Cambridge A Level Mathematics 9709 — 2020 Feb/March Paper 2 · Variant 2

2 (a) Find the quotient when 4x3 17x2 9x is divided by x2 5x 6, and show that the remainder is 18. + + + + [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 4x3 17x2 9x 0. [3] + + −18 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) + Obtain quotient 4 3 x − A1 Confirm remainder is 18 A1 AG necessary detail needed 3 2(b) State or imply equation is 2 (4 3)( 5 6) 0 x x x − + + = B1FT Following their quotient from part (a) Attempt solution of cubic equation to find three real roots M1 Obtain 3 4 3, 2, − − A1 3

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3a 2 7 3 It is given that dx ln 2. 2x = Ôa −5 Find the value of the positive constant a. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Integrate to obtain ln(2 5) k x − *M1 For non-zero constant k Apply limits to obtain 7 2 ln(6 5) ln(2 5) ln a a − − − = A1 Apply subtraction law for logarithms *M1 OE Obtain equation 6 5 7 2 5 2 a a − = − A1 OE without logarithms Solve equation for a DM1 Obtain 25 2 a = A1 6

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5 (a) Sketch, on the same diagram, the graphs of y x 2k and y 2x , where k is a positive constant. = + = −3k Give, in terms of k, the coordinates of the points where each graph meets the axes. [3] (b) Find, in terms of k, the coordinates of each of the two points where the graphs intersect. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find, in terms of k, the largest value of t satisfying the inequality 2t . [2] 2k + ≥ 2t+1 −3k ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Draw two V-shaped graphs with one vertex on negative x-axis and one vertex on positive x-axis M1 Draw correct graphs related correctly to each other A1 State correct coordinates 3 2 2 , 2 , , 3 k k k k − A1 Either given on axes or stated separately 3 5(b) State or imply non-modulus equation 2 2 ( 2 ) (2 3 ) x k x k + = − or pair of linear equations B1 Attempt solution of 3-term quadratic equation or pair of linear equations M1 Obtain 1 3 , 5 x k x k = = A1 Obtain 7 3 , 7 y k y k = = A1 If A0A0, award A1 for one pair of correct coordinates 4 5(c) Relate 2t to larger value of x from part (b) M1 Apply logarithms to obtain ln(5 ) ln2 k t = A1 OE such as 10 10 log (5 ) log 2 k or 2 log (5 ) k 2

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6 A curve has equation y x3e0.2x where x At the point P on the curve, the gradient of the curve is 15. = ≥0. _ 75e−0.2x (a) Show that the x-coordinate of P satisfies the equation x . [4] 15 x = + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use the equation in part (a) to show by calculation that the x-coordinate of P lies between 1.7 and 1.8. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) Differentiate using the product rule *M1 Obtain 2 0.2 3 0.2 3 e 0.2 e x x x x + A1 OE Equate first derivative to 15 and rearrange to ... x = DM1 Confirm 0.2 75e 15 x x x − = + A1 AG – necessary detail needed 4 6(b) Consider sign of 0.2 75e 15 x x x − − + or equivalent for 1.7 and 1.8 M1 Obtain 0.08... − and 0.03... or equivalents and justify conclusion A1 2 6(c) Use iterative process correctly at least once M1 Answer required to exactly 4 sf Obtain final answer 1.771 A1 Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [1.7705, 1.7715] A1 3

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7 y A x O The diagram shows part of the curve with equation y 4 sin2x 8 sin x 3, = + + where x is measured in radians. The curve crosses the x-axis at the point A and the shaded region is bounded by the curve and the lines x 0 and y 0. = = (a) Find the exact x-coordinate of A. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact gradient of the curve at A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the exact area of the shaded region. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Solve equation 0 y = to find value of x M1 Obtain 7 6 π A1 2 7(b) Attempt first derivative using chain rule M1 OE Obtain d 8sin cos 8cos d y x x x x = + A1 OE Substitute value from part (a) to find gradient 2 3 − A1 Or exact equivalent 3 7(c) Express integrand in the form 1 2 3 cos2 sin k k x k x + + *M1 Obtain correct 5 2cos2 8sin x x − + A1 OE. Allow unsimplified Integrate to obtain 5 sin2 8cos x x x − − A1 Apply limits 0 and their value from part (a) correctly DM1 Obtain 35 7 3 8 6 2 π + + or exact equivalent A1 5

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