Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 3 · Variant 1

1 The polynomial 4x 3 + ax 2 + 5x + b , where a and b are constants, is denoted by p ( )x . It is given that ( 2x + 1) is a factor of p ( )x . When p ( )x is divided by ( x - 4) the remainder is equal to 3 times the remainder when p ( )x is divided by ( x - 2) . Find the values of a and b. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Substitute x = − 12 and equate the result to zero M1 a Obtain a correct equation, e.g. − 84 + a4 − 52 + b = 0 A1 ( 4 + b = 3) Any equivalent form. Substitute x = 2 and x = 4 and use p ( 4 ) = 3p ( 2 ) M1 If using long division, M1 is for correct use of two constant remainders. Condone if 3 is on the wrong side. Obtain a correct equation, e.g. 3 ( 32 + 4 a + 10 + b ) = 256 + 16 a + 20 + b A1 ( − 2 a + b = 75 ) Any equivalent form. Obtain a = −32 and b = 11 A1 5

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3 2 Find the exact value of x 2 ln 3x dx . Give your answer in the form a ln b + c , where a and c are rational y 1 and b is an integer. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 3 2 M1* x dx Integrate to obtain px ln3 x + q  1 3 1 2 A1 Or unsimplified equivalent. Obtain x ln3 x x dx 3 −3 1 3 1 3 A1 Complete integration to obtain x ln3 x − x 3 9 Use correct limits correctly in an expression of the form rx 3 ln3 x + sx 3 DM1 9ln9 − 3 − 13 ln3 + 19 An exact expression for their integral. 53 26 A1 Or 2-term equivalent. Obtain ln3 − 3 9 5

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3 The equation of a curve is ln ( x + y) = 3x 2 y . Find the gradient of the curve at the point ( 1 , 0) . [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 1+ ddyx B1 State or imply as the derivative of ln ( x + y ) x + y 2 dy 2 B1 State or imply 6 xy + 3 x as the derivative of 3x y dx d y M1 Having the correct form for at least one of the above. Substitute (1, 0 ) and solve for d x dy 1 A1 Obtain = or 0.5 dx 2 Alternative Method for Question 3: x 2 y dy B1 Rewrite as x + y = 3e and state or imply 1 + as the derivative of the LHS dx  2 dy  3 x 2 y x 2 y B1 State or imply  6 xy + 3 x  e as the derivative of 3e  dx  d y M1 Having the correct form for at least one of the above. Substitute (1, 0 ) and solve for d x dy 1 A1 Obtain = or 0.5 dx 2 4

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4 (a) Show that sec 4 i - tan 4 i / 1 + 2 tan 2 i . 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(b) Hence or otherwise solve the equation sec 4 2 a - tan 4 2 a = 2 tan 2 2 a sec 2 2 a for 0° 1 a 1 180° . 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Mark scheme: 4(b) Form an equation in tan2. M1 2 2 2 1 + 2tan 2= 2tan 2 1 + tan 2 ( ) 4 2 2 2 4 cos 2+ 2sin 2cos 2= 2sin 2 Or multiply through by cos 2 to form an equation in sin2or cos2  sin 4 2+ 2sin 2 2−=1 0 or cos 4 2− 4cos 2 2+ 2 = 0 Solve for tan2 or equivalent M1  1   tan2=      2  Obtain one correct solution for α, e.g. 20.0 ( 30.. )  A1 Obtain a second correct value for α, e.g. 70.0  ( 69.9698.. ) A1 Obtain solutions 110 (110.0) and 160 (160.0) for α, and no others in range A1 5

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5 (a) By sketching a suitable pair of graphs, show that the equation 2 + e -0 .2 x = ln ( 1 + x) has only one root. [2] (b) Show by calculation that this root lies between 7 and 9. 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(c) Use the iterative formula x = exp `2 + e -0 .2 x nj - 1 n + 1 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [exp(x) is an alternative notation for ex.] [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = 2 + e− 0.2 x B1 y 1 For the sketches: y=2+exp(- 5x) y=ln(1+x) Correct curvature Intersections with the y-axis approximately correct 3 Horizontal asymptote approximately correct – need not draw in 2 Allow scale not marked and implied by their sketch O x Sketch a second relevant graph, e.g. y = ln (1 + x ) and justify the given B1 statement 2 5(b) Calculate the value of a relevant expression or values of a relevant pair of M1 expressions at x = 7 and x = 9 Complete the argument correctly with correct calculated values A1 E.g. 2.079  2.246 and 2.302  2.165, or 0.167  0 and −0.137  0. 2 5(c) Use the iterative process correctly at least once M1 I.e., obtain one value and substitute that value back into the formula. Obtain final answer 8.03 A1 Show sufficient iterations to at least 4 decimal places to justify 8.03 to 2 A1 E.g. decimal places, or show that there is a sign change in the interval 7, 8.4555, 7.8846, 8.0849, 8.0115, 8.0380, 8.0283, 8.0318 ( 8.025,8.035 ) 8, 8.0421, 8.0268, 8.0324, 9, 7.7172, 8.1490, 7.9887, 8.0463, 8.0253, 8.0329 3

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6 y R 3 x O r M 4 The diagram shows the curve y = sin 2 x ( 1 + sin 2x) , for 0 G x G 3 r , and its minimum point M. The 4 shaded region bounded by the curve that lies above the x-axis and the x-axis itself is denoted by R. (a) Given that the x-coordinate of M lies in the interval 1 r 1 x 1 3 r , find the exact coordinates 2 4 of M. 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(b) Find the exact area of the region R. 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Mark scheme: 6(a) Correct use of product rule to differentiate *M1 2cos2 x (1 + sin2 x ) + sin2 x  2cos2 x , or 2cos 2 x − 2sin 2 x + 8sin x cos 3 x − 8cos x sin 3 x. All terms needed but could have errors in the coefficients. Obtain 2cos2 x + 4cos2 x sin2 x A1 OE 1 1 Equate derivative to zero and solve for 2x or x DM1 2cos2 x (1 + 2sin2 x ) = 0  x = 2 sin −1 ( − 2 ) Condone if they only consider cos 2x = 0. 7 1 A1 Mark degrees as a misread. Obtain x = π, y = − The Q asks for an exact answer. 12 4 4 6(b) Use correct double angle formula to integrate M1 1 − cos4 x sin2 x + dx  2 Or use integration by parts and correct double angle formula 1  1 + cos4 x Or − cos2 x (1 + sin2 x ) + dx. 2  2 1 x 1 A1 1 1 x 1 Obtain − cos2 x + − sin4 x ( +C ) Or − cos2 x − cos2 x sin2 x + + sin4 x ( + C ) 2 2 8 2 2 2 8 1 M1 1 1 1 + π − 0 + − 0 + 0 Use limits 0 and π correctly in a solution containing p cos2 x and q sin4 x 2 4 2 2 1 A1 Obtain π + 1 4 4

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7 2 5x + 8x + 5 Let f ( x) = . ( 1 + 2 x)( 2 + x 2 ) (a) Express f ( )x in partial fractions. 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(b) Hence find the coefficient of x3 in the expansion of f ( )x . 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Mark scheme: 7(a) A Bx + C B1 State or imply the form + 1 + 2 x 2 + x 2 Use a correct method to find a constant M1 Obtain one of A = 1, B = 2 and C = 3 A1 Obtain a second value A1 Obtain the third value A1 5 7(b) −1. − 2. − 3 3 B1 FT Correct term in 3x or coefficient of 3x in the expansion of State ( 2 x ) or −8 3! −1 A (1 + 2 x ) . Any equivalent form. 2 M1 Do not need to deal with 2-1 at this stage. Use a correct method to obtain the coefficient of x in the expansion of −1 2 2 x 2 . 2 + x or the coefficient of x in the expansion of ( − 1 ) ( 1 + 2 ) 1 1 −B x 3 or −B A1 FT Follow their B (and C). Obtain ( Bx + C ) − 2 2 x 2 or 4 4 Obtain final answer − 8 12 or −8 12 x3 A1 Or simplified equivalent. Ignore additional terms for other powers of x. 4

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1 8 (a) Given that z = 1 + y i and that y is a real number, express in the form a + bi , where a and b are z functions of y. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 2 2 1 (b) Show that ba - l + b = , where a and b are the functions of y found in part (a). 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[3] (d) The complex number z is such that Re ( z ) = 1. Use your answer to part (b) to give a geometrical description of the locus of 1. 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Mark scheme: 8(a) Multiply numerator and denominator by 1 − iy M1 OE 1 − y A1 OE Obtain + i 1 + y 2 1 + y 2 2 8(b) 2 M1 2 2  1  2  1 1  −( ) y  Express  a −  + b in terms of y and expand the bracket  2 −  +  2   2   1 + y 2   1 + y    2 A1FT Follow their answer from (a) provided it gives an  1 2 1  y expression in y. Obtain − + +  2 1 + y 2  4  1 + y 2 1 + y 2 ( ) ) 2 ) 2  (  ( Obtain 1 from full and correct working A1 AG 4 3 8(c) Show a vertical straight line through 1 + 0i B1 Im(z) 1 2i Show a circle centre 12 + 0i B1 1 1 Show a circle with radius 12 and centre not at the origin B1 2 Re(z) 1 - 2i 3 8(d) circle centre 12 + 0i with radius 12 B1 OE Condone inclusion of the origin. 1

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9 The position vector of point A relative to the origin O is OA = 8i - 5j + 6k . The line l passes through A and is parallel to the vector 2i + j + 4k . (a) State a vector equation for l. 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(b) The position vector of point B relative to the origin O is OB =- t i + 4 tj + 3t k, where t is a constant. The line l also passes through B. Find the value of t. 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(c) The line m has vector equation r = 5 i - j + 2k + n ( a i - j + 3 k ) . The acute angle between the 1 directions of l and m is i, where cos i = . 6 Find the possible values of a. 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Mark scheme: 9(a) Use a correct method to form a vector equation M1 Allow in column vectors. Obtain r = 8i − 5 j + 6k + ( 2 i + j + 4k ) A1 Need r = … 2 9(b) State the position vector of a point on l in component form B1 FT Follow their equation ( 8 + 2) i + ( −+5 ) j + ( 6 + 4) k . Or at least 2 correct components seen Might see the correct equation for the first time in (b). Equate to −+ti 4tj + 3tk and solve for t M1 Obtain t = −2 A1 3 9(c) Evaluate the scalar product of a pair of relevant vectors M1 ( 2i + j + 4k )  ( ai −+j 3k ) = 2 a + 11 OE, SOI Complete the process for finding the cosine of  *M1 Divide the scalar product by the product of the moduli and equate to cos. 2 a + 11 1 A1 OE Obtain =  21 10 + a 2 6 Form a 3-term quadratic equation in a and solve for a DM1 a 2 + 88a + 172 = 0 OE Obtain a = −2, a = −86 A1 Correct only (both values). 5

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10 h m 4 m A large cylindrical tank is used to store water. The base of the tank is a circle of radius 4 metres. At time t minutes, the depth of the water in the tank is h metres. There is a tap at the bottom of the tank. When the tap is open, water flows out of the tank at a rate proportional to the square root of the volume of water in the tank. d h (a) Show that =- m h , where m is a positive constant. 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(b) At time t = 0 the tap is opened. It is given that h = 4 when t = 0 and that h = 2.25 when t = 20. Solve the differential equation to obtain an expression for t in terms of h, and hence find the time taken to empty the tank. 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Mark scheme: 10(a) dV dV dh B1 SOI =  k V or = 16π dt dt dt Correct use of chain rule and V = 16πh M1 dV dh OE, e.g. = 16π . dt dt dh dV dV − k 16πh A1 Any equivalent form in terms of h. Obtain =  = dt dt dh 16π  k  k A1 Obtain given answer from full and correct working. = −   h = − h since is constant  4 π  4 π 4 10(b) Separate variables correctly and commence integration *M1 1 dt OE  dh =  − h Obtain −t = 2 h ( +C ) A1 Use the boundary conditions in an equation containing pt and q h to form DM1 OE, e.g. 0 = 4 + C or − 20= 2  1.5 + C. an equation in and/or C Use the boundary conditions in an equation containing pt and q h to form a DM1 1 C = −4, = second equation in and/or C and solve 20 t OE, e.g. − = 2 h − 4. 20 Hence t = 80 − 40 h A1 Must be seen. Time to empty the tank is 80 minutes A1 6

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