Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 3 · Variant 2

1 1 Expand ( 9 - 3)x 2 in ascending powers of x, up to and including the term in x2, simplifying the coefficients. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................. ............................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 1 Obtain a correct unsimplified version of the x or x 2 term of the expansion of M1 1 1 1 2  12 1   E.g. − x or − x 2 or 1  1  2 2 3 2 9 ( 9 −x3 ) 2 or  1 − x  −1 1 −1 −3  3  1 2 1 2  2 2 9 ( −3 x ) or 9 ( −3 x ) 2 . 2 2 Not for symbolic coefficients in the form n C r . State correct first term 3 B1 1 1 2 A1 A1 A1 for each term correct. Obtain the next two terms − x − x Do not ISW. 2 24 1 1 2 SC M1A1 for 1 − x − x seen on its own or 6 72 as a factor. 4

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2 (a) By sketching a suitable pair of graphs, show that the equation cot 2x = sec x has exactly one root in the interval 0 1 x 1 1 r . [2] 2 (b) Show that if a sequence of real values given by the iterative formula 1 -1 x = tan ( cos x ) n + 1 2 n converges, then it converges to the root in part (a). [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 2(a) Sketch a relevant graph, e.g. y = cot 2x B1 Alt: use tan2xand cosx . And only one root in π range. (also cross at 2 ) Sketch a second relevant graph on the same axes, e.g. y = sec x and justify the given B1 Need to mark intersection with a dot, a cross, or statement say roots at points of intersection, OE. 2 2(b) 1 −1 B1 Should see tan2 x = cos x before the given State x = tan ( cos x ) 2 conclusion. and rearrange to the given equation cot 2x = sec x 1 −1 Or rearrange cot 2x = sec x to x = tan ( cos x ) 2 and state iterative formula 1 −1 xn +1 = tan ( cos xn ) . Note: If using the alternative approach in (a), can stop at tan2 x = cos x 2 1

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3 The square roots of 6 - 8i can be expressed in the Cartesian form x + yi , where x and y are real and exact. By first forming a quartic equation in x or y, find the square roots of 6 - 8i in exact Cartesian form. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 Square x + iy and equate real and imaginary parts to 6 and −8 respectively *M1 Condone +8 in place of -8 and/or 2i = 1. Obtain equations x2 – y2 = 6 and 2xy = –8 A1 OE Eliminate one variable and find an equation in the other (from 2 equations each in 2 DM1 Condone a slip but not seriously incorrect unknowns) algebra, e.g. use of x = −4 y is M0. Obtain x4 – 6x2 – 16 = 0 or y4 + 6y2 – 16 = 0 A1 Accept 3-term equivalents e.g. x 4 = 6 x 2 + 16. Condone missing ‘= 0’ if implied by subsequent work. Obtain answers  2 2 − 2i or exact equivalents A1 Allow if values of x and y stated separately but ( ) the pairing is clear. Ignore additional correct solutions for x and y not real, but A0 if any additional incorrect answers. 5

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4 Solve the equation 5 x = 5 x + 2 - 10 . Give your answer correct to 3 decimal places. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 Use laws of indices correctly and solve for 5x *M1 x 5 E.g. obtain 5 = OE. 12 Allow for y = … if they have previously stated y = 5 x . Could be implied if they have a correct simplified equation in 5x, e.g. 12  5 x = 5. Use a correct method for solving an equation of the form 5 x = a , where a > 0 DM1  10  Allow x ln5 = ln   .  24  Obtain answer – 0.544 A1 CWO. If no working shown, 0/3. Note: 3 dp required. 3

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5 (a) The complex number u is given by ( cos 1 r + isin 1 r) 4 u = 7 7 . cos 1 r - isin 1 r 7 7 Find the exact value of arg u. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The complex numbers u and u* are plotted on an Argand diagram. Describe the single geometrical transformation that maps u onto u* and state the exact value of arg u*. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 5(a) 4   M1 SOI and/or − 4  1  4 1 7 7 Allow −− π  or − π 7  7  7 7 Note: Many multiply top and bottom by the conjugate, which is fine, but to score the M1 they need to state or imply the argument of a complex number. 5 A1 Do not accept degrees. Obtain arg u = π 7 2 5(b) Reflection (in the) real axis B1 Correct non-contradictory statement. Condone x–axis or horizontal axis. Need ‘reflection’. Not ‘mirror’, ‘flip’. 5 B1FT FT their exact (a). Accept 2π− their exact (a). arg u* = − π 7 Accept an ‘exact’ expression in place of an exact value. Need to see a value or an expression. Do not accept arg u* = − arg u without a value seen. 2

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6 ln y (3.40, 8.27) (0.50, 2.24) O x The variables x and y satisfy the equation ay = bx , where a and b are constants. The graph of lny against x is a straight line passing through the points (0.50, 2.24) and (3.40, 8.27), as shown in the diagram. Find the values of a and b. Give each value correct to 1 significant figure. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 6 Form a pair of equations in a and b *M1 Condone sign slips but must be using the given coordinates correctly. ln a + 8.27 = 3.4ln b e.g.  ln a + 2.24 = 0.5ln b ae 2.24 = b 0.5 or  8.27 3.4  ae = b Carry out a correct method for finding lna or lnb or a or b DM1 Condone sign slip. Obtain value a = 0.3 A1 (0.30109…) Obtain value b = 8 A1 (7.99895…) Allow A0A1 if both values ‘correct’ but not rounded to 1 sf. Allow 4/4 for 0.3 y = 8 x with correct working shown. 6 Alternative Method for Question 6: Carry out a correct method for finding lnb or b *M1 Condone sign slips but must be using the given (Need to link the gradient to lnb at some point) coordinates correctly. 8.27 − 2.24 ln b = ( = 2.079.... ) 3.4 − 0.5 Obtain value b = 8 A1 Correct method to find  ln aor a DM1 Condone sign slip ( ln a = − 1.200... ) . Obtain value a = 0.3 A1 Allow A0A1 if both values ‘correct’ but not rounded to 1 sf. Allow 4/4 for 0.3 y = 8 x with correct working shown. 4

More questions on Logarithmic and exponential functions

7 (a) Show that the equation tan 3 x + 2 tan 2x - tan x = 0 may be expressed as tan 4 x - 2 tan 2 x - 3 = 0 for tanx ! 0 . 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(b) Hence solve the equation tan 3 2 i + 2 tan 4i - tan 2i = 0 for 0 1 i 1 r . Give your answers in exact form. 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Mark scheme: 7(a) Use correct double angle formula to obtain an equation in tan x M1 3 2  2tan x e.g. tan x + − tan x ( = 0 ) . 2 1 − tan x Allow if the correct formula is quoted but then they lose the 2 from the numerator when they use the formula. Obtain a correct equation in tan x in any form without fractions A1 E.g. tan 3 x − tan 5 x + 4tan x − tan x + tan 3 x ( = 0 ) . Condone if ‘= 0’ is missing here. Reduce to the given answer of tan 4 x − 2tan 2 x −=3 0 correctly A1 Obtain given answer from correct working but condone if never mention tan x  0. Condone the right terms in a different order ‘Show that’ so each line must be correct. 3 tan2=  37(b) A complete correct method to solve the equation to obtain a value for  M1 ( ) Allow if they make a slip in copying the equation but do have a complete method to obtain a value of . M0 if they get a value for 2 but never halve it. 1 1 2 5 A1 Obtain two of ( =) , , and  6 3 3 6 1 1 2 5 A1 Exact, ignore any answers outside interval Obtain the other two of ( =) , , and  and no others in the interval 6 3 3 6 2 1 4 2 Accept  for and  for  6 3 6 3  Do not need to see  = (from tan2= 0). 2 3

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9 With respect to the origin O, the points A, B and C have position vectors given by 2 0 - 3 OA = f 1p, OB = f 4 p and OC = f- 2p. - 3 1 2 (a) The point D is such that ABCD is a trapezium with DC = 3AB . Find the position vector of D. 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(b) The diagonals of the trapezium intersect at the point P. Find the position vector of P. 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(c) Using a scalar product, calculate angle ABC. 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Mark scheme: 9(a) Use a correct method to find OD M1 E.g. OC + 3 OA − OB = ( ) (–3i – 2j + 2k) + 3((2i + j – 3k) – (4j + k)) AB = −2i + 3 j + 4k ( ) Accept column vectors throughout. Obtain position vector of D is 3i – 11j – 10k A1 Accept coordinates. 2 9(b) Carry out correct method for finding a vector equation for AC or BD *M1 E.g. 2i + j – 3k + λ (5i + 3j – 5k) or 4j + k + µ (3i – 15j – 11k). Condone missing r = … Both diagonal equations correct. A1ft Seen or implied. Follow their D. Condone missing r = … Equate at least two pairs of corresponding components and solve for λ or for µ DM1 Dependent on using relevant lines and two different parameters. 1 1 A1 The values will depend on the directions of their Obtain λ = – or µ = lines 4 4 3 1 7 A1 OE Obtain position vector of P is i + j – k Accept coordinates. 4 4 4 Do not ISW. 9(b) Alternative Method for Question 9(b): State or imply AC = 5i − 3 j + 5k B1 FT Or BD = 3i − 15 j − 11k Follow their D if used. Identify similar triangles with ratio 1 : 3 M1 1 M1 Must be correct fraction. Use similar triangles to obtain OP , e.g. OP = OA + AC 4 3 1 7 A2 OE Obtain position vector of P is i + j – k Allow A1A0 if any two values are correct. 4 4 4 5 9(c) Find direction vector BA = 2i – 3j – 4k and BC = –3i – 6j + k or equivalent B1FT Or AB and CB . FT if using an incorrect AB from earlier work. Carry out correct process for evaluating the scalar product of two relevant vectors M1 Allow if one is going in the negative direction, e.g. AB and BC . Using the correct process for the moduli, divide their scalar product by the product M1 Independent of the first M1. of their moduli and evaluate the inverse cosine of the result to obtain an angle For their two vectors −1 8 = cos = ... 29 46 Obtain answer 77.3° (or 1.35 radians) A1 77.347… Correctly rounded to more than 3 sf or AWRT 77.3. 4

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10 A balloon in the shape of a sphere has volume V and radius r. Air is pumped into the balloon at a constant rate of 40r starting when time t = 0 and r = 0 . At the same time, air begins to flow out of the balloon at a rate of 0.8rr . The balloon remains a sphere at all times. (a) Show that r and t satisfy the differential equation dr 50 - r = . [3] td 5r 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the quotient and remainder when 5r 2 is divided by 50- r . 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(c) Solve the differential equation in part (a), obtaining an expression for t in terms of r. 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(d) Find the value of t when the radius of the balloon is 12. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 10(a) dV B1 Need a complete correct statement seen or Obtain = 40π − 0.8πr or equivalent implied. dt dV 2 dV 2 dr B1 Need a complete correct statement seen or Obtain = 4πr or equivalent e.g. = 4 r implied. dr dt dt Use the chain rule to obtain given answer (including the derivative) B1 dr 50 − r dr 40 − 0.8r Allow if = follows = dt 5r 2 dt 4 r 2 without further explanation (π already cancelled) and no incorrect statements seen. 3 10(b) Commence division and reach quotient of the form M1 Allow M1 if divide by r − 50 to obtain –5r ± 250 5r  250 . or 5r2 = (50 – r)(Ar + B) + C and reach A = –5 and B = ± 250 Obtain quotient –5r – 250 A1 Do not need to state which is quotient and which is remainder. However, if clearly muddled, then M1A1A0 for both expressions correct. Obtain remainder 12 500 A1 Note: 12 500 following division by r – 50 is correct and scores this A1 ISW. SC B1 only for correct use of remainder theorem to obtain correct remainder. 3 10(c) Prepare to integrate e.g. separate variables correctly B1FT 2 5 r 1d t d r =  2  50 − r d t 5 r  12500  Condone missing dr, dt or missing integral Or express in the form =  = − ( 5 r + 250 ) +  dr 50 − r  50 − r  signs, but not both. Follow their division in (b) if substitute before separating. Obtain term t DB1 A 2 M1 C Obtain terms r + Br − Cln ( 50 − r ) From their Ar + B + in (b) where 2 50 − r ABC ≠ 0. Allow a single slip in the coefficients. 5 2 A1FT FT their (b), provided of the correct form. Obtain terms − r − 250r − 12500ln(50 − r ) 2 Use t = 0, r = 0 to evaluate a constant or as limits in a solution containing terms of M1 the form r2, r, ln(50 – r) and t 5 2 A1 OE Obtain final answer t = − r − 250r − 12500ln(50 − r ) + 12500ln50 Must be t = ….. 2 Allow with 12500ln50 = 48900 or better. 6 10(d) Obtain t = 70.5 B1 May be more accurate (70.4605…). 1

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2 e 2 x 11 Let f ( )x = . e 2 x - 3 e x + 2 (a) Find f l ( x) and hence find the exact coordinates of the stationary point of the curve with equation y = f ( )x . 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Mark scheme: 11(a) Use correct quotient rule NB the question asks for f x( ) so need complete form M1 Or correct product rule. 4e 2 x (e 2 x − 3e x + 2) − 2e 2 x (2e 2 x − 3e x ) A1 Obtain correct derivative in any form, e.g. 2 e 2 x − 3e x + 2 ( ) Equate their derivative to zero *M1 Can be implied by numerator equated to zero for quotient rule. 8 = 6ex ( ) Solve for x to obtain x = lna DM1 a positive. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept equivalent exact forms, e.g. x = ln . 6 11(a) Alternative Method for Question 11(a) Complete method to express f ( x ) in partial fractions M1 As far as p + q + r with values for p, e x − 2 e x − 1  8 2  q and r  2 + x − x  .  e − 2 e − 1  u = e x Allow in u ( ) . s e x t e x *M1 Note: the question requires f’(x) so if they have Differentiate to obtain f  ( x ) = 2 + 2 substituted for ex, they will also need chain rule. e x − 2 e x − 1 ( ) ( ) − 8e x 2e x A1 From correct work. Obtain f  ( x ) = 2 + 2 e x − 2 e x − 1 ( ) ( ) Equate derivative to zero and solve for x to obtain x = lna DM1 Must follow correctly to give a positive value of a. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept x = ln , or equivalent. 6 5 11(b) du x B1 State or imply = e dx 2u B1 Correct expression in u. Obtain  u du or equivalent 2 − 3u + 2 Condone missing du or missing integral but not both.  2  8 2 + Or1   − du Allow FT if using their partial fractions from   u  u ( u − 2 ) u ( u − 1) (a). A B B1 FT Complete reduction to partial fractions. State or imply partial fractions of the form + Correct form for their integrand. u − 1 u − 2 C D E 2u − 3 3 2u − 3 F G Or1 + + Or2 + = + + u u − 2 u − 1 u 2 − 3u + 2 u 2 − 3u + 2 u 2 − 3u + 2 u − 2 u − 1 Use a correct method for finding a constant M1 Available if they have incorrect form. −2 4 A1 Obtain correct + u − 1 u − 2 2u − 3 3 3 Or2 + − u 2 − 3u + 2 u − 2 u − 1 Integrate to obtain a ln (u – 1) + b ln (u – 2) or equivalent *M1 M0 if they have additional terms that do not cancel out. Obtain correct –2 ln (u – 1) + 4 ln (u – 2) or equivalent A1FT FT values of their partial fraction coefficients. Correctly use limits u = 5 and 3 in an expression of the form a ln (u – 1) + b ln (u – 2) DM1 or x = ln 5 and ln 3 in an expression of the form a ln (ex – 1) + b ln (ex – 2) 81 A1 Accept ln 20.25. Obtain ln 4 9

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