Mathematics 9709/32 October/November 2019: question paper and mark scheme

Mark scheme

1 Remove logarithms and state 4 − 3x = e1.2 , or equivalent B1 Accept 4 − 3x = 3.32 ( 01169.....) 3 s.f. or better Use correct method to solve an equation of the form 3x = a , where M1 ( 3 = 0.67988..) x a > 0. Complete method to x =… If using log3 the subscript can be implied Obtain answer x = – 0.351 only A1 CAO must be to 3 d.p. 3

Mark scheme

2 Use correct quotient rule or correct product rule M1 Obtain correct derivative in any form A1 dy −2e = −2 x (1 − x ) + 2 xe 2 −2 x dx (1 − x ) 2 2 Equate derivative to zero and obtain a 3 term quadratic in x M1 Obtain a correct 3-term equation A1 From correct work only e.g. 2 x 2 + 2 x − 2 = 0 or x 2 + x = 1 Solve and obtain x = 0.618 only A1 From correct work only 5

Mark scheme

3 Commence division and reach partial quotient x 2 + kx M1 Obtain correct quotient x 2 + 2 x − 1 A1 Set their linear remainder equal to 2x + 3 and solve for a or for b M1 Remainder = ( a + 3) x + ( b − 1) Obtain answer a = – 1 A1 Obtain answer b = 4 A1 Alternative method for question 3 ( )( ) State x 4 + 3 x 3 + ax + b = x 2 + x − 1 x 2 + Ax + B + 2 x + 3 and form M1 e.g. 3 = 1 + A and 0 = −1 + A + B and solve two equations in A and B Obtain A = 2, B = −1 A1 Form and solve equations for a or b M1 e.g. a = B − A + 2, b = − B + 3 Obtain answer a = – 1 A1 Obtain answer b = 4 A1 5 3 Alternative method for question 3 −1 ± 5 M1 Allow for correct use of either root in exact or decimal form. Use remainder theorem with x = 2 a a 5 9 5 A1 Expand brackets and obtain exact equation for either root. Obtain − ± +b = ∓ Accept exact equivalent. 2 2 2 2 Solve simultaneous equations for a or b M1 Obtain answer a = – 1 from exact working A1 Obtain answer b = 4 from exact working A1 5

Mark scheme

4(i) State R = 7 B1 Use correct trig formulae to find α M1 1 1 6 e.g. tan α = , sin α = , or cos α = 6 7 7 Obtain α = 22.208 ° A1 ISW 3 4(ii)  2  B1FT 49.107 ° to 3 d.p. B1 can be implied by correct answer(s) later. Evaluate sin −1   to at least 1 d.p. The FT is on their R  7 SC: allow B1 for a correct alternative equation e.g. 3tan 2 θ − 2 6 tan θ + 1 = 0 Use correct method to find a value of θ in the interval M1 Must get to θ Obtain answer, e.g. 13.4 ° A1 Accept correct over-specified answers. 13.449…, 54.3425… Obtain second answer, e.g. 54.3 ° and no extras in the given A1 Ignore answers outside the given interval. interval 4

Mark scheme

5 dy B1 State 4 xy + 2 x 2 , or equivalent, as derivative of 2x 2 y dx dy B1 State y 2 + 2 xy , or equivalent, as derivative of xy 2 dx dy *M1 dy y 2 − 4 xy Equate attempted derivative of LHS to zero and set equal to = 2 dx dx 2 x − 2 xy zero (or set numerator equal to zero) Reject y = 0 B1 Allow from y 2 − kxy = 0 Obtain y = 4x A1 OE from correct numerator. ISW Obtain an equation in y (or in x) and solve for y (or for x) in terms DM1 y3 y3 of a 8 x3 − 16 x3 = a3 or − = a3 8 4 Obtain y = – 2a A1 With no errors seen 7 5 Alternative method for question 5 a3 M1 Correct use of function of a function and implicit differentiation Rewrite as y = and differentiate 2 x 2 − xy Obtain correct derivative (in any form) A1  dy  −a3  4 x − y − x  dy  dx  = ( ) 2 dx 2 2 x − xy dy *M1 set equal to zero (or set numerator equal to zero) dx Obtain 4x – y = 0 A1 Confirm 2 x 2 − xy ≠ 0 B1 x = 0 and 2x = y both give a = 0 Obtain an equation in y (or in x) and solve for y (or for x) DM1 y3 y3 8 x3 − 16 x3 = a3 or − = a3 8 4 Obtain y = – 2a A1 With no errors seen 7

Mark scheme

6 1 1 B1 Or equivalent integrands. ∫ Separate variables correctly to obtain x + 2 ∫ dx = cot θ dθ 2 Integral signs SOI Obtain term ln ( x + 2 ) B1 Modulus signs not needed. 1 M1 Obtain term of the form k ln sin θ 2 1 A1 Obtain term 2ln sin θ 2 1 M1 Reach C = an expression or a decimal value Use x = 1, θ = π to evaluate a constant, or as limits, in an 3 ( expression containing p ln ( x + 2 ) and q ln sin 1 θ 2 ) Obtain correct solution in any form A1 ln12 = 2.4849….. Accept constant to at least 3 s.f. 1 1 e.g. ln ( x + 2 ) = 2 ln sin θ + ln12 Accept with ln3 − 2ln 2 2 Remove logarithms and use correct double angle formula M1  x + 2 1 − cosθ  Need correct algebraic process.  =   12 2  Obtain answer x = 4 − 6cosθ A1 8

Mark scheme

7(a) Substitute and obtain a correct horizontal equation in x and y in any B1 zz * +iz − 2 z* = 0 ⇒ form x 2 + y 2 + ix − y − 2 x + 2iy = 0 Allow if still includes brackets and/or i 2 Use i 2 = −1 and equate real and imaginary parts to zero OE *M1 For their horizontal equation Obtain two correct equations A1 Allow ix + 2iy = 0 e.g. x 2 + y 2 − y − 2 x = 0 and x + 2y = 0 Solve for x or for y DM1 6 3 A1 1 Obtain answer − i and no other OE, condone ( 6 − 3i ) 5 5 5 5 7(b)(i) Show a circle with centre 2i and radius 2 B1 Show horizontal line y = 3 – in first and second quadrant B1 Im(z) P 3i 2i Re(z) SC: For clearly labelled axes not in the conventional directions, allow B1 for a fully ‘correct’ diagram. 2 7(b)(ii) Carry out a complete method for finding the argument. (Not by measuring the sketch) M1 ( z = 3 + 3i ) Must show working if using 1.7 in place of 3 . 1 A1 SC: Allow B2 for 60° with no working Obtain answer π (or 60°) 3 2

Mark scheme

8(i) A Bx + C B1 State or imply the form + 2 2x −1 x +2 Use a correct method for finding a constant M1 Obtain one of A = 4, B = – 1, C = 0 A1 Obtain a second value A1 Obtain the third value A1 5 8(ii) Integrate and obtain term 2ln ( 2 x − 1) B1FT 1 The FT is on A. Aln ( 2 x − 1) 2 ( Integrate and obtain term of the form k ln x 2 + 2 ) *M1 From nx 2 x +2 A1FT The FT is on B 1 ( Obtain term − ln x 2 + 2 2 ) Substitute limits correctly in an integral of the form DM1 2ln 9 ( −2ln1) − 1 ln 27 + 1 ln 3 a ln ( 2 x − 1) + b ln( x 2 + 2) , where ab ≠ 0 2 2 Obtain answer ln 27 after full and correct exact working A1 ISW 5

Mark scheme

9(i) Commence integration by parts, reaching *M1 1 1 ∫ ax sin x − b sin x dx 3 3 1 1 A1 ∫ Obtain 3 x sin x − 3 sin x dx 3 3 1 1 A1 Complete integration and obtain 3 x sin x + 9 cos x 3 3 Substitute limits correctly and equate result to 3 in an integral of the DM1 a a 3 = 3a sin + 9 cos ( −0 ) − 9 1 1 3 3 form pxsin x + qcos x 3 3 a A1 With sufficient evidence to show how they reach the given 4 − 3cos equation Obtain a = 3 correctly a sin 3 5 9(ii) Calculate values at a = 2.5 and a = 3 of a relevant expression or pair M1 2.5 < 2.679 and 3 > 2.827 of expressions. If using 2.679 and 2.827 must be linked explicitly to 2.5 and 3. Solving f(a) = 0, f(2.5) = 0.179. and f(3) = –0.173 or if f ( a ) = a sin 1 a + 3cos 1 a − 4 ⇒ f (2.5) = −0.13..,f (3) = 0.145... 3 3 Complete the argument correctly with correct calculated values A1 Accept values to 1 sf. or better 2 9(iii) 4 − 3cos 13 an M1 Use the iterative process an +1 = an +1 correctly at least sin 13 an once Show sufficient iterations to at least 5 d.p. to justify 2.736 to 3d.p., A1 or show a sign change in the interval (2.7355, 2.7365) Obtain final answer 2.736 A1 3

Mark scheme

10(i) Express general point of l in component form B1 e.g. (1 + λ, 3 – 2λ, – 2 + 3λ) Substitute in equation of p and solve for λ M1 5 5 2 A1 OE Obtain final answer i + j from λ = Accept 1.67i + 1.67 j or better 3 3 3 3 10(ii) Use correct method to evaluate a scalar product of relevant vectors M1 e.g. (i – 2j + 3k).(2i + j – 3k) Using the correct process for calculating the moduli, divide the M1 9 scalar product by the product of the moduli and evaluate the inverse sin θ = 14 sine or cosine of the result Obtain answer 40.0 ° or 0.698 radians A1 AWRT 3 Alternative method for question 10(ii) Use correct method to evaluate a vector product of relevant vectors M1 e.g. (i – 2j + 3k)x(2i + j – 3k) Using the correct process for calculating the moduli, divide the M1 115 modulus of the vector product by the product of the moduli of the cosθ = 14 two vectors and evaluate the inverse sine or cosine of the result Obtain answer 40.0 ° or 0.698 radians A1 AWRT 3 10(iii) State a – 2b + 3c = 0 or 2a + b – 3c = 0 B1 Obtain two relevant equations and solve for one ratio, M1 Could use 2a + b – 3c = 0 and e.g. a : b  a + 3b − 2c = d  5 5  3 a + 3 b = d i.e. use two points on the line rather than the direction of the line. The second M1 is not scored until they solve for d. Obtain a : b : c = 3 : 9 : 5 A1 OE Substitute a, b, c and a relevant point in the plane equation and M1 Using their calculated normal and a relevant point evaluate d Obtain answer 3x + 9y + 5z = 20 A1 OE Alternative method for question 10(iii) Attempt to calculate vector product of relevant vectors, M1 e.g. (i – 2j + 3k) × (2i + j – 3k) Obtain two correct components A1 Obtain correct answer, e.g. 3i + 9j + 5k A1 Use the product and a relevant point to find d M1 Using their calculated normal and a relevant point Obtain answer 3x + 9y + 5z = 20, or equivalent A1 OE 10(iii) Alternative method for question 10(iii) Attempt to form a 2-parameter equation with relevant vectors M1 State a correct equation A1 e.g. r = i + 3j – 2k + λ(i – 2j + 3k) + µ(2i + j – 3k) State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain answer 3x + 9y + 5z = 2 A1 OE 5