Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 2 · Variant 1

1 (i) Solve the equation 9x 3x 2 . [3] −2 = + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence, using logarithms, solve the equation 2 , giving your answer correct to 3 significant figures. 3y+2 −2 = 3y+1 + [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) State or imply non-modular equation 2 2 (9 2) (3 2) x x − = + or pair of linear equations B1 Attempt solution of quadratic equation or of 2 linear equations M1 Obtain 0 and 2 3 A1 SC: B1 for one correct solution 3 1(ii) Apply logarithms and use power law for 3y k = where 0 k > M1 Must be using their answers to part (i) Obtain 0.369 − A1 2

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7 6 2 Show dx ln 125. [5] 2x 1 = thatÔ1 + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Integrate to obtain form ln(2 1) k M1 Obtain correct 3ln(2 1) x + A1 Use subtraction law of logarithms correctly M1 Dependent on first M1 Use power law of logarithms correctly M1 Dependent on first M1 Confirm ln125 A1 5

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3 Solve the equation 3 cosec for [5] sec21 = 1 0Å < 1 < 180Å. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 State 2 1 3 sin cos θ θ = or 2 3 1 tan sin θ θ + = B1 Produce quadratic equation in sinθ M1 Dependent on B1 Solve 3-term quadratic equation to find value between –1 and 1 for sinθ M1 Dependent on first M1 Obtain 1 6 sin ( 1 37) θ = −+ and hence 57.9 A1 Obtain 122.1 and no others between 0 and 180 A1 5

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5 A curve has parametric equations x t ln t 1 , y 3te2t. = + + = (i) Find the equation of the tangent to the curve at the origin. 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(ii) Find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places. 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Mark scheme: 5(i) Use product rule to differentiate y obtaining 2 2 1 2 e e t t k k t + M1 Obtain correct 2 2 3e 6 e t t t + A1 State derivative of x is 1 1 1 t + + B1 Use d d d / d d d y y x x t t = with 0 t = to find gradient M1 Obtain 3 2 y x = or equivalent A1 5 Question Answer Marks Guidance 5(ii) Equate d d y x or d dt y to zero and solve for t M1 Allow full marks if correct solution is obtained but d d x t is incorrect Obtain 1 2 t = − A1 Obtain 1.19 x = − A1 Obtain 0.55 y = − A1 4

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6 y R x 0 O The diagram shows the curve with equation 1 for 0 y The region R is 3 cos2 12x ! = / + ≤x ≤0. bounded by the curve, the axes and the line x = 0. (i) Use the trapezium rule with two intervals to find an approximation to the area of R, giving your answer correct to 3 significant figures. 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(ii) The region R is rotated completely about the x-axis. Without using a calculator, find the exact volume of the solid produced. 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Mark scheme: 6(i) Use y values 2, 2.5 , 1 or equivalents B1 Use correct formula, or equivalent, with 1 2 h π = and three y values M1 Obtain 1 1 2 2 (2 2 2.5 1) π × + + or equivalent and hence 4.84 A1 3 Question Answer Marks Guidance 6(ii) State or imply volume is 2 1 2 (1 3cos ) d x x π + ∫ B1 Allow if π appears later; condone omission of dx Use appropriate identity to express integrand in form 1 2 cos k k x + M1 Obtain 5 3 2 2 ( cos ) d x x π + ∫ or 5 3 2 2 ( cos ) d x x + ∫ A1 Condone omission of dx Integrate to obtain 5 3 2 2 ( sin ) x x π + or 5 3 2 2 sin x x + A1 Obtain 2 5 2 π with no errors seen A1 5

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