Mathematics 9709/22 February/March 2023: question paper and mark scheme

Mark scheme

1 Express integrand as 2sec2 12 x − 2 B1 Integrate to obtain form a tan 12 x + bx M1 where ab ≠ 0 . Obtain correct 4 tan 12 x − 2 x A1 Obtain 4 − π A1 4

Mark scheme

2 tan θ − tan 60 3 B1 OE involving tanθ only. State = 1 + tan θ tan 60 tan θ Attempt arrangement of equation to quadratic form M1 Condone sign errors in first step and retention of tan 60 . Obtain tan 2 θ − 4 3 tan θ − 3 = 0 A1 OE involving decimals. Solve 3-term quadratic equation to obtain at least one value of θ M1 Obtain −22.2 and 82.2 A1 Or greater accuracy; and no others between –90 and 90. 5

Mark scheme

3(a) Substitute x = −2 and equate to zero *M1 Substitute x = 3 and equate to 35 *M1 Obtain −8a − 4a − 2a + b = 0 and 27a − 9a + 3a + b = 35 A1 Solve a pair of relevant simultaneous linear equations to find a or DM1 Dependent at least one M mark. b Obtain a = 1 and b = 14 A1 5 3(b) Divide by x + 2 at least as far as the x term M1 Obtain [( x + 2)] ( x2 − 3x + 7) A1 Conclude with reference to −2 , and discriminant is 9 − 28 and A1 OE hence no root 3

Mark scheme

4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw approximately correct graph of y = 3x − 3 with greater B1 Crossing x-axis between origin and vertex of first graph. gradient 2 4(b) Attempt solution of linear equation where signs of 2x and 3x are M1 different Solve −2x + 11 = 3x − 3 to obtain x = 145 A1 OE Conclude x > 145 A1 OE Alternative method for Question 4(b) Attempt solution of 3-term equation (2 x − 11) 2 = (3 x − 3) 2 to M1 Or equivalent inequality. obtain at least one value of x Obtain at least x = 145 A1 OE Conclude x > 145 A1 OE 3 4(c) Attempt value of N (maybe non-integer at this stage) using M1 logarithms and their answer to part (b). Conclude with single integer 17 A1 2

Mark scheme

5(a) Integrate to obtain form k1 ln(1 + 2 x) + k2 ln x *M1 k1 ≠ 0, k2 ≠ 0 . Obtain correct 2ln(1 + 2 x) + 3ln x A1 Use limits correctly and equate to ln10 DM1 Apply relevant logarithm properties correctly and arrange as far as DM1 a3 = ... Confirm given result a = 3 90(1 + 2a)−2 with sufficient detail A1 AG 5 5(b) Use iteration process correctly at least once M1 Need to see 1.6848 . Obtain final answer 1.68 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign A1 change in the interval [1.675, 1.685] 3

Mark scheme

6(a) Differentiate using quotient rule M1 OE (e x + 2) 8e2 x − (4e2 x + 9) e x A1 OE Obtain (e x + 2)2 Substitute x = 0 in first derivative and attempt evaluation M1 Obtain 119 A1 4 6(b) Equate first derivative to zero and attempt factorisation or M1 equivalent Solve a three-term quadratic equation in e x to obtain e x = ... M1 (2e x + 9)(2e x − 1) = 0 . Obtain x-coordinate ln 12 or − ln 2 A1 Obtain y-coordinate 4 A1 4

Mark scheme

7(a) Use identity sin 2t = 2sin t cos t B1 Attempt solution of y = 0 for cost M1 Obtain cost = 23 A1 Or exact equivalent. 3 7(b) dx dy *M1 Differentiate to obtain at least one of and correct dt dt dy 6cos 2t − 4cos t A1 Obtain = dx k sec2 t dy DM1 Using correct identities. Attempt to express in terms of cost dx dy 6(2cos 2 t − 1)cos 2 t − 4cos3 t A1 OE Obtain = dx k 4 7(c) dy *M1 Substitute value from part (a) in expression for involving k dx and cost Equate to − 109 and solve for k DM1 Obtain k = 43 A1 3