Mathematics 9709/33 May/June 2023: question paper and mark scheme

Mark scheme

1 Use exponentials or law for the logarithm of a product, quotient or power M1* eln(5+ x ) = e5+ln x insufficient. x+5 Need e.g. ln x ( ) = 5 or ln ( x + 5 ) = ln e + ln x 5 or ln(x + 5) = ln(e5x) or x + 5 = e5 + lnx or x + 5 = e5 elnx and others. Correctly remove logarithms DM1 Obtain a correct equation in x A1 x+5 5 e.g. = e (or 148.4…) or x + 5 = xe5. x Obtain 0.034 A1 CAO Final answer must be 3d.p. 4

Mark scheme

2 Divide to obtain quotient 2 x 2 ± 2 x + k (k ≠ 0) M1 Obtain result in answer column, together with a linear polynomial or a constant as remainder. If correct: Obtain [quotient] 2 x 2 − 2 x − 4 A1 Allow unless quotient and remainder interchanged, then A0 A1. Obtain [remainder] 10 x − 15 A1 Allow (x2 + x + 3)(2x2 – 2x – 4) + 10x – 15. Alternative Method for Question 2 ( )( ) Expand x 2 + x + 3 Ax 2 + Bx + C + ( Dx + E ) and reach A = 2, B = ± 2, M1 Solve all 3 equations for A, B and C, allow sign errors in establishing equations and in solving. C=k If correct, A = 2, A + B = 0, 3A + B + C = 0, 3B + C + D = 0, 3C + E = − 27. Obtain result in answer column, together with a linear polynomial or a constant as remainder. Obtain [quotient] 2 x 2 − 2 x − 4 A1 Allow unless quotient and remainder interchanged, then A0 A1. Obtain [remainder] 10 x − 15 A1 Allow (x2 + x + 3) (2x2 – 2x – 4) + 10x – 15. 3

Mark scheme

3 Show a circle with centre 3 + i B1 Must be some evidence of scale on both axes or centre stated as 3 + i or (3, 1). Show a circle with radius 3 and centre not at the origin B1 Must be some evidence that radius = 3 or stated r = 3 Show the line y = 2 B1 Line y = 2 can be represented by 2 or correct dashes. Shade the correct region B1 Line and circle must be correct. Im 4 Scales may be replaced by dashes on axes for all marks. Correct figure, with no scale on either axis then allow 1/3 and the B1 for correct shaded region Max 2/4. If B0 above for line but relatively correct position then 2 B1 for correct shaded region Max 3/4. 1 Re and Im axes interchanged but clearly labelled, allow 3 Re SCB1 for centre and radius of circle correct and SCB1 for line and shading correct Max 2/4.

Mark scheme

4 dy B1 Ignore left side throughout dx/dt, dy/dt, dx, dy State = 1 − 2sin θ dθ dy but must see for final A1. dx Use correct quotient rule, or product rule if rewrite x as cos θ ( 2 – sin θ ) − 1 M1 Incorrect formula seen M0 A0 otherwise BOD. dx − ( 2 − sin θ ) sin θ + cos θ A1 − sin θ (2 – sin θ)–1 – cos θ (2 – sin θ )–2(− cos θ) or 2 Obtain = o.e. equivalent. dθ ( 2 − sin θ ) 2 dy dy dx M1 dy Use = ÷ 1 − 2sin θ dx dθ dθ = (1 − 2sin θ ) ÷ . dx ( 2 − sin θ ) 2 Allow M1 even if errors in both derivatives. dy A1 AG – must see working in above cell to gain final A1. = ( 2 − sin θ ) . 2 Obtain Allow cos2 θ + sin2 θ = 1 to be implied. dx x instead of θ or missing θ more than twice on right side then A0 final mark. 5

Mark scheme

5(a) Use correct product rule M1 d 2 d (x )cos(3x) + x2 (cos 3x). dx dx Obtain correct derivative in any form A1 e.g. 2 x cos3 x − 3 x 2 sin 3 x . 1 2 A1 1 2 Equate derivative to zero and obtain a = tan −1 . AG Condone a = tan −1 . 3 3a 3 3a Must at least reach expression 2x = 3x2 tan(3x) or better before final answer to gain A1. Final answer must be in terms of a. Can work with x and switch to a at very end. 2 2 Look for a or x in working not immediately 3 3 corrected or as penultimate line A0. 3 5(b) 1 2 M1 Degrees 0/3. Use the iterative process an +1 = tan −1 correctly at least twice during 3 3an successive iterations in the numerous iterations Obtain final answer 0.36 A1 Must be 2d.p. Show sufficient iterations to 4 or more d.p. to justify 0.36 to 2 d.p. or show A1 Allow small errors in 4th d.p. there is a sign change in the interval ( 0.355, 0.365 ) Allow errors at start if self corrects later. 0.5 0.4 0.3 0.2 0.1 π/6 π/12 3 0.3091 0.3435 0.3826 0.4264 0.4740 0.3017 0.3989 0.3789 0.3650 0.3499 0.3339 0.3176 0.3820 0.3439 0.3513 0.3566 0.3625 0.3688 0.3754 0.3502 0.3649 0.3619 0.3599 0.3576 0.3552 0.3526 0.3624 0.3567 0.3578 0.3604 0.3614 0.3576 0.3580

Mark scheme

6(a) Expand cos ( x − 60° ) correctly and evaluate 3 cos x + 2 cos(x − 60°) B1 Need to see A cos x + B sin x with A and B correct to obtain 4cos x + 3 sin x or unsimplified coefficients A may be 4 or 3 + 2 cos 60 and B may be 3 or 2 sin 60. State R = 19 [R cos α = 4 R sin α = 3] B1 FT Follow through their 4 and 3 . If coefficients are 3 and 2 then B0. R = 19 from R = 4.36 B0 but 4.36 seen after 19 ISW. Use correct trig formulae for their expansion to find α M1 If sin α = 3 cos α = 4 seen then M0 A0. 3 4 3 If tan α = 23.41° M0 A0 but can recover if α = 23.41° e.g. α = tan−1 or cos−1 or sin−1 4 19 19 seen later. 2 3 α = tan−1 M1 (α = 33.69° ) but α = tan−1 M0 3 2 Obtain α = 23.41° A1 Allow if x instead of α. 4 6(b) 2.5 B1 FT SOI [55.0°]. cos −1 R Follow through their 19 . Use a correct method to find a value of 2θ (not x) in the interval. M1 2.5 Allow sign error in moving α to right side 2θ = cos −1 + 23.41° R 2.5 or 2θ = 360° − cos −1 + 23.41° R with R substituted. Obtain one correct answer e.g. 39.2° A1 If working for M1 not seen then M1 implied by 39.2° or 164.2° Must be at least 1d.p. Obtain second correct answer e.g. 164.2° and no others in the interval A1 Must be at least 1d.p. Ignore answers outside the given interval. 4

Mark scheme

7(a) du B1 SOI = − sin x dx Use double angle formula and substitute for x and dx throughout the integral M1 All x’s must be removed, can be coefficient errors provided 2 seen in working. A1 Limits may be omitted, or left as 0 and π, during the Obtain ± 2ue 2u du change of variable stage. 1 A1 AG Must see x = 0, u = 1 and x = π, u = −1. 2ue du from correct working 2u Justify new limits and obtain −1 Inequalities alone e.g. 0 ⩽ x ⩽ π and 1 ⩽ u ⩽ −1 or –1 ⩽ u ⩽ 1 for limits are insufficient A0 If sign in expression and order of limits incorrect then A0. If negative sign is present in the integrand then this can be removed and limits introduced in correct order in a single step. 4 7(b) Commence integration and reach aue 2u + b e 2u du, where ab ≠ 0, b < 0 M1* Condone dx. Complete integration and obtain ue 2u − 12 e 2u A1 1 1 OE Allow (2 u e 2u ) − e 2u . 2 2 Use correct limits correctly in c ue 2u + d e2u having integrated twice DM1 1 and −1 for u, 0 and π for x e.g. ce2 + de2 − (− ce−2 + de−2). Not decimals. or in c cos x e2 cos x + d e2cos x Allow one sign error at most in going from c ue 2u + d e2u or c cos x e2 cos x + d e2 cos x to ce2 + de2 − (− ce−2 + de−2). [e2 − ½ e2 − (− e−2 – ½ e−2)] Complete reversal of sign by converting back to cos x and not making x = 0 upper limit is DM0 A0. 1 2 3 −2 A1 ISW Obtain e + e 2 2 e4 + 3 Or equivalent 2-term expression e.g. or 2e 2 1 2 3 e + 2 . 2 e 4

Mark scheme

8 Separate the variables correctly B1 y+4 1 2 dy = dx . y +4 x Obtain ln x B1 ( ) Split the fraction and integrate to obtain p ln y 2 + 4 or q tan −1 2y correctly *M1 Only following subdivision into y 2 + 2 4 . y +4 y +4 If no subdivision seen then both terms ( ) y p ln y 2 + 4 and q tan −1 must be present. 2 Obtain 12 ln y 2 + 4 ( ) A1 Obtain 2 tan −1 2y A1 ( ) Use 4, 2 3 in an expression containing at least 2 of DM1 2π Allow one sign or arithmetic error e.g. . 3 ( ) a ln x, b ln y 2 + 4 and c tan −1 2y to obtain constant of integration ( ) May use 4, 2 3 and (x, 2) as limits to find x for the final 3 marks. Correct solution (any form) A1 However solution not asked for so allow 1 ( y ) e.g. ln y 2 + 4 + 2 tan −1 = ln x + 2 2 2π 3 1 2 1 ln8 + 2 tan −1 1 = ln x + 2 tan −1 3 + ln16 − ln 4 . 2 1 ( y ) 1 or ln y 2 + 4 + 2 tan −1 = ln x + 2 tan −1 3 + ln16 − ln 4 2 2 2 1 − π 1 − π 8 A1 ISW Must remove ln so x = e (ln2√2 − π/6) A0. 0.516 Obtain 8e 6 or 1.68 or more accurate or 2 2e 6 or 1 or e π e6 8 Alternative method for first *M1 A1 A1 y y *M1 Allow sign error. p ( y + 4 ) tan −1 − tan −1 dy 2 2 1 y y y y A1 ( y + 4 ) tan −1 − tan −1 + 2 dy 2 2 2 2 y +4 ( Obtain 2 tan −1 2y + 12 ln y 2 + 4 ) A1 8

Mark scheme

9(a) Perform scalar product of direction vectors and set result equal to zero M1 2c + 6 + 4 = 0 . Use P to find the value of λ M1 3 − 2λ = 7 λ = −2 [a + λ c = 4, b + 4 λ = − 2]. Equation for line l may contain – λ instead of + λ leading to λ = 2 all marks available. Obtain c = −5 or b = 6 A1 a = −6, b = 6 and c = –5 all correct A1 4 SC1: Use P to find the value of λ M1 Substitute λ = –2 into point P, so a − 2c = 4, and put μ = − 1 and λ = − 1 into l so a − c = − 1, then solve to obtain a = −6 , b = 6 and c = –5. All 3 values correct A1. Max 2/4. ρ ρ 9(b) Find PQ (or QP) for a general point Q on m B1 −3 + 2μ = ± ((1 + 2 μ , 2 − 3 μ , 3 + μ ) – (a + λ c, 3 − 2 λ , b + 4 λ )) ρ ρ PQ or QP = ± −5 − 3μ 5 + μ Could be their a, b, c and λ values provided M1 M1 gained in (a). Allow expression in answer column. ρ ρ Equate the scalar product of PQ (or QP) and a direction vector for m to M1* ( 2 ( −3 + 2μ ) − 3( −5 − 3μ ) + ( 5 + μ ) ) = 0 . zero and obtain an equation in μ ρ ρ ρ Allow PQ = OQ + OP sign problem. Solve and obtain μ = −1 A1 PQ2 = ( −3 + 2 μ ) 2 + ( −5 −3μ )2 + ( 5 + μ )2. [= 14( μ + 1)2 + 45]. Min when μ = − 1 or by differentiation. ρ ρ Obtain OQ = −i + 5 j + 2k or PQ = −5i − 2 j + 4k A1 The working may be in (a) provided at least this result is Must be labelled correctly used in (b). Carry out a method to find the position vector of R DM1 ρ ρ 5 ρ ρ ρ 3 ρ e.g. Use OR = OP + PQ or OR = OQ + PQ or 2 2 ρ 5 ρ 3 ρ ρ ρ ρ ρ Alternative method for DM1 OR = OQ − OP or 2QR =2( OR − OQ ) = 3 PQ ρ ρ ρ ρ 2 2 OR = (4, 7, − 2) + t (− 5, − 2, 4) QR = OR − OQ ρ where OR = (x, y, z). 9 3 ρ Solve QR 2 = | PQ | 2 or QR = | PQ | t = 2.5 PQ used in all these approaches, may be incorrect, must 4 2 ρ ρ be in the correct direction, i.e. not using QP for PQ . 9(b) 17 A1 Accept coordinates. Obtain − i + 2 j + 8k from correct working 2 17 4 16 Don’t accept − i + j + k . 2 2 2 6 SC2 Equate lines, attempt to find μ = −1 or λ = −1 M1* ρ OQ = −i + 5 j + 2k A1. ρ Attempt to find OQ using other parameter value DM1. ρ OQ = −i + 5 j + 2k therefore intersect A1. Then use main scheme for the final DM1 A1. First DM1 A1 are available if they show the 3 coordinates are consistent for the 2 parameter values ρ instead of attempting to find OQ using the other parameter value and then showing intersection

Mark scheme

10(a) A B C B1 A Dx + E State or imply the form + + Alternative form: + . 1 + 2 x 3 − x ( 3 − x )2 1 + 2 x ( 3 − x )2 Use a correct method to find a constant M1 Incorrect format for partial fractions: Allow M1 and a possible A1 if obtain one of these correct values. Max 2/5 Allow M1 even if multiply up by (1 + 2x)(3 – x)3. Obtain one of A = 2, B = 2 and C = −3 A1 Alternative form: obtain one of A = 2, D = −2 and E = 3 . Obtain a second value A1 Obtain the third value A1 Do not need to substitute values back into original form. 5 A B Cx + D If + + B0 but M1 A1 for A, A1 for B 1 + 2 x 3 − x ( 3 − x )2 and A1 for C and D. If C = 0 then recovers B1 from above. 10(b) Use a correct method to obtain the first two terms of one of the unsimplified M1 (1 + 2 x ) −1 = 1 + (−1)(2x) + … expansions −1 −1 −2 1 (1 + 2 x )−1 , 1 − x , 1 − x ( 3 − x ) , ( 3 − x ) 1 1 −1 −2 1 − x =1 + (-1)(-x/3) +… 3 3 3 −2 1 1 − x = 1 + (−2)(−x/3) + ... 3 ( 3 − x ) = 3−1 + (−1)3−2 (−x) … −1 ( 3 − x ) = 3−2 + (−2)3−3(−x) + … −2 Obtain the correct unsimplified expansions up to the term in x 2 for each A1 FT Follow through on their A, B, C partial fraction A(1 + (−1)(2x) + ((−1)(−2)/2)(2x)2 + …) A1 FT B (1 + (−1)(−x/3) + ((−1)(−2)/2)(−x/3)2 + …) If correct, should be working with 3 A1 FT 2 2 3 2 −2 x + 3 C + − or + (1 + (−2)(−x/3) + ((−2)(−3)/2)(−x/3)2 + …). 1 + 2x 3 − x (3 − x) 2 1 + 2 x ( 3 − x )2 32 Must be their coefficients from (a) but may be unsimplified expansions for FT marks. If correct, expect to see 2(1 − 2x + (2x)2 ) or 2 − 4 x + 8 x 2 2 2 x x 2 2 2 2 (1 + + ) or + x + x 3 3 3 3 9 27 2 1 2x x 1 2 x2 − (1 + + (3) ) or − − x − . 3 3 3 3 9 9 Obtain final answer 7 − 4 x + 215 x 2 A1 Accept 2 1 − 4 x + 7 26 x 2 . No ISW. 3 27 3 27 10(b) Alternative Method for Question 10(b) A Dx + E M1* For the first two terms of an expanded partial fraction, For the form + 1 + 2x (3 − x ) 2 following their A, D, E. Obtain the correct unsimplified expansions up to the term in x 2 for each A1FT A(1 + (−1)(2x) + ((−1)(−2)/2)(2x)2 + …) + partial fraction 1 2 A1FT (Dx + E) 32 (1 + (−2)(−x/3) + ((−2)(−3)/2)(−x/3) + ) 2(1 − 2x + (2x)2 + …) 2 −2 x + 3 2x x + (1 + + (3) + …). 32 3 3 Multiply out fully DM1 Provided DE ≠ 0 . Ignore cubic terms and above. Allow error in one term but all terms must be present. If correct, expect to see 2 4 2 1 2 1 2 − 4 x + 8x2 − x − x + + x + x2 9 27 3 9 9 Obtain final answer 73 − 4 x + 215 x2 A1 Accept 2 1 − 4 x + 7 26 x 2 . No ISW 27 3 27 10(b) Alternative Method for Question 10(b): Maclaurin’s Series Correct derivatives for A(1 + 2x)−1, B(3 – x)−1 and C(3 – x)−2 B1 FT (−1)(2)A(1 + 2x)−2, (−1)(−1)B(3 – x)−2 and (−2)(−1)C(3 – x)−3 One of following (−2)(2)(−1)(2)A(1 +2x)−3, (−2)(-1)(−1)(−1)B(3 − x)−3 and B1 FT (−3)(−1)(−2)(−1)C(3 – x)−4 All correct B1 FT x2 M1 Substitute in f (0) + xf '(0) + f ''(0) 2 Obtain final answer 7 − 4 x + 215 x 2 A1 Accept 2 1 − 4 x + 7 26 x 2 . No ISW 3 27 3 27 5

Mark scheme

11(a) Multiply numerator and denominator by ( 3 − ai ) M1 Must perform complete multiplications but need not simplify i2. Can have errors but no term duplicated or missing. ( ( 5a – 2i )( 3 – ai ) 13a − i 5a + 6 = 2 ) M0 M1 A0 9 − a2 9 − a2 No working so unsure if denominator multiplied by 3 – ai M1 M1 A0 Use i 2 = −1 at least once and separate real and imaginary parts M1 Obtain ( 13a − i 5a 2 + 6 ) or 13a − 5a i − 6i 2 A1 OE If 15a – 2a = 13a seen later award this A1. 9 + a2 9 + a2 Use arg z to form equation in a M1 Allow expression given in answer column 2 5a + 6 π 13a π or 5a 2 + 6 = ±13a − = ± tan ± or − 2 = ± tan ± or use − (x ± xi) = (13a – i(5a2 + 6))/(9 + a2) 13a 4 5a + 6 4 and eliminate x so 5a2 + 6 = ± 13a M1. 5a 2 + 6 π −1 13a π or tan −1 − = ± or tan − 2 =± 13a 4 5a + 6 4 Obtain a = 2 A1 3 Need to reject a = or ignore it in future work. 5 3 May not see second root, but if present, must be . 5 Obtain z = 2 − 2i only A1 Allow z = − 2i + 2. 11(a) Alternative Method 1 for the first four marks arg z = arg (5a – 2i) – arg (3 + ai) M1 −2 a M1 Allow one sign error in second M1. = tan −1 − tan −1 5a 3 = tan −1 −2 a −2 a − / 1 + 5a 3 5a 3 5a 2 + 6 A1 −1 13a = tan −1 − or tan − 2 13a 5a + 6 5a 2 + 6 M1 5a 2 + 6 π −1 13a π ± = tan −1 − or tan − 2 Equate their tan −1 − to ± . 4 13a 5a + 6 13a 4 Then as original scheme for final 2 marks. Alternative Method 2 for the first four marks (x + iy)(3+ ai) = 5a – 2i M1 A1 3x – ay = 5a and ax + 3y = − 2 2 5a M1 x = ± y Find x or y in terms of a, e.g. x = or x = 3− a 3+ a 2 2 M1 Then as original scheme for final 2 marks. Substitute in other equation, for example 3 + a = 5a 3− a 3− a 6 11(b) ( ) State arg z 3 = − 34 π or evaluate from z = b – bi or from – 2b3(1 + i) B1 If 2 different values given award B0. Do not ISW. Complete method to obtain r from their z M1 ( x +y ) . 3 z3 = 2 2 3 If z correct, may see z 3 = 22 + ( −2 ) 2 or z 3 = ( −16 )2 + ( −16 )2 . r = 16 2 A1 CAO A1 if z = 2 – 2i obtained correctly. or z = used with a = 2 found correctly, otherwise A0XP. 3 − πi May see arg and r given in a final answer i.e. 16 2e 4 . Allow this form for arg and r to collect full marks, even if i missing. Ignore answers outside the given interval. If 2 different values given award A0. 3