Cambridge A Level Mathematics 9709 — 2023 May/June Paper 3 · Variant 3

2 Find the quotient and remainder when 2x4 is divided by x2 x 3. [3] −27 + + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Divide to obtain quotient 2 2 2   x 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 2 2 4   x x A1 Allow unless quotient and remainder interchanged, then A0 A1. Obtain [remainder] 10 15  x A1 Allow (x2 + x + 3)(2x2 – 2x – 4) + 10x – 15. Alternative Method for Question 2 Expand     2 2 3       x x Ax Bx C Dx E and reach A = 2, B = ± 2, C = k M1 Solve all 3 equations for A, B and C, allow sign errors in establishing equations and in solving. 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 2 2 4   x x A1 Allow unless quotient and remainder interchanged, then A0 A1. Obtain [remainder] 10 15  x A1 Allow (x2 + x + 3) (2x2 – 2x – 4) + 10x – 15. 3

More questions on Quadratics

3 On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z and z z . [4] −3 −i ≤3 ≥ −4i

Mark scheme: 3 Show a circle with centre 3 i 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 2  y B1 Line y = 2 can be represented by 2 or correct dashes. Shade the correct region B1 Line and circle must be correct. 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 B1 for correct shaded region Max 3/4. Re and Im axes interchanged but clearly labelled, allow SCB1 for centre and radius of circle correct and SCB1 for line and shading correct Max 2/4. 2 Im Re 1 3

More questions on Complex numbers

5 y M x O a 6π1 The diagram shows the part of the curve y x2 cos 3x for 0 and its maximum point M, where = ≤x ≤16π, x a. = @ A 1 2 (a) Show that a satisfies the equation a = 3 tan−1 3a . 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(b) Use an iterative formula based on the equation in (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 5(a) Use correct product rule M1 d dx (x2)cos(3x) + x2 d dx (cos 3x). Obtain correct derivative in any form A1 e.g. 2 2 cos3 3 sin3  x x x x . Equate derivative to zero and obtain 1 1 2 tan . 3 3        a a A1 AG Condone 1 1 2 tan 3 3   a a . 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. Look for 2 3 a or 2 3 x in working not immediately corrected or as penultimate line A0. 3 5(b) Use the iterative process 1 1 1 2 tan 3 3          n n a a correctly at least twice during successive iterations in the numerous iterations M1 Degrees 0/3. 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 there is a sign change in the interval   0.355, 0.365 A1 Allow small errors in 4th d.p. Allow errors at start if self corrects later. 0.5 0.4 0.3 0.2 0.1 /6 /12 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 3

More questions on Differentiation

7 (a) Use the substitution u cosx to show that = 1 π cos x sin 2x e2 dx 2ue2u du. [4] Ó 0 = Ó −1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ π cos x(b) Hence find the exact value of sin 2x e2 dx. 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Mark scheme: 7(a) d sin d  u x x B1 SOI 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. Obtain 2 2 e d  u u u A1 Limits may be omitted, or left as 0 and , during the change of variable stage. Justify new limits and obtain 1 2 12 e d  u u u from correct working A1 AG Must see x = 0, u = 1 and x = , u = −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 Question Answer Marks Guidance 7(b) Commence integration and reach 2 2 e e d ,  u u au b u where 0,  ab 0  b M1* Condone dx. Complete integration and obtain 2 2 1 2 e e  u u u A1 OE Allow (2 2 2 1 1 e ) e 2 2  u u u . Use correct limits correctly in c 2 e  u u d e2u having integrated twice or in c cos x e2 cos x + d e2cos x DM1 1 and −1 for u, 0 and π for x e.g. ce2 + de2 − (− ce−2 + de−2). Not decimals. Allow one sign error at most in going from c 2 e  u u 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. Obtain 2 2 1 3 e e 2 2   A1 ISW Or equivalent 2-term expression e.g. 4 2 e 3 2e  or 2 2 1 3 e 2 e        . 4

More questions on Integration

9 The lines l and m have equations l : r ai 3j bk ci 4k , = + + + , −2j + m : r i 2j 3k 2i k . = + + + - −3j + Relative to the origin O, the position vector of the point P is 4i 7j + −2k. (a) Given that l is perpendicular to m and that P lies on l, find the values of the constants a, b and c. 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(b) The perpendicular from P meets line m at Q. The point R lies on PQ extended, with PQ : QR 2 : 3. = Find the position vector of R. 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Mark scheme: 9(a) Perform scalar product of direction vectors and set result equal to zero M1 2 6 4 0 c Use P to find the value of  M1 3 2 7 2       [a + c = 4, b + 4  = − 2]. Equation for line l may contain – instead of + leading to  = 2 all marks available. Obtain 5  c or b = 6 A1 a = −6, b = 6 and c = –5 all correct A1 4 SC1: Use P to find the value of  M1 Substitute  = –2 into point P, so a − 2c = 4, and put  = − 1 and  = − 1 into l so a − c = − 1, then solve to obtain 6  a , 6  b and c = –5. All 3 values correct A1. Max 2/4. Question Answer Marks Guidance 9(b) Find  PQ (or )  QP for a general point Q on m = ± ((1 + 2 , 2 − 3 , 3 + ) – (a + c, 3 − 2, b + 4 )) B1 3 2 or 5 3 5                             PQ QP Could be their a, b, c and  values provided M1 M1 gained in (a). Allow expression in answer column. Equate the scalar product of  PQ (or )  QP and a direction vector for m to zero and obtain an equation in  M1*        2 3 2 3 5 3 5 0          . Allow  PQ =  OQ +  OP sign problem. Solve and obtain 1  A1 PQ2 =   3 2  2 + ( 5 3  )2 + (5   )2. [= 14(  + 1)2 + 45]. Min when  = − 1 or by differentiation. Obtain 5 2    OQ i j k or 5 2 4     PQ i j k Must be labelled correctly A1 The working may be in (a) provided at least this result is used in (b). Carry out a method to find the position vector of R Alternative method for DM1  OR = (4, 7, − 2) + t (− 5, − 2, 4)  QR =  OR −  OQ Solve QR 2 = 9 4 | | PQ 2 or QR = 3 2 | | PQ t = 2.5 DM1 e.g. Use 5 2      OR OP PQ or 3 2      OR OQ PQ or 5 3 2 2      OR OQ OP or 2  QR =2( )    OR OQ = 3  PQ where  OR = (x, y, z).  PQ used in all these approaches, may be incorrect, must be in the correct direction, i.e. not using  QP for  PQ . Question Answer Marks Guidance 9(b) Obtain 17 2 8 2    i j k from correct working A1 Accept coordinates. Don’t accept 17 4 16 2 2 2    i j k . 6 SC2 Equate lines, attempt to find = −1 or  = −1 M1* 5 2    OQ i j k A1. Attempt to find  OQ using other parameter value DM1. 5 2    OQ i j k 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 instead of attempting to find  OQ using the other parameter value and then showing intersection

More questions on Vectors

5a11 The complex number z is defined by z −2i , where a is an integer. It is given that arg z = 3 ai = −14π. + (a) Find the value of a and hence express z in the form x iy, where x and y are real. 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........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 0 and Give the simplified exact values of(b) Express z3 in the form rei1, where r > −π < 1 ≤π. r and [3] 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 11(a) 3 i a M1 Must perform complete multiplications but need not simplify i2. Can have errors but no term duplicated or missing.    2 5 – 2i 3 – i 9  a a a   2 2 13 5 6 9     a i a a M0 M1 A0 No working so unsure if denominator multiplied by 3 – ai M1 M1 A0 Use 2i 1 at least once and separate real and imaginary parts M1 Obtain   2 2 13 i 5 6 9    a a a or 2 2 13 5 i 6i 9    a a a A1 OE If 15a – 2a = 13a seen later award this A1. Use arg z to form equation in a 2 2 5 6 π 13 π tan or tan 13 4 4 5 6                     a a a a or 2 1 1 2 5 6 π 13 π tan or tan 13 4 4 5 6                     a a a a M1 Allow expression given in answer column or 2 5 6 13   a a or use − (x ± xi) = (13a – i(5a2 + 6))/(9 + a2) and eliminate x so 5a2 + 6 = ± 13a M1. Obtain 2  a A1 Need to reject a = 3 5 or ignore it in future work. May not see second root, but if present, must be 3 5 . Obtain 2 2i   z only A1 Allow z = − 2i + 2. Question Answer Marks Guidance 11(a) Alternative Method 1 for the first four marks arg z = arg (5a – 2i) – arg (3 + ai) M1 = 1 1 2 tan tan 5 3                 a a 1 tan 2 2 / 1 5 3 5 3                            a a a a M1 Allow one sign error in second M1. 2 1 1 2 5 6 13 tan or tan 13 5 6                    a a a a A1 π 4  = 2 1 5 6 tan 13         a a 1 2 13 or tan 5 6         a a M1 Equate their 2 1 5 6 tan 13         a a to π 4  . Then as original scheme for final 2 marks. Alternative Method 2 for the first four marks (x + iy)(3+ ai) = 5a – 2i 3x – ay = 5a and ax + 3y = − 2 M1 A1 x = ± y Find x or y in terms of a, e.g. 2 3   x a or 5 3   a x a M1 Substitute in other equation, for example 2 2 3 5 3 3                 a a a a M1 Then as original scheme for final 2 marks. 6 Question Answer Marks Guidance 11(b) State   3 3 4 arg  z 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   3 3 2 2   z x y . If z correct, may see   3 2 3 2 2 2         z or 3z =     2 2 16 16   . 16 2  r A1 CAO A1 if z = 2 – 2i obtained correctly. or z = used with a = 2 found correctly, otherwise A0XP. May see arg and r given in a final answer i.e. 3 i 4 16 2e   . 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

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