Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 2 · Variant 1

1 Solve the inequality 2x x. [4] −5 > ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Solve 2 x −=5 x to obtain x = 5 B1 Attempt solution of linear equation where signs of 2x and x are different M1 Obtain x = 5 A1 3 Conclude x  53, x  5 A1 Must be 2 separate inequalities.  5  5,  ) . Allow equivalents  −,   (  3  Alternative method for question 1 State or imply non-modulus equation (2 x − 5) 2 = x 2 B1 Attempt solution of 3-term quadratic equation M1 Obtain 5 and 5 A1 3 Conclude x  53, x  5 A1 Must be 2 separate inequalities.  5  5,  ) . Allow equivalents  −,   (  3  4

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2 Use logarithms to solve the equation giving your answer correct to 3 significant figures. 14e−2x = 5x+1, [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Apply logarithms correctly to both sides and apply power law at least once *M1 Obtain ln14 − 2 x = ( x + 1)ln5 A1 OE with x no longer part of a power. Attempt solution of linear equation DM1 Must have ln14 − ln5 = x ( 2 + ln5 ) . Obtain 0.285 A1 4

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4 (a) By sketching a suitable pair of graphs on the same diagram, show that the equation 2x x5 e−1 = has exactly one real root. [2] 5? 2xn to determine the root correct to 4 significant figures. Give (b) Use the iterative formula the result of each iterationxn+1to 6=significante−1 figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) − 12 x B1 with some curve in second quadrant as well as first. Draw approximately correct sketch of y = e Draw approximately correct sketch of y = x 5 and confirm one root B1 with some curve in third quadrant as well as first. Alternative method for question 4(a) Draw approximately correct sketch of y = 5ln x or y = ln x 5 B1 x B1 Must have intersection in the 4th quadrant. Draw approximately correct sketch of y = − and confirm one root 2 2 4(b) Use iteration process correctly at least once M1 Obtain final answer 0.9128 A1 answer required to exactly 4 s.f. Show sufficient iterations to 6 s.f. to justify answer or show sign change in A1 the interval [0.91275, 0.91285] 3

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7 y B A x O 3 2 ln x The diagram shows the curve with equation y 3x 1. The curve crosses the x-axis at the point A = and has a maximum point B. The shaded region is bounded+ by the curve and the lines x 3 and y 0. = = (a) Find the gradient of the curve at A. 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(b) Show by calculation that the x-coordinate of B lies between 3.0 and 3.1. 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(c) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places. 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Mark scheme: 7(a) Differentiate using quotient rule (or product rule) M1 2 A1 OE (3 x + 1) − 6ln x x Obtain (3 x + 1) 2 Substitute x = 1 to obtain 1 A1 OE 2 3 7(b) Equate numerator of first derivative to zero M1 May be implied. 2 M1 OE Consider sign of (3 x + 1) − 6ln x for 3.0 and 3.1 x Obtain 0.074… and –0.14… or equivalents and justify conclusion A1 AG – necessary detail needed. 0.00075 and – 0.001275 . 3 7(c) Use y-values [0], 2 ln2 or 0.1980 and 2 ln3 or 0.2197 B1 7 10 Use correct formula, or equivalent, with h = 1 M1 Obtain 0.31 A1 3

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