Mathematics 9709/33 May/June 2024: question paper and mark scheme

Mark scheme

1 Use law of the logarithm of product or quotient on each side *B1 Allow logs to any base, as well as decimals, throughout. ln 83 + ln 8−6x and ln 4 + ln 5−2x. 83 Allow for ln and ln 86x – ln 52x. 4 (3 − 6x) ln 8 and ln 4 + ln 5−2x gains next DB1 as well. Use law of logarithm of a power involving x on ONE side, DB1 SC If *B0 DB0, then allow B1 (1/4) for a correct logarithm law e.g. ln 83 + (−)6x ln 8 or (3 − 6x) ln 8 or ( 9 − 18x ) ln 2 seen anywhere. or ln 4 − 2x ln 5 Obtain a correct linear equation in x, e.g. B1 If in decimals, allow small errors in 2nd and 3rd dp. ( 3 − 6 x ) ln 8 = ( 9 − 18 x ) ln2 = ln 4 − 2 x ln 5 Obtain answer x = 0.524 B1 3dp required. No working scores 0/4 marks. After *B1 DB1 to correct answer with no more log working seen, then SC B1 for x = 0.524. Maximum 3/4 possible. Alternative Method for Question 1 Use laws of indices to get to a = b ±2x or c ± x in a correct form so now (B2) (83/4) and (5/83)−2x or (52/86)−x opposite sides or only ONE log power law required (4/83) and (83/5)–2x or (86/52)–x opposite sides 83 83 (B1) −2x ln (5/83) or 2x ln (83/5) or x ln (86/52) or – x ln (52/86). Obtain a correct linear equation in x, e.g. ln = 2 x ln SC: If B0 then allow B1 (1/4) for a correct term seen anywhere. 4 5 If in decimals, allow small errors in 2nd and 3rd dp. Obtain answer x = 0.524 (B1) 3dp required. No working scores 0/4 marks. From the first line to correct answer with no log working seen, then B2 and SC B1 for x = 0.524. Maximum 3/4 possible. 1 Alternative Method 2 for Question 1 Use laws of indices to get to any correct form with indices (*B1) Allow 27 – 18x and 5–2x on opposite sides combined so now TWO log power laws are required or 29 – 18x and 22 – 4.64x on opposite sides. Use law of logarithm of a power involving x on ONE side, (DB1) e.g. (7 – 18x) ln 2 or ( 9 − 18 x ) ln2 or –2x ln 5 or (2 – 4.64x) ln 2 e.g. (7 – 18x) ln 2 = ln 5–2x or ln 27 – 18x = − 2x ln 5 or … SC: If *B0 DB0 then allow B1 (1/4) for a correct term seen anywhere. Allow 7 – 18x ln 2 or 9 − 18 x ln2 E.g. any term in *B1 shown above. Obtain a correct linear equation in x, (B1) If in decimals, allow small errors in 2nd and 3rd dp. e.g. (7 – 18x) ln 2 = − 2x ln 5 or (9 – 18x) ln 2 = (2 – 4.64x) ln 2 Obtain answer x = 0.524 (B1) 3dp required. No working scores 0/4 marks. From the first line to correct answer with no log working seen, then *B1 and SC B1 for x = 0.524. Maximum 2/4 possible. 4

Mark scheme

2 Use correct product rule M1 ae2xsin2x + e2xbcos2x. Need a or b = 2. cos 2x may be 1 – 2 sin2x or … Allow M1 if only error is ex instead of e2x in one of terms, then maximum 1/5. Obtain correct derivative 2e 2 x sin2 x + 2e 2 x cos2 x ( ) A1 OE, e.g. 4e 2 x sinx cos x + 2e 2 x cos 2 x − sin 2 x . Equate derivative of the form ae2xsin2x + e2xbcos2x to 0 and solve M1 Obtain 2x = tan−1(− their b/their a) OE. for 2x or x using a correct method Allow one slip in rearranging. Note may have substituted for sin2x and/or cos2x Allow degrees. Variety of other methods available, such as solving quadratic equation in sin x or tan x e.g. tan² x – 2tan x – 1 = 0 leading to x = tan-1(1 + √2). Obtain x = 83 π only or exact equivalent A1 CWO 67.5° gets A0. π Ignore any answers outside interval 0 ⩽ x ⩽ . 2 1 π 3 A1 CWO, ISW. Obtain y = 2e 4 only or exact simplified equivalent 3 3π 2 Not sin πe 4 . 4 π Ignore any answers using x outside interval 0 ⩽ x ⩽ . 2 5

Mark scheme

3 Square x + iy obtaining three terms when simplified and equate real M1 Having used i2 = −1. and imaginary parts to 24 and −7 respectively Obtain equations x2 – y2 = 24 and 2xy = –7 A1 Allow 2xyi = –7i. Eliminate one variable by correct method and find a horizontal M1 All powers of x or y are positive and are in the numerator. equation in the other Obtain 4x4 – 96x2 – 49 = 0 or 4y4 + 96y2 – 49 = 0 A1 or 3-term equivalents 7 2 2 7 2 2 A1 7 2 2 7 2 2 Obtain answers − i and − + i or exact equivalents E.g. ± − i , but not ± μ i or 2 2 2 2 2 2 2 2 and no others 7 2 2 ± μ i . 2 2 Allow coordinates or x =…, y =… paired correctly. ISW converting to different form. Must simplify 49. 5

Mark scheme

4 1 B1 Allow ln k + ln y = cx lne State or imply that ln k + ln y = cx or ln y = cx + ln etc. k Carry out a completely correct method for finding ln k or c M1 Equations must have been formulated correctly. Obtain value c = 0.80 A1 AWRT Allow 0.8 for 0.80. Not a fraction. Accept in the equation ky = ecx. Obtain value k = 6.5 A1 AWRT Not a fraction. Accept in the equation ky = ecx.

Mark scheme

5 B C B1 State or imply the form A + + x −1 2x + 1 Use a correct method for finding a constant M1 Correct appropriate method. Obtain one of A = 3, B = 2 and C = –3 A1 Obtain a second value A1 Obtain a third value A1 Alternative Method for Question 5 Divide numerator by denominator to reach A = 3 (M1) ax + b May be implied by 3 [+] with a and b not both 0. ( x − 1)( 2 x + 1) x+5 (A1) Obtain 3 + ( x − 1)( 2 x + 1) D E (B1) State or imply the form + x −1 2x + 1 Obtain one of D = 2 and E = − 3 (A1) Obtain a second value (A1) 5

Mark scheme

6(a) Show a circle centre (4, 3) B1 Note full circle is not required but must show centre and include Allow dashes for coordinates on axes relevant arc. Show a circle with radius 2. B1FT FT centre not at the origin. Can be implied by at least two of the points (2, 3), (6, 3), (4, 1) and (4, 5) being correct Point representing (2, 1) B1 Half-line or ‘correct’ full line extending into the third quadrant implies point (2, 1). Show a half-line at their (2, 1) at an angle of 13 π , cutting top of B1FT FT the point (±2, ±1) or (±1, ±2). circle between x = 3 and x = 5 Shade the correct region B1 Needs correct half-line or “correct” full line extending into the third quadrant AND correct circle 5 6(b) Carry out a correct method for finding the greatest value of arg z in M1 E.g. sin−1(2/√(25)) + tan−1(3/4) or the correct region in (a) sin−1(2/√(25)) + sin−1(3/5). Or, e.g., substitute y = kx in circle equation, solve when 6 + 21 discriminant = 0, to get tan−1 . 6 Obtain answer 1.06, or 1.05 or 1.055 or 1.056 or 60.4° or 60.5° A1 The marks in (b) are available even if errors in (a). No working seen scores 0/2 marks. 2

Mark scheme

7(a) Show 8 × (–7)3 + 54 × (–7)2 – 17 × (–7) – 21 = 0 B1 No errors allowed. This is sufficient if no errors seen. [ – 2744 + 2646 +119 – 21 = 0] Correct division: 8x2 −2x −3 . Or complete division of 8x3 + 54x2 – 17x – 21 by x + 7 to get x + 7 8x3 + 54x2 − 17x − 21 quotient 8x2 – 2x – 3 and remainder of 0 8x3 + 56x2 . − 2x2 −17x Or state (x + 7)(8x2 – 2x – 3) is sufficient − 2x2 −14x . Factors must be stated again in (b) to collect marks there − 3x − 21 − 3x − 21 1 7(b) Commence division and reach partial quotient of the form 8x2 ± 2x M1 Condone no visible working. or 8x3 + 54x2 – 17x – 21 = (x + 7)(Ax2 + Bx + C) and reach A = 8 and B = ± 2 or C = –3 Obtain quotient 8x2 – 2x – 3 with no errors seen A1 Division can terminate with 0 or −3x – 21 stated once or twice. Stating (x + 7)(8x2 – 2x – 3) is sufficient The working of division and finding quotient may be seen in (a) but results required here to collect marks. 2 7(c) −1 M1 (x + 7) (8x2 − 2x – 3) = (x + 7)(4x – 3)(2x + 1) = 0 Solve quadratic from (b) to obtain a value for 𝜃 = cos −1 2 2 ± 4 + 96 1 3 x = cosθ = = − and . 3 16 2 4 or cos −1 4 Obtain one answer, e.g. 𝜃 = 120° A1 Obtain three further answers, e.g. 𝜃 = 240°, 41.4° and 318.6° A1 Accept more accurate answers. (condone 319°) and no others in the interval Answers in radians, maximum 2/3. 3

Mark scheme

8(a) State R = 12 or exact equivalent B1 ISW Use trig formula to find α M1 ± 3 3 −1 1 Allow α = 30° or tan −1 or cos ± −1 or sin ± 3 2 2 3 Allow M1 if – tan −1 etc. 3 NB: If cosα = 3 and sinα = 3 seen then M0 A0. 1 A1 3 Obtain α = π CWO, so A0 if from tan −1 − . 6 3 3 8(b) Express integral in the form A sec 2 ( 2 x + ...) dx or B1FT FT α from (a). A sec 2 ( 2 x − ...) dx Integrate and reach B tan ( 2 x + ...) or B tan ( 2 x − ...) B1FT FT α from (a). Where B = A or 2A or 0.5A. 1 B1FT OE Obtain tan ( 2 x + ...) FT α from (a). 8 1 1 1 Allow as × . 8 4 2 Coefficient must be correct. Use limits of x = 0 and x = 121 π in the correct order in expression of M1 Allow with tan still present. FT α from (a). π form B tan ( 2 x ± ...) so B tan + ... −B tan (...) 3 1 1 6 SC: B1 OE after tan 2 x + π with no working. 12 8 6 π or B tan − ... −B tan ( −...) 6 1 1 A1 1 1 1 3 −1 Obtain answer 121 3 or or ( 3 – )= needs simplifying. 4 3 48 8 3 8 3 or single term exact equivalent 5 1 Note: allow all marks in (b) even if α = π found by an incorrect 6 method in (a).

Mark scheme

9(a) dV k dV 1 B1 Obtain = [ ± ] or = [±] dt t dt kt dV B1 Obtain = 20 x − 3x 2 dx Correct use of chain rule involving k M1 dV dV dx Use = × . dt dx dt dV dV Expressions for and must be seen to get M1. dt dx dx k A1 If this expression is first seen with numerical values, allow A1 Obtain = [±] or equivalent, dt ( t 20 x − 3x 2 ) when their value of k is substituted back into the general expression. 1 1 dx 20 A1 dx −1 Use t = , x = and = − to obtain given answer which = AG 10 2 dt 37 ( dt 2t 20 x − 3 x 2 ) must be stated 20 k k Need to at least see − = if dx 20 37 1 3 t =− needed to score final A1 10 − dt 37 10 4 20 −k k or − = if − in working for correct k. 37 1 3 t 10 − 10 4 dx 20 = seen anywhere, then A0. dt 37 5 9(b) Separate variables correctly & integrate at least one side correctly B1 Obtain terms 10x2 – x3 B1 May see –10x2 + x3 if negative sign moved across or e.g. 20x2 – 2x3 if 2 moved across. 20 x 2 3x3 Allow − . 2 3 Obtain term ln t with ‘correct’ coefficient from their separation of B1FT FT sign and position of 2 from their separation but B0 if error from a later manipulation. variables, for example a ln t for . t 1 1 M1 Allow numerical and sign errors and decimals. Use t = , x = to evaluate a constant or as limits in a solution Allow if exponentiate before substitution, even if exponentiation 10 2 containing terms of the form x2, x3 and ln t (or ln 2t) done incorrectly, allow for c or ec. Obtain correct answer in any form, for example A1 ln 2t 19 ln 0.2 10 x 2 − x3 = − + + ln t 19 ln 0.1 2 8 2 10 x 2 − x3 = − + + 2 8 2 ln t or 10 x 2 − x3 = − + 2.5 − 0.125 − 1.15 2 Allow 1.14 to 1.16 for 1.15 and allow 2.44 to 2.46 for 2.45 2 x3 − 20 x 2 + 19 A1 ISW Obtain answer t = 101 e 4 or equivalent Need t =……… 19 2 x3 + 19 0.1 e 4 1 4 2 x3 − 20 x2 E.g. 19 , 2 , e e . 20 x 2 − 2 x3 − 10e 20 x 10 e 4 3 2 Allow decimals, allow 2.44 to 2.46 for 2.45, e.g. e2 x − 20 x + 2.45 . 1 ln A0 if e 10 present in final answer. 6

Mark scheme

10(a) Carry out correct process for evaluating the scalar product of *M1 3 −1 3 1 direction vectors 4 . 2 or 4 . −2 a 2 a −2 3(–1) + 4(2) + 2a or –3 + 8 + 2a or 5 + 2a. Allow one slip in unsimplified form. Using the correct process for the moduli, divide the scalar product *M1 *M1 marks independent of each other, so *M0 *M1 for failure to 2 use both direction vectors, but must be using scalar product and by the product of the moduli and equate to ± , same 2 vectors throughout. 2 2 2 2 or equate the scalar product to the product of the moduli and ± Allow or − throughout question. 2 2 2 5 + 2a 2 A1 5 + 2a 2 State a correct equation in any form, e.g. = [±] = [ ±] OE 3 25 + a 2 2 9 + 16 + a 2 1+ 4 + 4 2 Allow unsimplified as in guidance 2 E.g. 5 + 2a = [ ± ] 9 + 16 + a 2 1 + 4 + 4 2 If moduli initially correct but later has errors, award A1 when using 2 2 2 or ± or − . 2 2 2 Form a quadratic equation in a with 3 or more terms all on one side DM1 Must square (5 + 2a) to get 3 terms and must remove square roots and solve for a. from both terms on other side. DM1 depends on BOTH *M1 9 25 + 20a + 4a2 = (25 + a2) 2 a2 − 40a + 175 = 0 hence (a – 5)(a – 35) = 0. 10(a) Obtain a = 5 and a = 35 A2 A1 for each, working not needed if quadratic correct. 6 10(b) Express general point of at least one line correctly in component B1 Often the third point on the line occurs after M1 A1 is gained. form, 1 + 3λ –3 – µ i.e. 1 + 4λ or –1 + 2 µ 2 a + aλ 4 + 2µ Equate at least two pairs of corresponding components and solve M1 If solve for a first, they must have a complete method to eliminate for λ or µ or a both λ and µ. If using a to solve for λ or for µ, a must have been found from a valid method. Obtain λ = –1 or µ = –1 A1 Obtain a = 2 A1 Obtain position vector of the point of intersection is –2i – 3j + 2k A1 Accept coordinates, row or column, but not (–2i,– 3j,+ 2k) or Two different answers for point of intersection scores A0 even if −2i one is correct –3j but ISW after correct form seen. 2k 5