Mathematics 9709/32 October/November 2022: question paper and mark scheme

Mark scheme

1 Use law of the logarithm of a power or product M1 Ignoring the 3 or the 5 is not a misread. Obtain a correct linear equation in any form, A1 Condone invisible brackets if they are used correctly e.g. ( 3x − 1) ln 2 = ln 5 + (1 − x ) ln 3 later. Solve for x M1 Get as far as x = … Condone minor slips in the processing e.g. sign errors and losing a term that had been there, but award M0 for a fundamental error e.g. 3x ln 2 + x ln3 = 3x ln 6 or ignoring the 3 or the 5 completely. Condone working in decimals. ln 30 A1 Do not ISW Obtain final answer x = ln 24 Alternative method for question 1 Use laws of indices to split at least one exponential term M1 23 x x x e.g. or an arrangement with 8 and/or 3 . 2 Obtain 24 x = 30 A1 Or equivalent e.g. 3x8 x = 30 not for 3x 23 x = 30 (need two factors with the same index). Solve for x M1 Get as far as x = … ln 30 A1 Do not ISW Obtain final answer x = ln 24 4

Mark scheme

2(a) 3 M1 Or divide by 2 x + 3 and set constant remainder equal to Substitute x = − and equate result to zero 2 zero. ( ) ( ) Or state 2 x3 − x 2 + a = ( 2 x + 3) x 2 + px + q , compare coefficients and solve for p or q. Obtain a = 9 A1 2 2(b) Commence division by ( 2 x + 3) reaching a partial quotient x + kx *M1 The M1 is earned if inspection reaches an unknown 2 factor: x 2 + Bx + C and an equation in B and/or C, or an unknown factor Ax 2 + Bx + 3 and an equation in A and/or B. ( Obtain factorisation ( 2 x + 3) x 2 − 2 x + 3 ) A1 Allow if the correct quotient seen. Correct factors seen in (a) and quoted or used here scores M1A1. Show that x 2 − 2 x + 3 is always positive, or 2 x3 − x 2 + 9 only intersects the DM1 Must use their quadratic factor. x-axis once 3 SC If M0, allow B1 if state x < − and no error seen 2 3 A1 State final answer x <− from correct work 2 4

Mark scheme

3 Use correct product rule on given expression *M1 Obtain correct derivative in any form A1 e.g. cos x sin2x + 2sin x cos2x Use correct double angle formulae to express derivative in terms of sin x and *M1 cos x Equate derivative to zero and obtain an equation in one trig variable DM1 dependent on the 2 previous M Marks. Obtain 3sin 2 x = 2 , 3cos2 x = 1 or tan 2 x = 2 A1 OE Solve and obtain x = 0.955 A1 3 sf only. Final answer in degrees is A0. Ignore any attempt to find the corresponding value of y. Alternative method for the first three marks Use correct double angle formula to obtain y = 2cos x − 2cos3 x *M1 or y = 2sin 2 x cos x Use chain rule and / or product rule *M1 Obtain derivative y ′ = −2sin x + 6sin x cos 2 x A1 y ′ = −2sin 3 x + 4sin x cos 2 x Alternative method for the second and third M marks Equate derivative to zero and obtain an equation in tan x and tan 2x *M1 Use correct double angle formula to obtain an equation in tan x DM1 6

Mark scheme

4(a) State R = 17 B1 Allow if working from an incorrect expansion but not from decimals. Use correct trig formulae to find α (Correct expansion and correct expression for M1 NB: cosα = 4 and sin α = 1 scores M0A0. trig ratio for α ) M0 for incorrect expansion of cos ( x − α ) M1 for correct expression for trig ratio for α and no errors seen. Obtain α = 14.04° A1 2 d.p. required Allow M1A1 for correct answer with no working shown. Correct answer from incorrect working (e.g. tan −1 ( − 14 ) ) is awarded M0A0. 180° − tan −1 ( − 14 ) is awarded M1 3 4(b) 3 B1 FT FT their R. Accept awrt 43.3° or awrt 316.7° Evaluate cos−1 to at least 1 d.p. (43.31 38°) Can be implied by subsequent working. 17 Use correct method to find a value of x in the interval M1 Must be working with 2x and their α . Obtain answer, e.g. 14.6 ° A1 Accept overspecified answers but they need to be correct. (14.6388…. and 151.3249…). Use a correct method to find a second answer in the interval M1 Must be working with 2x, their α and 360° − their 43.3 . Obtain second answer in the interval, e.g. 151.3 ° , and no other in the interval A1 Ignore answers outside the given interval. Treat answers in radians (0.255… and 2.64…) as a misread. 5

Mark scheme

5(a) Use quadratic formula, or completing the square (( z − 3i ) − 3 = 0) 2 M1 and use i 2 = −1 to find a root Obtain a root, e.g. 3 + 3i A1 Or exact 2 term equivalent e.g. 6i2 + 12 ISW. 2 Obtain the other root, e.g. − 3 + 3i A1 Or exact 2 term equivalent ISW. 3 5(b) Show points representing the roots correctly B1 FT 2 roots consistent with their (a) and with no errors seen on the diagram. B0 if they only have one root or more than 2 roots Must match their scale and 1 < 3 < 2 Linear scales seen or implied. Need some indication of scale (numbers or dashes). Scales along an axis must be approximately consistent but scales may be different on the 2 axes. 1 5(c) State modulus of either root is 2 3 , or simplified exact equivalent B1 FT ISW if converted to decimal . Ignore modulus of second root if seen. Follow their root(s) not on either axis (from (a) or (b)). Find the argument of one of their roots – get as far as tan −1 (...) M1 SOI but must be correct for their root. 1 2 A1 Must obtain values. Allow degrees. Obtain correct arguments π and π , or simplified exact equivalents 3 3 3 5(d) Give a complete justification that the correct triangle is equilateral B1 Check their diagram in (b). Possible justifications: 3 equal sides, or all angles equal π π to 3 , or isosceles and an angle of 3 . 1

Mark scheme

6(a) State or imply AB or AC correctly in component form B1 ( AB = 2i − 2j + k, AC = 4i − 3k ) . Using the correct process with relevant vectors to evaluate the scalar product M1 or BA.CA ( 8 − 3 = 5 ) . AB. AC , M0 for AB.CA . Using the correct process for the moduli, divide their scalar product by the M1 5 product of their moduli to obtain cosθ or θ 9 25 Independent of the first M1. 1 A1 ISW. Need to see a value for cosθ . Obtain answer 3 Accept 155 or 0.333 ( cos−1 13 alone is not sufficient) 4 6(b) Use correct method to find an exact value for the sine of angle BAC from their (a) M1 ( 1− ) 1 9 2 A1 Obtain answer 2 , or equivalent 3 Use correct area formula to find the area of triangle ABC with their versions of relevant vectors M1 ( 1 2 ) ( 9 25 × their sin θ or 12 9 25 × sin cos −1 13 ) Obtain answer 5 2 or 50 A1 Only ISW Alternative method 1 for question 6(b) Use correct method to find the perpendicular distance from A to BC (or B to AC M1 2 + 2λ 2 or C to AB) −2 + 2λ . 2 = 0 λ = 1 1 − 4λ −4 6 Obtain 13 75 A1 ( i − j+ k ) 7 3 5 3 1 3 Use correct area formula to find the area of triangle ABC M1 ( × their 24 × their 1 2 1 3 75 ) The length they use for their base must be found correctly. Obtain answer 5 2 or 50 A1 6(b) Alternative method 2 for question 6(b) Correct method to find the semi-perimeter M1 Obtain 4 + 6 A1 Correct application of Hero’s (Heron’s) formula M1 ( 4 + 6 )(1 + 6 )( −1 + 6 )( 4 − 6 ) Obtain answer 5 2 or 50 A1 4

Mark scheme

7(a) Show sufficient working to justify the given statement B1 e.g. see 2cot θ × − cos ec2θ in the working or express in terms of sin θ and cosθ and use quotient rule to obtain the given result. Solution must have θ present throughout and must reach the given answer. 1 7(b) Separate variables correctly B1 tan 2 θ 2 cot θ xdx = − dθ sin θ 2 sin 2 θ Check for relevant working in (a) Condone incorrect notation e.g. missing dx. Need either the integral sign or the dx, dθ. 1 2 B1 Obtain term x 2 Obtain terms tan θ + cot 2θ 2cot θ 2cos θ Alternative: dθ = B1 + B1 1 dθ = − ( +C ) sin θ 2 sin θ3 sin 2 θ 1 M1 Need to have 3 terms. Constant of correct form. Form an equation for the constant of integration, or use limits x = 2, θ = π , in 4 a solution with at least two correctly obtained terms of the form ax 2 , btanθ and ccot 2θ , where abc ≠ 0 1 2 A1 or 1 x2 = tan θ + cosec2θ − 1 State correct solution in any form, e.g. x = tan θ + cot 2θ 2 2 If everything else is correct, allow a correct final answer to imply this A1. 1 A1 18+ 2 3 Substitute θ = π and obtain answer x = 2.67 2.6748… 3 If see a correctly rounded value ISW. 6 7

Mark scheme

8(a) State (a =) π 2 B1 Allow 32400, 180 2 . Accept ( x = ) π 2 . 1 8(b) State or imply dx =2u du or equivalent B1 e.g. du = 1 dx 2 x Incorrect statements e.g. du = 2 1 x is B0. Substitute for x and dx throughout the integral M1 Obtain 2u sin u du A1 Allow with missing du. Commence integration of ku sin u du by parts and reach *M1 ku cos u ± k cos u du Obtain integral −ku cos u + k sin u A1 Substitute limits u = 0 and u = their a , a ≠ 0 , a in radians DM1 −2π cos π + 2sin π ( +0 − 2sin 0 ) or x = 0 and their a Need limits stated but condone if zeros not shown in in the complete integral substitution. Obtain answer 2π A1 7

Mark scheme

9(a) State or imply angle AOC = π − 2θ B1 Might be seen on the printed diagram. Use correct formulae for the area of a sector and triangle, or of a segment, and M1 1 2 r 2 ( π − 2θ ) − 12 r 2 sin ( π − 2θ ) or find the area of the shaded region 1 2 πr 2 − 12 r 2 ( 2θ ) + 12 r 2 sin ( π − 2θ ) M0 if subtraction the wrong way round. 1 2 A1 e.g. 1 πr 2 = 1 r 2 ( π − 2θ ) − 1 r 2 sin ( π − 2θ ) . Equate to πr and obtain a correct equation in any form 6 2 2 6 Obtain θ = 13 ( π − 1.5sin 2θ ) correctly A1 AG Condone if state / imply sin ( π − 2θ ) = sin 2θ . 4 9(b) Evaluate a relevant expression or pair of expressions at θ = 0.5 and θ = 0.7 M1 Allow work on a smaller interval. Need to evaluate for both limits, with at least one correct. When using x = f ( x ) embedded values are not sufficient e.g. f(0.5) …. is accepted but 1 3 (π − 1.5sin 2 × 0.5 ) = ... is not. Complete the argument correctly with correct calculated values A1 e.g. 0.5 < 0.626, 0.7 > 0.554 or 0.126 > 0, − 0.146 < 0 If using pairs then the pairing must be clear. Need to see the inequalities or an appropriate comment. Need to see values calculated to at least 2 sf. 2 9(c) 1 M1 i.e obtain one value and use that value to obtain a second Use the iterative process θn +1 = (π − 1.5sin 2θn ) correctly at least once value. Must be working in radians. 3 Obtain final answer 0.586 A1 Show sufficient iterations to 5 d.p. to justify 0.586 to 3 d.p., or show there is a A1 0.5, 0.62646, 0.57225, 0.59195, 0.58416, 0.58715, sign change in the interval (0.5855, 0.5865). e.g. 0.58599, 0.58644 0.6, 0.58118, 0.58833, 0.58553, 0.58661, 0.58619, 0.58636 0.7, 0.55447, 0.59958, 0.58133, 0.58827, 0.58556, 0.58661, 0.58620, 0.58636 Allow working to more than 5 dp, but not less. 3

Mark scheme

10(a) A Bx + C B1 State or imply the form + 1+ x 2 + x2 Use a correct method for finding a constant M1 Obtain one of A = 2, B = – 1 and C = 0 A1 SC: A maximum of M1A1 is available for obtaining A = 2 after scoring B0. Obtain a second value A1 Obtain the third value A1 5 10(b) Integrate and obtain term 2ln (1+ x ) B1 FT A ln (1 + x ) ( ) Integrate and obtain term of the form k ln 2 + x2 from an integral of the *M1 Ignore any separate working relating to C ≠ 0. correct form 1 ( Obtain term − ln 2 + x2 ) A1 FT B 2 ln (2 + x ) 2 2 Substitute limits in an integral containing terms of the form DM1 Ignore working relating to C ≠ 0. ( ) a ln (1 + x ) + b ln 2 + x 2 , where ab ≠ 0 ( 2ln5 − 2ln1 − ln18 + ln 2) 1 2 1 2 Dependent on the first M1. Must be subtracting the correct way round. Must have an exact substitution. 25 A1 ISW Obtain answer ln 3 Any exact equivalent e.g. ln 25 2 with no number in 18 front of the logarithm. 5