Mathematical Methods SEE marking guide

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Mathematical Methods SEE marking guide

External assessment 2021

SEE 1: Short response (54 marks)

SEE 2 Paper 1: Technology-free (55 marks) SEE 2 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 Unit 3 Topic 2 2. comprehend mathematical concepts and techniques drawn from Unit 3 Topic 2

3. communicate using mathematical, statistical and everyday language and conventions 4. evaluate the reasonableness of solutions

5. justify procedures and decisions by explaining mathematical reasoning

6. solve problems by applying mathematical concepts and techniques drawn from Unit 3 Topic 2.

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Mathematical Methods SEE marking guide External assessment 2021

Queensland Curriculum & Assessment Authority

Purpose

This marking guide:

• provides a tool for calibrating external assessment markers to ensure reliability of results

• indicates the correlation, for each question, between mark allocation and qualities at each level of the mark range

• 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 ‘0’

will be recorded.

Where no response to a question has been made, a mark of ‘N’ will be recorded.

Allowing for FT marks — refers to ‘follow through’, 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 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 — 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.

Throughout this task, accept equivalent decimal values for all calculations and transcriptions from a calculator.

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SEE 1

External assessment m arking guide

Q Sample response The response:

1a) • accurately plots the given data points [1 mark]

• correctly labels the axes and scales for the scatterplot [1 mark]

(4)

Queensland Curriculum & Assessment Authority 1b) For the general shape of a trigonometric function, the curve

repeats for a particular period and oscillates between a maximum and minimum value.

For the general shape of an exponential function, the curve increases at an increasing rate over time.

For the general shape of a logarithmic model

𝑓𝑓(𝑥𝑥) = log𝑎𝑎(𝑥𝑥),𝑎𝑎> 1, the graph is close to the 𝑦𝑦-axis for 𝑥𝑥< 1 and increases slowly for positive 𝑥𝑥 to a limiting value.

The graph of the trigonometric model changes concavity over time.

The tangent to the graph of the logarithmic function is above the curve and therefore concave down.

The tangent to the graph of the exponential function is below the curve and therefore concave up.

• describes the general shape of a trigonometric function [1 mark]

• describes the general shape of an exponential function [1 mark]

• describes the general shape of a logarithmic function [1 mark]

• explains the concept of concavity for the trigonometric function [1 mark]

• explains the concept of concavity for the logarithmic function [1 mark]

• explains the concept of concavity for the exponential function [1 mark]

1c) The trigonometric model does not represent the population of Yellowhead birds over time, as the population is increasing initially and would not reach a maximum and then reduce (assumption).

Assuming the population continues to increase, the graph of the exponential function is observed to best fit the data.

• states an appropriate first assumption [1 mark]

• states an appropriate second assumption [1 mark]

• makes a suitable selection [1 mark]

2 Let 𝑥𝑥= the number of years since 2000, i.e. in 2001, 𝑥𝑥= 1. Let 𝑦𝑦= the population of Yellowhead birds at the beginning of each year.

Using statistics functionality on GDC to determine exponential function, 𝑦𝑦= 18.715𝑒𝑒^{0.3878𝑥𝑥}

• correctly defines variables [1 mark]

• states the method used to determine the model [1 mark]

• correctly determines the model [1 mark]

(5)

3 • provides a relevant table showing 𝑥𝑥 value (or 𝑡𝑡 value), actual, predicted and residual [1 mark]

• correctly calculates the predicted values for each year using the model from Question 2 [1 mark]

• calculates residuals for all data values [1 mark]

(6)

Queensland Curriculum & Assessment Authority The data values generated by the model are all decimal values,

which do not authentically indicate the population of Yellowhead birds over time.

The data value generated by the model for 𝑡𝑡= 9 is much greater than the actual data value.

• correctly labels the axes and selects appropriate scales for residual plot [1 mark]

• accurately plots the points [1 mark]

• evaluates the reasonableness of the model by considering the first observation from the residual plot [1 mark]

• evaluates the reasonableness of the model by considering the second observation from the residual plot [1 mark]

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4 𝑛𝑛(𝑡𝑡) =_{1+𝑏𝑏𝑒𝑒}^𝐿𝐿−𝑘𝑘𝑘𝑘=𝐿𝐿(1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘})⁻¹ Using the chain rule

𝑛𝑛^′(𝑡𝑡) =−𝐿𝐿(1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘})⁻²×−𝑏𝑏𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘} =𝐿𝐿𝑏𝑏𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}× (1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘})⁻²

= ^{𝐿𝐿𝑏𝑏𝑘𝑘𝑒𝑒}^{−𝑘𝑘𝑘𝑘}

�1+𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}�²

Using the quotient rule and chain rule

𝑛𝑛^′′(𝑡𝑡) =^{−�1+𝑏𝑏𝑒𝑒}^{−𝑘𝑘𝑘𝑘}^�²^{×𝐿𝐿𝑏𝑏𝑘𝑘}²^𝑒𝑒^{−𝑘𝑘𝑘𝑘}_{�1+𝑏𝑏𝑒𝑒}^{−𝐿𝐿𝑏𝑏𝑘𝑘𝑒𝑒}_{−𝑘𝑘𝑘𝑘}^{−𝑘𝑘𝑘𝑘}_�4^{×2�1+𝑏𝑏𝑒𝑒}^{−𝑘𝑘𝑘𝑘}^{�×−𝑏𝑏𝑘𝑘𝑒𝑒}^{−𝑘𝑘𝑘𝑘}

=^{−𝐿𝐿𝑏𝑏𝑘𝑘𝑒𝑒}^{−𝑘𝑘𝑘𝑘}^{×�1+𝑏𝑏𝑒𝑒}^{−𝑘𝑘𝑘𝑘}��𝑘𝑘�1+𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}�−2𝑏𝑏𝑘𝑘𝑒𝑒^{−𝑘𝑘𝑘𝑘}�

�1+𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}�⁴

Point of inflection 𝑛𝑛^′′(𝑡𝑡) = 0

0 = −𝐿𝐿𝑏𝑏𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}× (1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘})(𝑏𝑏(1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘})−2𝑏𝑏𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}) Using the null factor rule

∴ 𝐿𝐿𝑏𝑏𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘} = 0 (not possible if 𝐿𝐿,𝑏𝑏, and 𝑏𝑏 are all positive real numbers)

or 1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}= 0

∴ 𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}=−1 (not possible if 𝑏𝑏 is a positive real number) or 𝑏𝑏(1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘})−2𝑏𝑏𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}= 0

∴ 𝑏𝑏(1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}) = 2𝑏𝑏𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}

∴1 +𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}= 2𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}

∴1 = 𝑏𝑏𝑒𝑒^{−𝑘𝑘𝑘𝑘}

• correctly determines the first derivative [1 mark]

• correctly determines the second derivative [1 mark]

• determines factorised equation using point of inflection definition [1 mark]

• evaluates the reasonableness of solutions by stating that the first two factors cannot equal 0 [1 mark]

(8)

∴ 𝑡𝑡=^ln(𝑏𝑏)_𝑘𝑘 (as shown) Substituting into 𝑛𝑛(𝑡𝑡)

𝑛𝑛(𝑡𝑡) = ^𝐿𝐿

1+𝑏𝑏𝑒𝑒^{�−𝑘𝑘×}^ln(𝑏𝑏)^{𝑘𝑘 �}

𝑛𝑛(𝑡𝑡) =_{1+𝑏𝑏𝑒𝑒}^𝐿𝐿−ln(𝑏𝑏)

𝑛𝑛(𝑡𝑡) =_1+𝑏𝑏×^𝐿𝐿 1 𝑏𝑏

𝑛𝑛(𝑡𝑡) =^𝐿𝐿₂ (as shown)

• generates 𝑥𝑥-ordinate of point of inflection [1 mark]

• generates 𝑦𝑦-ordinate of point of inflection [1 mark]

• shows logical organisation

communicating key steps, including appropriate mathematical

vocabulary, symbols and conventions [1 mark]

5a) The graph is increasing over time.

The growth rate (slope of the graph) is slow at the start, increases in the middle section of the graph and then begins to decrease at the end.

• correctly describes a feature of the shape of the graph [1 mark]

• correctly describes a second feature of the shape of the graph [1 mark]

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5b) • provides a relevant table showing 𝑥𝑥 value (or 𝑡𝑡 value), actual, predicted and residual [1 mark]

• correctly calculates the predicted values for each year using the model from 5a) [1 mark]

• calculates residuals for all data values [1 mark]

(10)

Queensland Curriculum & Assessment Authority The data values generated by the model are all decimal

values, which do not authentically indicate the population of Yellowhead birds.

The model produces a result that is close to the real-world value (either above or below) for all times, e.g. in 2016, the model generates a value of 2475, and the actual population is 2457.

• correctly labels the axes and selects appropriate scales for residual plot [1 mark]

• accurately plots the points [1 mark]

• evaluates the reasonableness of the model by considering the first finding from the residual plot [1 mark]

• evaluates the reasonableness of the model by considering the second finding from the residual plot [1 mark]

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5c) The logistics model does level off at a value; therefore, this model does predict a carrying capacity.

The exponential model increases at a faster rate, never levelling off; therefore, this model does not predict a carrying capacity.

The ecologist’s model produces results that are closer to the real-world data, with a smaller maximum residual.

The exponential model produces reasonable results for the data values in Stimulus 1 but using the model to compare with real-world data from Stimulus 2 produces results that do not compare, e.g. for 2016, the predicted population is almost 7000 birds more than the actual population.

The more valid model is the ecologist’s model from 5a).

• documents a strength or limitation for the logistics model based on the carrying capacity [1 mark]

• documents a strength or limitation for the model from Question 2 based on the carrying capacity [1 mark]

• documents a different strength or limitation for one of the models [1 mark]

• documents another different strength or limitation for one of the models [1 mark]

• states that the logistic model is more valid [1 mark]

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Queensland Curriculum & Assessment Authority Page 11 of 12

6a) Determine the carrying capacity

Using the GDC, the function levels off at 2910.

Determine 99% of the carrying capacity 𝑛𝑛(𝑡𝑡) = 99

100× 2910

= 2880.9

Find 𝑡𝑡

2880.9 = 2910 1 + 205𝑒𝑒^{−0.4415𝑘𝑘}

Using the GDC to find the intersection point, 𝑡𝑡= 22.465

Determine the derivative at 𝑡𝑡= 22.465

Using the GDC to find the derivative when 𝑡𝑡= 22.465 𝑛𝑛(22.465) = 12.717

The growth rate will be 12.717 birds per year.

• provides evidence of method used [1 mark]

• correctly determines the carrying capacity [1 mark]

• determines 99% of carrying capacity [1 mark]

• determines 𝑡𝑡 value [1 mark]

• determines the rate [1 mark]

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6b) Determine the maximum population growth rate From the GDC, maximum occurs at (12.057, 321.191)

Determine the target growth rate 𝑛𝑛^′(𝑡𝑡) =1

2 × 321.191

= 160.5955

Determine 𝑡𝑡 when 𝑛𝑛^′(t) = 160.5955

Using the GDC to find the intersection points 𝑡𝑡= 8.064 and 𝑡𝑡= 16.049

Therefore, the population is experiencing its ‘high growth phase’ from early 2008 until early 2016.

• provides evidence of method used [1 mark]

• correctly determines the maximum growth rate [1 mark]

• determines target growth rate [1 mark]

• determines both 𝑡𝑡 values [1 mark]

• determines when the population is experiencing high growth [1 mark]

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Marking guide

Multiple choice

SEE 2 Paper 1: Technology-free (55 marks)

Question Response

1 A

2 B

3 C

4 C

5 D

6 A

7 C

8 B

9 B

10 C

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11a) Let 𝑢𝑢=𝑒𝑒^𝑥𝑥+ 1 Using the chain rule

𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑=3𝑒𝑒^𝑥𝑥(𝑒𝑒^𝑥𝑥+ 1)²

• correctly identifies the use of the chain rule [1 mark]

• correctly determines the derivative [1 mark]

11b) Using the quotient rule 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑=𝑑𝑑²cos(𝑑𝑑)−sin(𝑑𝑑)2𝑑𝑑 (𝑑𝑑²)² 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑=𝑑𝑑cos(𝑑𝑑)−2 sin(𝑑𝑑) 𝑑𝑑³

• correctly identifies the use of the quotient rule [1 mark]

• provides derivative in simplest form [1 mark]

12a) Changing from log to index form 5𝑑𝑑+ 7 = 32

5𝑑𝑑= 25 𝑑𝑑= 5

• correctly establishes the linear equation [1 mark]

• determines 𝑑𝑑 [1 mark]

12b) Using addition log law

log10(𝑑𝑑²−9) = log10(9𝑑𝑑 −29) Equating and rearranging

• correctly applies the log law [1 mark]

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Queensland Curriculum & Assessment Authority 13a) Solving simultaneously

𝑑𝑑²= 4𝑑𝑑

Rearranging and factorising 𝑑𝑑(𝑑𝑑 −4) = 0

∴ 𝑑𝑑= 0 and 𝑑𝑑= 4

• correctly uses the simultaneous procedure [1 mark]

• correctly determines both the 𝑑𝑑-intercept ordinates [1 mark]

13b) Area = ∫₀⁴(4𝑑𝑑 − 𝑑𝑑²)𝑑𝑑𝑑𝑑

= 2𝑑𝑑²−𝑑𝑑³ 3�4

0

=�2 × 4²−4³ 3� −0

=³²₃ square units

• correctly determines the integral [1 mark]

• substitutes limits into integral [1 mark]

• determines area [1 mark]

14a) 𝑓𝑓^′(𝑑𝑑) =_{(3𝑥𝑥+4)}³ • correctly determines the derivative [1 mark]

14b) 𝑑𝑑-intercept (𝑑𝑑= 0) 0 = ln(3𝑑𝑑+ 4) 3𝑑𝑑+ 4 = 1 𝑑𝑑=−1

• correctly determines the linear equation [1 mark]

• correctly determines the x-intercept [1 mark]

14c) 𝑓𝑓′(−1)

= 3

−3 + 4

= 3 • determines the gradient at the 𝑑𝑑-intercept [1 mark]

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15a)

• correctly labels the given angle and sides in the isosceles triangle [1 mark]

15b) Area =¹₂× 4 × 4 × sin 120°

=1

2 × 4 × 4 ×

√3 2 Area =4√3 cm²

• establishes expression for the area [1 mark]

• correctly determines the exact value of sine [1 mark]

• determines area in simplest form [1 mark]

16 𝑑𝑑^′=−𝑒𝑒^2−𝑥𝑥

Gradient of tangent at (1, 𝑒𝑒) 𝑑𝑑^′(1) =−𝑒𝑒

Equation of the tangent (𝑑𝑑 − 𝑒𝑒) =−𝑒𝑒(𝑑𝑑 −1)

𝑑𝑑-intercept (2,0) y-intercept (0,2𝑒𝑒) Area of triangle = ¹

2× 2 × 2𝑒𝑒 Area of triangle = 2𝑒𝑒units²

• determines equation of tangent [1 mark]

• determines x- and y-intercepts [1 mark]

• determines area of triangle [1 mark]

(18)

Queensland Curriculum & Assessment Authority 17 Using binomial distribution

𝑛𝑛= 5

𝑝𝑝=³₅ and 𝑞𝑞=²₅

𝑃𝑃(catches the bus on 1 day)

=�5 1 � �3

5�¹�2 5�⁴

=3 ×¹⁶₅₄=₆₂₅⁴⁸

• correctly determines the values for 𝑝𝑝^and𝑛𝑛^{[1 mark]}

• establishes expression for the required probability [1 mark]

• determines probability [1 mark]

18 � 𝑓𝑓(𝑑𝑑)𝑑𝑑𝑑𝑑=13

6

0

−1

� 𝑓𝑓(𝑑𝑑)𝑑𝑑𝑑𝑑=43 6

1

0 𝑎𝑎𝑥𝑥³

3 +^{𝑏𝑏𝑥𝑥}₂²+ 4𝑑𝑑+𝑐𝑐� 0

−1 =¹³⁶ ⁽ⁱ⁾

𝑎𝑎𝑥𝑥³

3 +^{𝑏𝑏𝑥𝑥}₂²+ 4𝑑𝑑+𝑐𝑐 �1

0 =⁴³⁶⁽ⁱⁱ⁾ From (i)

0− �−𝑎𝑎 3 +𝑏𝑏

2−4�=13 2𝑎𝑎 −3𝑏𝑏=−11 (A) 6 From (ii)

2𝑎𝑎+ 3𝑏𝑏= 19 (B) (A)−(B)

−6𝑏𝑏=−30

𝑏𝑏= 5 Sub into (A) 2𝑎𝑎= 4 𝑎𝑎= 2

∴ 𝑓𝑓(𝑑𝑑) = 2𝑑𝑑²+ 5𝑑𝑑+ 4

• correctly identifies the use of integrals [1 mark]

• correctly determines the two equations in the two unknowns 𝑎𝑎^and𝑏𝑏^{[1 mark]}

• determines 𝑏𝑏^{[1 mark]}

• determines 𝑎𝑎^{[1 mark]}

(19)

19 0.05 = 2�^{𝑝𝑝�(1−𝑝𝑝�)}_𝑛𝑛 rearranging 𝑛𝑛=²²×𝑝𝑝�(1−𝑝𝑝�)

(0.05)²

The largest sample size will result when 𝑝𝑝̂(1− 𝑝𝑝̂) is maximised in the numerator, therefore generating largest 𝑛𝑛^value.

Maximum occurs at 𝑝𝑝̂= 0.5 Substituting

𝑛𝑛=2²× 0.5(0.5) (0.05)² 𝑛𝑛=400

•correctly establishes an equation in 𝑝𝑝̂^and𝑛𝑛^using given information [1 mark]

•correctly rearranges to make 𝑛𝑛 the subject [1 mark]

•correctly verifies the firm’s decision to use 𝑝𝑝̂= 0.5 using mathematical reasoning [1 mark]

•determines sample size 𝑛𝑛^{[1 mark]}

(20)

Queensland Curriculum & Assessment Authority 20 Method 1

To determine intervals 𝑃𝑃^′(𝑡𝑡) =𝑡𝑡+ 2𝑡𝑡× ln(3𝑡𝑡) Critical points 𝑃𝑃^′(𝑡𝑡) = 0 0 =𝑡𝑡(1 + 2 ln(3𝑡𝑡)) 𝑡𝑡= 0 (reject)

1 + 2 ln(3𝑡𝑡) = 0→ 𝑡𝑡= 1 3√𝑒𝑒

To determine the nature of 𝑡𝑡=_3√𝑒𝑒¹

𝑃𝑃^′′(𝑡𝑡) = 3 + 2 ln(3𝑡𝑡) 𝑃𝑃^′′� 1

3√𝑒𝑒�= 3 + 2 ln� 1

√𝑒𝑒� 𝑃𝑃^′′� 1

3√𝑒𝑒�= 2

∴ the point is a minimum.

Therefore the population decreases for 0 <𝑡𝑡<_3√𝑒𝑒¹ but increases for 𝑡𝑡>_3√𝑒𝑒¹

•correctly determines 𝑃𝑃′(𝑡𝑡) [1 mark]

•correctly determines the rejected solution for 𝑡𝑡 [1 mark]

•determines 𝑡𝑡-ordinate of critical point [1 mark]

•determines value of 𝑃𝑃′′(𝑡𝑡) at critical point [1 mark]

•determines nature of critical point [1 mark]

•communicates when the population is increasing and when it is decreasing [1 mark]

•shows logical organisation communicating key steps [1 mark]

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20 Method 2

To determine intervals 𝑃𝑃^′(𝑡𝑡) =𝑡𝑡+2𝑡𝑡× ln(3𝑡𝑡) 𝑃𝑃′(𝑡𝑡) =𝑡𝑡(1 + 2 ln(3𝑡𝑡))

Population is increasing when 𝑃𝑃^′(𝑡𝑡) > 0 𝑡𝑡(1 + 2 ln(3𝑡𝑡)) > 0

Given 𝑡𝑡> 0

∴(1 + 2 ln(3𝑡𝑡)) > 0 ln(3𝑡𝑡) >−1 2 𝑡𝑡> 1

∴ the population decreases for 3√𝑒𝑒 0 <𝑡𝑡<_3√𝑒𝑒¹

•correctly determines 𝑃𝑃′(𝑡𝑡) [1 mark]

•correctly identifies the time interval required to determine increasing population [1 mark]

•identifies relevance of domain [1 mark]

•establishes inequality in 𝑡𝑡 [1 mark]

•determines interval when population is increasing [1 mark]

•determines interval when population is decreasing [1 mark]

•shows logical organisation communicating key steps [1 mark]

(22)

Marking guide

Multiple choice

SEE 2 Paper 2: Technology-active (55 marks)

Question Response

1 A

2 A

3 D

4 C

5 D

6 A

7 B

8 D

9 D

10 A

(23)

11a) 𝑥𝑥-intercepts (𝑦𝑦= 0) 0 =𝑒𝑒^𝑥𝑥sin(𝑥𝑥) Using null factor rule 𝑒𝑒^𝑥𝑥 ≠0

sin(𝑥𝑥) = 0

∴ 𝑥𝑥= 0, 𝜋𝜋, 2𝜋𝜋

• correctly generates the equation required [1 mark]

• correctly determines the three 𝑥𝑥-intercepts [1 mark]

11b)

� 𝑒𝑒^𝑥𝑥sin(𝑥𝑥)𝑑𝑑𝑥𝑥+�� 𝑒𝑒^2𝜋𝜋 ^𝑥𝑥sin(𝑥𝑥)𝑑𝑑𝑥𝑥

𝜋𝜋 �

𝜋𝜋 0

• determines expression for integral above the 𝑥𝑥^-axis [1 mark]

• determines expression for integral below the 𝑥𝑥^-axis (including absolute value brackets) [1 mark]

11c) Area enclosed (using GDC)

291 square units • determines area to the nearest unit [1 mark]

12a)

𝑠𝑠(𝑡𝑡)=1

6sin�6𝑡𝑡+𝜋𝜋

2�+ 2𝑡𝑡+𝑐𝑐 Substituting (0, 0)

0 =1 6sin�𝜋𝜋

2�+𝑐𝑐 𝑐𝑐=−1

6 𝑠𝑠(𝑡𝑡) =1

6sin�6𝑡𝑡+𝜋𝜋

2�+ 2𝑡𝑡+−1 6

• correctly determines the indefinite integral 𝑠𝑠(𝑡𝑡) [1 mark]

• correctly determines the displacement function [1 mark]

𝑠𝑠(3)− 𝑠𝑠(0) •

(24)

Queensland Curriculum & Assessment Authority 13a) Using GDC

� 3

2(1− 𝑥𝑥²)𝑑𝑑𝑥𝑥= 1

1

0 • correctly establishes the definite integral equated to 1

[1 mark]

13b) 𝑃𝑃(𝑋𝑋< 0.25)

=� 3

2(1− 𝑥𝑥²)𝑑𝑑𝑥𝑥

0.25

=0.3670

• correctly establishes the definite integral [1 mark]

• correctly determines the probability [1 mark]

13c) Mean =∫₀¹³₂𝑥𝑥(1− 𝑥𝑥²)𝑑𝑑𝑥𝑥

=0.375

Variance =∫₀¹³₂(𝑥𝑥 −0.375)²(1− 𝑥𝑥²)𝑑𝑑𝑥𝑥

=0.059 tonnes²

• correctly establishes the definite integral [1 mark]

• correctly establishes the mean [1 mark]

• establishes definite integral [1 mark]

• determines variance [1 mark]

(25)

14a) Using GDC

𝑃𝑃(student height under 180 cm)

=0.841 • correctly determines the probability [1 mark]

14b)

𝑥𝑥= 195.806

∴ minimum height is 196 cm

• correctly uses an appropriate mathematical representation [1 mark]

• correctly determines the lowest decimal height [1 mark]

• determines lowest whole height [1 mark]

14c)

𝑧𝑧School A=196 − 165 15 𝑧𝑧School A=31

15 31

15 < 3

∴ student in School B ranked higher

• determines 𝑧𝑧-score for student in school A [1 mark]

• provides statement to justify decision [1 mark]

• determines higher ranked student [1 mark]

(26)

Current ranking

= log10(50 × 100²)

= 5.69897

∴ increased ranking is 6.69897 6.69897 = log₁₀(50ℎ²) ℎ= 316.228

∴ the website requires an additional 217 hits to increase their ranking by 1.

• correctly determines the website’s increased ranking [1 mark]

• determines number of hits for increased ranking [1 mark]

• provides reasonable solution for number of additional hits [1 mark]

• shows logical organisation communicating key steps [1 mark]

(27)

15 Method 2

𝑅𝑅_{𝑜𝑜𝑜𝑜𝑜𝑜}= log₁₀(50ℎ²)

𝑅𝑅𝑜𝑜𝑜𝑜𝑜𝑜= log10(50) + 2 log10(ℎ) Let 𝑘𝑘= number of additional hits 𝑅𝑅𝑛𝑛𝑛𝑛𝑛𝑛= log10(50) + 2 log10(ℎ+𝑘𝑘) 𝑅𝑅_{𝑛𝑛𝑛𝑛𝑛𝑛}=𝑅𝑅_{𝑜𝑜𝑜𝑜𝑜𝑜}+ 1

log10(50) + 2 log10(ℎ+𝑘𝑘) = log10(50) + 2 log₁₀(ℎ) + 1

2 log10(ℎ+𝑘𝑘) = 2 log10(ℎ) + 1 2 log₁₀(ℎ+𝑘𝑘)−2 log₁₀(ℎ) = 1 log10(ℎ+𝑘𝑘)

ℎ = ¹

(ℎ+𝑘𝑘) 2

ℎ =√10 Currently ℎ= 100 𝑘𝑘= 216.228

∴ the website requires an additional 217 hits to increase their ranking by 1.

• correctly determines the equation using increased ranking of 1 [1 mark]

• determines number of additional hits for increased ranking [1 mark]

• provides reasonable solution for number of additional hits [1 mark]

• shows logical organisation communicating key steps [1 mark]

(28)

16

Distance from A to the tower (𝑏𝑏): tan 28° =60

𝑏𝑏

Distance from B to the tower (𝑎𝑎): tan 35° =60

𝑎𝑎

𝑏𝑏= 112.84 m and 𝑎𝑎= 85.69 m

Distance between A and B (𝑐𝑐) 𝑐𝑐²= 85.69²+ 112.84²– 2 ×

85.69 × 112.84 × cos 110°

𝑐𝑐= 163.37 m

The distance between A and B is 163^m.

• correctly establishes the angles using given bearings [1 mark]

• correctly determines the distance 𝑎𝑎 and 𝑏𝑏 [1 mark]

• determines distance between A and B [1 mark]

• shows logical organisation communicating key steps [1 mark]

(29)

17 Models are periodic.

Rabbits: 𝑅𝑅(𝑡𝑡) = 11 + 3.5 cos�^𝜋𝜋₆𝑡𝑡�

Foxes: 𝐹𝐹(𝑡𝑡) = 9 + 2 sin�^𝜋𝜋₆𝑡𝑡�

Total population of foxes and rabbits 𝑇𝑇(𝑡𝑡) = 20 +3.5 cos�^𝜋𝜋₆𝑡𝑡�+ 2 sin�^𝜋𝜋₆𝑡𝑡�

Graphing 𝑇𝑇(𝑡𝑡)

Greatest total population of foxes and rabbits is 24 031.

Jane’s claim is not correct.

The maximum total population occurs on one occasion in the year, but does not exceed 25 000 (maximum is

• correctly determines the models for the two populations [1 mark]

• determines model for total population of foxes and rabbits [1 mark]

• uses an appropriate mathematical representation [1 mark]

• evaluates reasonableness of the claim [1 mark]

(30)

18 Graphing 𝑃𝑃(𝑡𝑡) (dotted line) and 𝑃𝑃′(𝑡𝑡) (solid line)

The population is increasing most rapidly at the maximum value of 𝑃𝑃^′(𝑡𝑡)

à t = 1.386

∴ P(1.386)~49.993

There are approximately 50 000.

• correctly identifies the conditions for the most rapid increase [1 mark]

• determines when population growing the fastest [1 mark]

• determines population at this time [1 mark]

(31)

19 Method 1

If 𝑋𝑋 denotes the number of minutes that elapse between the placement and delivery of the order, then 𝑋𝑋 can take any value in the interval 100≤ 𝑋𝑋 ≤180.

Since the width is 80, the height of the density function must be ¹

80 for the total area under the density function to equal 1.

The probability density function is:

𝑓𝑓(𝑥𝑥) =₈₀¹,100≤ 𝑥𝑥 ≤180 Using the given rules 𝐸𝐸(𝑋𝑋) =180 + 100

2 Var(𝑋𝑋) =80²

12 = 1600

3

∴Mean = 140 and standard deviation =⁴⁰_√3 Required probability

𝑃𝑃(140−_√3⁴⁰<𝑋𝑋< 140 +_√3⁴⁰)

=

∫

^{140 +} ₈₀¹

𝑑𝑑𝑥𝑥

40

√3 140 − ⁴⁰_√3

= 0.57735

• correctly determines the probability density function [1 mark]

• correctly determines the mean and the standard deviation [1 mark]

• establishes definite integral to represent probability [1 mark]

(32)

Graphing the probability density function:

Using the given rules 𝐸𝐸(𝑋𝑋) =180 + 100

2 Var(𝑋𝑋) =80²

12 =1600 3

∴Mean = 140 and standard deviation =_√3⁴⁰ Identifying required probability as an area

Required probability =₈₀¹ × 2 ×⁴⁰_√3 Required probability =_√3¹

• correctly identifies the probability density function graphically [1 mark]

• correctly determines the mean and the standard deviation [1 mark]

• uses appropriate graphical representation [1 mark]

(33)

20 The quadratic has real roots when 𝑏𝑏²−4𝑎𝑎𝑐𝑐≥0

∴9 – 8𝐵𝐵 ≥0

∴9 ≥8𝐵𝐵

∴9

8 ≥ 𝐵𝐵

Using the standard normal distribution (the given distribution)

𝑃𝑃(𝐵𝐵 ≤9 8)

= 0.8697

• correctly identifies the need to use the discriminant [1 mark]

• correctly determines the range of values for 𝐵𝐵 [1 mark]