Mathematics 9709/32 May/June 2022: question paper and mark scheme

Mark scheme

1 Use law of the logarithm of a product, power or quotient M1 e2 x + 3 ( 2 x + ln 3) or a law of indices (on an expression that is relevant to the question) e.g. ln(e2 x + 3) − ln 3 = ln or e = e 2 x eln 3 3 State a correct equation without logs (in any form) A1 e.g. 3 + e 2 x = 3e 2 x Carry out correct method to solve an equation of the form e 2 x = a , where M1 Allow for x = 1 ln 3 . 2 2 a > 0, or for solving e x = b (b > 0) if they have already taken the square M1 can be implied by correct answer. root Obtain answer x = 0.203 A1 CAO. The question requires 3 d.p. Answer only with no working shown is 0/4. 4

Mark scheme

2 Use correct double-angle formula to obtain an equation in cos θ ( ) M1 e.g. 3 2cos 2 θ − 1 = 3cos θ + 2 Obtain 6cos 2θ − 3cos θ − 5 = 0 , or 3-term equivalent A1 M1 A0 is scored if they use any correct formula for cos 2θ and make a subsequent error. Solve a 3-term quadratic in cos θ for θ ( ) M1 As far as θ = cos −1 3− 129 if quadratic correct. 12 Obtain a correct answer, e.g. 134.1 ° A1 Accept greater accuracy e.g. 134.1456, 225.8544. Obtain a second answer, e.g. 225.9 ° and no other in 0°, 360° A1 FT Treat answers in radians (2.34 and 3.94) as a misread. Ignore answers outside [ 0°, 360°] . The FT is for 360° minus the first answer. Special Ruling: If they have an incorrect quadratic that leads legitimately to 4 solutions for θ , allow FT for 360° minus an answer in ( 0°,180° ) . More than 4 solutions is maximum M1 A0 M1 A0 A0. If their equation should have 4 solutions and the candidate only gives 3 solutions then M1 A0 M1 A0 A0. Mis-read leading to a quadratic with 4 solutions could score maximum M1 A0 M1 A1 A1 or M1 A0 M1 A1 A0 if extra/missing solution. 5

Mark scheme

3 1 M1 Or divide by 2x–1 and equate constant remainder to zero. Substitute x = , equate result to zero 2 Obtain a correct simplified equation A1 1 1 1 e.g. a + + b + 3 = 0 or a + 4b = −26 8 4 2 Substitute x = – 2, equate result to 5 M1 Or divide by x+2 and equate constant remainder to 5. Obtain a correct simplified equation A1 e.g. −8a + 4 − 2b + 3 = 5 or 8a + 2b = 2 Obtain a = 2 and b = – 7 A1 WWW 5

Mark scheme

4 Use the correct product rule and then the chain rule to differentiate either M1 e.g. two terms with one part of cos3 x or sin x dy cos3 x cos x = p cos 2 x sin x sin x + q . dx sin x Obtain correct derivative in any form A1 A1 A1 for each correct term substituted in the complete dy 3 cos x cos x derivative. e.g. = −3cos 2 x sin x sin x + dx 2 sin x Equate their derivative to zero and obtain a horizontal equation with positive M1 e.g. −3cos 2 x sin 2 x + 1 cos 4 x = 0 2 integer powers of sin x and/or cos x from an equation including sin x or 1 using sensible algebra. sin x Use correct formula(s) to express their equation/derivative in terms of one M1 Can be awarded before the previous M1. trigonometric function May involve more than one trigonometric term. Obtain 7cos 2 x = 6 , 7sin 2 x = 1 , or 6tan 2 x = 1 , or equivalent, and obtain A1 CAO. The question asks for 3 sf. answer x = 0.388 Ignore additional answers outside ( 0, π2 ) . 22.2° is A0. 6

Mark scheme

5(a) Sketch a relevant graph, e.g. y = ln x B1 2 Sketch a second relevant graph, e.g. y = 3x − x2 , and B1 justify the given statement by marking the root on the sketch or by use of a suitable comment 2 ln( x) : sketch should imply y-axis is an asymptote. Through (1, 0) if marked. Correct shape. 3x − x 2 : Symmetrical. Through (0, 0) and (3, 0) if marked. If ln(x) correct accept parabola for +ve y only. If ln(x) incorrect then need parabola in 3 quadrants. 2 5(b) Calculate the values of a relevant expression or pair of expressions at x = 2 M1 Allow for a smaller interval. and x = 2.8 At least one value correct if comparing with 0. If using pairs then the pairing must be clear. Complete the argument correctly with correct calculated values A1 e.g. 0.693 < 2 and 1.03 > 0.56 or 1.307 > 0, −0.47 < 0 using 3 x − ln x 0.304 > 0, −0.085 < 0 . Need to have calculated values to at least 2 sf. 2 5(c) Use the iterative process correctly at least once M1 Obtain final answer 2.63 A1 Show sufficient iterations to at least 4 dp to justify 2.63 to 2 dp A1 SC Allow M1 A1 A0 to a candidate who starts at a point or show there is a sign change in the interval (2.625, 2.635) in the interval and reaches a premature conclusion 3

Mark scheme

6(a) Correct separation of variables B1 e dy = xe dx Condone missing integral signs. −y −x Obtain term −e− y B1 Commence integration by parts and reach ± xe − x ± e − x dx *M1 M0 if clearly using differentiation of a product. Complete integration and obtain − xe − x − e − x A1 Use x = 0 and y = 0 to evaluate a constant or as limits in a solution DM1 Must see working for this. In a correct solution they containing or derived from terms ae − y , bxe − x and ce − x , where abc ≠ 0 should have −e − y + C = − xe − x − e − x or equivalent. If they take logarithms before finding the constant, the constant must be of the right form. Correct solution in any form A1 e.g. −e − y = − xe − x − e − x Must follow from correct working A0 if constant of integration ignored or assumed to be zero. ( Obtain final answer y = − ln ( x + 1) e − x ) from correct working A1 ex OE e.g. y = x − ln ( x + 1) , y = ln . x +1 A0 if constant of integration ignored or assumed to be zero. 7 6(b) Obtain answer ( y = )1 − ln 2 B1 Must follow from at least 6 or7 obtained in part 6(a). 1

Mark scheme

7(a) dy B1 State or imply 6 xy + 3x2 as derivative of 3x 2 y dx State or imply 3y 2 dy as derivative of y3 ( ) B1 Allow B1 B1 for 3 x 2 dx + 6 xydx + 3 x 2 dy − 3 y 2 dy [ = 0 ] dx Equate attempted derivative of left-hand side to zero and solve to obtain an M1 Allow if zero implied by subsequent working. dy dy equation with as subject Allow if recover from an extra = at the beginning dx dx of the left-hand side. dy x 2 + 2 xy A1 AG Accept y ′ for dy . Obtain = 2 2 correctly dx dx y −x 4 7(b) Equate numerator to zero *M1 Must be using the given derivative. Obtain x = – 2y, or equivalent A1 An equation with x or y as the subject SOI. Use x3 + 3x 2 y − y 3 = 3 to obtain an equation in x or y DM1 −8 y 3 + 12 y 3 − y3 = 3 or x 3 − 32 x 3 + 18 x 3 = 3 or any equivalent form (do not need to evaluate powers). Obtain the point (– 2, 1) and no others from solving their cubic equation A1 Allow if each component stated separately. ISW. State the point (0, − 3 3 ), or equivalent from correct work B1 Accept (0, 3 −3 ), or (0, – 1.44) (-1.44225). Allow if each component stated separately. ISW. 5

Mark scheme

8(a) A Bx + C B1 State or imply the form + 2 3x − 1 x +3 Use a correct method for finding a constant M1 Obtain one of A = 1, B = 0 and C = 3 from correct working A1 A maximum of M1 A1 is available after B0. Obtain a second value from correct working A1 Obtain the third value from correct working A1 5 8(b) 1 B1 FT 1 1 Integrate and obtain term ln ( 3 x − 1) OE e.g. ln( x − ) . The FT is on the value of A. 3 3 3 x M1 Obtain term of the form ktan −1 3 x A1 FT OE. The FT is on the value of C. Obtain term 3tan −1 3 Substitute correct limits in an integral of the form M1 Must be subtracted the right way round. x a ln ( 3x − 1) + ktan −1 , where ak ≠ 0 , and evaluate trigonometry ( ln 8 − ln 2 + 3 × − 3 × ) 1 3 1 3 π 3 π 6 3 Angles should be in radians. Condone angles as decimals. 2 3π A1 1 π Obtain answer ln 2 + from correct working in part 8(b) Or exact 2-term equivalent e.g. ln 4 + ISW. 3 6 3 2 3 5

Mark scheme

9(a) Express general point of l or m in component form, i.e. B1 ( −1 + 2λ , 3 − λ,4 − λ ) or ( 5 + aμ , 4 + bμ ,3 + μ ) Equate components and eliminate either λ or μ M1 e.g. μ = 2 , λ = −1−b , μ = −4 , λ = a + 6 1−b 1−b 2+ a a+2 Eliminate the other parameter or obtain a second expression in the first M1 λ and μ are not required to be the subject of the equations. Show intermediate steps to obtain 2b − a = 4 A1 AG Alternative method for question 9(a) Express general point of l or m in component form, i.e. B1 ( −1 + 2λ , 3 − λ,4 − λ ) or ( 5 + aμ , 4 + bμ ,3 + μ ) Express a or b in terms of λ and μ M1 2λ − 6 −1 − λ a= , b= μ μ Use λ = 1 − μ M1 Obtain 2b − a = 4 A1 AG 4 9(b) Using the correct process equate the scalar product of the direction vectors *M1 ( 2i − j − k ) . ( ai + bj + k ) = 0 SOI. to zero Obtain 2a − b − 1 = 0 A1 OE e.g. 2 ( 2b − 4 ) − b − 1 = 0 Solve simultaneous equations for a or for b DM1 Obtain a = 2, b = 3 A1 4 9(c) Substitute found values in component equations and solve for λ or for µ M1 Obtain answer 3i + j + 2k from either λ = 2 or µ = – 1 A1 Accept as coordinates or equivalent. 2

Mark scheme

10(a) Substitute x = −1 + 7i in the equation and attempt expansions of x 2 and *M1 3 x Use i 2 = −1 correctly at least once and solve for k DM1 ( ) ( ) ( 2 20 − 4 7i + 3 −6 − 2 7i + 14 −1 + ) 7i + k = 0 Obtain answer k = – 8 A1 SC B1 only for those who show no working for the cube and square and obtain answer k = – 8. Alternative method for question 10(a) ( ) Attempt division by x + 1 − 7i as far as 2 x2 + z1 x + ... *M1 See division on next page. Use i 2 = −1 correctly at least once and obtain 2x 2 + z1 x + z2 + remainder DM1 Obtain answer k = – 8 A1 3 10(b) State answer −1 − 7i B1 Can be seen simply stated on its own, or in a list of roots. Allow if stated clearly in part 10(a). Carry out a method for finding a quadratic factor with zeros −1 + 7i and ( ( M1 Or state x − −1 + 7i ) ) ( x − ( −1 − 7i ) ) ( 2 x − p ) −1 − 7i Obtain x 2 + 2 x + 8 ( )( A1 Or obtain −1 + 7i −1 − 7i p = −8 ) Or obtain ( −1 + 7i ) + ( −1 − 7i ) + = − p 3 2 2 1 A1 Needs to follow from the working. Obtain root x = , or equivalent, via division or inspection 2 4 10(c) Show a circle with centre −1 + 7i B1 Im Show circle with radius 2 and centre not at the origin B1 7 There needs to be some evidence of scale e.g. radius marked or a scale on the axes -1 Re If the scales are very different from each other then B1 for centre in the correct position and B1 for an ellipse. If there is more than one circle the max score is B1. 2 10(d) Carry out a complete method for calculating the maximum value of arg z for M1 e.g. π + tan −1 1 + π Can be implied by 155.7° . 2 7 4 correct circle Obtain answer 2.72 radians A1 CAO. The question requires radians. 2