Cambridge A Level Mathematics 9709 — 2009 May/June Paper 3 · Variant 1

2 y 1 x O 4p 2 for 0 The diagram shows the curve y = √(1 + tan2x) ≤x ≤14π. (i) Use the trapezium rule with three intervals to estimate the value of 14π 2 dx, ã 0 √(1 + tan2x) giving your answer correct to 2 decimal places. [3] (ii) The estimate found in part (i) is denoted by E. Explain, without further calculation, whether another estimate found using the trapezium rule with six intervals would be greater than E or less than E. [1]

Mark scheme: 2 (i) State or imply 3 of the 4 ordinates 1, 1.069389..., 1.290994..., 1.732050... B1 Use correct formula, or equivalent, with h = 121 π and four ordinates M1 Obtain answer 0.98 with no errors seen A1 3 [Accept h = 0.26 but not h = 15 when awarding the M1] [SR: if only 5 and/or 3 are given, and decimals are not seen, the B1 is available] 3 [SR: solutions with 2 or 4 intervals can score only the M1 for a correct expression] (ii) Justify statement that the second estimate would be less than E B1 1

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3 (i) Prove the identity cosec 2θ cot 2θ θ. [3] + ≡cot (ii) Hence solve the equation cosec 2θ cot 2θ 2, for [2] + = 0◦≤θ ≤360◦.

Mark scheme: 3 (i) Use cot A = 1/tan A or cos A/sin A and/or cosec A = 1/sin A on at least two terms M1 Use a correct double angle formula or the sin(A – B) formula at least once M1 Obtain given result A1 3 (ii) Solve cot θ = 2 for θ and obtain answer 26.6° B1 Obtain answer 206.6° and no others in the given range B1 √ 2 [Ignore answers outside the given range. Treat answers given in radians as a misread] 3

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4 The equation x3 0 has one real root. −2x −2 = (i) Show by calculation that this root lies between x 1 and x 2. [2] = = (ii) Prove that, if a sequence of values given by the iterative formula 2x3 2 n + xn+1 = 3x2 n −2 converges, then it converges to this root. [2] (iii) Use this iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 4 (i) Compare signs of x3 – 2x – 2 when x = 1 and x = 2, or equivalent M1 Complete the argument with correct calculations A1 2 (ii) State or imply the equation x = (2x3 + 2) / (3x2 – 2) B1 Rearrange this in the form x 3 – 2x – 2 = 0, or work vice versa B1 2 (iii) Use the iterative formula correctly at least once with xn > 0 M1 Obtain final answer 1.77 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change In the interval (1.765, 1.775) A1 3 GCE A/AS LEVEL – May/June 2009 9709 03 2 ( ) 2

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2 5 When 3, where a is a constant, is expanded in ascending powers of x, the coefficient (1 + 2x)(1 + ax) of the term in x is zero. (i) Find the value of a. [3] 2 (ii) When a has this value, find the term in x3 in the expansion of 3, simplifying the (1 + 2x)(1 + ax) coefficient. [4]

Mark scheme: ax B15 (i) State correct first two terms of the expansion of (1 + ax )3 , i.e. 1+ 23 2 Form an expression for the coefficient of x in the expansion of (1 + 2 x )(1 + ax )3 and equate it to zero M1 Obtain a = –3 A1 3 2 (ii) Obtain correct unsimplified terms in x2 and x3 in the expansion of (1 − 3 x )3 2 or (1 + ax )3 B1√ + B1 √ Carry out multiplication by 1 + 2x obtaining two terms in x3 M1 Obtain final answer − 103 x 3 , or equivalent A1 4  2  [Symbolic binomial coefficients, e.g.  3  , are not acceptable for the B marks in (i) or (ii)]  1  dx 2 dy 2

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6 The parametric equations of a curve are x a sin3t, = a cos3t, y = where a is a positive constant and 0 t 12π. < < dy (i) Express in terms of t. [3] dx (ii) Show that the equation of the tangent to the curve at the point with parameter t is x sin t y cos t a sin t cos t. + = [3] (iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY is always equal to a. [2]

Mark scheme: dx 2 dy 2 6 (i) EITHER State = −3a cos t sin t or = 3a sin t cos t , or equivalent B1 d t d t dy dy dx Use = ÷ M1 dx dt dt 2 − 13 2 − 13 23 23 OR State 3 x d x or 3 y d y as differentials of x or y respectively, or equivalent B1 dy Obtain in terms of t, having taken the differential of a constant to be zero M1 dx dy Obtain in any correct form A1 3 dx (ii) Form the equation of the tangent M1 Obtain the equation in any correct form A1 Obtain the given answer A1 3 (iii) State the x-coordinate of X or the y-coordinate of Y in any correct form B1 Obtain the given answer with no errors seen B1 2 GCE A/AS LEVEL – May/June 2009 9709 03

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7 (i) Solve the equation 0, giving your answers in the form x iy, where x and y ß2 + (2 √3)iß −4 = + are real. [3] (ii) Sketch an Argand diagram showing the points representing the roots. [1] (iii) Find the modulus and argument of each root. [3] (iv) Show that the origin and the points representing the roots are the vertices of an equilateral triangle. [1]

Mark scheme: 7 (i) Use quadratic formula, or completing the square, or the substitution z = x + iy to find a root, using i2 = –1 M1 Obtain a root, e.g. 1 – i3 A1 Obtain the other root, e.g. –1 – i3 A1 3 (ii) Represent both roots on an Argand diagram in relatively correct positions B1 √ 1 (iii) State modulus of both roots is 2 B1 √ State argument of 1 – i3 is –60° (or 300°, − 13 π , − 53 π ) B1 √ State argument of –1 – i3 is –120° (or 240°, − 23 π , − 43 π ) B1 √ 3 (iv) Give a complete justification of the statement B1 1 [The A marks in (i) are for the final versions of the roots. Allow (±2 – 2 i3 )/2 as final answer. The remaining marks are only available for roots such that xy ≠ 0.] [Treat answers to (iii) in polar form as a misread] A B C

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100 8 (i) Express x2(10 −x) in partial fractions. [4] (ii) Given that x 1 when t 0, solve the differential equation = = dx 1 −x), dt = 100x2(10 obtaining an expression for t in terms of x. [6]

Mark scheme: A B C 8 (i) State or imply the form + 2 + B1 x x 10 − x Use any relevant method to determine a constant M1 Obtain one of the values A = 1, B = 10, C = 1 A1 Obtain the remaining two values A1 4 Dx + E C [The form 2 + is acceptable and leads to D = 1, E = 10, C = 1] x 10 − x (ii) Separate variables and attempt integration of both sides M1 Obtain terms ln x, –10/x, –ln (10 – x), or equivalent A1√ + A1√ + A1 √ Evaluate a constant or use limits x = 1, t = 0 with a solution containing 3 of the terms kln x, l/x, mln (10 –x) and t, or equivalent M1  9 x  10 Obtain any correct expression for t, e.g. t = ln   − + 10 A1 6  10 − x  x adx [A separation of the form 2 = bdt is essential for the M1. The f.t. is on A, B, C] x (10 − x ) [If A or B (D or E) omitted from the form of fractions, give B0M1A0A0 in (i); M1A1√ A1√M1A0 in (ii)] GCE A/AS LEVEL – May/June 2009 9709 03

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9 The line l has equation r 4i 2j It is given that l lies in the plane with equation = + −k + t(2i −j −2k). 2x by 1, where b and c are constants. + + cß = (i) Find the values of b and c. [6] (ii) The point P has position vector 2j 4k. Show that the perpendicular distance from P to l is √5. + [5]

Mark scheme: 9 (i) EITHER Substitute coordinates of general point of l in equation of plane and equate constant terms, obtaining an equation in b and c M1 * Obtain a correct equation, e.g. 8 + 2b – c = 1 A1 Equate the coefficient of t to zero, obtaining an equation in b and c M1 * Obtain a correct equation, e.g. 4 – b – 2c = 0 A1 OR Substitute (4, 2, –1) in the plane equation M1 * Obtain a correct equation in b and c, e.g. 2b – c = –7 A1 EITHER Find a second point on l and obtain an equation in b and c M1 * Obtain a correct equation in b and c, e.g. b + 2c = 4 A1 OR Calculate scalar product of a direction vector for l and a vector normal for the plane and equate to zero M1 * Obtain a correct equation for b and c A1 Solve for b or for c M1(dep*) Obtain b = –2 and c = 3 A1 6 (ii) EITHER Find PQ for a point Q on l with parameter t, e.g. 4i – 5k + t(2i – j – 2k) B1 Calculate scalar product of PQ and a direction vector for l and equate to zero M1 Solve and obtain t = –2 A1 Carry out a complete method for finding the length of PQ M1 Obtain the given answer 5 correctly A1 OR 1 Calling (4, 2, –1) A, state AP (or PA ) in component form, e.g. 4i – 5k B1 Calculate vector product of AP and a direction vector for l, e.g. (4i – 5k) × (2i – j – 2k) M1 Obtain correct answer, e.g. –5i – 2j – 4k A1 Divide modulus of the product by that of the direction vector M1 Obtain the given answer correctly A1 OR 2 State AP (or PA ) in component form B1 Use a scalar product to find the projection of AP (or PA ) on l M1 18 Obtain correct answer in any form, e.g. A1 9 Use Pythagoras to find the perpendicular M1 Obtain the given answer correctly A1 OR 3 State AP (or PA ) in component form B1 Use a scalar product to find the cosine of PAQ M1 18 Obtain correct answer in any form, e.g. A1 41 ⋅ 9 Use trig to find the perpendicular M1 Obtain the given answer correctly A1 GCE A/AS LEVEL – May/June 2009 9709 03 OR 4 State AP (or PA ) in component form B1 Find a second point B on l and use the cosine rule in triangle APB to find the cosine of A, B or P, or use a vector product to find the area of APB M1 Obtain correct answer in any form A1 Use trig or area formula to find the perpendicular M1 Obtain the given answer correctly A1 OR 5 Find PQ for a point Q on l with parameter t, e.g. 4i – 5k + t(2i –j – 2k) B1 Use correct method to express PQ2 (or PQ) in terms of t M1 Obtain a correct expression in any form, e.g. (4 + 2t)2 + (–t)2 + (–5 – 2t)2 A1 Carry out a complete method for finding its minimum M1 Obtain the given answer correctly A1 5

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10 y M x O The diagram shows the curve y for x and its maximum point M. = x2√(1 −x2) ≥0 (i) Find the exact value of the x-coordinate of M. [4] (ii) Show, by means of the substitution x sin θ, that the area A of the shaded region between the = curve and the x-axis is given by 12π A 1 sin2 2θ dθ. = 4 ã 0 [3] (iii) Hence obtain the exact value of A. [4]

Mark scheme: 10 (i) EITHER Use product and chain rule M1 Obtain correct derivative in any form A1 OR Square and differentiate LHS by chain rule and RHS by product rule or as powers M1 Obtain correct result in any form A1 dy Set equal to zero and make reasonable attempt to solve for x ≠ 0 M1 dx Obtain answer x = 2 , or exact equivalent, correctly A1 4 3 dx (ii) State or imply dx = cos θ dθ or = cos θ B1 dθ Substitute for x and dx throughout the integral ∫ ydx M1 Obtain the given form correctly with no errors seen A1 3 (iii) Attempt integration and reach indefinite integral of the form aθ + bsin 4θ , where ab ≠ 0 M1* Obtain indefinite integral 18 θ − 321 sin 4θ , or equivalent A1 Substitute limits correctly M1(dep*) Obtain exact answer 161 π A1 4 [Working to carry out the change of limits is needed for the A mark in (ii) but, if omitted, can be earned retrospectively if it is seen in part (iii)]

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