Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 3 · Variant 3

16 1 Expand in ascending powers of x, up to and including the term in x2, simplifying the (2 + x)2 coefficients. [4]

Mark scheme: 1 Either Obtain correct unsimplified version of x or x2 term in expansion of M1 1 (2 + x)–2 or (1 + x)–2 2 Correct first term 4 from correct work B1 Obtain –4x A1 Obtain + 3x2 A1 Or Differentiate and evaluate f(0) and f΄ (0) where f΄ (x) = k(2+x)–3 M1 State correct first term 4 B1 Obtain –4x A1 Obtain + 3x2 A1 [4]

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e2x 2 The equation of a curve is y . Show that the gradient of the curve at the point for which = 1 e2x 9 + x ln 3 is 50. [4] =

Mark scheme: 2 Use correct quotient or product rule or equivalent M1 2 x 2 x 2 x 2 x (1 + e ).2e − e .2e Obtain 2 x 2 or equivalent A1 (1 + e ) Substitute x = ln 3 into attempt at first derivative and show use of relevant logarithm M1 property at least once in a correct context Confirm given answer 509 legitimately A1 [4]

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α 0 and Give the value3 (i) Express 8 cos θ 15 sin θ in the form R where R + cos(θ −α), > < 90◦. 0◦< of α correct to 2 decimal places. [3] θ 15 sin θ 12, giving all solutions in the interval (ii) Hence solve the equation 8 cos θ + = < 360◦. 0◦< [4]

Mark scheme: 3 (i) State or imply R =17 B1 Use correct trigonometric formula to find α M1 Obtain 61.93° with no errors seen A1 [3] 12 (ii) Evaluate cos–1 R ( = 45.099) M1 Obtain answer 107.0° A1 Carry out correct method for second answer M1 Obtain answer 16.8° and no others between 0° and 360° A1 [4] GCE AS/A LEVEL – October/November 2011 9709 33

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4 During an experiment, the number of organisms present at time t days is denoted by N, where N is treated as a continuous variable. It is given that dN dt = 1.2e−0.02tN0.5. When t 0, the number of organisms present is 100. = (i) Find an expression for N in terms of t. [6] (ii) State what happens to the number of organisms present after a long time. [1]

Mark scheme: 4 (i) Separate variables and attempt integration on both sides M1* Obtain 2N0.5 on left-hand side or equivalent A1 Obtain –60e–0.02t on right-hand side or equivalent A1 Use 0 and 100 to evaluate a constant or as limits in a solution containing terms DM1* aNo.5 and be –0.02t Obtain 2N0.5 = –60e–0.02t + 80 or equivalent A1 Conclude with N = (40 – 30e–0.02t)2 or equivalent A1 [6] (ii) State number approaches 1600 or equivalent, following expression of form B1√ [1] (c + de–0.02t)n (i) Eith

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a 5 It is given that x ln x dx 22, where a is a constant greater than 1. ã 1 = r 87 (i) Show that a . [5] = 2 ln a −1 (ii) Use an iterative formula based on the equation in part (i) to find the value of a correct to 2 decimal places. Use an initial value of 6 and give the result of each iteration to 4 decimal places. [3]

Mark scheme: 5 (i) Either Use integration by parts and reach an expression kx2 lnx ± n ∫ x2. 1x dx M1 Obtain 12 x2 ln x – ∫ 12 x dx or equivalent A1 Obtain 12 x2 ln x – 14 x2 A1 Or Use Integration by parts and reach an expression kx(xlnx – x) ± m ∫ xlnx – xdx M1 Obtain I = (x2 lnx – x2) – I + ∫ xdx A1 Obtain 12 x2 ln x – 14 x2 A1 Substitute limits correctly and equate to 22, having integrated twice DM1* 87 Rearrange and confirm given equation a = A1 [5] 2ln a − 1 (ii) Use iterative process correctly at least once M1 Obtain final answer 5.86 A1 Show sufficient iterations to 4 d.p. to justify 5.86 or show a sign change in the A1 interval (5.855, 5.865) (6 → 5.8030 → 5.8795 → 5.8491 → 5.8611 → 5.8564) [3] GCE AS/A LEVEL – October/November 2011 9709 33 2 3

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6 The complex number w is defined by w i. = −1 + (i) Find the modulus and argument of w2 and w3, showing your working. [4] (ii) The points in an Argand diagram representing w and w2 are the ends of a diameter of a circle. Find the equation of the circle, giving your answer in the form k. [4] |ß −(a + bi)| =

Mark scheme: 6 (i) Use correct method for finding modulus of their w2 or w3 or both M1 Obtain │w2│ = 2 and │w3│= 2 2 or equivalent A1 Use correct method for finding argument of their w2 or w3 or both M1 Obtain arg(w2) = – 12 π or 32 π and arg(w3) = 14 π A1ft [4] (ii) Obtain centre – 12 – 12 i (their w2) B1ft Calculate the diameter or radius using │w–w2│ w21 or right-angled triangle M1 or cosine rule or equivalent Obtain radius 12 10 or equivalent A1 Obtain │z + 12 + 12 i│ = 12 10 or equivalent A1ft [4] (i) S b i 1 d

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7 The polynomial is defined by p(x) ax3 4x p(x) = −x2 + −a, where a is a constant. It is given that is a factor of (2x −1) p(x). (i) Find the value of a and hence factorise [4] p(x). 8x (ii) When a has the value found in part (i), express −13 in partial fractions. [5] p(x)

Mark scheme: 7 (i) Substitute x = 12 and equate to zero or divide by (2x –1), reach a2 x2 + kx + ...and equate remainder to zero or by inspection reach a2 x2 + bx + c and an equation in b/c or by inspection reach Ax2 + Bx + a and an equation in A/B M1 Obtain a = 2 A1 Attempt to find quadratic factor by division or inspection or equivalent M1 Obtain (2x –1)(x2 +2) A1cwo [4] A Bx + C (ii) State or imply form + 2 , following factors from part (i) B1√ 2 x − 1 x + 2 Use relevant method to find a constant M1 Obtain A = –4, following factors from part (i) A1√ Obtain B = 2 A1 Obtain C = 5 A1 GCE AS/A LEVEL – October/November 2011 9709 33 2 2

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8 y x O The diagram shows the curve with parametric equations x sin t cos t, y sin3t cos3t, = + = + t 54π. for 14π < < dy (i) Show that sin t cos t. [3] dx = −3 (ii) Find the gradient of the curve at the origin. [2] (iii) Find the values of t for which the gradient of the curve is 1, giving your answers correct to 2 significant figures. [4]

Mark scheme: 8 (i) Differentiate y to obtain 3sin2 t cos t – 3cos2 t sin t o.e. B1 dy dy dt Use = / M1 dx dt dx Obtain given result –3sin t cos t A1cwo [3] 3 (ii) Identify parameter at origin as t = 4π B1 3 3 Use t = 4π to obtain 2 B1 [2] (iii) Rewrite equation as equation in one trig variable B1 e.g. sin2t = − 23 , 9 sin4 x – 9 sin2 x + 1 = 0, tan2 x + 3 tan x + 1 = 0 Find at least one value of t from equation of form sin 2t = k o.e. M1 Obtain 1.9 A1 Obtain 2.8 and no others A1 [4] GCE AS/A LEVEL – October/November 2011 9709 33

More questions on Differentiation

9 The line l has equation r 1 ! λ 3 !, where a is a constant. The plane p has equation = + 4 −2 2x 10. −2y + ß = (i) Given that l does not lie in p, show that l is parallel to p. [2] (ii) Find the value of a for which l lies in p. [2] (iii) It is now given that the distance between l and p is 6. Find the possible values of a. [5]

Mark scheme: 9 (i) Calculate scalar product of direction of l and normal to p M1 Obtain 4 x 2 + 3 × (–2) + (–2) × l = 0 and conclude accordingly A1 [2] (ii) Substitute (a, 1, 4) in equation of p and solve for a M1 Obtain a = 4 A1 [2] (iii) Either Attempt use of formula for perpendicular distance using (a, 1, 4) M1 2 a − 2 + 4 − 10 Obtain at least = 6 A1 4 + 4 + 1 Obtain a = 13 A1 2 a − 8 Attempt solution of = −6 M1 3 Obtain a = –5 A1 Or M1 Form equation of parallel plane and substitute (a, 1, 4) = 6 A1 Obtain 2 a+3 2 − 103 Obtain a = 13 A1 Solve 2 a+3 2 − 103 = −6 M1 Obtain a = –5 A1 Or State a vector from a pt on the plane to (a, 1, 4) e.g. B1  a − 5   a      1 or 1         4 −6     Calculate the component of this vector in the direction of the unit M1  a − 5  2  1    normal and equate to 6 : 3  1 . −2  = 6    4 1    Obtain a = 13 A1  a − 5  2  1    Solve 3  1 . −2  = −6 M1    4 1    Obtain a = – 5 A1 GCE AS/A LEVEL – October/November 2011 9709 33 Or a  2     B1 State or imply perpendicular line r = 1 + µ −2       4 1    Substitute components for p and solve for µ M1 8 − 2 a Obtain µ = A1 9 Equate distance between (a, 1, 4) and foot of perpendicular to ±6 M1 3(8 − 2 a ) Obtain = ±6 or equivalent and hence –5 and 13 A1 [5] 9 du 2

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10 (i) Use the substitution u tan x to show that, for n = ≠−1, 14π 1 dx ã 0 (tann+2x + tannx) = n 1. [4] + (ii) Hence find the exact value of 14π (a) dx, [3] ã 0 (sec4x −sec2x) 14π (b) 5 tan7x 5 tan5x dx. [3] ã 0 (tan9x + + + tan3x)

Mark scheme: du 10 (i) State or imply = sec 2 x B1 dx Express integrand in terms of u and du M1 u n +1 Integrate to obtain or equivalent A1 n + 1 1 Substitute correct limits correctly to confirm given result A1 [4] n + 1 (ii) (a) Use sec2 x =1 + tan2 x twice M1 Obtain integrand tan4 x + tan2 x A1 Apply result from part (i) to obtain 13 A1 [3] Or M1 Use sec2 x = 1 + tan2 x and the substitution from (i) Obtain ∫ u2du A1 Apply limits correctly and obtain 13 A1 (b) Arrange, perhaps implied, integrand to B1 t9 + t7 + 4(t7 + t5) + t5 + t3 Attempt application of result from part (i) at least twice M1 1 4 1 25 Obtain + + and hence 24 or exact equivalent A1 [3] 8 6 4

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