Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 2 · Variant 2

1 Solve the equation 7 cot 3 cosec for [3] 1 = 1 0Å < 1 < 90Å. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Use cos cot sin θ θ θ = and 1 cosec sin θ θ = B1 SOI Simplify to obtain cos k θ = where 0 1 k < < M1 Obtain 3 cos 7 θ = and hence 64.6 θ = and no other solutions in the range A1 Alternative method for question 1 Use identity 2 2 cosec 1 cot = + θ θ B1 Simplify to obtain 1 tan k θ = or 2 sin k θ = where 2 0 1 k < < M1 Obtain 1 tan 40 3 θ = or 1 sin 40 7 θ = and hence 64.6 θ = and no other solutions in the range A1 3

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8 23x+22 Given that 5, find the value of 23x and hence, using logarithms, find the value of x correct + 23x = to 4 significant figures.−7 [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Use 3 2 3 2 4 2 x x × Solve equation for 3 2 x M1 Obtain 3 2 43 x = A1 Apply logarithms and use power law for 3 2 x k = where 0 k > M1 Obtain 1.809 A1 AWRT 5

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. . 3 (a) Sketch, on a single diagram, the graphs of y 12x and y 32x 2a, where a is a positive constant. = −a = −1 [2] (b) Find the coordinates of the point of intersection of the two graphs. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ . . (c) Deduce the solution of the inequality 12x 32x 2a. [1] −a > −1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw straight line graph correctly positioned with greater gradient B1 2 3(b) Solve linear equation with signs of 1 2 x and 3 2 x different or solve non-modulus equation 2 2 1 3 1 2 2 2 x a x a     − = −         to obtain x = M1 Obtain 3 4 x a = A1 Obtain 5 8 y a = A1 And no other point 3 Question Answer Marks Guidance 3(c) State 3 4 x a < B1 FT Following their (single) x-coordinate from part (b) 1

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4 y x O 14 x The diagram shows the curve with equation y The shaded region is bounded by the curve −2 = x2 8. and the lines x 14 and y 0. + = = dy (a) Find and hence determine the exact x-coordinates of the stationary points. [4] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give the answer correct to 2 significant figures. 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Mark scheme: 4(a) Differentiate using quotient rule (or product rule) *M1 Obtain 2 2 2 ( 8) 2 ( 2) ( 8) x x x x + − − + A1 OE Equate first derivative to zero and attempt solution to get x = … DM1 Obtain 2 12 ± or exact equivalents A1 4 4(b) Use y values 4 8 12 (0), , , 44 108 204 or decimal equivalents B1 Decimal equivalents need to be to at least 2 decimal places Use correct formula, or equivalent, with 4 h = M1 Obtain 4 8 12 2 0 2 2 44 108 204   + × + × +     or equivalent and hence 0.78 A1 3

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