Mathematics 9709/32 October/November 2020: question paper and mark scheme

Mark scheme

1 State that 1 + e −3 x =e 2 B1 With no errors seen to that point Use correct method to solve an equation of the form e −3 x = a , where M1 (e −3 x ) = 6.389... Evidence of method must be seen. a > 0, for x or equivalent Obtain answer x = – 0.618 only A1 Must be 3 decimal places Alternative method for question 1 State that 1 + e −3 x = e 2 B1 Rearrange to obtain an expression for e x and solve an equation of the M1 1 ex = 3 form e x = a , where a > 0, or equivalent e2 − 1 Obtain answer x = – 0.618 only A1 Must be 3 decimal places 3

Mark scheme

2(a) 2 3 State a correct unsimplified version of the x or x or x term M1 For the given expression State correct first two terms 1 + 2x A1 40 3 A1 + A1 One mark for each correct term. ISW Accept 13 13 Obtain the next two terms −4 x 2 + x 3 The question asks for simplified coefficients, so candidates should cancel fractions. 4 2(b) 1 B1 OE. Strict inequality State answer x < 6 1

Mark scheme

3(a) State or imply y log 2 = log 3 − 2 x log 3 B1 Accept y ln 2 = (1 − 2 x ) ln 3 State that the graph of y against x has an equation which is linear in x B1 Correct equation. Need a clear statement/comparison with and y, or is of the form ay = bx + c matching linear form. 2ln 3 B1 Must be exact. Clear indication that the gradient is – 2log ln 2 k , log 1 3 Any equivalent e.g. − 2 log 2 9 k 3 3(b) Substitute y = 3x in an equation involving logarithms and solve for x M1 ln 3 A1 Allow M1A1 for the correct answer following decimals Obtain answer x = ln 72 2

Mark scheme

4(a) Use correct tan ( A + B ) formula and obtain an equation in tan θ M1 tan θ + tan 60° 2 e.g. = 1 − tan θ tan 60° tan θ Use tan 60° = 3 and obtain a correct horizontal equation in any form ( ) ( A1 e.g. tan θ tan θ + 3 = 2 1 − 3 tan θ ) Reduce to tan 2 θ + 3 3 tan θ − 2 = 0 correctly A1 AG 3 4(b) Solve the given quadratic to obtain a value for θ M1 −3 3 ± 35 tan θ = = 0.3599, − 5.556 2 Obtain one correct answer e.g. θ = 19.8 ° A1 Accept 1d.p. or better. If over-specified must be correct. 19.797...., 100.2029... Obtain second correct answer θ = 100.2° and no others in the given A1 Ignore answers outside the given interval. interval 3

Mark scheme

5(a) dx dy B1 CWO, AEF. State = sec 2 θ or = −2sin θ cos θ dθ dθ dy dy dx M1 Use = ÷ dx dθ dθ dy A1 AG Obtain = −2sin θ cos3 θ from correct working dx Alternative method for question 5(a) Convert to Cartesian form and differentiate M1 1 y= 1 + x2 dy −2 x A1 OE = ( dx 1 + x 2 2 ) dy A1 AG Obtain = −2sin θ cos3 θ from correct working dx 3 5(b) d M1 Condone incorrect naming of the derivative Use correct product rule to obtain dθ ( ±2cos3 θ sin θ ) For work done in correct context Obtain correct derivative in any form ( A1 e.g. ± −2cos 4θ + 6sin 2θ cos 2θ ) Equate derivative to zero and obtain an equation in one trig ratio A1 e.g. 3 tan 2θ = 1 , or 4 sin 2θ = 1 or 4 cos 2θ = 3 1 A1 Obtain answer x = – Or − 3 3 3 Alternative method for question 5(b) d2 y M1 Use correct quotient rule to obtain dx 2 Obtain correct derivative in any form A1 ( ) + 2 × 2 x × 2 x (1 + x ) 2 −2 1 + x 2 2 (1 + x ) 2 4 Equate derivative to zero and obtain an equation in x2 A1 e.g. 6x2 = 2 1 A1 Obtain answer x = – 3 4

Mark scheme

6(a) Multiply numerator and denominator by 1 + i, or equivalent M1 Must multiply out Obtain numerator 6 + 8i or denominator 2 A1 Obtain final answer u = 3 + 4i A1 Alternative method for question 6(a) Multiply out (1 − i )( x + iy ) = 7 + i and compare real and imaginary parts M1 Obtain x + y = 7 or y − x = 1 A1 Obtain final answer u = 3 + 4i A1 3 6(b) Show the point A representing u in a relatively correct position B1 FT The FT is on xy ≠ 0. Show the other two points B and C in relatively correct positions: B1 A approximately equal distance above / below real axis 4 2 B 5 C 2 Take the position of A as a guide to ‘scale’ if axes not marked 2 6(c) 1 B1 ArgC State or imply arg(1 – i) = − π 4 Substitute exact arguments in arg(7 + i) – arg(1 – i) = arg u M1 Must see a statement about the relationship between the Args e.g. ArgA = ArgB − ArgC or equivalent exact method 4 1 1 A1 Obtain given answer correctly from their u = k ( 3 + 4i ) Obtain tan −1 = tan −1 + π correctly 3 7 4 3

Mark scheme

7(a) Correct separation of variables B1 sec 2 x dx = e dt 2 −3t Needs correct structure 1 B1 Obtain term − e−3t 3 Obtain term of the form k tan 2 x M1 From correct working 1 A1 Obtain term tan 2 x 2 Use x = 0, t = 0 to evaluate a constant, or as limits in a solution M1 containing terms of the form a tan 2 x and be −3t , where ab ≠ 0 Obtain correct solution in any form A1 1 1 1 e.g. tan 2 x = − e−3t + 2 3 3 1 −1 2 A1 Obtain final answer x = 2 ( tan 1 − e −3t 3 ) 7 7(b) 1 −1 2 B1 FT Correct value. Accept x → 0.294 State that x approaches tan 2 3 The FT is dependent on letting e −3t → 0 in a solution containing e −3t . 1

Mark scheme

8(a) 2 2 B1 Or equivalent seen or implied ρ ρ Obtain AB = −2 and CD = 1 −4 1 Use the correct process for calculating the modulus of both vectors to M1 AB = 24, CD = 6 obtain AB and CD Using exact values, verify that AB = 2CD A1 Obtain given statement from correct work Allow from BA = 2DC, OE 3 8(b) Use the correct process to calculate the scalar product of the relevant M1 2 2 2 4 ρ ρ vectors (their AB and CD ) − 2 and 1 or −2 and 2 −4 1 −4 2 Divide the scalar product by the product of the moduli and evaluate the M1 inverse cosine of the result Obtain answer 99.6 ° (or 1.74 radians) or better A1 Do not ISW if go on to subtract from 180° (99.594…, 1.738...) Accept 260.4° 3 ρ ρ 8(c) State correct vector equations for AB and CD in any form, B1ft Follow their AB and CD 2 2 1 2 4 2 3 2 e.g. ( r = ) 1 + λ −2 and ( r = ) 1 + μ 1 Alternative: ( r = ) −1 + λ −2 and ( r = ) 2 + μ 1 5 −4 2 1 1 −4 3 1 Equate at least two pairs of components of their lines and solve for λ or M1 for µ Obtain correct pair of values from correct equations A1 Alternatives when taking A or B as point on line A λ µ B λ µ 17 7 17 7 ij − 16 1 3 ≠ ij − 76 − 23 ≠ 3 3 3 3 ik 1 2 1 0≠2 ik − 12 0 0≠2 3 1 jk 2 -3 5 ≠ −5 jk 2 -4 5 ≠ −5 Verify that all three equations are not satisfied and that the lines do not A1 CWO with conclusion intersect 17 7 17 7 e.g. ≠ or = is inconsistent or equivalent 3 3 3 3 4

Mark scheme

9(a) A Bx +C B1 State or imply the form + 2 3x + 2 x +4 Use a correct method for finding a constant M1 Obtain one of A = 3, B = –1, C = 3 A1 Obtain a second value A1 Obtain the third value A1 5 9(b) Integrate and obtain ln ( 3x + 2 ) ... B1 FT The FT is on A ( State a term of the form k ln x 2 + 4 . ) M1 From λx x + 4dx 2 1 A1 FT The FT is on B ( ) ... − ln x 2 + 4 ... 2 3 x B1 FT The FT is on C ... + tan −1 2 2 Substitute limits correctly in an integral with at least two terms of the M1 Using terms that have been obtained correctly from x completed integrals ( ) form a ln ( 3 x + 2 ) , b ln x 2 + 4 and c tan −1 , and subtract in correct 2 order 3 3 A1 Obtain answer ln 2 + π , or exact 2-term equivalent 2 8 6

Mark scheme

10(a) Use correct product rule M1 Obtain correct derivative in any form A1 dy 1 e.g. = cos x − x sin x . Accept in a or in x dx 2 x 1 A1 Obtain given answer from correct working. Equate derivative to zero and obtain tan a = The question says ‘show that ..’ so there should be an 2a intermediate step e.g. cos x = 2 x sin x . 1 Allow tan x = 2x 3 10(b) Use the iterative process correctly at least once (get one value and go on M1 Must be working in radians to use it in a second use of the formula) Degrees gives 1, 12.6039, 5.4133, ... M0 Obtain final answer 3.29 A1 Clear conclusion Show sufficient iterations to at least 4 d.p.to justify 3.29, or show there A1 3, 3.3067, 3.2917, 3.2923 is a sign change in the interval (3.285, 3.295) Allow more than 4d.p. Condone truncation. 3 ( x cos x ) dx 10(c) 2 B1 1 State or imply the indefinite integral for the volume is π [If π omitted, or 2π or π used, give B0 and follow 2 through. 4/6 available] Use correct cos 2A formula, commence integration by parts and reach *M1 x2 x 1 x(ax + b sin 2 x) ± ax + b sin 2 xdx Alternative: + sin 2 x − 4 4 4sin 2 xdx 1 1 1 1 A1 Obtain x( x + sin 2 x) − x + sin 2 x dx , or equivalent 2 4 2 4 Complete integration and obtain A1 OE 1 2 1 1 x + x sin 2 x + cos 2 x 4 4 8 1 DM1 π π2 1 1 Substitute limits x = 0 and x = π , having integrated twice +0− −0−0− 2 2 8 4 4 1 A1 CAO Obtain answer 16 ( ) π π 2 − 4 , or exact equivalent 6