Cambridge A Level Mathematics 9709 — 2021 May/June Paper 2 · Variant 1

1 Solve the inequality 3x 4x 5 . [4] −7 < + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State or imply non-modulus inequality 2 2 (3 7) (4 5) − < + x x or corresponding equation or pair of linear equations B1 Attempt solution of 3-term quadratic equation/inequality or of two linear equations M1 Obtain critical values 12 − and 2 7 A1 May be seen in a number line. State answer 2 12, 7 < − > x x or ( ) 2 . 12 , 7   −∞− ∪ ∞     or ( ) 2 . 12 , , 7   −∞− ∞     A1 OE 2 12 7 − > > x or similar would get A0 Mark the final answer. 4

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4 A curve has parametric equations x ln 2t 6 t, y t ln t. = + −ln = (a) Find the value of t at the point P on the curve for which x ln 4. 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(b) Find the exact gradient of the curve at P. 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Mark scheme: 4(a) Equate x to ln4 and use relevant logarithm property M1 Obtain equation with no logarithm present, 2 6 4 + = t t A1 OE Obtain 3 = t A1 3 4(b) Obtain d 2 1 d 2 6 = − + x t t t B1 Use product rule to find d d y t M1 Obtain 1 ln + × t t t A1 Divide to obtain d d y x using their d d y t and d d x t correctly DM1 Must have at least B1 or M1. Do not condone incorrect inverting of terms unless a correct statement is seen initially. Obtain 6(ln3 1) − + A1 or exact equivalents 5

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5 y M x O 3x 2 The diagram shows the curve with equation y . The curve has a minimum point M. + ln x = dy 3x 2 (a) Find an expression for and show that the x-coordinate of M satisfies the equation x . + dx 3 ln x = [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use the equation in part (a) to show by calculation that the x-coordinate of M lies between 3 and 4. 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(c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 5 significant figures. Give the result of each iteration to 7 significant figures. 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Mark scheme: 5(a) Use quotient rule (or equivalent) to find first derivative M1 Obtain 2 1 3ln (3 2) d d (ln ) − + = x x y x x x A1 OE Equate first derivative to zero and confirm 3 2 3ln + = x x x A1 AG 3 5(b) Consider 3 2 3ln + −x x x or equivalent for values 3 and 4 M1 M0 if using d d y x Obtain 0.33... − and 0.63... or equivalents and justify conclusion A1 AG 2 5(c) Use iteration process correctly at least once M1 Obtain final answer 3.3223 A1 Answer required to exactly 5 s.f. Show sufficient iterations to 7 s.f. to justify answer or show sign change in the interval [3.32225, 3.32235] A1 3

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4 6 6 (a) Use the trapezium rule with three intervals to find an approximation to dx. Give your 1 x Ô1 + answer correct to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 4 1 (b) Find the exact value of Ó 1 2e 2x−2 dx. 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(c) y y = 2e 2x−21 6 y = 1 + x x O 1 4 6 1 The diagram shows the curves y and y 2e 2x−2 which meet at a point with x-coordinate 4. 1 x = = The shaded region is bounded by the+ two curves and the line x 1. = Use your answers to parts (a) and (b) to find an approximation to the area of the shaded region. Give your answer correct to 3 significant figures. 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(d) State, with a reason, whether your answer to part (c) is an over-estimate or under-estimate of the exact area of the shaded region. 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Mark scheme: 6(a) Use y-values 6 6 3, , , 2 1 2 1 3 + + or 3, 6 2 6, 3 3 3, 2 − − or 3, 2.4853..., 2.19615..., 2 B1 Use correct formula, OE, with 1 = h M1 Allow 3 separate trapezia of width 1. Obtain 7.1814 A1 AWRT 3 6(b) Integrate to obtain form 1 2 2 e − x k M1 1 ≠ k If 2 = k , integration must be implied by use of square bracket notation or by substitution of limits. Obtain correct 1 2 2 4e − x A1 or exact equivalents Obtain 3 2 4 4e − − A1 or exact equivalents 3 6(c) Evaluate answer to part (a) minus answer to part (b) M1 Obtain 4.07 A1 2 6(d) State over-estimate with reference to top of each trapezium being above the [first] curve, or clear equivalent, e.g. concave up so over- estimate or convex down so over-estimate. B1 1

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