Mathematics 9709/32 October/November 2021: question paper and mark scheme

Mark scheme

1 Use law of the logarithm of a product, a quotient or power ( ) *M1 e.g. ln 7 x = x ln 7 Obtain a correct linear equation in any form A1 e.g. ln 3 + (1 − x ) ln 2 = x ln 7 Solve a linear equation for x DM1 ln 6 A1 Maximum 3 out of 4 available if final answer not in Obtain answer x = required form e.g. 0.67… ln14 ISW once correct answer seen. Alternative method for Question 1 21− x = 2 × 2− x *M1 OE 6 = 2 x 7 x = 14 x A1 Use law of the logarithm of a power to solve for x DM1 Must be a linear power. Allow x = ln14 ( 6 ) . ln 6 A1 ISW once correct answer seen. Obtain answer x = ln14 4

Mark scheme

2 State or imply non-modular inequality ( 3x − a ) > 22 ( x + 2a ) , or 2 2 B1 Need 22 seen or implied. corresponding quadratic equation, or pair of linear equations or linear inequalities Make reasonable attempt to solve a 3-term quadratic, or solve two linear equations for x in terms of a M1 ( 5 x − 22ax − 15a = 0 ) 2 2 3 A1 OE Obtain critical values x = 5a and x = − a and no others Accept incorrect inequalities with correct critical values. 5 Must state 2 values i.e. a ±c b is not sufficient. 3 A1 Do not condone ⩾ for > or ⩽ for < in the final answer. State final answer x > 5a, x < − a 5 5a < x < − 53 a is A0, ‘and’ is A0. Alternative method for Question 2 Obtain critical value x = 5a from a graphical method, or by solving a linear B1 equation or linear inequality 3 B2 Maximum 2 marks if more than 2 critical values. Obtain critical value x = − a similarly 5 3 B1 Do not condone ≥ for > or ≤ for < in the final answer. State final answer x > 5a , x < − a 5 5a < x < − 53 a is B0, ‘and’ is B0. 4

Mark scheme

3(a) Substitute for u and w and state correct conjugate of one side B1 Express the other side without conjugates and confirm ( u + w) = u* + w* * B1 Given answer. Needs explicit reference to conjugate of both sides. 2 3(b) Substitute and remove conjugates to obtain a correct equation in x and y B1 e.g. x + 2 − ( y + 1) i + ( 2 + i )( x + iy ) = 0 Use i 2 = −1 and equate real and imaginary parts to zero M1 Obtain two correct equations in x and y A1 e.g. 3x – y + 2 = 0 and x + y – 1 = 0. Allow xi + yi − i = 0 . 1 5 A1 Allow for real and imaginary parts stated separately. Solve and obtain answer z = − + i 4 4 4

Mark scheme

4 B C B1 Dx + E + F and P + Qx + R are also valid. State or imply the form A + + 2x −1 x−3 2x − 1 x − 3 2x −1 x − 3 Use a correct method for finding a constant M1 Obtain one of A = 2, B = – 3 and C = 2 A1 Allow maximum M1A1 for one or more ‘correct’ values after B0. Obtain a second value A1 Obtain the third value A1 Alternative method for Question 4 Divide numerator by denominator M1 A1 Obtain 2 + ( 2 x −1)( x −3) 2 + x+7 Px + Q ( 2 x −1)( x −3) D E B1 State or imply the form + 2x −1 x−3 Obtain one of D = – 3 and E = 2 A1 Obtain a second value A1 5

Mark scheme

5(a) Show circle with centre 3 + 2i B1 Show circle with radius 1. Must match their scales: if scales not identical B1 should have an ellipse. 2i Show line y = 2 in at least the diameter of a circle in the first quadrant B1 O 3 Shade the correct region in a correct diagram B1 4 5(b) Identify the correct point B1 Carry out a correct method for finding the argument M1 e.g. arg x = tan −1 2 + sin −1 1 3 13 Exact working required. Obtain answer 49.8° A1 Or better. 0.869 radians scores B1M1A0. 3 Special Case 1: B1M0 for 45° if they have shaded the wrong half of the circle. Special Case 2: 3 out of 3 available if they identify the correct point on the correct circle and it is consistent with their shading.

Mark scheme

6(a) State correct expansion of sin ( 3 x + 2 x ) or sin ( 3 x − 2 x ) B1 1 M1 Substitute expansions in ( sin 5 x + sin x ) , or equivalent 2 Simplify and obtain 1 ( sin 5 x + sin x ) = sin 3 x cos 2 x A1 Obtain the given identity correctly. 2 3 6(b) 1 1 B1 Obtain integral − cos5 x − cos x , or equivalent 10 2 Substitute limits correctly in an expression of the form p cos5x + q cos x M1 Correct limits and subtracted the right way around. Do not need values of trig functions for M1. Maximum one slip. ( Obtain 15 3 − 2 ) A1 Substitute values and obtain the given answer following full, correct and exact working. 3

Mark scheme

7 Separate variables correctly B1 1 y d y = 4 xe d x −2 x 2 1 1 B1 OE y dy = − y 2 Commence the other integration and reach axe−2 x + b e−2 x dx M1 1 1 A1 SOI (might have taken out factor of 4) Obtain −2 xe−2 x + 2 e−2 x dx or − xe −2 x + e −2 x dx 2 2 Complete integration and obtain −2 xe −2 x − e −2 x A1 Evaluate a constant or use x = 0 and y = 1 as limits in a solution containing M1 p terms of the form , qxe −2 x , re −2 x , or equivalent. y e2 x A1 ISW Obtain y = , or equivalent expression for y 2x + 1 7

Mark scheme

8(a) Expand the square and equate to 1 B1 Use correct double angle formula M1 4 Need to see or sin 2θ = 2sin θ cosθ stated. 2 Obtain cos 4 θ + sin 4 θ = 1 − 12 sin 2 2θ A1 Obtain the given result correctly. 3 8(b) Use the identity and carry out a method for finding a root M1 (1 − 12 sin 2θ = 95 ) 2 Obtain answer 35.3° A1 Must be correct if overspecified: 35.264... Obtain a second answer, e.g. 54.7° A1 FT [e.g 90° − their 35.3°] Do not FT if mixing degrees and radians. Obtain the remaining answers, e.g. 144.7° and 125.3° and no others in the A1 FT [e.g.180° − .. and 180° − ..] given interval Ignore answers outside the given interval. Treat answers in radians as a misread. ( 0.615, 0.955, 2.19, 2.53) Do not FT if mixing degrees and radians. 4

Mark scheme

9(a) State correct derivative of ye 2 x with respect to x B1 dy 2 ye2 x + e2 x dx State correct derivative of y 2 e x with respect to x B1 dy 2 ye x + y 2e x dx dy M1 Equate attempted derivative of the LHS to zero and solve for dx dy 2 ye x − y 2 A1 Obtain the given answer correctly. Obtain = dx 2 y − ex Condone multiplication by −1 and cancelling of ex −1 without comment. Alternative method for Question 9(a) 2 d 2x dy B1 Other rearrangements are possible e.g. Rearrange as y = ( ) e − ye x = 2e 2 x − ye x − e x d 2 −x dy ( ) 2x e − ye x dx dx y = 2e −2 x + y 2 e − x y e = 2 ye − x − y 2e− x dx dx dy 2 dy B1 dy dy =− × 2e 2 x − ye x − e x = −4e − x + 2 ye − x − y 2e− x ( ) dx dx 2 dx e2 x − ye x dx dy M1 Solve for dx dy 2 ye x − y 2 A1 Obtain the given answer correctly. Obtain = dx 2 y − ex 4 9(b) Equate denominator to zero and substitute for y or for e x in the equation of *M1 the curve Obtain equation of the form ae3 x = b or cy 3 = d DM1 ( e = 8, y = 1) SOI 3x 3 Obtain x = ln 2 A1 Accept 1 ln 8 ISW 3 Obtain y = 1 A1 4

Mark scheme

10(a) Obtain direction vector −i + j + 2k , or equivalent B1 Accept answers as column vectors throughout. Use a correct method to form a vector equation M1 State answer r = i + 2j – k + λ (– i + j + 2k), or equivalent correct form A1 0 1 x e.g. r = 3 + μ −1 Allow y for r. 1 −2 z 3 10(b) Use a correct method to find the position vector of C M1 1− 3 e.g. OC = OA + AC = 2 + 3 −1 + 6 Obtain answer – 2i + 5j +5k, or equivalent A1 Accept as coordinates. 2 ρ 10(c) State OP in component form B1 FT Form an equation in λ by equating the modulus of OP to 14 , or equivalent M1 Simplify and obtain 3λ 2 − λ − 4 = 0 , or equivalent A1 3λ 2 + λ − 4 = 0 if using i − j − 2k in (a). 3μ 2 + 5 μ − 2 = 0 if using − i + j + 2k in (a) and OB. Solve a 3-term quadratic and find a position vector M1 (λ = −1, 43 or λ = 1, − 43 or μ = 13 , −2 or μ = − 13 , 2) 1 10 5 A1 Accept as coordinates. Obtain answers 2i + j – 3k and − i + j + k , or equivalent 3 3 3 5

Mark scheme

11(a) Use chain rule M1 Allow if not starting with the correct index. Obtain correct derivative in any form A1 dy sec 2 x e.g. = dx 2 tan x Use correct Pythagoras to obtain correct derivative in terms of tan x A1 dy 1 + tan 2 x e.g. = dx 2 tan x dy B1 Confirm the given statement from correct work. Use a correct derivative to obtain = 1 when x = 14 π dx Should see at least 22 = 1 . 4 11(b) Equate answer to part (a) to 1 and obtain a quartic equation in t or tan x *M1 ( ) = 4 tan x . 2 At least as far as 1 + tan 2 x Obtain correct answer, i.e. t 4 + 2t 2 − 4t + 1 = 0 A1 Or equivalent horizontal form. Commence division by t – 1 DM1 As far as t 3 + t 2 + …by long division or inspection. Allow verification by multiplying given answer by t – 1. Obtain the given answer A1 4 11(c) Use the iterative process correctly with the given formula at least once M1 Obtain one value and use that to obtain the next. Must be working in radians. Obtain final answer a = 0.29 A1 Show sufficient iterations to 4 d.p. to justify 0.29 to 2 d.p., or show there is a A1 e.g. 0.3, 0.2854, 0.2894, 0.2883, ..... sign change in (0.285, 0.295) 0.4, 0.2436, 0.2984, 0.2841, 0.2883, 0.2871, ... 0.5, 0.1776, 0.3103, 0.2805, 0.2893, 0.2868, ... 3