Cambridge A Level Mathematics 9709 — 2008 May/June Paper 3 · Variant 1

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

Mark scheme: 1 EITHER State or imply non-modular inequality( x − 2 )2 > (3(2 x + 1))2 , or corresponding quadratic equation, or pair of linear equations ( x − 2 ) = ±3(2 x + )1 B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = −1 and x = − 1 A1 7 State answer − 1 < x < − 1 A1 7 OR Obtain the critical value x = −1 from a graphical method, or by inspection, or by solving a linear equation or inequality B1 Obtain the critical value x = − 1 similarly B2 7 State answer − 1 < x < − 1 B1 [4] 7 [Do not condone ≤ for <; accept − 5 and − 0.14 for − 1 .] 35 7 2

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2 Solve, correct to 3 significant figures, the equation ex + e2x = e3x. [5]

Mark scheme: 2 EITHER State or imply e x + 1 = e 2 x , or 1 + e − x = e x , or equivalent B1 Solve this equation as a quadratic in u = e x , or in e x , obtaining one or two M1 roots Obtain root 1 1 + 5 ) , or decimal in [1.61, 1.62] A1 2 ( Use correct method for finding x from a positive root M1 Obtain x = 0.481 and no other answer A1 [For the solution 0.481 with no working, award B3 (for 0.48 give B2). However a suitable statement can earn the first B1 in addition, giving a maximum of 4/5 (or 3/5) in such cases.] n 1 + e x ) or OR State an appropriate iterative formula, e.g. nx +1 = 12 ln ( n e x n + e 2 x ) B1 nx +1 = 13 ln ( Use the iterative formula correctly at least once M1 Obtain final answer 0.481 A1 Show sufficient iterations to justify its accuracy to 3 d.p., or show there is a sign change in the value of a relevant function in the interval (0.4805, 0.4815) A1 Show that the equation has no other root A1 [5]

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3 N D C a r x M A B 3a In the diagram, ABCD is a rectangle with AB = 3a and AD = a. A circular arc, with centre A and radius r, joins points M and N on AB and CD respectively. The angle MAN is x radians. The perimeter of the sector AMN is equal to half the perimeter of the rectangle. (i) Show that x satisfies the equation sin x = 14(2 + x). [3] (ii) This equation has only one root in the interval 0 < x < 12π. Use the iterative formula 2 + xn xn+1 = sin−1 , 4 with initial value x1 = 0.8, to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 3 (i) State or imply r = a cosec x, or equivalent B1 Using perimeters, obtain a correct equation in x, e.g. 2a cosec x + ax cosec x = 4a, or 2r + rx = 4a B1 Deduce the given form of equation correctly B1 [3] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 0.76 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the value of sin x − 1 (2 + x ) in the interval (0.755, 0.765) A1 [3] 4

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4 (i) Show that the equation tan(30◦+ θ) = 2 tan(60◦−θ) can be written in the form tan2θ + (6 √3) tan θ −5 = 0. [4] (ii) Hence, or otherwise, solve the equation tan(30◦+ θ) = 2 tan(60◦−θ), for 0◦≤θ ≤180◦. [3]

Mark scheme: 4 (i) Use tan(A ± B) formula correctly at least once to obtain an equation in tan θ M1 Obtain a correct horizontal equation in any form A1 Use correct exact values of tan 30° and tan 60° throughout M1 Obtain the given equation correctly A1 [4] GCE A/AS LEVEL – May/June 2008 9709 03 (ii) Make reasonable attempt to solve the given quadratic in tan θ M1 Obtain answer θ = 24.7° A1 Obtain answer θ = 95.3° and no others in the given range A1 [3] [Ignore answers outside the given range.] [Treat answers in radians as MR and deduct one mark from the marks for the angles.]

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5 The variable complex number is given by = 2 cos θ + i(1 −2 sin θ), where θ takes all values in the interval −π < θ ≤π. (i) Show that | −i| = 2, for all values of θ. Hence sketch, in an Argand diagram, the locus of the point representing . [3] 1 (ii) Prove that the real part of is constant for −π < θ < π. [4] + 2 −i

Mark scheme: 5 (i) Find modulus of 2cosθ − 2isinθ and show it is equal to 2 B1 Show a circle with centre at the point representing i B1 Show a circle with radius 2 B1 [3] (ii) Substitute for z and multiply numerator and denominator by the conjugate of z + 2 – i, or equivalent M1 Obtain correct real denominator in any form A1 Identify and obtain correct unsimplified real part in terms of cosθ, e.g. (2cosθ + 2)/(8cosθ + 8) A1 1 State that real part equals A1 [4] 4 dy 2 2

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6 The equation of a curve is xy(x + y) = 2a3, where a is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the x-axis, and find the coordinates of this point. [8]

Mark scheme: dy 26 EITHER State x + 2 xy , or equivalent, as derivative of x2y B1 dx dy 2 State y + 2 xy , or equivalent, as derivative of xy2 B1 dx dy OR State xy 1( + ) , or equivalent, as a term in an attempt to apply the product dx rule B1 dy State ( y + x )( x + y ) , or equivalent, in an attempt to apply the product rule dx B1 dy Equate attempted derivative of LHS to zero and set equal to zero dx M1 Obtain a horizontal equation, e.g. y2 = –2xy, or y = −2x, or equivalent A1√ Explicitly reject y = 0 as a possibility A1 Obtain an equation in x (or in y) M1 Obtain x = a A1 Obtain y = −2a only A1 [8] [The first M1 is dependent on at least one B mark having been earned.] [SR: for an attempt using (x + y) = 2a3 / xy, the B marks are given for the correct derivatives of the two sides of the equation, and the M1 for setting dy equal to zero.] dx [SR: for an attempt which begins by expressing y in terms of x, give M1A1 dy for a reasonable attempt at differentiation, M1A1√ for setting equal to dx zero and obtaining an equation free of surds, A1 for solving and obtaining x = a; then M1 for obtaining an equation for y, A1 for y = −2a and A1 for finding and rejecting y = a as a possibility.] GCE A/AS LEVEL – May/June 2008 9709 03 B C

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x2 + 3x + 3 7 Let f(x) ≡ (x + 1)(x + 3). (i) Express f(x) in partial fractions. [5] 3 (ii) Hence show that f(x) dx = 3 −1 ln 2. [4] 2 0

Mark scheme: B C 7 (i) State or imply the form A + + B1 x + 1 x + 3 State or obtain A = 1 B1 Use correct method for finding B or C M1 Obtain B = 1 A1 2 Obtain C = − 3 A1 [5] 2 (ii) Obtain integral x + 1 ln ( x + 1) − 3 ln ( x + 3 ) B2√ 2 2 [Award B1√ if only one error. The f.t. is on A, B, C.] Substitute limits correctly M1 Obtain given answer following full and exact working A1 [4] [SR: if A omitted, only M1 in part (i) is available, then in part (ii) B1√ for each correct integral and M1.] y d y

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8 y P x O T N 12 In the diagram the tangent to a curve at a general point P with coordinates (x, y) meets the x-axis at T. The point N on the x-axis is such that PN is perpendicular to the x-axis. The curve is such that, for all values of x in the interval 0 < x < 12π, the area of triangle PTN is equal to tan x, where x is in radians. PN (i) Using the fact that the gradient of the curve at P is , show that TN dy 1 = 2y2 cot x. [3] dx (ii) Given that y = 2 when x = 16π, solve this differential equation to find the equation of the curve, expressing y in terms of x. [6]

Mark scheme: y d y 8 (i) State = , or equivalent B1 TN d x dy Express area of PTN in terms of y and , and equate to tan x M1 dx Obtain given relation correctly A1 [3] (ii) Separate variables correctly B1 2 Integrate and obtain term − , or equivalent B1 y Integrate and obtain term ln(sin x), or equivalent B1 Evaluate a constant or use limits y = 2, x = 1 π in a solution containing a term of the 6 form a/y or bln(sin x) M1 2 Obtain correct solution in any form, e.g. − = ln (2 sin x ) − 1 A1 y Rearrange as y = 2 / (1 − ln (2 sin x )) , or equivalent A1 [6] [Allow decimals, e.g. as in a solution y = 2 / (3.0 − ln (sin x )) .]

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9 y M R x O The diagram shows the curve y = e−12x√(1 + 2x) and its maximum point M. The shaded region between the curve and the axes is denoted by R. (i) Find the x-coordinate of M. [4] (ii) Find by integration the volume of the solid obtained when R is rotated completely about the x-axis. Give your answer in terms of π and e. [6]

Mark scheme: 9 (i) Either use correct product or quotient rule, or square both sides, use correct product rule and make a reasonable attempt at applying the chain rule M1 Obtain correct result of differentiation in any form A1 Set derivative equal to zero and solve for x M1 Obtain x = 1 only, correctly A1 [4] 2 + 2 x )d x B1 (ii) State or imply the indefinite integral for the volume is π ∫ e − x (1 x M1 Integrate by parts and reach ± e − x (1 + 2 x ) ± ∫ 2e − x d x , or equivalent A1 Obtain − e − x (1 + 2 x ) + ∫ 2e − x d Complete integration correctly, obtaining − e − x (1 + 2 x ) − 2e − x , or equivalent A1 Use limits x = − 1 and x = 0 correctly, having integrated twice M1 2 Obtain exact answer π (2 e − 3) , or equivalent A1 [6] [If π omitted initially or 2π or π/2 used, give B0 and then follow through.] GCE A/AS LEVEL – May/June 2008 9709 03

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10 The points A and B have position vectors, relative to the origin O, given by −−→ −−→ OA = i + 2j + 3k and OB = 2i + j + 3k. The line l has vector equation r = (1 −2t)i + (5 + t)j + (2 −t)k. (i) Show that l does not intersect the line passing through A and B. [4] (ii) The point P lies on l and is such that angle PAB is equal to 60◦. Given that the position vector of P is (1 −2t)i + (5 + t)j + (2 −t)k, show that 3t2 + 7t + 2 = 0. Hence find the only possible position vector of P. [6]

Mark scheme: 10 (i) State a vector equation for the line through A and B, e.g. r = i + 2j + 3k + s(i − j) B1 Equate at least two pairs of components of general points on AB and l, and solve for s or for t M1 Obtain correct answer for s or t, e.g. s = −6, 2, −2 when t = 3, −1, −1 respectively A1 Verify that all three component equations are not satisfied A1 [4] (ii) State or imply a direction vector for AP has components (−2t, 3 + t, −1−t), or equivalent B1 AP. AB State or imply cos 60° equals M1* AP . AB Carry out correct processes for expanding the scalar product and expressing the product of the moduli in terms of t, in order to obtain an equation in t in any form M1(dep*) Obtain the given equation 3t 2 + 7t + 2 = 0 correctly A1 Solve the quadratic and use a root to find a position vector for P M1 Obtain position vector 5i + 3j + 4k from t = −2, having rejected the root t = − 1 for 3 a valid reason A1 [6]

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