Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 2 · Variant 1

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

Mark scheme: 1 EITHER: State or imply non-modular inequality ( x + 1)2 > ( x − 4 )2 , or corresponding equation or pair of linear equations M1 3 Obtain critical value A1 2 3 State correct answer x > A1 2 OR: State a correct linear equation for the critical value, e.g. x + 1 = − x + 4 , or corresponding correct linear inequality, e.g. x + 1 > − ( x − 4 ) M1 3 Obtain critical value A1 2 3 State correct answer x > A1 [3] 2

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giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 5x = 22x+1, [4]

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain x log 5 = (2 x + 1) log 2 , or equivalent A1 Solve for x, via correct manipulative technique(s) M1(dep*) Obtain answer x = .311 . Allow x ∈ [.3 10, .3 11] A1 [4] 1 2

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1 3 Show that 2e dx 12e2 2. [5] ã = + −3 0 (ex + 1)2

Mark scheme: 1 2 x 3 Integrate and obtain e term B1 2 Obtain 2e x term B1 Obtain x B1 Use limits correctly, allow use of limits x = 1 and x = 0 into an incorrect form M1 Obtain given answer A1 [5] S. R. Feeding limits into original integrand, 0/5 dx 1 dy 2

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4 The parametric equations of a curve are 9 x 1 y t , for t 2. = + ln(t −2), = + t > dy (i) Show that (t2 −9)(t −2) . [3] dx = t2 (ii) Find the coordinates of the only point on the curve at which the gradient is equal to 0. [3]

Mark scheme: dx 1 dy − 24 (i) State = or = 1 − 9t B1 dt t − 2 dt dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [3] (ii) Equate derivative to zero and solve for t M1 State or imply that t = 3 is admissible c.w.o., and note t = –3, 2 cases A1 Obtain coordinates (1, 6) and no others A1 [3]

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5 Solve the equation 8 cot θ 2 cosec2θ, giving all solutions in the interval [6] + = 0◦≤θ ≤360◦.

Mark scheme: 5 Use correct trig identity to obtain a quadratic in cot θ or tan θ M1 Solve the quadratic correctly A1 1 2 Obtain tan θ = or − A1√ 2 3 Obtain answer 26.6° or 146.3° A1 Carry out correct method for second answer from either root M1 Obtain remaining 3 answers from 26.6°, 146.3°, 206.6°, 326.3° and no others in the range A1 [6] [Ignore answers outside the given range] GCE A/AS LEVEL – October/November 2010 9709 21 6

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6 6 The curve with equation y intersects the line y x 1 at the point P. = x2 = + (i) Verify by calculation that the x-coordinate of P lies between 1.4 and 1.6. [2] (ii) Show that the x-coordinate of P satisfies the equation q 6 x . = x 1 [2] + (iii) Use the iterative formula r 6 xn+1 = , xn 1 + with initial value x1 1.5, to determine the x-coordinate of P correct to 2 decimal places. Give = the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 6 (i) Consider sign of 2 −x − 1 at x = 1.4 and x = 1.6, or equivalent M1 x Complete the argument correctly with appropriate calculations A1 [2] 6 (ii) State 2 = x + 1 B1 x Rearrange equation to given equation or vice versa B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.54 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.535, 1.545) B1 [3]

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7 The polynomial 3x3 2x2 ax b, where a and b are constants, is denoted by It is given that + + + p(x). is a factor of and that when is divided by the remainder is 10. (x −1) p(x), p(x) (x −2) (i) Find the values of a and b. [5] (ii) When a and b have these values, solve the equation 0. [4] p(x) =

Mark scheme: 7 (i) Substitute x = 1, equate to zero and obtain a correct equation in any form B1 Substitute x = 2 and equate to 10 M1 Obtain a correct equation in any form A1 Solve a relevant pair of equations for a or for b M1 Obtain a = –17 and b = 12 A1 [5] (ii) At any stage, state that x = 1 is a solution B1 EITHER: Attempt division by x – 1 and reach a partial quotient of 3 x 2 + 5 x M1 Obtain quotient 3 x 2 + 5 x − 12 A1 4 Obtain solutions x = –3 and x = A1 3 OR: Obtain solution x = –3 by trial and error or inspection B1 4 Obtain solution x = B2 3 [If an attempt at the quadratic factor is made by inspection, the M1 is earned if it reaches an unknown factor of 3 x 2 + 5 x + λ and an equation in λ ] [4]

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8 y Q p x O The diagram shows the curve y x sin x, for 0 The point Q 12π, 12π lies on the curve. = ≤x ≤π. (i) Show that the normal to the curve at Q passes through the point [5] (π, 0). d (ii) Find x cos [2] dx(sin −x x). 12π (iii) Hence evaluate x sin x dx. [3] ã 0

Mark scheme: 8 (i) Use product rule M1 Obtain correct derivative in any form A1 1 Substitute x = π , and obtain gradient of –1 for normal A1√ 2 from y ′ = sin x − x cos x ONLY  1 1  Show that line through  π , π  with gradient –1 passes through (π , 0 ) M1  2 2  A1 [5] (ii) Differentiate sin x and use product rule to differentiate xcos x M1 Obtain xsin x , or equivalent A1 [2] (iii) State that integral is sin x − x cos x (+ c ) B1 π Substitute limits 0 and correctly M1 2 Obtain answer 1 A1 [3] S. R. Feeding limits into original integrand, 0/3

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