|,

Mark scheme: 1 EITHER: State or imply non-modular inequality ( x + 3a ) 2 > ( 2( x − 2 a )) 2 , or corresponding quadratic equation, or pair of linear equations ( x + 3a ) = ± 2 ( x − 2 a ) B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = 13 a and x = 7a A1 State answer 13 a < x < 7 a A1 OR: Obtain the critical value x = 7a from a graphical method, or by inspection, or by solving a linear equation or inequality B1 Obtain the critical value x = 13 a similarly B2 State answer 13 a < x < 7 a B1 [4] [Do not condone Y for <; accept 0.33 for 13 .]

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2 Solve the equation sin θ 2 cos 2θ 1, = + giving all solutions in the interval [6] 0◦≤θ ≤360◦.

Mark scheme: 2 Use correct cos 2A formula and obtain an equation in sin θ M1 Obtain 4 sin 2 θ + sin θ − 3 = 0 , or equivalent A1 Make reasonable attempt to solve a 3-term quadratic in sin θ M1 Obtain answer 48.6° A1 Obtain answer 131.4° and no others in the given range A1 √ Obtain answer 270° and no others in the given range A1 [6] [Treat the giving of answers in radians as a misread. Ignore answers outside the given range.]

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3 The variables x and y satisfy the equation xny C, where n and C are constants. When x 1.10, y 5.20, and when x 3.20, y 1.05. = = = = = (i) Find the values of n and C. [5] (ii) Explain why the graph of ln y against ln x is a straight line. [1]

Mark scheme: 3 (i) EITHER: State or imply n ln x + ln y = ln C B1 Substitute x- and y-values and solve for n M1 Obtain n = 1.50 A1 Solve for C M1 Obtain C = 6.00 A1 OR: Obtain two correct equations by substituting x- and y-values in x n y = C B1 Solve for n M1 Obtain n = 1.50 A1 Solve for C M1 Obtain C = 6.00 A1 [5] (ii) State that the graph of ln y against ln x has equation nln x + ln y = ln C which is linear in ln y and ln x, or has equation of the form nX + Y = ln C, where X = ln x and Y = ln y, and is thus a straight line B1 [1]

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4 (i) Using the expansions of and prove that cos(3x −x) cos(3x + x), 1 2x 3x sin x. 2(cos −cos 4x) ≡sin [3] (ii) Hence show that 13π sin 3x sin x dx 18 √3. 1 ã 6π = [3]

Mark scheme: 4 (i) State correct expansion of cos(3x – x) or cos(3x + x) B1 Substitute expansions in 12 (cos 2 x − cos 4 x ) , or equivalent M1 Simplify and obtain the given identity correctly A1 [3] (ii) Obtain integral 14 sin 2 x − 18 sin 4 x B1 Substitute limits correctly in an integral of the form a sin 2 x + b sin 4 x M1 Obtain given answer following full, correct and exact working A1 [3] GCE AS/A LEVEL – May/June 2010 9709 31

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5 Given that y 0 when x 1, solve the differential equation = = dy xy y2 4, dx = + obtaining an expression for y2 in terms of x. [6]

Mark scheme: 5 Separate variables correctly B1 Integrate and obtain term ln x B1 Integrate and obtain term 12 ln( y 2 + 4 ) B1 Evaluate a constant or use limits y = 0, x = 1 in a solution containing aln x and bln(y2 + 4) M1 Obtain correct solution in any form, e.g. 12 ln( y 2 + 4) = ln x + 12 ln 4 A1 Rearrange as y 2 = 4( x 2 − )1 , or equivalent A1 [6] 2 2

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6 C r x rad A O B The diagram shows a semicircle ACB with centre O and radius r. The angle BOC is x radians. The area of the shaded segment is a quarter of the area of the semicircle. (i) Show that x satisfies the equation x 34π x. = −sin [3] (ii) This equation has one root. Verify by calculation that the root lies between 1.3 and 1.5. [2] (iii) Use the iterative formula xn 34π xn+1 = −sin to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Using the formulae 12 r 2θ and 12 r 2 sin θ , or equivalent, form an equation M1 Obtain a correct equation in r and x and/or x/2 in any form A1 Obtain the given equation correctly A1 [3] (ii) Consider the sign of x − ( 34 π − sin x ) at x = 1.3 and x = 1.5, or equivalent M1 Complete the argument with correct calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.38 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 (1.375, 1.385) A1 [3]

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7 The complex number 2 2i is denoted by u. + (i) Find the modulus and argument of u. [2] (ii) Sketch an Argand diagram showing the points representing the complex numbers 1, i and u. Shade the region whose points represent the complex numbers which satisfy both the inequalities and ß [4] |ß −1| ≤|ß −i| |ß −u| ≤1. (iii) Using your diagram, calculate the value of for the point in this region for which arg is least. |ß| ß [3]

Mark scheme: 7 (i) Obtain modulus 8 B1 Obtain argument 14 π or 45° B1 [2] (ii) Show 1, i and u in relatively correct positions on an Argand diagram B1 Show the perpendicular bisector of the line joining 1 and i B1 Show a circle with centre u and radius 1 B1 Shade the correct region B1 [4] (iii) State or imply relevance of the appropriate tangent from O to the circle B1 √ Carry out complete strategy for finding z for the critical point M1 Obtain answer 7 A1 [3] A B

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2 8 (i) Express (x + 1)(x + 3) in partial fractions. [2] (ii) Using your answer to part (i), show that 2 2 1 1 1 1 x 1 x 3 ≡ − + + [2] (x + 1)(x + 3) (x + 1)2 + + (x + 3)2. 1 4 7 3 (iii) Hence show that dx 12 2. [5] ä 0 = −ln (x + 1)2(x + 3)2

Mark scheme: A B 8 (i) State or imply the form + and use a relevant method to find A or B M1 x + 1 x + 3 Obtain A = 1, B = −1 A1 [2] (ii) Square the result of part (i) and substitute the fractions of part (i) M1 Obtain the given answer correctly A1 [2] 1 1 (iii) Integrate and obtain − − ln ( x + 1) + ln( x + 3) − B3 x + 1 x + 3 Substitute limits correctly in an integral containing at least two terms of the correct form M1 Obtain given answer following full and exact working A1 [5] GCE AS/A LEVEL – May/June 2010 9709 31

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9 y P x –1 O 1 r 1 The diagram shows the curve y . −x 1 x = + 1 dy (i) By first differentiating obtain an expression for in terms of x. Hence show that the −x 1 dx [5] gradient of the normal to +thex,curve at the point is (x, y) (1 + x) √(1 −x2). (ii) The gradient of the normal to the curve has its maximum value at the point P shown in the diagram. Find, by differentiation, the x-coordinate of P. [4]

Mark scheme: 9 (i) Use quotient or product rule to differentiate (1 – x)/(1 + x) M1 Obtain correct derivative in any form A1 dy Use chain rule to find M1 dx Obtain a correct expression in any form A1 Obtain the gradient of the normal in the given form correctly A1 [5] (ii) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain x = 12 A1 [4]

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10 The lines l and m have vector equations r i j k and r 4i 6j k 2j = + + + s(i −j + 2k) = + + + t(2i + + k) respectively. (i) Show that l and m intersect. [4] (ii) Calculate the acute angle between the lines. [3] (iii) Find the equation of the plane containing l and m, giving your answer in the form ax by d. + + cß =[5]

Mark scheme: 10 (i) Express general point of l or m in component form, e.g. (1 + s, 1 – s, 1 + 2s) or (4 + 2t, 6 + 2t, 1 + t) B1 Equate at least two corresponding pairs of components and solve for s or t M1 Obtain s = −1 or t = −2 A1 Verify that all three component equations are satisfied A1 [4] (ii) Carry out correct process for evaluating the scalar product of the direction vectors of l and m M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 74.2° (or 1.30 radians) A1 [3] (iii) EITHER: Use scalar product to obtain a – b + 2c = 0 and 2a + 2b + c = 0 B1 Solve and obtain one ratio, e.g. a : b M1 Obtain a : b : c = 5 : −3 : −4, or equivalent A1 Substitute coordinates of a relevant point and values for a, b and c in general equation of plane and evaluate d M1 Obtain answer 5x – 3y – 4z = −2, or equivalent A1 OR 1: Using two points on l and one on m, or vice versa, state three equations in a, b, c and d B1 Solve and obtain one ratio, e.g. a : b M1 Obtain a ratio of three of the unknowns, e.g. a : b : c = −5 : 3 : 4 A1 Use coordinates of a relevant point and found ratio to find the fourth unknown, e.g. d M1 Obtain answer –5x + 3y + 4z = 2, or equivalent A1 OR 2: Form a correct 2-parameter equation for the plane, e.g. r = i + j + k + λ(i – j + 2k) + µ(2i + 2j + k) B1 State three equations in x, y, z, λ and µ M1 State three correct equations A1 Eliminate λ and µ M1 Obtain answer 5x – 3y – 4z = −2, or equivalent A1 OR 3: Attempt to calculate vector product of direction vectors of l and m M1 Obtain two correct components of the product A1 Obtain correct product, e.g. –5i + 3j + 4k A1 Form a plane equation and use coordinates of a relevant point to calculate d M1 Obtain answer –5x + 3y + 4z = 2, or equivalent A1 [5]

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