Cambridge A Level Mathematics 9709 — 2022 Feb/March Paper 2 · Variant 2

1 Solve the equation 5x 4x 9 . [3] −2 = + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Solve 5 2 4 9 − = + x x to obtain 11 = x Attempt solution of linear equation where signs of 5x and 4x are different M1 Obtain final value 7 9 = − x A1 Alternative method for question 1 State or imply non-modulus equation 2 2 (5 2) (4 9) − = + x x B1 Attempt solution of 3-term quadratic equation M1 Obtain x = 7 9 − and x =11 A1 3

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3 The variables x and y satisfy the equation y 32aax, where a is a constant. The graph of ln y against x is a straight line with gradient 0.239. = (a) Find the value of a correct to 3 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the value of x when y 36. Give your answer correct to 3 significant figures. [2] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) State or imply equation is 2 ln ln3 ln = + a y x a B1 Equate gradient of line involving a to 0.239 M1 Obtain ln 0.239 = a and hence 1.27 = a A1 3 3(b) Substitute 36 = y in ln ... = y equation and solve for x M1 Or substitute in original equation with necessary manipulation Obtain 3.32 A1 2

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4 (a) Show that sin cot 1. [3] 21 1 −cos 21  ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the exact value of sin 1 cot 1 [2] 6π 12π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the smallest positive value of (in radians) satisfying the equation 1 sin cot cos 1. [2] 21 1 −3 21 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Use at least two of sin 2 2sin cos θ θ θ = , 2 2 cos2 cos sin θ θ θ = − , cos cot sin θ θ θ = B1 OE Express LHS in terms of sinθ and cosθ only and attempt valid simplification M1 Obtain 2 2 cos sin θ θ + or equivalent and hence 1 A1 AG – necessary detail needed 3 4(b) Substitute 1 12 θ = π and show or imply 1 1 1 6 12 6 sin cot 1 cos π π = + π M1 Obtain 1 2 1 3 + or exact equivalent A1 2 Question Answer Marks Guidance 4(c) Use the identity from part (a) to obtain 2cos2 0 θ − = or equivalent M1 Or alternative starting again, using valid simplification and reaching single trigonometric function. Obtain 1 4 θ = π A1 2

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