Mathematics 9709/23 October/November 2021: question paper and mark scheme

Mark scheme

1 Integrate to obtain 2e 2 x B1 Integrate to obtain 2e− x B1 Apply limits correctly to integral of the form k1e2 x + k2e− x M1 k1 ≠ 4 . Condone one error. Obtain 2e 4 − 2e A1 or exact equivalent. 4

Mark scheme

2(a) Draw V-shaped graph with vertex on positive x-axis B1 Must be straight lines. Draw straight line through origin with positive gradient greater than gradient B1 Must have the first B1. of first graph, together with a V shaped graph 2 2(b) Solve linear equation with signs of 3x and x different or solve non-modulus M1 equation (3x) 2 = ( x − 3)2 Obtain x = 34 A1 Obtain y = 94 A1 And no other point. 3 2(c) State x < 34 B1 FT Following their (single) x-coordinate from part (b). 1

Mark scheme

3 State or imply equation is y ln a = ln k + ln x B1 1 M1 Or eliminate lnk from simultaneous equations. Equate gradient of line to ln a 1 2.64 A1 AWRT Obtain = or equivalent and hence a = 1.8 ln a 1.55 Substitute appropriate values to find lnk M1 Obtain ln k = 2.7... and hence k = 15 A1 AWRT Alternative method for question 3 a 6.36 = ke1.03 and a 9 = ke 2.58 B1 For both. ( Elimination of k to obtain an equation in a only a 2.64 = e1.55 ) M1 Must have previous B1. Use of a correct method to obtain a M1 Allow for a = e0.59 . a = 1.8 A1 k = 15 A1 5

Mark scheme

4(a) Attempt use of product rule to differentiate xe 2 x M1 Obtain e 2 x + 2 xe 2 x − 5e − x A1 Equate first derivative to zero and multiply by e x to obtain an equation M1 involving e3 x Obtain e3x (1 + 2 x) = 5 or equivalent A1 Confirm given result x = 13 ln 5 − 13 ln(1 + 2 x) with sufficient detail A1 AG 5 4(b) Use iteration process correctly at least once M1 Need 0.35 and 2 correct values. Obtain final answer 0.357 A1 Answer required to exactly 3sf. Allow recovery. Show sufficient iterations to 5sf to justify answer or show sign change in A1 interval [0.3565, 0.3575] 3

Mark scheme

5(a) dx 2 B1 Obtain = dt 2t + 3 dy M1 Use quotient rule, or equivalent, to find dt dy 2(2t + 3) − 2(2t − 3) A1 OE Obtain = dt (2t + 3) 2 dy 6 A1 AG Divide to confirm = dx 2t + 3 4 5(b) Attempt to find value of t corresponding to x = 0 M1 Obtain t = −1 and hence gradient is 6 A1 2 5(c) Attempt to find value of t corresponding to y = 0 M1 Obtain t = 32 and hence gradient is 1 A1 2

Mark scheme

6(a) Substitute x = −3 , equate to zero and attempt solution for a M1 Allow attempt at synthetic division, must be a complete method. Allow one sign error carried through. Allow attempt at algebraic long division, must be complete with the remainder equated to zero. Obtain a = 13 A1 2 6(b) Substitute x = 2 , equate to 40 and attempt solution for b M1 Obtain b = 9 A1 2 6(c) Identify x + 3 as factor of f ( x) − g( x) B1 May be implied by synthetic division. If working backwards from solutions from a calculator then B0 M0. Attempt, by division or equivalent, to find quadratic factor M1 k ( x + 3)( 2 x + 1)( 2 x − 1) where k ≠ 1 gets B1 M1. Obtain ( x + 3)(2 x − 1)(2 x + 1) A1 3 6(d) Attempt correct process to find at least 1 value from cosec θ = k where M1 Allow for 199.5o or −19.5o . k < −1 Obtain 3.48 or 5.94 A1 Obtain a second correct solution A1 And no others within the range. 3

Mark scheme

7(a) State cos2θ cosθ − sin 2θ sinθ B1 Attempt correct relevant identities to express in terms of cosθ only M1 M0 if moving terms from side to side. Confirm 4cos3 θ − 3cosθ with sufficient detail A1 AG 3 7(b) Use identity with θ = 185 π M1 Obtain 12 cos 56 π and hence − 14 3 A1 2 7(c) Express integrand in form k1 (cos3x + 3cos x) + k2 (cos9 x + 3cos3 x) M1 Obtain correct integrand 9cos x − cos9x A1 OE (allow unsimplified). Integrate to obtain form k3 sin x + k4 sin 9 x M1 Obtain correct 9sin x − 19 sin 9 x A1 Now simplified; condone missing ... + c . 4