Mathematics 9709/31 October/November 2024: question paper and mark scheme

Mark scheme

1 Substitute x = − 12 and equate the result to zero M1 Obtain a correct equation, e.g. − 84 + a4 − 52 + b = 0 A1 ( a4 + 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 + 4a + 10 + b ) = 256 + 16a + 20 + b A1 ( −2a + b = 75 ) Any equivalent form. Obtain a = −32 and b = 11 A1 5 2 Integrate to obtain px3 ln 3x + q x 2 dx M1* 1 3 1 2 A1 Or unsimplified equivalent. Obtain 3 x ln 3x − 3 x dx 1 3 1 A1 Complete integration to obtain x ln 3x − x3 3 9 Use correct limits correctly in an expression of the form rx3 ln 3x + sx3 DM1 9ln 9 − 3 − 13 ln 3 + 19 An exact expression for their integral. 53 26 A1 Or 2-term equivalent. Obtain ln 3 − 3 9 5

Mark scheme

3 1 + ddyx B1 State or imply as the derivative of ln ( x + y ) x+ y dy B1 State or imply 6 xy + 3x 2 as the derivative of 3x 2 y dx dy M1 Having the correct form for at least one of the above. Substitute (1, 0 ) and solve for dx dy 1 A1 Obtain = or 0.5 dx 2 Alternative Method for Question 3: 2 dy B1 Rewrite as x + y = e3x y and state or imply 1 + as the derivative of the LHS dx dy 2 2 B1 State or imply 6 xy + 3x 2 e3 x y as the derivative of e3 x y dx dy M1 Having the correct form for at least one of the above. Substitute (1, 0 ) and solve for dx dy 1 A1 Obtain = or 0.5 dx 2 4

Mark scheme

4(a) Factorise M1 (sec θ − tan θ )(sec θ + tan θ ) 2 2 2 2 Or obtain an expression in cos 2 θ or sec 2 θ 2 Or −1 + , OE. cos2 θ Use of 1 + tan 2 θ = sec2 θ (anywhere) M1 The 2 method marks can appear in either order, in which ( ) − tan θ . 2 case the 2nd M1 is for expanding 1 + tan 2 θ 4 ( ) Obtain 1× 1 + tan 2 θ + tan 2 θ = 1 + 2 tan 2 θ A1 Obtain given answer from full and correct working. 3 4(b) Form an equation in tan2α. M1 ( 1 + 2 tan 2 2α = 2 tan 2 2α 1 + tan 2 2α ) Or multiply through by cos 4 2α to form an equation in sin 2α or cos2α cos4 2α + 2sin 2 2α cos2 2α = 2sin 2 2α sin 4 2α + 2sin 2 2α − 1 = 0 or cos 4 2α − 4cos 2 2α + 2 = 0 Solve for tan 2α or equivalent M1 1 tan 2α = ± 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

Mark scheme

5(a) Sketch a relevant graph, e.g. y = 2 + e − 0.2 x B1 y 1 y=2+exp(- x) For the sketches: 5 y=ln(1+x) Correct curvature 3 Intersections with the y-axis approximately correct Horizontal asymptote approximately correct – need not draw in 2 Allow scale not marked and implied by their sketch Sketch a second relevant graph, e.g. y = ln (1 + x ) and justify the given O B1 x 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

Mark scheme

6(a) Correct use of product rule to differentiate *M1 2cos 2 x (1 + sin 2 x ) + sin 2 x × 2cos 2 x, or 2cos2 x − 2sin 2 x + 8sin x cos3 x − 8cos x sin 3 x. All terms needed but could have errors in the coefficients. Obtain 2cos2x + 4cos2x sin2x A1 OE Equate derivative to zero and solve for 2x or x DM1 2cos2x (1 + 2sin 2 x ) = 0 x = 12 sin −1 ( − 12 ) 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 − cos 4 x sin 2x + 2 dx Or use integration by parts and correct double angle formula 1 + cos 4 x Or − 1 cos 2 x (1 + sin 2 x ) + dx. 2 2 1 x 1 A1 1 1 x 1 Obtain − cos 2 x + − sin 4 x ( +C ) Or − cos 2 x − cos 2 x sin 2 x + + sin 4 x ( +C ) 2 2 8 2 2 2 8 1 M1 1 + 1π−0+ 1 −0+0 Use limits 0 and π correctly in a solution containing p cos 2 x and q sin 4 x 2 4 2 2 1 A1 Obtain π +1 4 4

Mark scheme

7(a) A Bx + C B1 State or imply the form + 1 + 2 x 2 + x2 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 B1 FT Correct term in x 3 or coefficient of x 3 in the expansion of State ( 2 x )3 or −8 A(1 + 2 x ) . Any equivalent form. 3! −1 Use a correct method to obtain the coefficient of x 2 in the expansion of M1 Do not need to deal with 2-1 at this stage. ( 2 + x ) or the coefficient of x in the expansion of (1 + ) . −1 2 −1 2 x2 2 B 3 B A1 FT Follow their B (and C). Obtain ( Bx + C ) × 12 × − 12 x2 or − x 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

Mark scheme

8(a) Multiply numerator and denominator by 1 − y i M1 OE 1 −y A1 OE Obtain + i 1 + y 1 + y2 2 2 1 ( −) y 8(b) 1 2 M1 1 2 2 Express a − + b2 in terms of y and expand the bracket − + 2 1+ y 2 2 1 + y2 A1FT Follow their answer from (a) provided it gives an 1 2 1 y2 expression in y. Obtain − + + ( 1 + y2 2 ) ( 2 1 + y2 ) 4 1 + y 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 i 2 Show a circle centre 12 + 0i B1 1 1 B1 Re(z) Show a circle with radius 12 and centre not at the origin 2 1 - i 2 3 8(d) circle centre 12 + 0i with radius 12 B1 OE Condone inclusion of the origin. 1

Mark scheme

9(a) Use a correct method to form a vector equation M1 Allow in column vectors. Obtain r = 8i − 5 j + 6k + λ ( 2i + 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 + 2λ ) i + ( −5 + λ ) j + ( 6 + 4λ ) 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 ) = 2a + 11 OE, SOI Complete the process for finding the cosine of θ *M1 Divide the scalar product by the product of the moduli and equate to cosθ . 2a + 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

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 = −λ h since is constant 4 π 4 π 4 10(b) Separate variables correctly and commence integration *M1 1 h dh = −λdt OE Obtain −λt = 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 − 20λ = 2 ×1.5 + C. an equation in λ and/or C Use the boundary conditions in an equation containing pt and q h to form a DM1 1 C = −4, λ = second equation in λ 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