Mathematics 9709/32 May/June 2024: question paper and mark scheme

Mark scheme

1(a) y B1 Correct shape, roughly symmetrical. Both sections should be solid straight lines. If not drawn with a ruler the intention must be clear. Allow construction lines if dashed or clearly fainter. 2a marked on each axis (must be 2a, not just 2). 2a Needs to extend into negative x. If a is given a value, then B0. Ignore y = 2x – 3a if seen. O x 2a 1 1(b) Solve linear equation or inequality to obtain critical value B1 Ignore x = a if seen. x = 53 a or exact equivalent. Obtain x < 53 a or exact equivalent B1 Accept x < 10 a or ( −∞, 5 a ) . Must be strict inequality. 6 3 Need a clear final solution: x > a or x < a must be rejected if seen as part of the working. Rejection can be implied, e.g. if only the correct inequality is underlined. B0 B0 if a is given a value. Alternative Method for Question 1(b) Solve quadratic equation ( 2 x − 3a ) = ( x − 2a ) to obtain critical 2 2 (B1) ( 3x − 8ax + 5a = 0 ) 2 2 value x = 53 a or exact equivalent Ignore x = a if seen. Obtain x < 53 a or exact equivalent (B1) Accept x < 10 a or ( −∞, 5 a ) . Must be strict inequality. 6 3 Need a clear final solution: x > a or x < a must be rejected if seen as part of the working. Rejection can be implied, e.g. if only the correct inequality is underlined. B0 B0 if a is given a value. 2

Mark scheme

2 B C B1 Dx + E F P Qx + R State or imply the form A + + + and + are also valid. 2x + 3 x − 4 2x + 3 x − 4 2x + 3 x−4 Use a correct method for finding a constant M1 SC: If score B0, they can score M1 A1 for one correct constant. B0 M1 A0 available if they substitute two values to form simultaneous equations but get an incorrect answer, or they substitute one value and make an arithmetic error. Obtain one of A = 3, B = −2 and C = 4 A1 SC: If the horizontal equation is correct apart from an incorrect value for A, the other A marks may be available. Obtain a second value A1 SC: If denominator factorised as ( x + 3 ) ( x − 4 ) can score a 2 maximum of B0 M1 A1 A1 A0 for a split involving 3 terms. Obtain a third value A1 ISW Statement of the final split is not required. 2 Alternative method for Question 2 Divide numerator by denominator (M1) Px + Q (A1) 6 x + 20 Obtain 3 + 2 3 + 2 x − 5 x − 12 (2 x + 3)( x − 4) Px + Q D E (B1) Must deal with the 3 separately or include it correctly on both State or imply the form = + sides in their split. 2 x − 5 x − 12 2 2x + 3 x − 4 Obtain one of D = −2 and E = 4 (A1) SC: If denominator factorised as ( x + 3 ) ( x − 4 ) , then can score a 2 maximum of B0 M1 A1 A1 A0 for a split involving three terms. Obtain a second value (A1) ISW Statement of the final split is not required. 5

Mark scheme

3(a) Use logarithms to obtain a correct expression without powers e.g. B1 Could use logs to any base e.g. 2 y − 1 = ( x − y ) log a b. ( 2 y − 1) ln a = ( x − y ) ln b Do not condone missing brackets unless recovered later. Separate terms and factorise to obtain y ( 2ln a + ln b ) = x ln b + ln a B1 ln b ln a Or equivalent, e.g. y = x + or ln a b ln a 2b 2 y ( 2 + log a b ) = x log a b + 1. Clear explanation of linear form. B1 E.g. equation matches the linear form y = mx + c or py = qx + r. From correct work only. Condone if they compare with y = mx + c, but do not actually state that it must therefore be a straight line. Stating “this is a linear equation” without comparing to a relevant standard form scores B0. B0 if they have m = … and c = … correct but never actually mention y = mx + c. 3 3(b) Use a = b3 and log laws to simplify their equation M1 ln b ln b3 y = x + ln b7 ln b7 Denominator reduced to a single log term. 1 3 A1 x 3 x+3 Obtain y = x+ Accept y = + but not y = . 7 7 7 7 7 Alternative method for Question 3(b) Use a = b3 to obtain b 3( 2 y −1) = b x − y or equivalent (M1) Or log a b = 13 . 1 3 (A1) x 3 x+3 Obtain y = x+ Accept y = + but not y = . 7 7 7 7 7 2

Mark scheme

4 Substitute y = 1 and obtain e x = a , where a > 0 M1 Must come from a quadratic in e x . ( e2 x + e x − 6 = 0 ) Ignore any negative solution. Obtain e x = 2 only A1 Or equivalent e.g. x = ln 2. Condone x = 0.693….. d dy B1 dy State or imply dx ( ye2 x ) = 2 ye 2 x + e2 x dx Accept y ′ for dx . d 2 x dy B1 dy State or imply dx ( y e ) = y 2 e x + 2 ye x dx Accept y ′ for dx . Differentiate RHS of given equation to obtain zero (could be implied M1 Independent. dy dy dy by subsequent work), substitute for x and y and obtain = ... 2 × 4 + 4 dx + 2 + 4 dx = 0 dx dy NB: Could also rearrange to obtain then substitute. Both steps dx are needed for M1. 5 A1 Correct answer from correct working only Accept −810 . Obtain − or −1.25 4 -1.2499 is A0. 4 Alternative method for Question 4: Dividing through by e x Substitute y = 1 and obtain e x = a , where a > 0 (M1) Must come from a quadratic in e x . ( e2 x + e x − 6 = 0 ) Ignore any negative solution. Obtain e x = 2 only (A1) Or equivalent e.g. x = ln 2. Condone x = 0.693….. d 2 dy (B1) dy State or imply dx ( y ) = 2y dx Accept y ′ for dx . d dy (B1) dy State or imply dx ( ye x ) = ye x + e x dx Accept y ′ for dx . Differentiate RHS of given equation to obtain −6e − x , substitute for x (M1) Independent. dy dy dy 6 and y and obtain = ..... 1 × 2 + 2 dx + 2 dx = − 2 dx dy NB: Could also rearrange to obtain then substitute. Both steps dx are needed for M1. 5 (A1) Correct answer from correct working only Accept −810 . Obtain − or −1.25 4 -1.2499 is A0. 6

Mark scheme

5(a) Calculate the value of a relevant expression or values of a pair of M1 Allow if working with a smaller interval, e.g. (0.72, 0.78). expressions at x = 0.7 and x = 0.8 Need all relevant values but condone one error. Pairings must be clear for solutions involving four values (do not accept embedded values). M0 if working in degrees e.g -1.94…, 1.04… Complete the argument correctly with correct calculated values. A1 E.g. 4.95... > 4.26... and 4.06... < 4.49... (can be using the equation in the rubric or the equation in (b) or -0.439…< 0, 0.690…> 0 equivalent) 1.4 < 1.5029…, 1.6 > 1.449… 1.1 > 1, 0.86 < 1 0.0515 > 0, -0.075 < 0. Allow values rounded or truncated to 2sf. 2 5(b) State 2 x = ln ( 5 + cos3x ) and take exponential of both sides to B1 Given answer requires fully correct working or work vice versa. If working in reverse, must get to the iterative formula, including obtain e 2 x = 5 + cos3x subscripts. 1 5(c) Use the iterative process correctly at least once M1 M0 if working in degrees (e.g. values heading for 0.89….). Obtain final answer 0.740 A1 Show sufficient iterations to at least 5dp to justify 0.740 to 3dp, or A1 E.g. 0.7, 0.75150, 0.73719, 0.74105, 0.74000, 0.74028 show that there is a sign change in the interval ( 0.7395, 0.7405 ) 0.75, 0.73759, 0.74094, 0.74003, 0.74028 0.8, 0.72494, 0.74443, 0.73909, 0.74053, 0.74014, 0.74025 Allow recovery. Allow truncation or rounding and condone small differences in the final decimal place. 3

Mark scheme

6(a) Use correct product rule *M1 Or equivalent. Condone incorrect chain rule. M0 if a value is used for a (not equivalent work). Obtain correct derivative A1 dy E.g. = −axe− ax + e− ax dx Equate derivative to zero and solve for x DM1 Obtain x = 1a , y = a1e A1 ISW Or exact equivalent. 4 6(b) Use integration by parts to obtain pxe − ax + q e − ax dx *M1 Condone sign error in parts formula and omission of dx. M0 if a value is used for a (not equivalent work). Obtain − 1a xe − ax + 1a e − ax dx A1 OE Complete integration to obtain − a1 xe − ax − a12 e − ax A1 OE Correct use of limits 0 and a2 in an expression of the form DM1 ( e − e +0+ ) −2 a2 −2 1 a2 −2 1 a2 rxe − ax + se − ax Obtain a12 (1 − 3e −2 ) A1 ISW e2 − 3 Or simplified 2-term equivalent, e.g. . a 2e2 5

Mark scheme

7(a) Factorise LHS using difference of 2 squares *M1 ( ( cos θ − sin θ )( cos θ + sin θ )) 2 2 2 2 Simplify DM1 cos 2 θ + sin 2 θ = 1 must be seen or implied, e.g. ( cos2 θ − sin 2 θ )( cos2 θ + sin 2 θ ) = ( cos2 θ − sin 2 θ ) . Obtain cos 4 θ − sin 4 θ ≡ cos 2θ from correct working A1 AG Alternative Method for Question 7(a) Use of correct rearrangements of double angle formulae (*M1) 2 1 + cos 2θ 1 − cos 2θ 2 E.g. − 2 2 2 2 1 + cos 2θ cos 2θ − 1 Only condone − if correct expression for 2 2 sin θ seen. 2 Expand and simplify (DM1) Collect like terms. Condone recovery from missing brackets. Obtain cos 4 θ − sin 4 θ ≡ cos 2θ from correct working (A1) AG Alternative Method 2 for Question 7(a) Correct use of Pythagoras ( ) ( ) ( (*M1) E.g. 1 − sin 2 θ 2 − sin 4 θ or cos 2 θ 1 − sin 2 θ − sin 2 θ 1 − cos 2 θ ) Expand and simplify (DM1) Condone recovery from missing brackets. Obtain cos 4 θ − sin 4 θ ≡ cos 2θ from correct working (A1) AG 3 7(b) Use part (a) and correct double angle formula to obtain expression M1 cos θ − sin θ + 4sin θ cos θ dθ = cos 2θ + sin 2θ dθ 4 4 2 2 2 involving sin 2 2θ dθ or cos 2 2θ dθ Allow BOD for 2sin 2 2θ if sin 2θ = 2sin θ cosθ seen. cos 2θ dθ = sin 2θ 1 2 B1 Seen or implied. Use of correct double angle formula on second part of the integral to M1 1 − cos 4θ obtain a form that can be integrated directly e.g. sin 2 2θ dθ = dθ 2 Obtain 12 θ − 18 sin 4θ A1 Condone a mixture of x and θ . Correct use of limits ± 8π in an expression of the form pθ + q sin 2θ + r sin 4θ and evaluate the trig M1 ( 2 ( × + − )) 1 2 1 2 π 16 1 8 Obtain 12 2 + 18 π − 14 A1 ISW Or exact equivalent from exact working. 6 ρ ρ

Mark scheme

8(a) Correct direction vector seen or implied ( BC = 3i + 3 j − 3k ) B1 Condone BC = −3i − 3 j + 3k. Use a correct method to form a vector equation M1 Allow for the RHS with no LHS. Obtain r = 5i + 2 j + λ ( i + j − k ) A1 ISW x Must have r = … or y = ... , not l1 = … z Or, equivalent vector form, e.g. r = 8i + 5 j − 3k + α ( i + j − k ) or r = 5i + 2 j + λ ( 3i + 3 j − 3k ) . Condone a column vector with i, j, k. 3 8(b) Use components to form two relevant equations in 2 unknowns B1FT 5 + λ −2 + 3μ For their l1 B0 if they use the same unknown for both lines. Two components of 2 + λ = 1 + μ seen or implied. −λ 4 − 2μ Solve 2 relevant equations in 2 unknowns for λ or μ M1 For their l1. ρ Obtain λ = 2 or μ = 3 A1 Or equivalent e.g. using BC as direction vector gives λ = 2 . 3 Obtain ( 7, 4, − 2 ) A1 Accept position vector. Condone a column vector with i, j, k. No need to check the third equation – the question implies that the SC: B1 M1 A1 A1 if one component of their line is incorrect but lines intersect. they do not use that component. 4 ρ 8(c) State AB = 72 + 12 + 42 = 66 ( ) B1 Or ( AB )2 = 66 Condone a sign error in AB. ρ State BD in component form B1 −7 + 3r −1 + r or equivalent. 4 − 2r AB = BD ( 3r − 7 ) + ( r − 1) + ( −2r + 4 ) = 66 M1 Or equivalent equation in one unknown for their AB and their 2 2 2 ρ ρ BD ≠ OD. (14r − 60r = 0 ) 2 If you never see a correct form and they go direct to 9r 2 + 49 + r 2 + 1..... then M0. r = 307 A1 Correct only. Ignore r = 0 if seen. ρ OD = 767 i + 377 j − 327 k A1 Must be a vector. ρ ρ Condone if also have OD = OA. 5

Mark scheme

9(a) ( ) ( State zω = −3 + 3 3 + 3 + 3 3 i ) B1 Or exact equivalent with real and imaginary parts collected. Need brackets around the coefficient of i. Allow for a =, b = stated correctly. 1 9(b) Obtain z = 2 B1 Obtain arg z = − π4 final answer B1 Obtain ω = 6 B1 Obtain arg ω = 2π 3 final answer B1 4 9(c) Im B (-3+3 3)+(3+3 3)i Note: The question does not require the diagram. If they use 5π 12 they need to demonstrate where it comes from. A Complex number equivalent to AB is 3 3 + 3i. -3+3 3i Re O Show OA = AB = 6, hence isosceles B1 One mark for ‘isosceles’ and one mark for ‘right angle’. There will be alternatives e.g. use of Pythagoras (ratio of lengths is AOB = arg ω − arg zω = − arg z = π4 hence third angle is a right B1 1:1: 2 ), expressing each number in “vector” form and using scalar product or explaining the effect of multiplying by 1 – i. angle 2 9(d) arg zω = arg z + arg ω ( = 23π − π4 = 512π ) M1 For showing correct use of their angles from part (b). Must demonstrate where 5π12 comes from. 3+3 3 M1 Correct method for their zω from part (a). arg zω = tan −1 Must link to point B on diagram or to arg zω. −3 + 3 3 3+3 3 3+3 3 Need to see tan −1 or tan θ = and not just −3 + 3 3 −3 + 3 3 1+ 3 tan −1 . −1 + 3 3 +1 A1 Obtain given answer from full and correct working. tan ( 125 π ) = 3 −1 3

Mark scheme

10(a) Use of correct chain rule (and correct quotient rule) and cos −3 θ M1 Obtain k × ( cosθ )−4 × sin θ or equivalent. dy A1 Obtain given answer from full and correct working (signs must be = −3 × − sin θ ( cosθ ) = 3sin θ sec4 θ −4 dθ ( ) shown), but condone ddθ sec 3 θ = .... and y '(θ ). Must be expressed in the given form 2 10(b) sin θ ( x + 3) B1 Or 3sin θ dθ = 3x + 9 Separate variables: dθ = 2 dx 2 dx. ( x + 9) cos θ 4 cos θ 4 x +9 Condone missing integral signs or missing dx or dθ , but not both. Obtain p sec3 θ ( + A ) B1 Correct form, p any constant but not 0. 3x + 9 3x 9 *M1 Might have one third of both sides. Use dx and obtain q ln ( x + 9 ) or 2 2 dx = 2 + 2 x +9 x +9 x +9 Alt: substitute x = 3tan φ to obtain q 1 + tan φ dφ ; condone if have r tan −1 3x ( +C ) . θ in place of φ in this method. Obtain q ln ( x 2 + 9 ) and r tan −1 3x ( +C ) DM1 Obtain q (φ μ ln ( cos φ ) ) OE. ( ) Obtain sec3 θ = 32 ln x 2 + 9 + 3 tan −1 3x ( +C ) or equivalent A1 Or might see a third of both sides. Must have 2 different variables. Use θ = 13 π , x = 3 in an equation including p sec3 θ , q ln ( x 2 + 9 ) M1 Or as limits in a definite integral. Limits for φ are 0 and 14 π . and r tan −1 3x to evaluate the constant of integration Obtain constant = 8 − 32 ln18 − 34 π A1 OE, e.g. 1.308… to at least 3sf. Obtain cosθ = 0.601 A1 Accept AWRT 0.601. 8