Mathematics Advanced

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1 A C E

EXAM

PAPER Student name: ______________________

PAPER 4

YEAR 12 YEARLY EXAMINATION

Mathematics Advanced

General

Instructions

Working time - 180 minutes

Write using black pen

NESA approved calculators may be used

A reference sheet is provided at the back of this paper

In questions 11-16, show relevant mathematical reasoning and/or calculations

Total marks:

100 Section I – 10 marks Attempt Questions 1-10

Allow about 15 minutes for this section Section II – 90 marks

Attempt questions 11-16

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2 10 marks

Attempt questions 1 - 10

Allow about 15 minutes for this section

Use the multiple-choice answer sheet for questions 1-10

1. If ! = 2$√$, which of the following is an expression for !"_!#? (A) 1 √$ (B) 1 2√$ (C) 2√$ (D) 3√$ 2. Evaluate 0 1$ $ % (A) –5 (B) 0 (C) 5 (D) $ 3.

Which of the following equations is likely to be the rule for the graph of the trigonometric function shown above?

(A) ! = 3 + 3sin 6π$ 49 (B) ! = 3 + 3cos 6π$ 49 (C) ! = 3 + 3sin 6$ 49 (D) ! = 3 + 3cos 6$ 49

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3

4. What values of x is the curve <($) = $&_{+ $}'_{concave down?}

(A) $ > 3 (B) $ < −3 (C) $ > −1 3 (D) $ < −1 3

5. The heights of a group of friends are normally distributed with a mean of 160 cm and a standard deviation of 15 cm. What percentage of the group are more than 190 cm tall? (A) 1%

(B) 2.5% (C) 5% (D) 95%

6. What is the domain and range of the function ! = 1 √$ − 9? (A) {$ ∶ $ ≥ 9} and {! ∶ ! > 0} (B) {$ ∶ $ > 9} and {! ∶ ! > 0} (C) {$ ∶ −∞ ≤ $ ≤ ∞} and {! ∶ −∞ ≤ ! ≤ ∞} (D) {$ ∶ −3 ≥ $ ≥ 3} and {! ∶ ! < 0} 7.

What is the correlation between the variables in this scatterplot? (A) Weak negative

(B) Weak Positive (C) Moderate negative (D) Moderate positive

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4 8. The derivative of L cos2$ with respect to x is: (A) −2L−4*

(sin2$ + 2cos2$)

(B) −L−4*

_{(sin2$ − 2cos2$)}

(C) 2L−4*

_{(sin2$ + 2cos2$)}

(D) L−4*

_{(sin2$ − 2cos2$)}

9. A runner started a new training program. On the first day he completed 30 laps of the running track. Every succeeding day he increased his training by 5 laps, until his daily schedule reached 105 laps. On which day did he complete 105 laps in a single day? (A) 15th_day

(B) 16th_day

(C) 20th_day

(D) 21st day

10. The probability density function for the continuous random variable X is:

<($) = M 3

4($'− 1) 1 < $ < 2 0 $ ≤ 1 or $ ≥ 2 The value of O(P ≤ 1.3) is closest to: (A) 0.07

(B) 0.30 (C) 0.43 (D) 0.93

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5 Section II

90 marks

Attempt questions 11 - 16

Allow about 2 hours and 45 minutes for this section Answer each question in the spaces provided.

Your responses should include relevant mathematical reasoning and/or calculations.

Question 11 (4 marks) Marks

The line ! = R$ is a tangent to the curve ! = L%.$#_{at a point A.}

(a) Sketch the line and the curve on the diagram below. 1

(b) Find the coordinates of A. 2

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6

Solve the equation (cos $ + 2)(2cos $ + 1) = 0 in the domain 0 ≤ $ ≤ 2π. 2

Question 13 (2 marks) Simplify 5 $ − 2− 2 $ − 3 2 Question 14 (2 marks)

What is the derivative of ! = 2sin3$ − 3tan$ at x = 0? 2

Question 15 (2 marks)

A class compared their assessment results to their head circumference. The correlation coefficient for these quantities was 0.2. What is the meaning of this correlation?

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7

The diagram below shows the graph of ! = $'_{− 2$ − 8}

(a) What are the coordinates of A? 1

(b) Find the area bounded by the x-axis and the curve ! = $'_{− 2$ − 8 between}

0 ≤ $ ≤ 6.

2

The second term of an arithmetic series is 37 and the sixth term is 17. What is the sum of the first ten terms?

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8

The scatterplot shows the number of sit-ups (s) and the number of push-ups (p) performed by ten students during a fitness test.

(a) Draw a line of best fit on the scatterplot. Find the gradient of this line. 2

(b) Alyssa was absent for the push-up test. Predict her push-up result if she scored 36 on the sit-up test.

1

(c) Calculate the value of the Pearson’s correlation coefficient. Answer correct to

two decimal places. 2

Find the period and amplitude for the graph 5! = sin 62$ −π

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9 A function <($) is defined by <($) = 4$&_{− 4$}'_.

(a) Find the stationary points for the curve ! = <($) and determine their nature. 3

(b) Sketch the graph of ! = <($) showing the stationary points. 2

(c) Show the point(s) at which ! = <($) cuts the x-axis. 1

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10

A particle moves along a straight line about a fixed point O so that its acceleration, a ms-2_{, at time t seconds, is given by:}

^ = 4cos 62_ +π 69

Initially the particle is moving to the right with a velocity of 1 ms-1_{from a position}

0.5√3 metres to the left of O.

(a) Find an expression for the velocity of the particle after t seconds. 2

(b) Find an expression for the position of the particle after t seconds. 2

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11

The table below shows the present value of a $1 annuity. Present value of $1 Period 2% 4% 6% 8% 10% 12% 1 0.98 0.96 0.94 0.93 0.91 0.89 2 1.94 1.89 1.83 1.78 1.74 1.69 3 2.88 2.78 2.67 2.58 2.49 2.40 4 3.81 3.63 3.47 3.31 3.17 3.04

(a) What would be the present value of a $6 000 per year annuity at 4% per

annum for 2 years, with interest compounding yearly? 1

(b) What is the value of an annuity that would provide a present value of $47 988

after 3 years at 8% per annum compound interest? 2

(c) An annuity of $1000 each six months is invested at 12% per annum,

compounded biannually for 2 years. What is the present value of the annuity? 1

The graph ! = <($) passes through the point (1, 4) and <′($) = 3$'_{− 2 .}

Find the expression for <($).

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12

On 1 June 2010, Samuel invested $20,000 in a bank account that paid interest at a fixed rate of 7% per annum, compounded annually

(a) How much is in the account after the payment of interest on 1 June 2020 if no additional deposits were made? Answer to the nearest cent.

1

(b) Samuel decided to deposit $2000 to his account on 1 June each year

beginning on 1 June 2011. How much is in his account on 1 June 2020 after the payment of interest, his deposit and the original investment? Answer to the nearest cent.

3

(c) Samuel’s friend Maya invested $20,000 in an account at another bank on 1 June 2010 and made no further deposits. On 1 June 2020, the balance of Maya’s account was $49,565. What was the annual rate of compound interest paid on Maya’s account? Answer correct to one decimal place.

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13

Find the solution to the equation cos 2$ = 0.5 in the domain −π ≤ $ ≤ π by sketching graphs.

3

Lola gained a standardised score (z-score) of 2.5 for a class test out of 100. (a) Describe Lola’s result in terms of mean and standard deviation of the class

test. 1

(b) The class test has a mean of 56% and a standard deviation of 9.5.

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14

The equation of least-squares line of best fit is given by y = mx + c where R = b.!

." and c = !d − R$̅

What is the y-intercept of the least-squares line of best fit given m = 0.6, $̅ = 50

and !d = 65 ? 1 Question 28 (3 marks) Differentiate (a) (L#_{− 2)}/ ₁ (b) 3$ sin2$ 2 Question 29 (2 marks)

The ages of the residents who live in Hudson Creek are normally distributed. The mean age is 56 years and the standard deviation is 14. What percentage of the residents are younger than 70?

2

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15

A continuous random variable X has a probability density function f given 3 <($) = fg$ + h 1 ≤ $ ≤ 6

0 elsewhere

where A and B are constants. The median of X is 3. Find the values of A and B.

The normal distribution below represents the mass of 400 students. It has a standard deviation of 10 kg. All measurements are in kilograms.

(a) What is the weight of a student with a z-score of –2? 1

(b) How many students have a mass in the region marked with an A? 1

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16

Molly is designing a garden bed for her backyard in the shape of a sector with a radius r and sector angle i. She has a total of 40 metres of garden edging materials to use as the perimeter of the garden bed.

(a) Show that the area A of the garden bed is given by the formula: 3 g =b

2(40 − 2b)

(b) What value of r gives the maximum value of A? Justify your answer. 2

(c) Find the maximum possible area that can be made using the 40 metres of edging material.

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17 The cross-section of a piece of artwork is shown below

The trapezoidal rule with 2 intervals was used to approximate the area of the artwork as 612 cm2. What is the value of x?

2

A radioactive isotope has a half-life of 163 days. Initially there was 10 mg of the isotope. The mass M(t) in milligrams of isotope, after t days, is given by:

2 j(_) = 10L012_{where k is a constant.}

Given that after 163 days only 5 mg of isotope remain, find the value of k. Answer correct to two significant figures.

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18

The rate of emission of carbon pollution C, in tonnes per year from a factory from 1st_{January 2014 is given by}

l = 500 − m_{1 + _}10 n

'

where _ is the time in years

(a) What was the rate of emission of carbon pollution C on 1st January 2014? 1

(b) What value does C approach as time passes? 1

(c) Draw a sketch of C as a function of t? 1

(d) Calculate the total amount of carbon pollution emitted from the factory from 1st_{January 2014 to 1}st_{January 2020? Answer correct to the nearest tonne.}

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19 The graph of ! = <($) is shown below.

Draw sketches of the following functions on the above number plane.

Clearly label each sketch. Indicate any asymptotes and intercepts with the axes.

(a) ! = <($ + 1) 2

(b) ! = 0.5<($) 2

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1

Worked solutions and marking guidelines

Section I Solution Criteria 1. _{! = 2$√$ = 2$ × $}!_"_{= 2$}#_" (! ($= 2 × 3 2× $ ! "= 3√$ 1 Mark: D 2. _{* ($ = * 1($}$ % $ % = [$]_%$_{= 5 − 0 = 5} 1 Mark: C 3. _{Amplitude = 3, Period =}2π 8 = π 4, Vertical shift = 3 ∴ ! = 3 + sin Jπ$ 4 K 1 Mark: A

4. Concave down when L′′($) < 0 L($) = $#_{+ $}"_{, L′($) = 3$}"_{+ 2$} L′′($) = 6$ + 2 6$ + 2 < 0 $ < −1 3 1 Mark: D 5. _{R =}$ − $̅ T = 190 − 160 15 = 2

\95% of scores have a z-score between –2 and 2. \5% ÷ 2 = 2.5% have a z-score greater than 2.

1 Mark: B 6. √$ − 9 ≠ 0 or $ ≠ 9 Also $ − 9 > 0 or $ > 9 lim &→( 1 √$ − 9= 0 Domain: {$ ∶ $ > 9} Range: {! ∶ ! > 0} 1 Mark: B

7. Correlation between –0.5 and –0.74. Moderate negative.

1 Mark: C

8. (

($(^)*&cos2$)($ = (^)*&× −2sin2$) + (cos2$ × −4^)*&) = −2^)*&_{(sin2$ + 2cos2$)}

1 Mark: A 9. 30, 35, 40,…,105. AP with a = 30, d = 5 and Tn = 105 _+ = ` + (a − 1)( 105 = 30 + (a − 1) × 5 105 = 30 + 5a − 5 5a = 80 a = 16 1 Mark: B 10. * 3 4 !·# ! ($"_{− 1)($ =}3 4b $# 3 − $c_! !.# =3 4bd 1.3# 3 − 1.3e − d 1# 3 − 1ec = 0.07425 ≈ 0.07 1 Mark: A

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2

11(a) Sample answer 1 mark: Correct

answer.

11(b) At A the y-values of the line and the curve coincide: ! = h$ = ^%.$&_①

At A the slopes of the line and the tangent to the curve coincide:

h = ( ($^%.$& = 0.5^%.$& ② Substitute ① into ②: h = 0.5h$ $ = 2 ! = ^%·$×"_{= ^} \Coordinates of A are (2, e) 2 Marks: Correct answer.

1 Mark: Finds one of the coordinates or shows some understanding. 11(c) At A h$ = ^%.$& h × 2 = ^%·$×" h = 0.5^ 1 Mark: Correct answer. 12 (cos $ + 2)(2 cos $ + 1) = 0 cos $ = −1 2 or cos $ = −2 $ =2π 3 or (No soln) In the domain 0 ≤ $ ≤ 2π $ =2π 3 , 4π 3 2 Marks: Correct answer.

1 Mark: Finds one solution or shows some understanding. 13 5 $ − 2− 2 $ − 3= 5($ − 3) ($ − 2)($ − 3)− 2($ − 2) ($ − 2)($ − 3) =5$ − 15 − 2$ + 4 ($ − 2)($ − 3) = 3$ − 11 ($ − 2)($ − 3) 2 Marks: Correct answer. 1 Mark: Finds a common denominator or shows some understanding. 14 ! = 2sin 3$ − 3tan$ !/_{= 2 × 3cos 3$ − 3 × sec}"_$ = 6cos 3$ − 3sec"_$ At $ = 0 !′ = 6 − 3 = 3 2 Marks: Correct answer.

1 Mark: Finds the derivative. 15 Assessment results increase as head circumference increases.

Low positive correlation. Not a strong relationship.

2 Marks: Correct answer.

1 Mark: Shows understanding

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3 $ − 2$ − 8 = 0

($ − 4)($ + 2) = 0 ∴ $ = 4 or $ = −2

The x value of A is positive (diagram) \Coordinates of A is (4, 0) 16(b) m* ($* "_{− 2$ − 8)($} % m + * ($0 "_{− 2$ − 8)($} * = nb$# 3 − $"− 8$c_% * n + b$# 3 − $"− 8$c_* 0 = m b4# 3 − 4"− 8 × 4c m + db 6# 3 − 6"− 8 × 6c − b 4# 3 − 4"− 8 × 4ce = 411 3 square units 2 Marks: Correct answer. 1 Mark: Calculates the primitive function or shows some understanding of the problem. 17 _{_}₊ _{= ` + (a − 1)(} _" = ` + ( = 37 ① _₀ = ` + 5( = 17 ② Equation ② − ① 4( = −20 ( = −5

Substitute ( = −5 into equation ① ` − 5 = 37 ` = 42 p₊=a 2[2` + (a − 1)(] =10 2 [2 × 42 + (10 − 1) × (−5)] = 195 3 Marks: Correct answer.

2 Marks: Finds the first term and the common

difference.

1 Mark: Finds two equations using the nth term of a AP or shows some understanding. 18(a) h =Rise Run= 50 50= 1 \Gradient is 1. 2 marks: Correct answer.

1 mark: Finds the line of best fit or shows some understanding.

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4

Alyssa should score 46 on the push-up test. answer.

18(c) _{Data: (0,10)(5,15)(10,25)(15,25)(20,25)(25,35)} (30,50)(35,45)(40,50)(45,50)(50,60) r = 0.968450. . . ≈ 0.97 2 marks: Correct answer. 1 mark: Finds a value of r close to 0.9. 19 _{5! = sin J2$ −}π 3K ! =1 5sin J2$ − π 3K Amplitude =1 5 Period =2π 2 = π 2 Marks: Correct answer. 1 Mark: Finds either amplitude or the period. 20(a) L($) = 4$#_{− 4$}" Stationary points L′($) = 0 L′($) = 12$"_{− 8$} 4$(3$ − 2) = 0 $ = 0, $ =2 3

\Stationary points are (0, 0) and ("_#, −!0_"1) L"($) = 24$ − 8 At (0, 0), L′′(0) = −8 < 0 Maxima At (2 3, − 16 27) L′′ x 2 3y = 8 > 0 Minima 3 Marks: Correct answer. 2 Marks: Finds both of the stationary points. 1 Mark: Finds one of the stationary points or recognises 12$"_{− 8$ = 0}_. 20(b) 2 Marks: Correct answer. 1 Mark: Makes some progress towards sketching the curve. 20(c) x-intercepts (y = 0) 4$#_{− 4$}"_{= 0} 4$"_{($ − 1)} $ = 0, $ = 1

\ The curve cuts the x-axis at x = 0 and x = 1.

1 Mark: Correct answer.

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5 21(a) _{| = * 4cos J2} +}π 6K (} = 2sin J2} +π 6K + ~ Initially t = 0 and v = 1 1 = 2sin J2 × 0 +π 6K + ~ ~ = 0 ∴ | = 2sin J2} +π 6K 2 Marks: Correct answer. 1 Mark: Integrates to find velocity function. 21(b) _{$ = * 2sin J2} +}π 6K (} = −cos J2} +π 6K + ~ Initially t = 0 and x = −0.5√3 −0.5√3 = −cos J2 × 0 +π 6K + ~ −√3 2 = − √3 2 + ~ ~ = 0 ∴ $ = −cos J2} +π 6K 2 Marks: Correct answer. 1 Mark: Integrates to find position function.

21(c) Particle changes direction when v = 0 0 = 2sin J2} +π 6K 2} +π 6= 0, π, 2π, … 2} = −π 6, 5π 6 , 11π 6 , … } =5π 12, 11π 12 , …

∴ Particle changes direction at } =5π 12

1 Mark: Uses

v = 0.

22(a) Intersection value is 1.89 (4% and 2 years) ÄÅ = 1.89 × 6000

= $11 340

1 mark: Correct answer.

22(b) Intersection value is 2.58 (8% and 3 years) Let the value of the annuity be x

47 988 = $ × 2.58

$ =47 988

2.58 = $18 600

\ Value of the annuity is $18 600 per year.

2 marks: Correct answer.

1 mark: Finds the intersection value.

22(c) Intersection value is 3.47 (6% and 4 years) ÄÅ = 3.47 × 1000

= $3 470

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6 L($) = $#_{− 2$ + ~}

Point (1, 4) satisfies the function. 4 = 1#_{− 2 × 1 + ~}

~ = 5

∴ L($) = $#_{− 2$ + 5}

answer.

1 Mark: Correctly integrates the first derivative. 24(a) ÉÅ = ÄÅ(1 + r)+ = 20 000(1 + 0.07)!% = 39 343.02715. . . ≈ $39 343.03 \Account balance is $39 343.03 1 Mark: Correct answer. 24(b) Ñ!%= 2000(1.07)2+ 2000(1.07)3+ ⋯ + 2000(1.07)% = 2000(1.072_{+ 1.07}3_{+ ⋯ + 1.07}!_{+ 1)} GP with ` = 1, r = 1.07 and n = 10 Ñ_!%= 2000 ×1[1.07!% − 1] 1.07 − 1 = 27 632.8959. . . ≈ $27 632.90 Account balance = $27 632.90 + $39 343.03 = $66 975.93 \ Account balance is $27 632.90 3 Marks: Correct answer.

2 Mark: Finds the amount of the annuity or makes significant progress. 1 Mark: Identifies a G.P. with 10 terms. 24(c) ÉÅ = ÄÅ(1 + r)+ 49 565 = 20 000 × (1 + r)!% (1 + r)!%_{= 2.47825} 1 + r = √2.47825!" r = √2.47825!" − 1 = 0.095000989. . . ≈ 9.5%

\Annual rate of compound interest is 9.5%.

1 Mark: Uses the future value interest formula with one correct value.

25 _{Draw graphs of}_{! = cos2$ and ! = 0.5}

Solutions are the points of intersection of the graphs

∴ $ =π 6, 5π 6 , − π 6, − 5π 6 3 Marks: Correct answer. 2 Marks: Makes significant progress towards the solution. 1 Mark: Draws one of the graphs or finds one of the solutions.

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7 26(b) _{R =}$ − $̅ T 2.5 =$ − 56 9.5 23.75 = $ − 56 $ = 79.75

\ Claire scored 79.75 in the class test.

1 mark: Correct answer. 27 á = !à − h$̅ = 65 − 0.6 × 50 = 35 \y-intercept is 35. 1 mark: Correct answer. 28(a) (

($(^&− 2)* = 4(^&− 2)#^& = 4^&_(^&_{− 2)}# 1 Mark: Correct answer. 28(b) ( ($x 3$ sin2$y = sin2$ × 3 − 3$ × 2cos2$ (sin2$)" =3sin2$ − 6$cos 2$ (sin2$)" 2 Marks: Correct answer. 1 Mark: Applies the quotient rule. 29 R =$ − $̅ T =70 − 56 14 = 1 Percentage = 50% +68% 2 = 84% 2 Marks: Correct answer. 1 mark: Calculates the z-score. \84% of scores have a z-score less than1.

30 ! = $ln$

!/_{= $ ×}1

$+ ln$ × 1 = 1 + ln$ When x = 1

!/_{= 1 + ln 1 = 1 (gradient of the tangent)} Gradient of the normal

h_!h_"= −1 h × 1 = −1 h = −1

When x = 1 then ! = 1 × ln 1 = 0 (1, 0) Equation of the normal

! − !_! = h($ − $_!) ! − 0 = −1($ − 1) $ + ! − 1 = 0

2 Marks: Finds the gradient of the normal.

1 Mark: Finds the derivative of the function.

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8 ∴ * (Ñ$ + â)($ = 1 0 ! bÑ$" 2 + â$c_! 0 = 1 35Ñ 2 + 5â = 1 35Ñ + 10â = 2 ① Also * (Ñ$ + â)($ =1 2 (Median 3) # ! bÑ$" 2 + â$c_! # =1 2 8Ñ + 2â =1 2② Multiply equation (2) by 5 40Ñ + 10â =5 2 ③ Equation ①–Equation ③ −5A = −1 2 or Ñ = 1 10 Substitute Ñ = ! !% in equation (1) 35 × 1 10+ 10â = 2 or â = − 3 20 ∴ Ñ = 1 10 and â = − 3 20 answer. 2 Marks: Makes significant progress towards the solution. 1 Mark: Finds one equation relating

A and B.

32(a) Students with a z-score of –2 is two standard deviations below the mean (70 − (2 × 10) = 50.

\Weight of the student is 50 kg.

32(b) 68% of scores have a z-score between -1 and 1 (or from 60 to 80)

Region A =68% 2 = 34% 1 mark: Correct answer. 32(c) R =$ − $̅ T = 100 − 70 10 = 3

Percentage of scores less than a z-score of 3 is 99.85% Number of students = 99.85% × 400 = 399.4 = 399

\There are 399 students with a mass less than 105 kg.

1 Mark: Finds the

z-score or shows some

understanding. 33(a) _{Area of a sector: Ñ =} å

360× πr" (need to eliminate å) Perimeter = r + r + å 360× 2πr 40 = 2r +2πrå 360 2πrå 360 = 40 − 2r å =360 2πr(40 − 2r) Ñ = å 360× πr"= πr" 360× å = πr" 360× 360 2πr(40 − 2r) =r 2(40 − 2r) 3 Marks: Correct answer. 2 Marks: Makes significant progress towards the solution. 1 Mark: Uses the perimeter to eliminate å or shows some understanding.

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9 (Ñ

(r = 20 − 2r ("_Ñ

(r" = −2

Maximum Ñ occurs when (Ñ

(r = 0 20 − 2r = 0 2r = 20 r = 10 Check if r = 10 is a maximum ("_Ñ (r" = −2 < 0

\r =10 gives the maximum value of A

1 Mark:

Differentiates the formula for A with respect to r.

33(c) _{Maximum area when}_r_{= 10}

Ñ =r 2(40 − 2r) =10 2 (40 − 2 × 10) = 100 m" 1 Mark: Correct answer.

34 _{Trapezoidal rule with 2 intervals.}

Ñ =ℎ 2[!%+ !"+ 2!!] 612 =18 2 [24 + 24 + 2 × $] 68 = 48 + 2$ 2$ = 20 $ = 10 cm 2 Marks: Correct answer.

1 Mark: Uses the trapezoidal rule with at least one correct value. 35 Initially t = 163 and M = 5 é(}) = 10^)45 5 = 10^)4×!0# 1 2= ^)4×!0# ln 0.5 = −163è è = ln 0.5 ÷ −163 = 0.004252436 … ≈ 0.0043 2 Marks: Correct answer. 1 Mark: Makes some progress towards the solution.

36(a) Initial calculation occurs on 1st_{January 2014 or}_t_{= 0}

~ = 500 − x_{1 + 0}10 y "

= 400 tonnes per year

\Rate of emission is 400 tonnes per year.

1 Mark: Correct answer.

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10 ~ = lim 5→( 500 − x 10 1 + }y x lim5→( 10 1 + }= 0y ≈ 500 tonnes per year

\C approaches 500 as time passes.

answer.

36(c) 1 Mark: Correct

answer.

36(d) Area under the curve represents the amount of carbon pollution. * 500 − x_{1 + }}10 y " 0 % (} = * 500 − 100(1 + })0 )"_(} % = [500} + 100(1 + }))!_] % 0 = [500 × 6 + 100(1 + 6))!_{− (100(1 + 0)})!_)] = 2914.2857. . . ≈ 2914 tonnes

\There was 2914 tonnes of carbon emitted from the factory.

1 Mark: Sets up the area under the curve.

37(a) 37(b)

1 Mark: Draws the general shape of the curve or shows some understanding.