Mathematical Methods marking guide
External assessment 2022
Paper 1: Technology-free (55 marks) Paper 2: Technology-active (55 marks)
Assessment objectives
This assessment instrument is used to determine student achievement in the following objectives:
1.
select, recall and use facts, rules, definitions and procedures drawn from Units 3 and 4
2.comprehend mathematical concepts and techniques drawn from Units 3 and 4
3.
communicate using mathematical, statistical and everyday language and conventions
4. evaluate the reasonableness of solutions5.
justify procedures and decisions by explaining mathematical reasoning
6. solve problems by applying mathematical concepts and techniques drawn from Units 3 and 4
Purpose
This marking guide:
β’ provides a tool for calibrating external assessment markers to ensure reliability of results
β’ indicates the correlation, for each question, between mark allocation and qualities at each level of the mark range
β’ informs schools and students about how marks are matched to qualities in student responses.
Mark allocation
Where a response does not meet any of the descriptors for a question or a criterion, a mark of β0β
will be recorded.
Where no response to a question has been made, a mark of βNβ will be recorded.
Allow FT mark/s β refers to βfollow throughβ, where an error in the prior section of working is used later in the response, a mark (or marks) for the rest of the response can still be awarded so long as it still demonstrates the correct conceptual understanding or skill in the rest of the response.
This mark may be implied by subsequent working β the full mathematical reasoning and/or working, as outlined in the sample response and associated mark, is not explicitly stated in the student response, but by virtue of subsequent working there is sufficient evidence to award the mark/s.
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Marking guide
Multiple choice
Paper 1: Technology-free (55 marks)
Question Response
1 A
2 D
3 C
4 C
5 D
6 B
7 B
8 A
9 A
10 D
Short response
Q Sample response The response:
11a) Change from log to index form and rearrange 2π₯π₯=ππ5
π₯π₯=ππ5 2
β’ correctly rearranges equation to remove ln [1 mark]
β’ correctly determines π₯π₯ [1 mark]
11b) Using log laws log4οΏ½4π₯π₯+ 16
π₯π₯2β2οΏ½= 1 (4π₯π₯+ 16) = 4(π₯π₯2β2) π₯π₯2β π₯π₯ β6 = 0
(π₯π₯ β3)(π₯π₯+ 2) = 0 π₯π₯=3,β2
β’ correctly applies the log law [1 mark]
β’ correctly determines quadratic equation to solve [1 mark]
β’ determines possible values for π₯π₯ [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
12a)
πΈπΈ(ππ) =οΏ½ πππππ₯π₯ππ
=
15 Γ 0 + 45 Γ 1=
45 β’ correctly determines the mean [1 mark]12b) ππππππ(ππ) =οΏ½ ππππ(π₯π₯ππβ ππ)2
=
15 ΓοΏ½
β45οΏ½
2+
45 ΓοΏ½
15οΏ½
2=
254 β’ correctly determines the variance [1 mark]12c) ππππππππππππππππ ππππππππππππππππππ = βππππππππππππππππ = 2
5 β’ determines the standard deviation [1 mark]
Q Sample response The response:
13a) ππβ²(π₯π₯)=6ππ2π₯π₯+1 β’ correctly determines the derivative [1 mark]
13b)
g(π₯π₯) =ln (π₯π₯)
Let π’π’= ln(π₯π₯) and πππ₯π₯ =π₯π₯
β΄πππππππ₯π₯=1π₯π₯ andπππππππ₯π₯= 1 gβ²(π₯π₯) =ππ πππππππ₯π₯ β ππ πππππππ₯π₯
ππ2
=
π₯π₯Γ1π₯π₯ βln(π₯π₯)Γ1π₯π₯2=
1βln(π₯π₯)π₯π₯2gβ²(ππ) =1βln(ππ) (ππ)2 = 0
β’ correctly identifies the use of the product or quotient rule [1 mark]
β’ correctly determines the derivative [1 mark]
β’ determines the derivative at the given value [1 mark]
13c) β(π₯π₯) =π₯π₯sin(π₯π₯)
Let π’π’=π₯π₯ and ππ= sin(π₯π₯) β΄πππππππ₯π₯= 1 and πππππππ₯π₯= cos(π₯π₯) ββ²(π₯π₯) =π’π’ππππ
πππ₯π₯+πππππ’π’
=π₯π₯Γ cos(π₯π₯) + sin(π₯π₯) Γ 1 πππ₯π₯
= π₯π₯cos(π₯π₯) + sin(π₯π₯) Let π’π’=π₯π₯ and ππ= cos(π₯π₯) β΄πππππππ₯π₯= 1 and πππππππ₯π₯=βsin(π₯π₯) ββ²β²(π₯π₯) =π’π’ππππ
πππ₯π₯+πππππ’π’
πππ₯π₯+ cos(π₯π₯)
=π₯π₯Γβsin(π₯π₯) + cos(π₯π₯) Γ 1 + cos(π₯π₯)
= 2 cos(π₯π₯)β π₯π₯sin(π₯π₯)
β’ correctly identifies the use of the product rule [1 mark]
β’ correctly determines the derivative [1 mark]
β’ correctly identifies the use of the product rule and that
ππ
πππ₯π₯οΏ½β(π₯π₯) +ππ(π₯π₯)οΏ½=πππ₯π₯ππ β(π₯π₯) +πππ₯π₯ππ ππ(π₯π₯) [1 mark]
β’ determines the second derivative [1 mark]
β’ simplifies the second derivative [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
14a) V =οΏ½0.25ππ0.25π‘π‘ ππππ =ππ0.25π‘π‘+ππ when ππ= 0,ππ= 0
β΄0 =ππ0.25Γ0+ππ
β΄ππ=β1
β΄ ππ=ππ0.25π‘π‘β1
β’ correctly determines the integral of the function ππ(ππ) [1 mark]
β’ determines the value of c [1 mark]
14b) ππ(8 ln(6))=ππ0.25Γ8 ln(6)β1 = 36β1
= 35 litres
β’ determines the simplified exponential term [1 mark]
β’ determines number of litres [1 mark]
14c) Using trapezoidal rule
Volume after 3 hours = 12οΏ½0.25+0.53+2(0.32+0.41)οΏ½
Volume after 3 hours =1.12 litres
β’ establishes expression for approximate number of litres of water in vessel after 3 hours [1 mark]
β’ determines approximate number of litres [1 mark]
Q Sample response The response:
15 The function is decreasing when ππβ²(π₯π₯) < 0 and concave up when ππβ²β²(π₯π₯) > 0
ππβ²(π₯π₯)= (π₯π₯ β4)πππ₯π₯ < 0 when π₯π₯< 4
ππβ²β²(π₯π₯) =(π₯π₯ β4)πππ₯π₯+πππ₯π₯=πππ₯π₯(π₯π₯ β3)> 0 when π₯π₯> 3 Therefore, the function is decreasing and concave up when 3 <π₯π₯< 4
β’ correctly describes conditions when the function is decreasing and concave up [1 mark]
β’ correctly determines the interval where ππ(π₯π₯) is decreasing [1 mark]
β’ correctly determines the interval where ππ(π₯π₯) is concave up [1 mark]
β’ determines interval when function is decreasing and concave up [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
16 The provided functions are a sine and a cosine function, both with negative amplitudes and the same period.
Therefore, the smooth line is π¦π¦β²=βππsin(πππ₯π₯) , and the dotted line is
π¦π¦β²β²= βπ΄π΄cos(πππ₯π₯), with ππ= 3 and π΄π΄= 4.5.
Therefore, the primary function must be a cosine function with the same period i.e. π¦π¦= 2 cosοΏ½32π₯π₯οΏ½
β’ correctly identifies the smooth curve as the first derivative [1 mark]
β’ provides mathematical reasoning for sketch [1 mark]
β’ sketches the function [1 mark]
Q Sample response The response:
17 Using first integral πΉπΉ(ππ)β πΉπΉ(ππ) = 117
ππ3β ππ3= 117 β¦ Equation I
Using second integral
(ππ β1)3β ππ3= 56 β¦ Equation II Equation I β Equation II
ππ3β(ππ β1)3= 61
ππ3β(ππ3β3ππ2+ 3ππ β1) = 61 3ππ2β3ππ β60 = 0
ππ2β ππ β20 = 0 (ππ β5)(ππ+ 4) = 0 ππ=β4, 5
Given ππ> 1
β΄ ππ= 5
β’ correctly establishes a formula for one of the integrals [1 mark]
β’ determines equation in ππ [1 mark]
β’ determines values of ππ [1 mark]
β’ evaluates the reasonableness of solutions [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
18 οΏ½ππ2π₯π₯ β2 πππ₯π₯= 0.36
1
π₯π₯2β2π₯π₯ οΏ½ππ1 = 0.36 (ππ2β2ππ)β(1β2) = 0.36 ππ2β2ππ+ 1 = 0.36 ππ2β2ππ+ 0.64 = 0
β΄ ππ=2Β±β4β4Γ1Γ0.64
2
β΄ ππ=2 Β±β1.44 2
β΄ ππ=2 Β± 1.2
β΄ ππ= 1.6 or 0.4 2 Given 1β€ π₯π₯ β€2
β΄ ππ= 1.6
β’ correctly determines the definite integral [1 mark]
β’ determines the quadratic equation [1 mark]
β’ determines values of ππ [1 mark]
β’ evaluates the reasonableness of solutions [1 mark]
Q Sample response The response:
19
Total area πππππ΄π΄ and ππππππ (π΄π΄ππ)(using area rule)
=12 Γ 20 Γ π₯π₯ Γ sin 30Β° +12 Γ (10β π₯π₯) Γ ππππ Γ sin 30Β°
Given πΆπΆπΆπΆπ΄π΄π΄π΄=πππΆπΆπππ΄π΄ β΄πΆπΆπΆπΆ20=10βπ₯π₯π₯π₯
β΄ππππ=20(10βπ₯π₯)π₯π₯
π΄π΄ππ=12Γ 20 Γπ₯π₯Γ sin 30Β° +12Γ (10β π₯π₯) Γ20(10βπ₯π₯)π₯π₯ Γ sin 30Β°
= 5π₯π₯+π₯π₯5(10β π₯π₯)2
π΄π΄ππ=10π₯π₯2β100π₯π₯+ 500 π₯π₯
β’ correctly uses all of the given information to draw a labelled diagram [1 mark]
β’ correctly establishes a formula for the total area [1 mark]
β’ correctly determines an expression for CP in terms of π₯π₯ [1 mark]
β’ determines simplified version of the formula for total area [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
Differentiate π΄π΄ππ
π΄π΄ππ= 10π₯π₯+ 500π₯π₯β1β100 π΄π΄β²ππ= 10β500π₯π₯β2 Let π΄π΄β²ππ= 0
β΄0 = 10β500π₯π₯β2
β΄500 π₯π₯2 = 10
β΄ π₯π₯= Β±β50 π₯π₯ is a positive length
β΄ π₯π₯=β50 Verifying using ππβ²β²(π₯π₯) ππβ²(π₯π₯) = 10β500π₯π₯β2 ππβ²β²(π₯π₯) =1000
π₯π₯3
ππβ²β²οΏ½β50οΏ½> 0β΄minimum
Therefore when π₯π₯=β50 the total area is minimised.
β’ determines an equation to solve for stationary points [1 mark]
β’ evaluates the reasonableness of solutions [1 mark]
β’ verifies solution [1 mark]
Marking guide
Multiple choice
Paper 2: Technology-active (55 marks)
Question Response
1 D
2 B
3 C
4 C
5 B
6 B
7 D
8 C
9 B
10 C
Note that question 3 has been updated to indicate option C as the correct answer.
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Short response
Q Sample response The response:
11a) Mean number of sales
=ππππ
= 25 Γ 0.2
= 5
β’ correctly substitutes into formula for mean [1 mark]
β’ correctly determines the mean [1 mark]
11b) Standard deviation of number of sales
=οΏ½ππππ(1β ππ)
=οΏ½25 Γ 0.2 Γ (1β0.2)
= 2
β’ correctly substitutes into formula for standard deviation [1 mark]
β’ correctly determines the standard deviation [1 mark]
11c) 1β οΏ½ππ0οΏ½0.200.8ππβ₯0.88 β΄1β0.8ππβ₯0.88
β΄0.8ππ β€0.12
β΄ ππ β₯log0.8(0.12)
β΄ ππ β₯9.50179
β΄ minimum number of customers would be 10.
β’ correctly determines the required inequation [1 mark]
β’ correctly determines the unknown in the inequation [1 mark]
β’ determines the minimum number of customers [1 mark]
Q Sample response The response:
12a)
ππ(π₯π₯< 12000) =0.3085
β’ correctly uses an appropriate mathematical representation [1 mark]
β’ correctly determines the probability [1 mark]
12b)
π₯π₯=14 561.38 kilometres
β’ correctly uses an appropriate mathematical representation [1 mark]
β’ correctly determines the value of π₯π₯ [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
13a) π€π€(2) = 8
β΄ ππ+ππsin(0) = 8
β΄ ππ= 8 π€π€(11) = 3
β΄8 +ππsinοΏ½3ππ 2οΏ½= 3
β΄ ππ Γ β1 =β5
β΄ ππ= 5
β’ correctly determines ππ [1 mark]
β’ correctly determines ππ [1 mark]
13b)
The rate of change at 8am is 5ππ6. Using sketch π‘π‘= 14
At 8 pm the rate is the same (for the first time).
β’ determines rate when π‘π‘= 2 [1 mark]
β’ determines first time when rate is the same as π‘π‘= 2 [1 mark]
Q Sample response The response:
14a) Using GDC to determine confidence interval associated with ππ= 200,ππΜ= 0.25,π§π§= 1.96 (0.19, 0.31)
β’ correctly identifies all of the information required to establish the confidence interval [1 mark]
β’ correctly determines the confidence interval [1 mark]
14b) Combining results ππ= 450,ππΜ=11
45 Using GDC (0.2047, 0.2842)
β’ correctly determines ππ and ππΜ for the combined sample [1 mark]
β’ determines confidence interval [1 mark]
14c) By combining the results, the sample size is increased and the confidence interval width is reduced.
The new sample statistic provides a better estimate for the population parameter.
β’ identifies changed width of confidence interval [1 mark]
β’ evaluates the reasonableness of Khadijaβs suggestion [1 mark]
14d) Using approximation to the normal distribution Mean = 0.24
Standard deviation= οΏ½0.24 Γ 0.76
200 = 0.0302 Using GDC
ππ(ππΜ> 0.30) =0.0235
β’ correctly determines the mean and standard deviation of the normal distribution [1 mark]
β’ determines the probability [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
15
π»π»= β= 5.2 πΏπΏ= 22Β° ππ=π₯π₯ πΆπΆ= ππ= 8 Using cosine rule
π₯π₯2= 82+ 5.22β2 Γ 8 Γ 5.2 Γ cos 22Β°
π₯π₯= 3.7280 km
Using sine rule 3.7280
sin 22Β° = 8 sinπΆπΆ πΆπΆ= 53.5019Β°
or πΆπΆ= 126.4981Β°
Bearing = 360 β126.4981Β°+ 30
β΄ walk back at a bearing of 263Β°30β² T
β’ correctly establishes the diagram using bearing and distance information [1 mark]
β’ correctly determines angle L [1 mark]
β’ determines unknown distance [1 mark]
β’ substitutes into appropriate rule [1 mark]
β’ determine obtuse angle at πΆπΆ [1 mark]
β’ determines direction needed to return to youth hostel using the obtuse angle for πΆπΆ [1 mark]
Q Sample response The response:
β’ shows logical organisation communicating key steps [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
16 οΏ½ ππ(4β π₯π₯2) 32 πππ₯π₯= 1
2
β2Using GDC (solving for equation) ππ= 3
ππ(β1β€ ππ β€1)
=β«β11 ππ(4βπ₯π₯32 2)
πππ₯π₯
= 0.6875
β’ correctly identifies required integral equation to solve [1 mark]
β’ correctly determines the value of ππ [1 mark]
β’ correctly identifies interval [1 mark]
β’ determines probability [1 mark]
Q Sample response The response:
17 Total displacement of the snail
β« ππ.ππππππ ππ π₯π₯π₯π₯(ππ+ππππ)π π ππ=ππππ.ππππππππ ππππ Velocity of the ant =β« ππ π π ππ
=ππππ+ππ πππππππππ₯π₯πππππππππππ₯π₯πππππ₯π₯ππ ππππππππ ππππ ππππ πππππππππ₯π₯= πππππππππ₯π₯πππππππππππ₯π₯πππππ₯π₯πππππ₯π₯ ππππππππ ππ ππππ πππππππππ₯π₯
β΄ β« ππππππππππππ +ππ=ππππ.ππππππππ
Solving numerically on GDC ππ=βππ.ππππππππ
Therefore, velocity of ant at ππ=ππππ
=ππΓππππ β ππ.ππππππππ
= 22.3477 cm minβ1 along the antβs path.
β’ correctly determines the total displacement of the snail [1 mark]
β’ establishes an equation linking the ant and the snail [1 mark]
β’ determines constant [1 mark]
β’ determines velocity of ant [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Q Sample response The response:
18
ππ(πΌπΌπΌπΌ β₯120) =0.1057 Using binomial distribution ππ= 9,ππ= 0.1057
Using GDC ππ(π₯π₯ β€2) =0.9391
β’ correctly determines the probability of IQ β₯120 [1 mark]
β’ correctly recognises context is suitable for modelling as a binomial [1 mark]
β’ determines the required probability [1 mark]
Q Sample response The response:
19 Number of flying foxes entering the region
=οΏ½42 sinοΏ½0.03π‘π‘ βππ 3οΏ½+ 71
=β1400 cosοΏ½0.03π‘π‘ βππ
3οΏ½+ 71π‘π‘+ππ1
Number of flying foxes leaving the region πΏπΏ(π‘π‘)
=οΏ½42 sinοΏ½0.04π‘π‘ βππ 3οΏ½+ 42
=β 42
0.04cosοΏ½0.04π‘π‘ βππ
3οΏ½+ 42π‘π‘+ππ2 Number of flying foxes in region ππ(π‘π‘)
=β1400 cosοΏ½0.03π‘π‘ βππ3οΏ½+ 71π‘π‘+ππ1
β οΏ½β1050 cosοΏ½0.04π‘π‘ βππ3οΏ½+ 42π‘π‘+ππ2οΏ½
Given number of flying foxes in the region at time π‘π‘= 0 is 100
ππ(π‘π‘) =β1400 cosοΏ½0.03π‘π‘ βππ3οΏ½+ 1050 cosοΏ½0.04π‘π‘ β
ππ
3οΏ½+ 29π‘π‘+ 275
Using GDC to sketch ππ(π‘π‘)
Using a GDC, the maximum point is identified (177.729, 7034.264).
Maximum of 7034 flying foxes in the region at 9: 58 pm.
β’ correctly determines the function to model the total number of flying foxes in the region [1 mark]
β’ uses an appropriate mathematical method to identify the maximum values [1 mark]
β’ determines the maximum number of flying foxes and time, i.e. the coordinates of the turning point [1 mark]
β’ states the number of flying foxes as a whole number and the time after 7 pm it occurs [1 mark]
Mathematical Methods marking guide
External assessment 2022 Queensland Curriculum & Assessment Authority
Β© State of Queensland (QCAA) 2022
Licence: https://creativecommons.org/licenses/by/4.0 | Copyright notice: www.qcaa.qld.edu.au/copyright β lists the full terms and conditions, which specify certain exceptions to the licence. | Attribution: Β© State of Queensland (QCAA) 2022
