Mathematics 9709/32 February/March 2023: question paper and mark scheme

Mark scheme

1 Use law of the logarithm of a quotient or express x as ln ex M1 x = ln[(2y – 3)/(y + 4)] or ln ex = ln(2y – 3) − ln (y + 4)]. 2y − 3 A1 Remove logarithms and obtain a correct equation e.g. e x = y+4 3 + 4e x A1 OE ISW Obtain answer y = 2 − ex 3

Mark scheme

2(a) Show correct half-lines from 1 + 2i, symmetrical about y = 2i B1 Im(z) π 5π (drawn between and ). 4 12 Show the line x = 3 extending in both quadrants. B1 2i Shade the correct region. B1 FT 1 3 Re(z) Allow dashes on axes as scale. FT If only error is one of following: FULL lines or x ≠ 3 or one sign error in 1 + 2i or angle outside tolerance or scale missing on one axis. SC No scale on either axis allow B1 FT for otherwise correct figure in correct position. 3 2(b) Carry out a complete method for finding the least value of arg z M1 e.g. − tan −1 ( 2 3 − 2) or tan ( −2 3 + 2) . −1 3 3 Obtain answer –0.454 3dp A1 SC B1 0.454 . 2

Mark scheme

3 Commence division and reach partial quotient 2x2 + (a ± 2)x M1 2x2 + (a + 2)x + a need 2x2 + (a ± 2)x (x – x + 1) 2x + ax + 0x2 + bx – 1 2 4 3 2x4 – 2x3 + 2x2 (a + 2)x3 – 2x2 + bx (a + 2)x3– (a + 2)x2 + (a + 2)x ax2 + (b – (a + 2))x – 1 ax2 – ax + a (b – 2)x – (1 +a ) 3x + 2 Working backwards from remainder: 2x2 + (….)x ± 3 M1 2x2 − x − 3 A1 Obtain correct quotient 2x2 + (a + 2)x + a A1 Allow sign error e.g. in b – 2. Set their linear remainder equal to part of “3x + 2” and solve for a or for b M1 Remainder = 3x + 2 = (b – 2)x – 1 – a. Allow for just equating x term or constant term. Obtain answer a = –3 A1 Obtain answer b = 5 A1 3 Alternative method for Question 3 ( )( ) State 2 x4 + ax3 + 0 x2 + bx − 1 = x 2 – x +1 2 x 2 + Ax + B + 3x + 2 and M1 e.g. 0 = B – A + 2 and –1 = B + 2. form and solve equation(s) to obtain A or B Obtain A = –1, B = –3 A1 Form and solve equations for a or for b M1 e.g. a = A – 2 or b = –B + A + 3. Obtain answer a = –3 A1 Obtain answer b = 5 A1 Alternative method for Question 3 1 ± −3 1± i 3 M1 Allow for correct use of a reasonable attempt at either root in Use remainder theorem with x = or x = exact or decimal form in the remainder theorem 2 2 −1 + −3 3 −1 − −3 x2 = x = − 1 x4 = . 2 2 b b −3 7 3 −3 A1 Expand brackets and obtain exact equation for either root. Obtain −a + ± −3 − 2 = ± or Accept exact equivalent. 2 2 2 2 b bi 3 7 3i 3 −a + ± i 3−2= ± 2 2 2 2 Solve simultaneous equations, or single equation, for a or for b M1 Obtain answer a = –3 from exact working A1 Obtain answer b = 5 from exact working A1 5

Mark scheme

4 Substitute z = x + iy and z* = x − iy to obtain a correct equation, horizontal B1 5(x + iy) – (x + iy)( x – iy )(1 + 2i) + (30 +10i)(1 + 2i) = 0 or with (1– 2i)/( 1 – 2i) seen, in x and y 5(x + iy)(1 – 2i)/[(1 + 2i)(1 – 2i)]– (x + iy)( x – iy )+ (30 +10i) =0 x − 2ix + iy + 2 y − x 2 − y 2 + 30 + 10i = 0 . Use i2 = –1 at least once and equate real and imaginary parts to zero *M1 OE For their horizontal equation. Obtain two correct equations A1 5x – (x2 + y2) + 10 = 0 e.g. x + 2y – x2 – y2 + 30 = 0 and –2x + y + 10 = 0 5y – 2(x2 + y2) +70 = 0 5y – 10x + 50 = 0 x + 2y – (x2 + y2) + 30 = 0 Allow –2ix + iy + 10i = 0. Solve quadratic equation for x or for y DM1 x2 – 9x + 18 = (x – 3)(x – 6) = 0 y2 + 2y – 8 = (y + 4)(y – 2) = 0 DM0 If x or y imaginary. Obtain answers 3 – 4i and 6 + 2i A1 5

Mark scheme

5(a) dx 2t B1 OE Obtain = e + 2te2t dt dy dy dt M1 dy 2t +1 Use = × = 2t . dx dt dx dx e (1 + 2t ) dy A1 AG Need to see e2t (1 + 2t) in denominator. Obtain the given answer = e −2t dx 3 5(b) 1 B1 Obtain x = −e−2 or − 2 and y = 3 at t = − 1 e 1 B1 Obtain gradient of normal = −e−2 or − 2 e x = 0 substituted into equation of normal or use of gradients to give 1 ( ) B1 Equation of normal y − 3 = −e−2 x − −e−2 . y = 3 − 4 with no errors AG e SC Decimals B0 B1 B0 − 0.135 . 3

Mark scheme

6(a) State R = 13 B1 Allow if √ (122 + (− 5)2) seen. Use correct trig formulae to find M1 cos(α) = 12 and sin(α) = 5 M0 α = tan-1(±5/12) = cos-1(±12/13) = sin-1(±5/13) However, sin(α)/cos(α) = 5/12 or − 5/12 with no error seen, or tan(α) = 5/12 or − 5/12 quoted then allow. Obtain α = 0.395 A1 CWO If negative sign seen when finding R then A0 here. If degrees 22.6 A0 MR. Only penalise degrees once in (a) and (b). Note α = 0.39479… 3 6(b) 6 B1FT SOI cos −1 1.0910… FT their incorrect R. R Use correct method to find a value of 2x in the interval M1 6 6 2x = cos −1 + α or 2π − cos −1 + α. R R Allow if cos(2x + 0.395) seen Obtain answer, e.g. x = 0.743 or 0.742 A1 42.5 or 42.6 degrees. Obtain second answer, e.g. x = 2.79 and no others in the interval A1 159.8, 159.9 or 160.0 degrees all possible depending whether using 3 dp or 4 dp. 4

Mark scheme

7(a) 1 2 B1 OE State or imply area of major sector = r (2π − x) 2 1 2 1 B1 OE State or imply area of shaded segment = r x − r 2 sin x r2 sin(x/2) cos(x/2) B0 until changed to (1/2)r2 sin x. 2 2 1 2 1 1 M1 OE State r (2π − x) = 3 r 2 x − r 2 sin x Area of major sector = 3 times (area of minor sector – area of 2 2 2 triangle). Allow r2 sin(x/2) cos(x/2). 3 1 A1 AG Allow rectified slip if before penultimate line. Obtain the given answer x = sin x + π after full and correct working 4 2 4 7(b) Calculate the values of a relevant expression or pair of expressions at x = M1 x= 2 x = 2.5 2 and x = 2.5 (3/4) sin x + (1/2)π 2.2(5277) 2.0(197) 2 < 2.2 or 2.3 2.5 > 2.0 x – (3/4) sin x – (1/2)π − 0.2(5277) < 0 + 0.4(803) > 0 or change of sign Attempt both values and one correct for M1. Complete the argument correctly with correct calculated values A1 Degrees award 0/2 2 7(c) Use the iterative formula correctly at least twice M1 Obtain final answer 2.18 A1 Show sufficient iterations to 4 d.p. to justify 2.18 to 2 d.p. or show there is A1 a sign change in the interval (2.175, 2.185) 2 2.25 2.5 2.2528 2.1543(5) 2.0196(5) 2.1530 2.1967 2.2465 2.1972 2.1786 2.1560 2.1784 2.1865 2.1960 2.1866 2.1831 2.1789 2.1830 2.1845 2.1863 2.1846 2.1831 2.1845 Degrees award 0/3 3

Mark scheme

8(a) Use the product rule correctly *M1 x3 d/dx(lnx) + d/dx(x3) lnx. Obtain the correct derivative in any form A1 x3 e.g. + 3 x 2 ln x . x Equate derivative to zero and solve exactly for x DM1 Reaching x = ea. 1 1 A1 ISW Obtain answer 3 , − or exact equivalent e 3e 4 8(b) Integrate by parts and reach ax 4 ln x + b ( x 4 / x)dx *M1 x4 1 A1 OE Obtain ln x − ( x 4 / x)dx 4 4 x4 x4 A1 OE Complete integration and obtain ln x − 4 16 1 DM1 Correct substitution [(1/4)ln1 or 0 − 1/16] – [(1/64)ln(1/2) – Use limits of x = and x = 1 in the correct order, having integrated twice (1/16)2] or minus this value CWO. 2 Allow omission of (1/4)ln1 or 0. 15 1 A1 Obtain answer − ln 2 or exact equivalent final answer 256 64 5

Mark scheme

9 Separate variables correctly and obtain e−3 y and sin 2 2x on the opposite B1 sides 1 B1 Obtain term − e −3 y 3 Use correct double angle formula for sin 2 2x = (1/2)[1 – cos 4x] M1 1 1 A1 Obtain terms x − sin 4 x oe 2 4 Use x = 0, y = 0 to evaluate a constant or as limits in a solution containing M1 terms of the form ax and b sin 4x and ce±3 y Obtain correct answer in any form A1 1 1 1 1 e.g. − e−3 y = x − sin 4 x − 3 2 4 3 Substitute x = 12 and obtain y = 0.175 or − 13 ln ( 14 + 83 sin 2) A1 OE ISW 7

Mark scheme

10(a) Carry out correct process for evaluating the scalar product of OA and OB M1 ± (3, − 1, 2).(1, 2, − 3) = ±(3 – 2 – 6) = [− 5]. Using the correct process for the moduli, divide the scalar product by the A1 product of the moduli and obtain cos-1{±(3 – 2 – 6)/[ √ (32 + (–1)2 +22) √ (12 + 22 + (–3)2)]} Obtain answer 110.9° or 1.94c A1 3 10(b) Use a correct method to form an equation for line through AB M1 Obtain r = 3i – j + 2k + μ1 (2i – 3j + 5k) A1 OE e.g. r = i + 2j – 3k + μ2 (–2i + 3j – 5k). Need r or (x, y, z). 2 10(c) Obtain a correct equation for line through CD B1 OE e.g. [r = ] 5i – 6j + 11k + λ2(–4i + 4j – 6k). e.g. [r = ] i – 2j + 5k + λ1(–4i + 4j – 6k) r can be omitted or another symbol used. Equate two pairs of components of general points on their l and their CD M1 and solve for λ or for μ Obtain e.g. λ1 = –2 or μ1 = 3 or λ2 = –1 or μ2 = − 4 A1 Obtain position vector 9i – 10j + 17k A1 Condone (9, –10, 17) but not (9i, – 10j, 17k). 4

Mark scheme

11(a) Ax + B C B1 State or imply the form + 4+ x 2 1+ x Use a correct method for finding a coefficient M1 (Ax + B)(1 + x) + C(4 + x2) = 5x2 + x + 11. Obtain one of A = 2, B = –1 and C = 3 A1 If error present in above still allow A1 for C. Obtain a second value A1 Obtain the third value A1 If A = 0 then max M1 A1 (for C). 5 A x ( ) 11(b) B B1FT + The FT is on A, B and C. Integrate and obtain terms ln 4 + x 2 + tan −1 + C ln (1 + x ) 2 2 2 B1FT + Ax + B B −1 x 2 B1FT Integral of 4 + x 2 = 2 tan 2 or (A/2)ln(4 + x ) only. B0FT unless clearly from single term. Substitute limits 0 and 2 correctly in an integral of the form M1 2 a ln ( 4 + 4 ) + btan −1 + c ln (1 + 2 ) − [ x ( ) a ln 4 + x 2 + btan −1 + c ln (1 + x ) , 2 2 a ln 4 + btan 0 + c ln (1)] . −1 where abc ≠ 0 Allow one sign or substitution error. Allow omission of btan −1 0 + c ln (1) . π A1 AG – work to combine or simplify at least 2 of ln terms is Obtain answer ln 54 − after full and correct working required CWO. 8 A0 if any non exact value(s) seen. 5