Cambridge A Level Mathematics 9709 — 2009 May/June Paper 2 · Variant 1

x 1 Given that use logarithms to find the value of correct to 3 significant figures. [3] y (1.25)x = (2.5)y,

Mark scheme: 1 Use logarithms to linearise an equation M1 x ln 5.2 Obtain = , or equivalent A1 y ln .125 Obtain answer 4.11 A1√ [3] 2 2

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2 Solve the inequality [4] |3x + 2| < |x|.

Mark scheme: 2 EITHER: State or imply non-modular inequality (3x + 2)2 < x2, or corresponding quadratic equation, or pair of linear equations 3x + 2 = ± x M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = –1 and x = – 1 A1 2 State answer –1 < x < – 1 A1 2 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 2 State answer –1 < x < – 1 B1 [4] 2

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3 y 1 x O 1 2 1 The diagram shows the curve y for values of x from 0 to 2. 1 = √x + (i) Use the trapezium rule with two intervals to estimate the value of 2 1 dx, 1 √x ä 0 + giving your answer correct to 2 decimal places. [3] (ii) 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 (i). [1]

Mark scheme: 3 (i) Show or imply correct ordinates 1, 0.5, 0.414213 ... B1 Use correct formula, or equivalent, with h = 1 and three ordinates M1 Obtain answer 1.21 with no errors seen A1 [3] (ii) Justify the statement that the rule gives an over-estimate B1 [1] dx

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4 The parametric equations of a curve are x 4 sin θ, y 3 cos 2θ, = = −2 1 dy where 2π θ 2π. Express in terms of θ, simplifying your answer as far as possible. [5] dx −1 < <

Mark scheme: dx 4 State = 4 cos θ B1 dθ dy State = 4 sin 2θ , or equivalent B1 dθ dy dy dx Use = ÷ M1 dx dθ dθ dy sin 2θ Obtain in any correct form, e.g. A1 dx cos θ Simplify and obtain answer 2 sinθ A1√ [5] [The f.t. is on gradients of the form k sin 2θ / cos θ, or equivalent.] 2 2 2 2

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5 Solve the equation sec x 4 tan2x, giving all solutions in the interval [6] = −2 0◦≤x ≤180◦.

Mark scheme: 5 Use tan2 x = sec2 x –1 or sin2 x = 1 – cos2 x M1 Obtain 3-term quadratic in sec x or cos x, e.g. 2sec2 x + sec x – 6 = 0 A1 Make reasonable solution attempt at a 3-term quadratic M1 Obtain sec x = 3 and sec x = –2, or equivalent A1 2 [or 6cos2 x – cos x – 2 = 0 cos x = 2 3 , −1 2 ] Obtain answer x = 48.2° A1 Obtain answer x = 120° and no others in the range A1 [6] [Ignore answers outside the given range.] GCE A/AS LEVEL – May/June 2009 9709 02

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6 The polynomial x3 ax2 bx 6, where a and b are constants, is denoted by It is given that is a factor of+ +and that+ when is divided by the remainderp(x).is 4. (x −2) p(x), p(x) (x −1) (i) Find the values of a and b. [5] (ii) When a and b have these values, find the other two linear factors of [3] p(x).

Mark scheme: 6 (i) Substitute x = 2, equate to zero and state a correct equation, e.g. 8 + 4a + 2b + 6 = 0 B1 Substitute x = 1 and equate to 4 M1 Obtain a correct equation. e.g. 1 + a + b + 6 = 4 A1 Solve for a or for b M1 Obtain a = –4 and b = 1 A1 [5] (ii) EITHER: Attempt division by x –2 reaching a partial quotient of x2 + kx M1 Obtain remainder quadratic factor x2 – 2x – 3 A1 State linear factors (x –3) and (x + 1) A1 OR: Obtain linear factor (x + 1) by inspection B1 Obtain factor (x –3) similarly B2 [3]

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7 y x O M The diagram shows the curve y xe2x and its minimum point M. = (i) Find the exact coordinates of M. [5] (ii) Show that the curve intersects the line y 20 at the point whose x-coordinate is the root of the equation = 1 20 x ln . 2 x = [1] (iii) Use the iterative formula 1 20 2 xn xn+1 = ln , with initial value x1 1.3, to calculate the root correct to 2 decimal places, giving the result of = each iteration to 4 decimal places. [3]

Mark scheme: 7 (i) Use product rule M1* Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1(dep*) Obtain answer x = − 1 correctly A1 2 Obtain y = –1/(2e) or exact equivalent A1 [5] (ii) Show that 20 = xe2x is equivalent to x = 1 ln(20 / x) or vice versa B1 [1] 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.35 A1 Show sufficient iterations to justify its accuracy to 2 d.p. A1 [3] 1

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8 (a) Find the equation of the tangent to the curve y at the point where x 1. [4] = ln(3x −2) = (b) (i) Find the value of the constant A such that 6x A 3x 3x ≡2 + [2] −2 −2. 6 6x 8 (ii) Hence show that dx 8 ln 2. [5] 3 3x ä = + 2 −2

Mark scheme: 8 (a) State derivative is k/(3x –2) where k = 3.1, or 1 M1 3 State correct derivative 3/(3x –2) A1 Form the equation of the tangent at the point where x = 1 M1 Obtain answer y = 3x –3, or equivalent A1 [4] (b) (i) Carry out a complete method for finding A M1 Obtain A = 4 A1 [2] (ii) Integrate and obtain term 2x B1 Obtain second term of the form aln(3x –2) M1 Obtain second term 4 ln(3x – 2) A1√ 3 Substitute limits correctly M1 Obtain given answer following full and correct working A1 [5]

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