Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 2 · Variant 2

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

Mark scheme: 1 Either Square both sides obtaining 3 terms on each side M1 Solve 3-term quadratic equation M1 Obtain − 54 and 6 A1 [3] Or Obtain value 6 from graphical method, inspection, linear equation, … B1 Obtain value − 54 similarly B2 [3] 3

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4 (a) Find the value of x satisfying the equation 2 ln x x ln 2. [5] −4 −ln = (b) Use logarithms to find the smallest integer satisfying the inequality 1.4y 1010. >

Mark scheme: 4 (a) Use power law to produce ln ( x − 4 )2 B1 Apply logarithm laws to produce equation without logarithms M1 Obtain ( x − 4 )2 = 2 x or equivalent A1 Solve 3-term quadratic equation DM1 Obtain (finally) x = 8 only A1 [5] (b) Apply logarithms and use power law (once) M1 ln 1010 Obtain or equivalent as part of inequality or equation A1 ln 4.1 Conclude with single integer 69 A1 [3]

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5 y M x O The diagram shows part of the curve y 2 cos x 2x = −cos and its maximum point M. The shaded region is bounded by the curve, the axes and the line through M parallel to the y-axis. (i) Find the exact value of the x-coordinate of M. [4] (ii) Find the exact value of the area of the shaded region. [4]

Mark scheme: 5 (i) Differentiate to obtain − 2 sin x + 2 sin 2 x or equivalent B1 Use sin 2 x = 2 sin x cos x or equivalent B1 Equate first derivative to zero and solve for x M1 Obtain 13 π A1 [4] (ii) Integrate to obtain form k1 sin x + k 2 sin 2 x M1 Obtain correct 2 sin x − 12 sin 2 x A1 Apply limits 0 and their answer from part (i) M1 Obtain 34 3 or exact equivalent A1 [4]

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