Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 2 · Variant 2

3 y A B x O 2 The diagram shows the curve with 2x. The points on the curve with equation y x-coordinates 0 and 2 are denoted and =The6e−1shaded region is enclosed by the curve, the line by A B respectively. through A parallel to the x-axis and the line through B parallel to the y-axis. (a) Find the exact gradient of the curve at B. 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(b) Find the exact area of the shaded region. 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Mark scheme: 3(a) − 12 x M1 For any non-zero k except 6. Differentiate to obtain form ke 1 3 A1 Substitute x = 2 to obtain − 3e− or − e 2 3(b) 2 x B1 OE Integrate to obtain −12e − 1 x M1 For any non-zero k except 6 1 Use limits 0 and 2 correctly to an integral of the form ke −, 2 retaining − or equivalent perhaps involving integration of 6 − 6e 2 x . exactness 1 12 A1 Subtract from 12 to obtain final answer 12e− or e 3

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4 (a) Sketch, on the same diagram, the graphs of y 3 and y 9 [2] = −x = −2x. (b) Solve the inequality 3 9 [3] −x > −2x. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use logarithms to solve the inequality 500. Give your answer in the form x a, where 23x−10 <figures. < [3] the value of a is given correct to 3 significant ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) List the integers that satisfy both of the inequalities 3 9 and 500. [1] −x > −2x 23x−10 < ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 9 − 2 x with steeper negative B1 Dependent on first B mark, appropriately positioned gradient with respect to first graph. 2 4(b) Solve linear equation or inequality with signs of x and 2x different M1 Obtain critical value 4 A1 Conclude x  4 only A1 6 must be discounted. Alternative Method for Question 4(b) State or imply non-modulus equation (or inequality) (3 − x ) 2 = (9 − 2 x ) 2 B1 Attempt solution of three-term quadratic equation (or inequality) M1 Dependent on previous B1. Conclude x  4 only A1 6 must be discounted. 3 4(c) State or imply (3 x − 10)ln2  ln500 B1 Or equivalent perhaps involving different logarithm base. Obtain critical value 6.32 B1 Obtain x  6.32 B1 Or greater accuracy. 3 4(d) State 5 and 6 only B1 1

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5 (a) Find the quotient when 6x3 is divided by 2x 1 , and show that the remainder is 6. −5x2 −24x −4 + [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find 7 6x3 dx, −5x2 −24x −4 2x 1 Ô2 + giving your answer in the form a ln b, where a and b are integers. 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Mark scheme: 5(a) Carry out division at least as far as 3x 2 + k1 x M1 OE (e.g. by inspection). Obtain quotient 3 x 2 − 4 x − 10 A1 Confirm given result of remainder is 6 with sufficient detail A1 AG SC If remainder = 6 shown using remainder theorem allow B1. 3 5(b) Integrate to obtain at least 3x and term of form k 2 ln(2 x + 1) *M1 ln term must be added. Obtain x 3 − 2 x 2 − 10 x + 3ln(2 x + 1) A1 Apply limits correctly to expression with four terms DM1 Apply appropriate logarithm properties correctly to obtain the form k3 ln a DM1 Obtain 195 + ln27 A1 5

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6 y B A x O The diagram shows the curve with parametric equations x 3 ln 2t , y 4t ln t. = −3 = The curve crosses the y-axis at the point A. At the point B, the gradient of the curve is 12. (a) Find the exact gradient of the curve at A. 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(b) Show that the value of the parameter t at B satisfies the equation 9 3 t [2] 1 ln t 2. = + + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use an iterative formula, based on the equation in (b), to find the value of t at B, giving your answer correct to 3 significant figures. Use an initial value of 5 and give the result of each iteration to 5 significant figures. 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Mark scheme: 6(a) dx 6 B1 Obtain = dt 2t − 3 d y M1 Allow unsimplified. Use product rule to find d t dy (4ln t + 4)(2t − 3) A1 OE Obtain = dx 6 Attempt to find t corresponding to point A using a complete and correct M1 method Obtain t = 2 and hence gradient is 23 ln 2 + 23 A1 Or exact equivalent. 5 6(b) d y k M1 Equate to 12 and attempt rearrangement to 2t −=3 d x 4ln t + 4 9 3 A1 AG Confirm given result t = + with sufficient detail 1 + ln t 2 2 6(c) Use iteration process correctly at least once M1 Need to see 4.9626 . Obtain final answer 4.96 A1 Answer required to exactly 3 s.f. Show sufficient iterations to 5 sf to justify answer or show sign change in A1 interval [4.955, 4.965] 3

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