Cambridge A Level Mathematics 9709 — 2005 May/June Paper 2 · Variant 1

3 The sequence of values given by the iterative formula 3xn 2 xn+1 = + , 4 x3 n with initial value x1 = 2, converges to α. (i) Use this iteration to calculate α correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3] (ii) State an equation which is satisfied by α and hence find the exact value of α. [2]

Mark scheme: 3 (i) Use the given iterative formula correctly at least once M1 Obtain final answer α = 1 . 68 A1 Show sufficient iterations to justify the answer to 2 dp B1 3 3 2 (ii) State equation, e.g. x = x + , in any correct form B1 4 x 3 Derive the exact answer α (or x) = 4 8 , or equivalent B1 2

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4 The polynomial x3 −x2 + ax + b is denoted by p(x). It is given that (x + 1) is a factor of p(x) and that when p(x) is divided by (x −2) the remainder is 12. (i) Find the values of a and b. [5] (ii) When a and b have these values, factorise p(x). [2]

Mark scheme: 4 (i) Substitute x = −1 and equate to zero obtaining e.g. (–1)3 – (–1)2 + a(–1) + b = 0 B1 Substitute x = 2 and equate to 12 M1 Obtain a correct 3-term equation A1 Solve a relevant pair of equations for a or b M1 Obtain a = 2 and b = 4 A1 5 (ii) Attempt division by x + 1 reaching a partial quotient of x 2 + kx , or similar stage M1 by inspection Obtain quadratic factor x 2 −x2 = 4 A1 2 [Ignore failure to repeat that x + 1 is a factor]

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1 dy 5 (i) By differentiating , show that if y = sec θ then = sec θ tan θ. [3] cos θ dθ (ii) The parametric equations of a curve are x = 1 + tan θ, y = sec θ, 1 dy for −12π < θ < 2π. Show that = sin θ. [3] dx (iii) Find the coordinates of the point on the curve at which the gradient of the curve is 12. [3]

Mark scheme: 5 (i) Differentiate using chain or quotient rule M1 Obtain derivative in any correct form A1 Obtain given answer correctly A1 3 (ii) State dx = sec2 θ, or equivalent B1 dθ d y d y d x Use = ÷ M1 d x d θ d θ Obtain given answer correctly A1 3 GCE AS LEVEL – JUNE 2005 9709 2 π (iii) State that θ = B1 6 1 Obtain x-coordinate 1 + , or equivalent B1 3 2 Obtain y-coordinate , or equivalent B1 3 3

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6 ln x The diagram shows the part of the curve y = for 0 < x ≤4. The curve cuts the x-axis at A and its x maximum point is M. (i) Write down the coordinates of A. [1] (ii) Show that the x-coordinate of M is e, and write down the y-coordinate of M in terms of e. [5] (iii) Use the trapezium rule with three intervals to estimate the value of 4 ln x dx, x 1 correct to 2 decimal places. [3] (iv) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii). [1]

Mark scheme: 6 (i) State coordinates (1, 0) B1 1 (ii) Use quotient or product rule M1 − ln x 1 Obtain correct derivative, e.g. + A1 x 2 x 2 Equate derivative to zero and solve for x M1 Obtain x = e A1 1 Obtain y = A1 5 e (iii) Show or imply correct coordinates 0, 0.34657..., 0.36620..., 0.34657,,, B1 Use correct formula, or equivalent, with h = 1 and four ordinates A1 Obtain answer 0.89 with no errors seen A1 3 (iv) Justify statement that the rule gives an under-estimate B1 1

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7 (i) By expanding sin(2x + x) and using double-angle formulae, show that sin 3x = 3 sin x −4 sin3x. [5] (ii) Hence show that 13π sin3x dx = 24.5 [5] 0

Mark scheme: 7 (i) Make relevant use of the sin(A + B) formula B1 Make relevant use of sin2A and cos2A formulae M1 Obtain a correct expression in terms of sin x and cos x A1 Use cos2 x = 1 – sin2 x to obtain an expression in terms of sin x M1(dep*) Obtain given answer correctly A1 5 3 1 (ii) Replace integrand by sin x – sin 3x, or equivalent B1 4 4 3 1 Integrate, obtaining − cos x + cos 3 x, or equivalent B1√ + B1√ 4 12 Use limits correctly M1 Obtain given answer correctly A1 5

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