Cambridge A Level Mathematics 9709 — 2019 May/June Paper 2 · Variant 1

2 (i) Solve the inequality 3x x 3 . [4] −5 < + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence find the greatest integer n satisfying the inequality 30.1n 3 . [2] 30.1n+1 −5 < + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) State or imply non-modular inequality 2 2 (3 5) ( 3) x x − < + or corresponding equation or pair of different linear equations/inequalities B1 SC: Allow B1 for 4 from only one linear inequality Attempt solution of 3-term quadratic equation/inequality or of two different linear equations/inequalities M1 For M1, must get as far as 2 critical values Obtain critical values 1 2 and 4 A1 State answer 1 2 4 x < < or equivalent A1 If given as 2 separate statements, condone omission of ‘and’ or ∩ but penalise inclusion of ‘or’ or ∪ 4 Question Answer Marks Guidance 2(ii) Attempt to find n (not necessarily an integer so far) from 0.1 3 n = or < their positive upper value from part (i) or 0.1 1 3 n+ = or < 3 × their positive upper value from part (i) M1 0/2 for trial and improvement Conclude 12 A1 2

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4 (a) Find tan2 3x dx. [3] Ó ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 e3x 4 (b) Find the exact value dx. Show all necessary working. [4] + ex ofÔ0 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Use identity 2 2 tan 3 sec 3 1 x x = − B1 Integrate to obtain form 1 2 tan3 k x k x + M1 Obtain correct 1 3 tan3x x c − + A1 3 4(b) Express integrand as 2 e 4e x x − + B1 Integrate to obtain form 2 3 4 e e x x k k − + M1 Obtain correct 2 1 2 e 4e x x − − A1 Use limits to obtain 2 1 7 1 2 2 e 4e− − + or similarly simplified equivalent A1 4

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7 (i) Show that 2 cosec cot [3] 21 1  cosec21. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence show that tan 4. [2] cosec215Å 15Å = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Solve the equation 2 cosec cot 1 cosec 1 12 for Show all necessary working. & 2& + 2& = −360Å < & < 360Å. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) State or imply 1 cosec2 2sin cos θ θ θ = B1 Attempt to express left-hand side in terms of sinθ and cosθ only M1 Simplify to confirm 2 cosec θ AG A1 3 7(ii) Use identity to express left-hand side in terms of sin30 or cosec30 M1 Obtain 2 sin30 or 2cosec30 and confirm 4 AG A1 2 7(iii) Solve quadratic equation of the form 2 cosec cosec 12 0 2 2 k φ φ + − = or *M1 Allow sign errors 2 12sin sin 0 2 2 k φ φ − − = correctly for 1 2 cosec φ or 1 2 sin φ to find two values of 1 2 sin φ or 1 2 cosec φ Obtain 1 1 1 2 4 3 sin , φ = − A1 Use correct process to find at least one correct value of φ from 1 1 1 2 4 3 sin , φ = ± ± DM1 Allow for any rounded or truncated value Obtain any two of –331.0, –29.0, 38.9, 321.1 A1 Allow greater accuracy Obtain all four values and no others between –360 and 360 A1 Allow greater accuracy 5

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