Cambridge A Level Mathematics 9709 — 2013 May/June Paper 2 · Variant 2

2 Solve the inequality x −8 > 2x −4 . [4]

Mark scheme: 2 Either State or imply non-modular inequality (x – 8)2 > (2x – 4)2, or corresponding equation or pair of linear equations M1 Make reasonable solution attempt at a quadratic, or solve two linear equations M1 Obtain critical values 4 and – 4 A1 State correct answer – 4 < x < 4 A1 Or Obtain one critical value, e.g. x = 4, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 Obtain the other critical value similarly B2 State correct answer – 4 < x < 4 B1 [4]

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4 The variables x and y satisfy the equation 5y+1 = 23x. (i) By taking logarithms, show that the graph of y against x is a straight line. [2] (ii) Find the exact value of the gradient of this line and state the coordinates of the point at which the line cuts the y-axis. [2]

Mark scheme: 4 (i) State or imply (y + 1) log 5 = 3x log 2 M1 State that this is of the form ay = bx + c and thus a straight line, or equivalent A1 [2] 3 ln 2 (ii) State gradient is , or equivalent, e.g. 3log52 B1 ln 5 State (0, –1) B1 [2] d y

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5 The equation of a curve is x2 −2x2y + 3y = 9. dy 2x −4xy (i) Show that = . [4] dx 2x2 −3 (ii) Find the equation of the normal to the curve at the point where x = 2, giving your answer in the form ax + by + c = 0. [4]

Mark scheme: d y 5 (i) State 3 as derivative of 3y, or equivalent B1 d x d y State 4xy + 2x2 as a derivative of 2x2y, or equivalent B1 d x d y Equate derivative of LHS to zero and solve for M1 d x Obtain given answer correctly A1 [4] (ii) Substitute x = 2 into given equation and solve for y M1 12 Obtain gradient = correctly A1 5 d y Form equation of the normal at their point, using negative recip of their M1 d x State correct equation of normal 5x + 12y + 2 = 0 or equivalent A1 [4]

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6 (i) By sketching a suitable pair of graphs, show that the equation 3ex = 8 −2x has only one root. [2] (ii) Verify by calculation that this root lies between x = 0.7 and x = 0.8. [2] (iii) Show that this root also satisfies the equation @8 −2x A x = ln . 3 @8 −2xn A to determine this root correct to 3 decimal places. (iv) Use the iterative formula xn+1 = ln 3 Give the result of each iteration to 5 decimal places. [3]

Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = 3ex or y = 8 – 2x B1 Sketch a second relevant graph and justify the given statement B1 [2] GCE AS LEVEL – May/June 2013 9709 22 (ii) Consider sign of 3ex – 8 + 2x at x = 0.7 and x = 0.8, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (f (0.7) = –0.559, f (0.8) = 0.277 or equivalent)  8 − 2 x  (iii) Show that given equation is equivalent to x = ln   , or vice versa B1 [1]  3  (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.768 A1 Show sufficient iterations to justify its accuracy to 3 d.p. xo = 0.7 xo = 0.75 xo = 0.8 0.78846 0.77319 0.75769 0.76129 0.76603 0.77082 0.76971 0.76825 0.76676 0.76711 0.76756 0.76802 0.76791 0.76763 0.76766 or show there is a sign change in the interval (0.7675, 0.7685) B1 [3] 2 1

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7 (a) Find the exact area of the region bounded by the curve y = 1 + e2x−1, the x-axis and the lines x = 1 and x = 2. [4] 2 (b) y M 1 x O 2p e2x The diagram shows the curve y = for 0 < x < 120, and its minimum point M. Find the sin 2x exact x-coordinate of M. [5]

Mark scheme: 7 (a) Obtain one term of form ke2x–1 with any non-zero k M1 1 Obtain correct integral x + e2x–1 A1 2 Substitute limits, giving exact values M1 1 Correct answer e3 + 1 A1 [4] 2 (b) Use product or quotient rule M1* Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1* dep Obtain tan 2x = 1 A1 π Obtain x = A1 [5] 8

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