Cambridge A Level Mathematics 9709 — 2016 Feb/March Paper 2 · Variant 2

2 Solve the inequality x 2x 3 . [4] −5 < +

Mark scheme: 2 Either State or imply non-modular inequality ( x − 5 ) 2 < ( 2 x + 3 ) 2 or corresponding pair of linear equations B1 Attempt solution of 3-term quadratic equation or of 2 linear equations M1 2 Obtain critical values −8 and A1 3 2 State answer x < − 8, x > A1 3 Or Obtain critical value −8 from graphical method, inspection, equation B1 2 Obtain critical value similarly B2 3 2 State answer x < − 8, x > B1 [4] 3

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4 The sequence of values given by the iterative formula , / n ! 12x2n + 4x−3 xn+1 = with initial value x1 1.5, converges to = !. (i) Use this iterative formula to find correct to 3 decimal places. Give the result of each iteration ! to 5 decimal places. [3] (ii) State an equation that is satisfied by and hence find the exact value of [2] ! !.

Mark scheme: 4 (i) Use the iterative formula correctly at least once M1 Obtain final answer 1.516 A1 Show sufficient iterations to justify accuracy to 3 dp or show sign change in interval (1.5155,1.5165) B1 [3] (ii) State equation x = 12 x 2 + 4 x −3 or equivalent B1 Obtain exact value 5 8 or 80.2 B1 [2]

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a 5 Given that dx 65, find the value of a correct to 3 decimal places. [5] = Ó0 6e2x+1

Mark scheme: 5 Obtain integral of form ke 2 x +1 M1 Obtain correct 3e 2 x +1 A1 Apply both limits correctly and rearrange at least to e 2 a +1 = ... M1 Use logarithms correctly to find a M1 Obtain 1.097 A1 [5]

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6 y M x O 120 The diagram shows the part of the curve y sin 2x for 0 and the stationary point M. = 3e−x ≤x ≤120, (i) Find the equation of the tangent to the curve at the origin. [4] (ii) Find the coordinates of M, giving each coordinate correct to 3 decimal places. [4]

Mark scheme: 6 (i) Use product rule to obtain expression of form k1e − x sin 2 x + k 2 e − x cos2 x M1 Obtain correct −3e − x sin2 x + 6e − x cos2 x A1 Substitute x = 0 in first derivative to obtain equation of form y = mx M1 Obtain y = 6 x or equivalent with no errors in solution A1 [4] (ii) Equate first derivative to zero and obtain tan2 x = k M1* Carry out correct process to find value of x dep M1* Obtain x = 0.554 A1 Obtain y = 1.543 A1 [4] dy

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7 The equation of a curve is 2x3 y3 24. + = dy (i) Express in terms of x and y, and show that the gradient of the curve is never positive. [4] dx (ii) Find the coordinates of the two points on the curve at which the gradient is [5] −2.

Mark scheme: dy 27 (i) State 3 y as derivative of 3y B1 dx d y Equate derivative of left-hand side to zero and solve for M1 d x d y 6 x 2 Obtain = − 2 or equivalent A1 d x 3 y Observe x 2 and y 2 never negative and conclude appropriately A1 [4] (ii) Equate first derivative to − 2 and rearrange to y 2 = x 2 or equivalent B1 Substitute in original equation to obtain at least one equation in 3x or 3y M1 Obtain 3 x 3 = 24 or x 3 = 24 or 3 y 3 = 24 or − y 3 = 24 A1 Obtain (2, 2) A1 Obtain ( 3 24, − 3 24) or (2.88, − 2.88) and no others A1 [5] cos x

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8 (i) Show that sin 2x cotx 2 cos2x. [2]  (ii) Using the identity in part (i), (a) find the least possible value of 3 sin 2x cotx 5 cos 2x 8 + + as x varies, [4] 160 (b) find the exact value of cosec 4x tan 2x dx. [5] Ó 1 80

Mark scheme: cos x 8 (i) State 2sin x cos x . B1 sin x Simplify to confirm 2cos 2 x B1 [2] (ii) (a) Use cos2 x = 2cos 2 x − 1 B1 Express in terms of cos x M1 Obtain 16cos 2 x + 3 or equivalent A1 State 3, following their expression of form a cos 2 x + b A1 [4] 1 2 (b) Obtain integrand as sec 2 x B1 2 Integrate to obtain form k tan2 x M1* 1 Obtain correct tan 2 x A1 4 Apply limits correctly dep M1* 1 1 Obtain 3 − or exact equivalent A1 [5] 4 4

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