Band 4 – Page 40

Calculus, 2ADV C4 SM-Bank 3

If `f^{′}(x)= (x^2)/(x^3 + 1)` and `f(1)= log_e 2,` find `f(x)`. (3 marks)

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`f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

Show Worked Solution

`f(x)`	`=int f^{′}(x)\ dx`
	`=int (x^2)/(x^3 + 1)\ dx`
	`= 1/3 int (3x^2)/(x^3 + 1)\ dx`
	`=1/3 log_e \|(x^3 + 1)\|+c`

`text(S)text(ince)\ \ f(1)= log_e 2,`

`1/3 log_e 2+c`	`= log_e 2`
`c`	`=2/3 log_e 2`

`:.\ f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

Calculus, MET1 2009 VCAA 2b

Evaluate `int_1^4(sqrt x + 1)\ dx`. (3 marks)

Show Answers Only

`23/3`

Show Worked Solution

`int_1^4 (x^(1/2) + 1)\ dx`

`= [2/3 x^(3/2) + x]_1^4`

`= (2/3 xx (sqrt4)^3 + 4) – (2/3 + 1)`

`=(2/3 xx 8 +4) – 5/3`

`= 23/3`

Calculus, MET1 2007 ADV 2aii

Let `y=xsinx.` Evaluate `dy/dx` for `x=pi`. (3 marks)

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`-pi`

Show Worked Solution

`y = x sin x`

`(dy)/(dx)`	`= x xx d/(dx) (sin x) + d/(dx) (x) xx sin x`
	`= x cos x + sin x`

`text(When)\ \ x = pi,`

`(dy)/(dx)`	`= pi xx cos pi + sin pi`
	`= pi (-1) + 0`
	`=-pi`

Functions, EXT1 F1 2020 MET1 6

`f(x) = 1/sqrt2 sqrtx`, where `x in [0,2]`

Find `f^(-1)(x)`, and state its domain. (2 marks)

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The graph of `y = f(x)`, where `x ∈ [0, 2]`, is shown on the axes below.

On the axes above, sketch the graph of `f^(-1)(x)` over its domain. Label the endpoints and point(s) of intersection with `f(x)`, giving their coordinates. (2 marks)

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`text(Domain) = [0, 1]`

`f^(-1)(x) = 2x^2`

Show Worked Solution

a. `text(Domain)\ \ f^(-1)(x)= text(Range)\ \ f(x)=[0,1]`

`y = 1/sqrt2 x`

`text(Inverse: swap)\ \ x ↔ y`

`x`	`= 1/sqrt2 sqrty`
`sqrty`	`= sqrt2 x`
`y`	`= 2x^2`

`:. f^(-1)(x) = 2x^2`

Calculus, MET1 2020 VCAA 6

Let `f:[0,2] -> R`, where `f(x) = 1/sqrt2 sqrtx`.

Find the domain and the rule for `f^(-1)`, the inverse function of `f`. (2 marks)

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The graph of `y = f(x)`, where `x ∈ [0, 2]`, is shown on the axes below.

On the axes above, sketch the graph of `f^(-1)` over its domain. Label the endpoints and point(s) of intersection with the function `f`, giving their coordinates. (2 marks)

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Find the total area of the two regions: one region bounded by the functions `f` and `f^(-1)`, and the other region bounded by `f, f^(-1)` and the line `x = 1`. Give your answer in the form `(a-bsqrtb)/6`, where `a, b ∈ ZZ^+`. (4 marks)

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`text(Domain) = [0, 1]`

`f^(-1)(x) = 2x^2`
`(5 + 2sqrt2)/6\ \ text(u²)`

Show Worked Solution

`text(Domain)\ \ f^(-1)(x)`

`= text(Range)\ \ f(x)=[0,1]`

`y = 1/sqrt2 x`

`text(Inverse: swap)\ \ x ↔ y`

`x`	`= 1/sqrt2 sqrty`
`sqrty`	`= sqrt2 x`
`y`	`= 2x^2`

`:. f^(-1)(x) = 2x^2`

`A`	`= int_0^(1/2) 1/sqrt2 sqrtx-2x^2 dx + int_(1/2)^1 2x^2-1/sqrt2 sqrtx\ dx`
	`= [sqrt2/3 x^(3/2)-2/3 x^3]_0^(1/2) + [2/3 x^3-sqrt2/3 x^(3/2)]_(1/2)^1`
	`= [sqrt2/3 (1/sqrt2)^3-2/3(1/2)^3] + [(2/3-sqrt2/3)-(2/24-sqrt2/3 · 1/(2sqrt2))]`
	`= (1/6-1/12) + 2/3-sqrt2/3-(1/12-1/6)`
	`= 1/12 + 2/3-sqrt2/3 + 1/12`
	`= (1 + 8-4sqrt2 + 1)/12`
	`= (5-2sqrt2)/6\ \ text(u²)`

Statistics, EXT1 S1 2020 MET1 5

For a certain population the probability of a person being born with the specific gene SPGE1 is `3/5`.

The probability of a person having this gene is independent of any other person in the population having this gene.

In a randomly selected group of four people, what is the probability that three or more people have the SPGE1 gene? (2 marks)

Show Answers Only

`297/625`

Show Worked Solution

`text(Let)\ \ X =\ text(number of people with gene)`

`X\ ~\ text(Bin) (4, 3/5)`

`P(X >= 3)`	`= P(X = 3) + P(X = 4)`
	`= \ ^4C_3(3/5)^3(2/5) + \ ^4C_4(3/5)^4`
	`= (4 xx 27 xx 2)/625 + 81/625`
	`= 297/625`

Probability, MET1 2020 VCAA 5

For a certain population the probability of a person being born with the specific gene SPGE1 is `3/5`.

The probability of a person having this gene is independent of any other person in the population having this gene.

In a randomly selected group of four people, what is the probability that three or more people have the SPGE1 gene? (2 marks)

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In a randomly selected group of four people, what is the probability that exactly two people have the SPGE1 gene, given that at least one of those people has the SPGE1 gene? Express your answer in the form `(a^3)/(b^4 - c^4)`, where, `a, b, c ∈ Z^+`. (2 marks)

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`297/625`
`(6^3)/(5^4 – 2^4)`

Show Worked Solution

a. `text(Let)\ \ X =\ text(number of people with gene)`

`X\ ~\ text(Bi) (4, 3/5)`

`P(X >= 3)`	`= P(X = 3) + P(X = 4)`
	`= \ ^4C_3(3/5)^3(2/5) + \ ^4C_4(3/5)^4`
	`= (4 xx 27 xx 2)/625 + 81/625`
	`= 297/625`

b.	`P(X = 2 \| X >= 1)`	`= (P(X = 2))/(1 – P(X = 0))`
		`= (\ ^4C_2(3/5)^2(2/5)^2)/(1 – (2/5)^4)`
		`= (6 xx (6^2)/(5^4))/((5^4 – 2^4)/(5^4))`
		`= (6^3)/(5^4 – 2^4)`

Probability, 2ADV S1 2012 MET1 2

A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`.

State the probability that at any given six-month service model X will require an air filter change without an oil change. (1 mark)

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The car manufacturer is developing a new model. The production goals are that the probability of model Y requiring an oil change at any given six-month service will be `m/(m + n)`, the probability of model Y requiring an air filter change will be `n/(m + n)` and the probability of model Y requiring both will be `1/(m + n)`, where `m, n ∈ Z^+`.

Determine `m` in terms of `n` if the probability of model Y requiring an air filter change without an oil change at any given six-month service is 0.05. (2 marks)

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`1/10`
`m = 19n – 20`

Show Worked Solution

	`text(Pr)(F ∩ O′)`	`= text(Pr)(F) – text(Pr)(F∩ O)`
		`= 3/20 – 1/20`
		`= 1/10`

`text(Pr)(F ∩ O′)`	`= n/(m + n) – 1/(m + n)`
`1/20`	`= (n – 1)/(m + n)`
`m + n`	`= 20n – 20`
`m`	`= 19n – 20`

Probability, MET1 2020 VCAA 2

State the probability that at any given six-month service model X will require an air filter change without an oil change. (1 mark)

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The car manufacturer is developing a new model. The production goals are that the probability of model Y requiring an oil change at any given six-month service will be `m/(m + n)`, the probability of model Y requiring an air filter change will be `n/(m + n)` and the probability of model Y requiring both will be `1/(m + n)`, where `m, n ∈ Z^+`.
Determine `m` in terms of `n` if the probability of model Y requiring an air filter change without an oil change at any given six-month service is 0.05. (2 marks)

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`1/10`
`m = 19n – 20`

Show Worked Solution

	`text(Pr)(F ∩ O′)`	`= text(Pr)(F) – text(Pr)(F∩ O)`
		`= 3/20 – 1/20`
		`= 1/10`

`text(Pr)(F ∩ O′)`	`= n/(m + n) – 1/(m + n)`
`1/20`	`= (n – 1)/(m + n)`
`m + n`	`= 20n – 20`
`m`	`= 19n – 20`

Algebra, MET1 2020 VCAA 4

Solve the equation `2 log_2(x + 5) - log_2(x + 9) = 1`. (3 marks)

Show Answers Only

`x = text{−1}`

Show Worked Solution

`2 log_2(x + 5) – log_2(x + 9)`	`= 1`
`log_2(x + 5)^2 – log_2(x + 9)`	`= 1`
`log_2(((x + 5)^2)/(x + 9))`	`= 1`
`((x + 5)^2)/(x + 9)`	`= 2`
`x^2 + 10x + 25`	`= 2x + 18`
`x^2 + 8x + 7`	`= 0`
`(x + 7)(x + 1)`	`= 0`

`:. x = −1\ \ \ \ (x != text{−7}\ \ text(as)\ \ x > text{−5})`

Trigonometry, 2ADV T3 2010 MET1 3

Shown below is part of the graph of a period of the function of the form `y = tan(ax + b)`.

Find the value of `a` and the value of `b`, where `a > 0` and `0 < b < 1`. (3 marks)

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`a = (7pi)/24, b = pi/24`

Show Worked Solution

`y = tan(ax + b)`

`text(Substitute)\ \ (1, sqrt3), (−1, −1)\ \ text(into equation:)`

`tan(a + b)`	`= sqrt3`
`tan(b – a)`	`= −1`

`a + b`	`= pi/3 \ …\ (1)`
`b – a`	`= −pi/4 \ …\ (2)`

`text{Add (1) + (2):}`

`2b`	`= pi/3 – pi/4`
`b`	`= pi/24`

`text{Substitute into (1):}`

`a + pi/24`	`= pi/3`
`a`	`= (7pi)/24`

Graphs, MET1 2020 VCAA 3

Shown below is part of the graph of a period of the function of the form `y = tan(ax + b)`.

The graph is continuous for `x = ∈ [-1, 1]`.

Find the value of `a` and the value of `b`, where `a > 0` and `0 < b < 1`. (3 marks)

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`a = (7pi)/24, b = pi/24`

Show Worked Solution

`y = tan(ax + b)`

`text(Substitute)\ \ (1, sqrt3), (-1,-1)\ \ text(into equation:)`

`tan(a + b)`	`= sqrt3`
`tan(b-a)`	`=-1`

`a + b`	`= pi/3 \ …\ (1)`
`b-1`	`=-pi/4 \ …\ (2)`

`text{Add (1) + (2):}`

`2b`	`= pi/3-pi/4`
`b`	`= pi/24`

`text{Substitute into (1):}`

`a + pi/24`	`= pi/3`
`a`	`= (7pi)/24`

Statistics, STD2 S1 2011 HSC 25a*

A study on the mobile phone usage of NSW high school students is to be conducted.

Data is to be gathered using a questionnaire.

The questionnaire begins with the three questions shown.

Classify the type of data that will be collected in Q2 of the questionnaire. (1 mark)

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Write a suitable question for this questionnaire that would provide discrete ordinal data. (1 mark)

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`text(Categorical, nominal)`
`text(How many outgoing calls do you make per day?)`

Show Worked Solution

i. `text(Categorical, nominal)`

ii. `text(How many outgoing calls do you make per day?)`

`text{(Ensure it can be answered with a numerical score.)}`

Proof, EXT2 P1 SM-Bank 14

Prove that `log_3 7` is irrational. (2 marks)

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`text(See Worked Solution)`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ log_3 7\ \ text(is rational.)`

`text(i.e.)\ \ log_3 7 = p/q\ text(where)\ \ p, q ∈ ZZ\ \ text(with no common factor except 1.)`

`log_3 7`	`= p/q`
`q log_3 7`	`= p`
`log_3 7^q`	`= p`
`7^q`	`= 3^p`

`text(S)text(ince 7 and 3 are prime numbers), 7^q != 3^p`

`:.\ text(Contradiction: integer values)\ \ p, q\ \ text(do not exist.)`

`:. log_3 7\ text(is irrational.)`

Mechanics, EXT2 M1 2018 SPEC1-N 1

A light inextensible string hangs over a frictionless pulley connecting masses of 3 kg and 7 kg, as shown below.

Draw all of the forces acting on the two masses on the diagram above. (1 mark)
Calculate the tension in the string. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`(21g)/5 \ text(N)`

Show Worked Solution

b.	`sum F = 7g – 3g = 4g`	` = (7 + 3)a`
	`a = (4g)/10= (2g)/5`
	`sum F_text(on 3kg) = T – 3g`	`= 3a`
	`T – 3g`	`= (6g)/5`
	`T`	`= 3g + (6g)/5`
		`= (21g)/5\ \ text(N)`

Mechanics, EXT2 M1 2012 SPEC2 20 MC

Particles of mass 3 kg and `m` kg are attached to the ends of a light inextensible string that passes over a smooth pulley, as shown.

If the acceleration of the 3 kg mass is `4.9\ text(m/s)^2` upwards, then

A. `m` = 4.5

B. `m` = 6.0

C. `m` = 9.0

D. `m` = 13.5

E. `m` = 18.0

Show Answers Only

`C`

Show Worked Solution

`sum F`	`=\ text(total mass × acceleration)`
`mg – 3g`	`= (m + 3) xx 4.9`
`9.8m-3xx9.8`	`= 4.9m + 14.7`
`:. m`	`= 9`

`=> C`

Mechanics, EXT2 M1 2015 SPEC2 19 MC

A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and `m` kg attached to the ends of the string.

If the acceleration of the 1 kg particle is 4.9 `text(ms)^(-2)` upwards, then `m` is equal to

Show Answers Only

`C`

Show Worked Solution

`T-(9.8 xx 1) = 4.9 xx 1`

`T=14.7`

`m xx 9.8-T`	`=m xx 4.9`
`4.9m`	`= 14.7`
`:. m`	`= 14.7/4.9`
	`= 3`

`=> C`

Mechanics, EXT2 M1 2014 SPEC2 20 MC

Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest.

Assuming the system remains connected, the speed of the 5 kg mass after two seconds is

A. 4.0 m/s

B. 4.9 m/s

C. 9.8 m/s

D. 19.6 m/s

Show Answers Only

`B`

Show Worked Solution

`sum F`	`= 5g – 3g = 2g`
	`=2 xx 9.8`
	`=19.6\ \ text(N)`
`(5+3)a`	`= 19.6`
`a`	`=2.45`

`u = 0, quad t = 2, quad a = 2.45`

`v`	`= u + at`
	`= 0 + 2.45 xx 2`
	`=4.9\ \ text(m/s)`

`=> B`

Complex Numbers, EXT2 N2 SM-Bank 1

Find all solutions for `z`, in exponential form, given `z^4 = -2 sqrt3 - 2 i`. (3 marks)

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`sqrt2 e^(-frac{i 17 pi}{24}) \ , \ sqrt2 e^(-frac{i 5 pi}{24}) \ , \ sqrt2 e^(frac{i 7 pi}{24}) \ , \ sqrt2 e^(frac{i 19 pi}{24})`

Show Worked Solution

`text{Convert} \ \ z^4\ \ text{to Mod/Arg:}`

`| z^4 | = sqrt{(2 sqrt3)^2 + 2^2} = 4`

`tan \ theta`	`= frac{2 sqrt3}{2} = sqrt3`
`theta`	`= frac{pi}{3}`

`text{arg} (z^4) = – (frac{pi}{2} + frac{pi}{3}) = – frac{5 pi}{6}`

`text{By De Moivre:}`

`| z | = root4 (4) = sqrt2`

`text{arg}(z) = -frac{5 pi}{24} + frac{ k pi}{2} \ , \ k = 0 , ± 1 , ± 2`

`k = 0:\ text{arg} (z)=-frac{5 pi}{24}`

`k = 1:\ text{arg} (z)= -frac{5 pi}{24} + frac{pi}{2} = frac{7 pi}{24}`

`k = – 1:\ text{arg} (z)= -frac{5 pi}{24} – frac{pi}{2} = frac{-17 pi}{24}`

`k = 2:\ text{arg} (z)= -frac{5 pi}{24} + pi = frac{19 pi}{24}`

`therefore \ text{Solutions are:} \ sqrt2 e^(- frac{i 17 pi}{24}) \ , \ sqrt2 e^(frac{ -i 5 pi}{24}) \ , \ sqrt2 e^(frac{i 7 pi}{24}) \ , \ sqrt2 e^(frac{i 19 pi}{24})`

Complex Numbers, EXT2 N2 SM-Bank 8

If `(x + iy)^3 = e^( - frac{i pi}{2}),\ \ x, y ∈ R`, find a solution in the form `x + i y, \ x, y ≠ 0`. (2 marks)

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`frac{sqrt3}{2} – i frac{1}{2}`

Show Worked Solution

`text{Let}`	`(x + i y)`	`= cos theta + i sin theta`
	`(x + i y)^3`	`= cos (3 theta) + isin (3 theta)`

`e^(- frac{i pi}{2}) = cos (frac{-pi}{2}) + i sin (frac{-pi}{2})`

COMMENT: Read the question fully and save time – only one solution is required here.

`text{Equating real parts:}`

`3 theta`	`= frac{-pi}{2}`
`theta`	`= frac{-pi}{6}`

`therefore \ x + i y`	`= cos (frac{-pi}{6}) + i sin (frac{-pi}{6})`
	`= frac{sqrt3}{2} – i frac{1}{2}`

Complex Numbers, EXT2 N1 SM-Bank 9

Let `z = sqrt3 - 3 i`

Express `z` in modulus-argument form. (2 marks)

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Find the smallest integer `n`, such that `z^n + (overset_z)^n = 0`. (3 marks)

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`2 sqrt3 text{cis} (frac{-pi}{3})`
`3`

Show Worked Solution

i.	`z`	`= sqrt3 – 3 i`
	`\|z\|`	`= sqrt((sqrt3)^2 + 3^2) = 2 sqrt3`

`tan theta`	`= frac{3}{sqrt3}=sqrt3`
`theta`	`= frac{pi}{3}`
`text{arg} (z)`	`= – frac{pi}{3}`

`therefore z = 2 sqrt3 \ text{cis} (frac{-pi}{3})`

ii. `z^n + (overset_z)^n = 0`

`[2 sqrt3 \ cos (frac{-pi}{3}) + i sin (frac{-pi}{3})]^n + [ 2 sqrt3 \ cos (frac{-pi}{3}) – i sin (frac{-pi}{3}) ]^n = 0`

`(2 sqrt3)^n [cos (frac{-n pi}{3}) + i sin (frac{-n pi}{3}) + cos (frac{-n pi}{3}) – i sin (frac{-n pi}{3}) = 0`

`2 \ cos (frac{-n pi}{3})`	`= 0`
`cos (frac{n pi}{3})`	`= 0`
`frac{n pi}{3}`	`= frac{pi}{2} + k pi \ , \ k = 0, ± 1, ± 2, …`
`frac{n}{3}`	`= frac{(2k + 1)}{2}`
`n`	`= frac{3 (2k + 1)}{2}`

`text{Numerator will always be odd ⇒ no solution exists}`

Complex Numbers, EXT2 N2 2004 HSC 2c

Sketch the region in the complex plane where the inequalities

`| z + overset_z | ≤ 1` and `| z - i | ≤ 1`

hold simultaneously. (3 marks)

Show Answers Only

Show Worked Solution

`| z + overset_z | ≤ 1 `

`\| x + i y + x – iy \|`	`≤ 1`
`\| 2x \|`	`≤ 1`
`\| x \|`	`≤ frac{1}{2}`

`| z – i | ≤ 1 \ => \ text{Circle, radius = 1 , centre (0, 1)`

Complex Numbers, EXT2 N2 2005 HSC 2c

Sketch the region on the Argand diagram where the inequalities

`| z - overset_z | < 2` and `| z - 1 | >=1`

hold simultaneously. (3 marks)

Show Answers Only

Show Worked Solution

`\| z – overset_z \|`	`< 2`
`\| x + i y – (x – i y) \|`	`< 2`
`\| 2 i y \|`	`< 2`
`\| y \|`	`< 1`

`| z – 1 | = 1 \ => \ text{Circle, radius = 1, centre (1, 0)}`

`:.\ text(Graph:)\ | z – overset_z |<2 \ ∩ \ | z – 1 | >= 1`

Complex Numbers, EXT2 N2 EQ-Bank 1

`z = sqrt2 e^((ipi)/15)` is a root of the equation `z^5 = alpha(1 + isqrt3), \ alpha ∈ R`.

Express `1 + isqrt3` in exponential form. (2 marks)

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Find the value of `alpha`. (1 mark)

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Find the other 4 roots of the equation in exponential form. (3 marks)

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`2e^((ipi)/3)`
`alpha = 2sqrt2`
`e^((i11pi)/15), e^(-(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)`

Show Worked Solution

i. `beta = 1 + isqrt3`

`|beta| = sqrt(1 + (sqrt3)^2) = 2`

`text(arg)(beta) = tan^(−1) (sqrt3/1) = pi/3`

`beta = 2e^((ipi)/3)`

ii.	`z`	`= sqrt2 e^((ipi)/15)`
	`z^5`	`= (sqrt2 e^((ipi)/15))^5`
		`= (sqrt2)^5 e^((ipi)/15 xx 5)`
		`= 4sqrt2 e^((ipi)/3)`

`:. alpha = 2sqrt2`

iii. `text(arg)(z^5) = pi/3 + 2kpi, \ \ k = 0, ±1, ±2, …`

`text(arg)(z) = pi/15 + (2kpi)/5`

`k = 1:\ text(arg)(z) = pi/15 + (2pi)/5 = (11pi)/15`

`k = text(−1):\ text(arg)(z) = pi/15 – (2pi)/5 = −pi/3`

`k = 2:\ text(arg)(z) = pi/15 + (4pi)/5 = (13pi)/15`

`k =text(−2):\ text(arg)(z) = pi/15 – (4pi)/5 = −(11pi)/15`

`:. 4\ text(other roots are:)`

`e^((i11pi)/15), e^(−(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)`

Complex Numbers, EXT2 N1 EQ-Bank 6

Given $z=\dfrac{-1-i \sqrt{3}}{1+i}$, calculate $z^2$ in exponential form. (3 marks)

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$2 e^{\small{\dfrac{i \pi}{6}}}$

Show Worked Solution

$z$	$=\dfrac{-1-i \sqrt{3}}{1+i} \times \dfrac{1-i}{1-i}$
	$=\dfrac{(-1-i \sqrt{3})(1-i)}{1-i^2}$
	$=\dfrac{-1+i-i \sqrt{3}+i^2 \sqrt{3}}{2}$
	$=\dfrac{-1-\sqrt{3}}{2}+i\left(\dfrac{1-\sqrt{3}}{2}\right)$

$\abs{z}$	$=\sqrt{\left(\dfrac{-1-\sqrt{3}}{2}\right)^2+\left(\dfrac{1-\sqrt{3}}{2}\right)^2}$
	$=\sqrt{\dfrac{1+2 \sqrt{3}+3+1-2 \sqrt{3}+3}{4}}$
	$=\sqrt{2}$

$\text{Find}\ \ \arg (z):$

$\tan \theta$	$=\dfrac{\frac{1-\sqrt{3}}{2}}{\frac{-1-\sqrt{3}}{2}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$
$\theta$	$=15^{\circ}=\dfrac{\pi}{12}$

$\therefore \arg (z)=-\dfrac{11 \pi}{12}$

$z$	$=\sqrt{2} \operatorname{cis}\left(-\dfrac{11 \pi}{12}\right)$
$z^2$	$=(\sqrt{2})^2 \operatorname{cis}\left(-\dfrac{11 \pi}{12} \times 2+2 \pi\right)$
	$=2 \operatorname{cis}\left(\dfrac{\pi}{6}\right)$
	$=2 e^{\small{\dfrac{i \pi}{6}}}$

Calculus, EXT2 C1 2005 HSC 1e

Let `t=tan(theta/2).`

Show that `(dt)/(d theta) = 1/2(1+t^2)` (1 mark)

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Show that `sin theta = (2t)/(1+t^2).` (2 marks)

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Use the substitution `t=tan(theta/2)` to find `int text(cosec)\ theta\ d theta.` (2 marks)

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Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`log_e | tan frac{theta}{2} | + c`

Show Worked Solution

i.	`t`	`= tan frac{theta}{2}`
	`frac{dt}{d theta}`	`= frac{1}{2} text{sec}^2 frac{theta}{2}`
		`= frac{1}{2} (1 + tan^2 frac{theta}{2})`
		`= frac{1}{2} (1 + t^2)`

ii. `text{Show} \ \ sin theta = frac{2t}{1 + t^2} :`

`sin theta`	`= 2 \ sin frac{theta}{2} cos frac{theta}{2}`
	`= 2 * frac{t}{sqrt(1 + t^2)} * frac{1}{sqrt(1 + t^2)}`
	`= frac{2t}{1 + t^2}`

iii. `int \ text{cosec} \ theta \ d theta`

`t = tan frac {theta}{2}`

`frac{dt}{d theta} = frac{1}{2} text{sec}^2 frac{theta}{2} \ , \ d theta = frac{2dt}{sec^2 frac{theta}{2}} = frac{2}{1 + t^2} dt`

`int \ text{cosec} \ theta\ d theta`	`= int frac{1 + t^2}{2t} xx frac{2}{1 + t^2} dt`
	`= int frac{1}{t}\ dt`
	`= log_e \| t \| + c`
	`= log_e \| tan frac{theta}{2} \| + c`

Complex Numbers, EXT2 N1 2008 HSC 2b

Write `frac{1 + i sqrt3}{1 + i}` in the form `x + iy`, where `x` and `y` are real. (2 marks)

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By expressing both `1 + i sqrt3` and `1 + i` in modulus-argument form, write `frac{1 + i sqrt3}{1 + i}` in modulus-argument form. (3 marks)

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Hence find `cos frac{pi}{12}` in surd form. (1 mark)

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By using the result of part (ii), or otherwise, calculate `(frac{1 + i sqrt3}{1 + i})^12`. (1 mark)

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`frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`
`sqrt2 (cos (frac{pi}{12}) + i sin (frac{pi}{12}))`
`frac{sqrt2 + sqrt6}{4}`
`-64`

Show Worked Solution

i.	`frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}`	`= frac{(1 + i sqrt3)(1 – i)}{1 – i^2}`
		`= frac{1 – i + i sqrt3 – sqrt3 i^2}{2}`
		`= frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`

ii. `z_1 = 1 + i sqrt3`

`| z_1 | = sqrt(1 + ( sqrt3)^2) = 2`

`text{arg} (z_1) = tan^-1 (sqrt3) = frac{pi}{3}`

`z_1 = 2 (cos frac{pi}{3} + i sin frac{pi}{3})`

`z_2 = 1 + i`

`| z_2 | = sqrt(1^2 + 1^2) = sqrt2`

`text{arg} (z_2) = tan^-1 (1) = frac{pi}{4}`

`z_2 = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`

`frac{1 + i sqrt3}{1 + i}`	`= frac{z_1}{z_2}`
	`= frac{2}{sqrt2} ( cos ( frac{pi}{3} – frac{pi}{4} ) + i sin ( frac{pi}{3} – frac{pi}{4} ) )`
	`= sqrt2 ( cos (frac{pi}{12}) + i sin (frac{pi}{12}) )`

iii. `text{Equating real parts of i and ii:}`

`sqrt2 cos (frac{pi}{12})`	`= frac{1 + sqrt3}{2}`
`cos(frac{pi}{12})`	`= frac{1 + sqrt3}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
	`= frac{sqrt2 + sqrt6}{4}`

iv.	`(frac{1 + i sqrt2}{1 + i})^12`	`= (sqrt2)^12 (cos (frac{pi}{12} xx 12) + i sin (frac{pi}{12} xx 12))`
		`= 64 (cos pi + i sin pi)`
		`= – 64`

Complex Numbers, EXT2 N1 EQ-Bank 3 MC

Which of the following is the complex number $-3-\sqrt{3}i$?

$12 e^{-\small{\dfrac{i 5 \pi}{6}}}$
$12 e^{\small{\dfrac{i \pi}{6}}}$
$2 \sqrt{3} e^{\small{-\dfrac{i 5 \pi}{6}}}$
$2 \sqrt{3} e^{\small{\dfrac{i \pi}{6}}}$

Show Answers Only

$C$

Show Worked Solution

$\abs{z}$	$=\sqrt{3^2+(\sqrt{3})^2}$
	$=\sqrt{12}$
	$=2 \sqrt{3}$

$\tan \theta$	$=\dfrac{3}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}=\sqrt{3}$
$\theta$	$=\dfrac{\pi}{3}$
$\arg (z)$	$=-\left(\dfrac{\pi}{3}+\dfrac{\pi}{2}\right)=-\dfrac{5 \pi}{6}$

$\text {In exponential form:}$

$z=2 \sqrt{3} e^{-\small{\dfrac{i5\pi}{6}}}$

$\Rightarrow C$

Complex Numbers, EXT2 N1 2020 HSC 9 MC

What is the maximum value of `|e^(i theta) - 2| + |e^(i theta) + 2|` for `0 ≤ theta ≤ 2 pi`?

`sqrt5`
`4`
`2 sqrt5`
`10`

Show Answers Only

`C`

Show Worked Solution

`e^(i theta) = a + ib \ \ text{where}\ \ a^2 + b^2 =1`

Mean mark 56%.

`X`	`= \|e^(i theta) – 2\| + \|e^(i theta) + 2\|`
	`= \| (a – 2) + ib \| + \| (a + 2) + i b \|`
	`= sqrt((a-2)^2 + b^2) + sqrt((a+2)^2 + b^2)`
	`= sqrt(a^2 – 4a + 4 + b^2) + sqrt(a^2 + 4a + 4 + b^2`
	`= sqrt(5 – 4a) + sqrt(5 + 4a)`

`frac{dX}{da}`	`= -frac{1}{2} xx 4 xx frac{1}{sqrt(5 – 4a)} + frac{1}{2} xx 4 xx frac{1}{sqrt(5 + 4a)}`
	`= frac{2}{sqrt(5 + 4a)} – frac{2}{sqrt(5 – 4a)}`

`text(When)\ \ frac{dX}{da} = 0:`

`5 + 4a`	`= 5 – 4a`
`a`	`= 0`

`:. X_text(max) = 2 sqrt5`

`=> \ C`

Mechanics, EXT2 M1 2020 HSC 16a

Two masses, `2m` kg and `4m` kg, are attached by a light string. The string is placed over a smooth pulley as shown.

The two masses are at rest before being released and `v` is the velocity of the larger mass at time `t` seconds after they are released.

The force due to air resistance on each mass has magnitude `kv`, where `k` is a positive constant.

Show that `frac{dv}{dt} = frac{gm-kv}{3m}`. (2 marks)

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Given that `v < frac{gm}{k}`, show that when `t = frac{3m}{k} ln 2`, the velocity of the larger mass is `frac{gm}{2k}`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

♦ Mean mark part (i) 37%.

`text{Taking} \ v \ text{downwards as positive.}`

`text{Forces acting on}\ 2m\ text{mass:}`

`kv + 2 mg-T = -2m * frac{dv}{dt}\ …\ (1)`

`text{Forces acting our} \ 4m \ text{mass:}`

`4mg-kv-T = 4m * frac{dv}{dt}\ …\ (2)`

`text{Subtract} \ \ (2)-(1)`

`2 mg-2 kv`	`= 6 m * frac{dv}{dt}`
`:. frac{dv}{dt}`	`= frac{2mg-2 kv}{6 m}`
	`= frac{gm-kv}{3m}`

ii.	`frac{dv}{dt}`	`= frac{gm-kv}{3m}`
	`frac{dt}{dv}`	`= frac{3m}{gm-kv}`
	`t`	`= int frac{3m}{gm-kv}\ dv`
		`= -frac{3m}{k} log_e \|gm-kv \| + c`

`text{When} \ \ t = 0, v = 0:`

`0`	`= -frac{3m}{k} log_e \ \| gm \| + c`
`c`	`= frac{3m}{k} log_e \ \| gm \| `
`t`	`= frac{3m}{k} log_e \ \| gm \| \-frac{3m}{k} log_e \ \| gm -kv \|`
	`= frac{3m}{k} log_e \ \| frac{mg}{gm-kv} \|`

`text{Find} \ \ v\ \ text{when} \ \ t = frac{3m}{k} log_e 2 :`

`frac{3m}{k} log_e 2`	`= frac{3m}{k} log_e \| frac{gm}{gm-kv} \|`
`2`	`= frac{gm}{gm-kv}`
`2gm-2kv`	`= gm`
`2kv`	`= gm`
`therefore \ v`	`= frac{gm}{2k}`

Proof, EXT2 P1 2020 HSC 14d

Prove that for any integer `n > 1, log_n (n + 1)` is irrational. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{See Worked Solutions}`

Show Worked Solution

`text{Proof by contradiction:}`

`text{Assume} \ log_n(n + 1) \ text{is rational}`

`therefore \ log_n (n + 1) = frac{p}{q} \ \ text{where} \ \ p,q ∈ ZZ \ text{with no common factor except 1}`

`n^(frac{p}{q}`	`= n + 1`
`n^p`	`= (n + 1)^q`

`text{Strategy 1}`

`n^p = (n + 1)^q \ \ text{when} \ \ p = q = 0\ \ text{only}`

`q ≠ 0`

`:.\ text{By contradiction}, log_n (n + 1) \ \ text{is irrational.}`

`text{Strategy 2}`

`n^p = (n + 1)^q`

`text{If} \ \ n\ \ text{is odd, LHS is odd and RHS is even.}`

`text{If} \ \ n\ \ text{is even LHS is even and RHS is odd.}`

`text{Statement is true for} \ \ p = q = 0 , text{but} \ \ q ≠ 0`

`therefore \ text{By contradiction,} \ log_n (n + 1) \ text{is irrational.}`

Vectors, EXT2 V1 2020 HSC 15b

The point `C` divides the interval `AB` so that `frac{CB}{AC} = frac{m}{n}`. The position vectors of `A` and `B` are `underset~a` and `underset~b` respectively, as shown in the diagram.

Show that `overset->(AC) = frac{n}{m + n} (underset~b - underset~a)`. (2 marks)

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Prove that `overset->(OC) = frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`. (1 mark)

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Let `OPQR` be a parallelogram with `overset->(OP) = underset~p` and `overset->(OR) = underset~r`. The point `S` is the midpoint of `QR` and `T` is the intersection of `PR` and `OS`, as shown in the diagram.

Show that `overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`. (3 marks)

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Using parts (ii) and (iii), or otherwise, prove that `T` is the point that divides the interval `PR` in the ratio 2 :1. (1 mark)

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`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

`frac{overset->(AC)}{overset->(AB)}`	`= frac{n}{m + n}`
`overset->(AC)`	`= frac{n}{m + n} * overset->(AB)`
	`= frac{n}{m + n} (underset~b – underset~a)`

ii.	`overset->(OC)`	`= overset->(OA) + overset->(AC)`
		`= underset~a + frac{n}{m + n} (underset~b – underset~a)`
		`= underset~a – frac{n}{m + n} underset~a + frac{n}{m + n} underset~b`
		`= (1 – frac{n}{m + n}) underset~a + frac{n}{m + n} underset~b`
		`= (frac{m + n – n}{m + n}) underset~a + frac{n}{m + n} underset~b`
		`= frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`

iii. `text{Show} \ \ overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`

♦ Mean mark part (iii) 50%.

`text{Consider} \ \ Delta PTO \ \ text{and} \ \ Delta RTS:`

`angle PTO = angle RTS \ (text{vertically opposite})`

`angle OPT = angle SRT \ (text{vertically opposite})`

`therefore \ Delta PTO \ text{|||} \ Delta RTS\ \ text{(equiangular)}`

`OT : TS = OP : SR = 2 : 1`

`(text{corresponding sides in the same ratio})`

`frac{overset->(OT)}{overset->(OS)}`	`= frac{2}{3}`
`overset->(OT)`	`= frac{2}{3} overset->(OS)`
	`= frac{2}{3} ( underset~r + frac{1}{2} underset~p)`
	`= frac{2}{3} underset~r + frac{1}{3} underset~p`

Mean mark part (iv) 51%.

iv. `text{Let} \ \ overset->(OT) \ text{divide} \ PR\ text{so that}\ \ frac{TR}{PT} = frac{m}{n}`

`text{Using part (ii):}`

`overset->(OT)`	`= frac{m}{m + n} underset~p + frac{n}{m + n} c`
`overset->(OT)`	`= frac{1}{3} underset~p + frac{2}{3} underset~r \ \ \ (text{part (iii)})`
`frac{m}{m + n}`	`= frac{1}{3} , frac{n}{m + n} = frac{2}{3}`

`=> \ m = 1 \ , \ n = 2`

`therefore \ T \ text{divides} \ PR \ text{in ratio 2 : 1}.`

Proof, EXT2 P1 2020 HSC 15a

In the set of integers, let `P` be the proposition:

'If `k + 1` is divisible by 3, then `k^3 + 1` divisible by 3.'

Prove that the proposition `P` is true. (2 marks)

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Write down the contrapositive of the proposition `P`. (1 mark)

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Write down the converse of the proposition `P` and state, with reasons, whether this converse is true or false. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

i. `text{Let} \ \ k + 1 = 3N, \ N∈ Z`

`=> k = 3N – 1`

`k^3 + 1`	`= (3N -1)^3 + 1`
	`= (3N)^3 + 3(3N)^2 (-1) + 3(3N)(-1)^2 + (-1)^3 + 1`
	`= 27N^3 – 27N^2 + 9N – 1 + 1`
	`= 3 (9N^3 – 9N^2 + 3N)`
	`= 3Q \ , \ Q ∈ Z`

`therefore \ text{If} \ \ k+ 1 \ \ text{is divisible by 3}, text{then} \ \ k^3 + 1 \ \ text{is divisible by 3.}`

ii. `text{Contrapositive}`

`text{If} \ \ k^3 + 1 \ \ text{is not divisible by 3, then}\ \ k + 1\ \ text{is not divisible by 3.}`

♦♦ Mean mark part (iii) 36%.

iii. `text{Converse:}`

`text{If} \ \ k^3 + 1\ \ text{is divisible by 3, then}\ \ k + 1\ \ text{is divisible by 3.}`

`text(Contrapositive of converse:)`

`text{If}\ \ k + 1\ \ text{is not divisible by 3, then}\ \ k^3 + 1\ \ text{is not divisible by 3.}`

`text(i.e.)\ \ k + 1 \ \ text{is not divisible by 3 when}\ \ k + 1 = 3Q + 1\ \ text{or}\ \ k + 1 = 3Q + 2, text{where}\ Q ∈ Z`

`text{If} \ \ k + 1`	`= 3Q + 1\ \ => \ k=3Q`
`k^3 + 1`	`= (3Q)^3 + 1`
	`= 27Q^3 + 1`
	`= 3(9Q^3) + 1`
	`= 3M + 1 \ \ (text{not divisible by 3,}\ M ∈ Z)`

`text{If} \ \ k + 1`	`= 3Q + 2\ \ => \ k=3Q+1`
`k^3 + 1`	`= (3Q + 1)^3 + 1`
	`= (3Q)^3 + 3(3Q)^2 + 3(3Q) + 1 + 1`
	`= 27Q^3 + 27Q^2 + 9Q + 2`
	`= 3(9Q^3 + 9Q^2 + 3Q) + 2`
	`= 3M + 2 \ (text{not divisible by 3,}\ M ∈ Z) `

`therefore \ text{By contrapositive, if}\ \ k^3 + 1\ \ text {is divisible by 3, k + 1 is divisible by 3.}`

Proof, EXT2 P2 2020 HSC 14c

Prove by mathematical induction that, for `n ≥ 2`,

`frac{1}{2^2} + frac{1}{3^2} + ... + frac{1}{n^2} < frac{n - 1}{n}` (4 marks)

--- 10 WORK AREA LINES (style=lined) ---

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`text{See Worked Solutions}`

Show Worked Solution

`text{Prove} \ frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{n^2} < frac{n – 1}{n}\ \ \ text(for)\ \ n>=2`

`text(If)\ \ n=2:`

`text{LHS} = frac{1}{2^2} = frac{1}{4}`

`text{RHS} = frac{2-1}{2} = frac{1}{2} > text{LHS}`

`therefore \ text{True for} \ \ n = 2`

`text{Assume true for} \ \ n =k`

`frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{k^2} < frac{k – 1}{k}`

`text{Prove true for} \ \ n = k + 1`

`frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{k^2} + frac{1}{(k + 1)^2}< frac{k}{k+1}`

`text{LHS}`	`= frac{k – 1}{k} + frac{1}{(k + 1}^2}`
	`= frac{(k – 1)(k + 1)^2 + k}{k (k + 1)^2}`
	`= frac{(k – 1)(k^2 + 2 k +1) + k}{k(k +1)^2}`
	`= frac{k^3 + 2k^2 + k – k^2 – 2k – 1 + k}{k(k + 1)^2}`
	`= frac{k^3 + k^2 -1}{k(k + 1)^2}`
	`= frac{k^2 (k + 1 – frac{1}{k^2})}{k(k + 1)^2}`
	`= frac{k}{k +1} xx frac{(k + 1 – frac{1}{k^2})}{k +1}`
	`< frac{k}{k +1} \ \ \ (text{since} \ frac{k + 1 – frac{1}{k^2}}{k +1} < 1 )`

`=> \ text{True for} \ n = k + 1`

`therefore \ text{S}text{ince true for} \ n = 2, \ text{by PMI, true integral} \ n ≥ 2.`

Mechanics, EXT2 M1 2020 HSC 14b

A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s², is modelled by

`a = 10 (1 - (kv)^2)`,

where `v` is the velocity of the particle in m/s and `k = 0.01`.

Find the velocity of the particle after 5 seconds. (4 marks)

Show Answers Only

`46.21 \ text{ms}^-1`

Show Worked Solution

`a`	`= 10 (1 – (kv)^2)`
`frac{dv}{dt}`	`= 10 – 10 xx 0.01^2 xx v^2`
	`= 10 – 0.001 v^2`
`frac{dt}{dv}`	`= frac{1}{10 – 0.001 \ v^2}`
	`= frac{1000}{10 \ 000 – v^2}`
`t`	`= int frac{1000}{100^2 – v^2}\ dv`

`text{Using partial fractions}:`

`frac{1}{100^2 – v^2}`	` = frac{A}{100 + v} + frac{B}{100 – v}`
`1`	`= A (100 – v) + B(100 + v)`

`text{If} \ \ v = 100 \ , \ 1 = 200 B \ => \ B = frac{1}{200}`

`text{If} \ \ v = -100 \ , \ 1 = 200 A \ => \ A = frac{1}{200}`

`t`	`= 1000 int frac{1}{200 (100 + v)}\ dv + 1000 int frac{1}{200(100 -v)}\ dv`
	`= 5 int frac{1}{100 + v}\ dv + 5 int frac{1}{100 – v}\ dv`
	`= 5 ln \ \| 100 + v \| – 5 ln \\|100 – v \| + c`
	`= 5 ln \ \| frac{100 + v}{100 – v} \| + c`

`text{When} \ \ t = 0 , \ v = 0`

`0 = 5 ln 1 + c \ => \ c = 0`

`:. t = 5 ln \ | frac{100 + v}{100 – v} |`

`text{Find} \ \ v \ \ text{when} \ \ t = 5 :`

`5`	`= 5 ln \| frac{100 + v}{100 – v} \|`
`1`	`= ln \| frac{100 + v}{100 – v} \| `
`e`	`= frac{100 + v}{100 – v}`
`100 e – ve`	`= 100 + v`
`v + ve`	`= 100 e – 100`
`v(1 + e)`	`= 100 e – 100`
`:. v`	`= frac{100 e – 100}{1 + e}`
	`= 46.21 \ text{ms}^-1 \ \ (text{2 d.p.})`

Complex Numbers, EXT2 N2 2020 HSC 14a

Let `z_1` be a complex number and let `z_2 = e^(frac{i pi}{3}) z_1`

The diagram shows points `A` and `B` which represent `z_1` and `z_2`, respectively, in the Argand plane.

Explain why triangle `OAB` is an equilateral triangle. (2 marks)

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Prove that `z_1 ^2 + z_2^2 = z_1 z_2`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

`text{Let}`	`z_1`	`= r(cos theta + i sin theta)`
	`z_2`	`= e^(i frac{pi}{3}) z_1`
		`= r (cos ( theta + frac{pi}{3} ) + i sin (theta + frac{pi}{3}))`

`| z_1 | = | z_2 | => OA = OB`

` angle AOB = frac{pi}{3} \ ( z_2 \ text{is a} \ frac{pi}{3} \ text{anti-clockwise rotation of} \ z_1 )`

`=> angle OBA = angle BAO = pi/3\ \ \ text{(angles opposite equal sides)}`

`therefore \ OAB \ text{is equilateral}`

ii.	`z_1`	`= z_2 e^(i frac{pi}{3})`
	`frac{z_1}{z_2}`	`= e^(i frac{pi}{3})`
	`(frac{z_1}{z_2})^3`	`= e^((3 xx i frac{pi}{3})) \ \ (text{by De Moivre})`
	`frac{z_1^3}{z_2^3}`	`= e^(i pi)`
	`z_1^3`	`= -z_2^3`
	`z_1^3 + z_2^3`	`= 0`

COMMENT: The identity to prove suggests the addition of 2 cubes is a possible strategy.

`(z_1 + z_2)(z_1^2 – z_1 z_2 + z_2^2)`	`= 0`
`z_1^2 – z_1 z_2 + z_2^2`	`= 0`
`z_1^2 + z_2^2`	`= z_1 z_2`

Mechanics, EXT2 M1 2020 HSC 13a

A particle is undergoing simple harmonic motion with period `frac{pi}{3}`. The central point of motion of the particle is at `x = sqrt(3)`. When `t = 0` the particle has its maximum displacement of `2 sqrt(3)` from the central point of motion.

Find an equation for the displacement, `x`, of the particle in terms of `t`. (3 marks)

Show Answers Only

`x = 2 sqrt(3) cos (6t) + sqrt(3)`

Show Worked Solution

`text{Period}`	`= frac{pi}{3}`
`frac{2 pi}{n}`	`= frac{pi}{3}`
`n`	`= 6`

`text{Amplitude} = 2 sqrt(3)`

`text{Centre of motion} = sqrt(3)`

`text{S} text{ince maximum displacement at}\ \ t = 0,`

`x = 2 sqrt(3) cos (6t) + sqrt(3)`

Mechanics, EXT2 M1 2020 HSC 12b

A particle is projected from the origin with initial velocity `u` m/s at an angle `theta` to the horizontal. The particle lands at `x = R` on the `x`-axis. The acceleration vector is given by `underset~a = ((0),(-g))`, where `g` is the acceleration due to gravity. (Do NOT prove this.)

Show that the position vector `underset~r (t)` of the particle is given by

`underset~r (t) = ((ut cos theta),(ut sin theta - frac{1}{2} g t^2))`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that the Cartesian equation of the path of flights is given by

`y = frac{-gx^2}{2u^2} (tan^2 theta - frac{2u^2}{gx} tan theta + 1)`. (3 marks)

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Given `u^2 > gR`, prove that there are 2 distinct values of `theta` for which the particle will land at `x = R`. (2 marks)

--- 7 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

i. `underset~a = ((0),(-g))`

`ddotx = 0`

`dotx = int ddotx \ dt = c`

`text{When} \ \ t = 0, \ dotx = u cos theta \ => \ c = u cos theta`

`=> dotx = u cos theta`

`x = int dotx \ dt =u t cos theta + c`

`text{When} \ \ t = 0 , \ x = 0, \ c = 0`

`therefore \ x = ut cos theta`

`ddoty = -g`

`doty = int – g \ dt = -g t + c`

`text{When} \ \ t = 0 , \ doty = u sin theta`

`=> doty = u sin theta – g t`

`y = int doty \ dt = ut sin theta – frac {1}{2} g t^2 + c`

`text{When} \ \ t = 0, \ y = 0 \ => \ c = 0`

`therefore y = ut sin theta – frac(1)(2) g t^2`

`:. underset~r = ((x),(y)) = ((ut cos theta),(ut sin theta – frac{1}{2} g t^2))`

ii. `x = ut \ cos theta`

`t = frac{x}{u \ cos theta}`

`text{Substitute into} \ y:`

`y`	`= u * frac{x}{u \ cos theta}\ sin theta – frac{1}{2} g ( frac{x}{u \ cos theta} )^2`
	`= x tan theta – frac{gx^2}{2 u^2 cos^2 theta}`
	`= frac{-gx^2}{2u^2} ( frac{1}{cos^2 theta} – frac{2u^2}{gx} tan theta )`
	`= frac{-gx^2}{2u^2} ( sec^2 theta – frac{2u^2}{gx} tan theta )`
	`= frac{-gx^2}{2u^2} ( tan^2 theta – frac{2u^2}{gx} tan theta + 1 )`

iii. `text{When} \ \ x = R \ , \ y = 0`

Mean mark part (iii) 52%.

`frac{-gR^2}{2 u^2}`	`( tan^2 theta – frac{2u^2}{gR} tan theta + 1 ) = 0`
	`tan^2 theta – frac{2u^2}{gR} tan theta + 1 = 0`
`Delta`	`= ( frac{-2u^2}{gR} )^2 – 4 * 1 * 1`
	`= frac{4u^2}{g^2 R^2} – 4`

	`u^2`	`> gR\ \ \ text{(given)}`
	`u^4`	`> g^2 R^2`
	`frac{u^4}{g^2 R^2}`	`> 1`
	`frac{4u^4}{g^2 R^2}`	`> 4`
	`frac{4u^4}{g^2 R^2} – 4`	`> 0`
	`Delta`	`> 0`

`therefore \ 2 \ text{distinct values of} \ \ theta\ \ text{satisfy} \ \ x = R.`

Mechanics, EXT2 M1 2020 HSC 12a

A 50-kilogram box is initially at rest. The box is pulled along the ground with a force of 200 newtons at an angle of 30° to the horizontal. The box experiences a resistive force of `0.3R` newtons, where `R` is the normal force, as shown in the diagram.

Take the acceleration `g` due to gravity to be 10m/s².

By resolving the forces vertically, show that `R =400`. (2 marks)

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Show that the net force horizontally is approximately 53.2 newtons. (2 marks)

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Find the velocity of the box after the first three seconds. (2 marks)

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Show Answers Only

`text{See Worked Solutions}`
`text{See Worked Solutions}`
`3.19 \ text{ms}^-1`

Show Worked Solution

`text{Resolving forces vertically:}`

`R + 200 \ sin 30^@`	`= 50g`
`R + 200 xx frac{1}{2}`	`= 50 xx 10`
`R + 100`	`= 500`
`therefore \ R`	`= 400 \ text(N)`

ii. `text{Resolving forces horizontally:}`

`text{Net Force}`	`= 200 \ cos 30^@ – 0.3 R`
	`= 200 xx frac{sqrt3}{2} – 0.3 xx 400`
	`= 100 sqrt3 – 120`
	`= 53.2 \ text{N (to 1 d. p.)}`

iii.	`F`	`=ma`
	`50 a`	`=100 sqrt300 – 120`
	`a`	`= frac{100 sqrt3 – 120}{50}\ text(ms)^(-2)`

`text{Initially} \ \ u = 0,`

`v`	`= u + at`
`v_(t=3)`	`= 0 + frac{100 sqrt3 – 120}{50} xx 3`
	`= 3.1923 \ …`
	`= 3.19 \ text{ms}^-1 \ text{(to 2 d.p.)}`

Mechanics, EXT2 M1 2020 HSC 5 MC

A particle undergoing simple harmonic motion has a maximum acceleration of 6 m/s² and a maximum velocity of 4 m/s.

What is the period of the motion?

`pi`
`frac{2pi}{3}`
`3pi`
`frac{4pi}{3}`

Show Answers Only

`D`

Show Worked Solution

`ddotx = -n^2 x`

Mean mark 57%.

`text{Find} \ n :`

`ddotx_text{max} = 6 \ \ text{occurs when} \ \ x = – a`

`6 = n^2 a\ …\ (1)`

`v^2 = n^2 (a^2 – x^2)`

`v_text{max} = 4 \ \ text{occurs when} \ \ x = 0`

`4^2`	`= n^2 (a^2-0)`
`16`	`= n^2 a^2`
`4`	`= n a\ …\ (2)`

`text{Substitute} \ \ na = 4 \ \ text{from (2) into (1):}`

`6`	`= 4n`
`n`	`= frac{3}{2}`

`therefore \ text{Period} = frac{2pi}{n} = frac{4pi}{3}`

`=> \ D`

Complex Numbers, EXT2 N1 2020 HSC 4 MC

The diagram shows the complex number `z` on the Argand diagram.

Which of the following diagrams best shows the position of `frac{z^2}{|z|}`?

Show Answers Only

`A`

Show Worked Solution

`text{Let} \ \ z = r\ text(cis)\ theta`

`z^2`	`= r^2 text(cis)\ (2 theta)`
`\|z\|`	`= r`
`therefore frac{z^2}{\|z\|}`	`= frac{r^2 \ text(cis)\ (2 theta)}{r}`
	`= r\ text(cis)\ (2 theta)`

`text{On Argand diagram, it lies on the dotted line`

`text{(modulus the same) with an argument that is}`

`text{doubled.}`

`=> \ A`

Probability, STD1 S2 2020 HSC 26

Barbara plays a game of chance, in which two unbiased six-sided dice are rolled. The score for the game is obtained by finding the difference between the two numbers rolled. For example, if Barbara rolls a 2 and a 5, the score is 3.

The table shows some of the scores.

Complete the six missing values in the table to show all possible scores for the game. (1 mark)
What is the probability that the score for a game is NOT 0? (2 marks)

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Show Answers Only

`frac{5}{6}`

Show Worked Solution

♦ Mean mark part (b) 47%.

b.	`Ptext{(not zero)}`	`= frac{text(numbers) ≠ 0}{text(total numbers)}`
		`= frac{30}{36}`
		`= frac{5}{6}`

$\text{Alternate solution (b)}$

b.	`Ptext{(not zero)}`	`= 1 – Ptext{(zero)}`
		`= 1 – frac{6}{36}`
		`= frac{5}{6}`

Statistics, STD1 S3 2020 HSC 22

A group of students sat a test at the end of term. The number of lessons each student missed during the term and their score on the test are shown on the scatterplot.

Describe the strength and direction of the linear association observed in this dataset. (2 marks)

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Calculate the range of the test scores for the students who missed no lessons. (1 mark)

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Draw a line of the best fit in the scatterplot above. (1 mark)
Meg did not sit the test. She missed five lessons.

Use the line of the best fit drawn in part (c) to estimate Meg's score on this test. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
John also did not sit the test and he missed 16 lessons.

Is it appropriate to use the line of the best fit to estimate his score on the test? Briefly explain your answer. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

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a. $\text{Strength : strong}$

$\text{Direction : negative} $

b. $\text{Range}\ = \text{high}-\text{low}\ = 100-80=20$

e. $\text{John’s missed days are too extreme and the LOBF is not}$

$\text{appropriate. The model would estimate a negative score for}$

$\text{John which is impossible.}$

Show Worked Solution

a. $\text{Strength : strong}$

$\text{Direction : negative} $

♦ Mean mark (a) 45%.
♦♦ Mean mark (b) 31%.

b. $\text{Range}\ = \text{high}-\text{low}\ = 100-80=20$

$\therefore\ \text{Meg’s estimated score = 40}$

e. $\text{John’s missed days are too extreme and the LOBF is not}$

$\text{appropriate. The model would estimate a negative score for}$

$\text{John which is impossible.}$

♦ Mean mark (e) 38%.

Algebra, STD1 A2 2020 HSC 20

The weight of a bundle of A4 paper (`W` kg) varies directly with the number of sheets (`N`) of A4 paper that the bundle contains.

This relationship is modelled by the formula `W = kN`, where `k` is a constant.

The weight of a bundle containing 500 sheets of A4 paper is 2.5 kilograms.

Show that the value of `k` is 0.005. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
A bundle of A4 paper has a weight of 1.2 kilograms. Calculate the number of sheets of A4 paper in the bundle. (2 marks)

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Show Answers Only

`text{Show Worked Solutions}`
`240 \ text{sheets}`

Show Worked Solution

a. `W = 2.5\ text{kg when} \ N = 500:`

`2.5`	`= k xx 500`
`therefore \ k`	`= frac{2.5}{500}`
	`= 0.005`

b. `text{Find}\ \ N \ \ text{when} \ \ W = 1.2\ text{kg:}`

♦ Mean mark 50%.

`1.2`	`= 0.005 xx N`
`therefore N`	`= frac{1.2}{0.005}`
	`= 240 \ text{sheets}`

Financial Maths, STD1 F1 2020 HSC 17

Matilda has a weekly net income of $ 510. She has created a budget where she allocates this income to rent, car expenses, personal expenses, phone, and rest to savings.

Her budget is shown below, with some details missing.

Matilda allocates 20% of her weekly net income to personal expenses.

How many weeks will it take Matilda to save $4930? (3 marks)

Show Answers Only

`85 \ text{weeks}`

Show Worked Solution

`text{Personal expenses} = 20text(%) xx 510 = $102`

`text{Savings}`	`= 510 – (115 + 210 + 102 + 25)`
	`= $58`

`therefore \ text{Weeks to save} \ $4930`

`= frac{4930}{58}`

`= 85 \ text{weeks}`

Algebra, STD1 A1 2020 HSC 16

Consider the equation `m = 6 - frac{3R}{2R - 5}`.

Find the value of `m` when `R = 10`. (2 marks)

Show Answers Only

`4`

Show Worked Solution

`m`	`= 6 – frac{3 xx 10}{2 xx 10 – 5}`
	`= 6 – frac{30}{15}`
	`= 6 – 2`
	`= 4`

Algebra, STD1 A3 2020 HSC 14

Adam travels on a straight road away from his home. His journey is shown in the distance – time graph.

Describe the journey in the first 4 minutes by referring to change in speed and distance travelled. (2 marks)

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After the 4 minutes shown on the graph. Adam rests for 2 minutes and then return home by travelling on the same road at a constant speed. Adam is away from home for a total of 10 minutes.

On the above, complete the distance-time using the information provided. (2 marks)

--- 0 WORK AREA LINES (style=lined) ---

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`text{Speed: Adam increases speed until approximately}\ \ t=2,`

`text{and then decreases speed until he stops when}\ \ t=4.`

`text{Distance travelled: Adam’s distance from home increases}`

`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

`text(distance from home remains the same.)`

Show Worked Solution

a. `text{Speed: Adam increases speed until approximately}\ \ t=2,`

♦ Mean mark part (a) 35%.

`text{and then decreases speed until he stops when}\ \ t=4.`

`text{Distance travelled: Adam’s distance from home increases}`

`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

`text(distance from home remains the same.)`

Mean mark part (b) 51%.

b. `text(S)text(ince Adam travels home at a constant speed, the graph is)`

`text{is a straight line and ends at (10, 0).}`

Measurement, STD1 M4 2020 HSC 12

Two painters each provide a quote for painting an area of 1500 square metres. Painter A charges $100 per 30 square metres. Painter B charges $80 per hour and bases their quote on painting 25 square metres per hour.

Calculate how much will be saved by choosing the cheaper quote. (3 marks)

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Show Answers Only

`$200`

Show Worked Solution

`text{Painter A cost:}`

`C_A = frac{1500}{30} xx 100 = $5000`

`text{Painter B cost:}`

`text(Hours)\ = frac{1500}{25} = 60`

`C_B = 60 xx 80 = $ 4800`

`therefore \ text{Savings by using Painter B}`

`= 5000 – 4800`

`= $200`

Statistics, STD1 S3 2020 HSC 4 MC

The table shows the average brain weight (in grams) and average body weight (in kilograms) of nine different mammals.

\begin{array} {|l|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Brain weight (g)} \rule[-1ex]{0pt}{0pt} & 0.7 & 0.4 & 1.9 & 2.4 & 3.5 & 4.3 & 5.3 & 6.2 & 7.8 \\
\hline
\rule{0pt}{2.5ex} \textit{Body weight (kg)} \rule[-1ex]{0pt}{0pt} & 0.02 &0.06 & 0.05 & 0.34 & 0.93 & 0.97 & 0.43 & 0.33 & 0.22 \\
\hline
\end{array}

Which of the following is the correct scatterplot for this dataset?

Show Answers Only

`C`

Show Worked Solution

`text{Consider data point} \ (1.9, 0.05)`

`→ \ text{Eliminate} \ A \ text{(too high)}`

`→ \ text{Eliminate} \ D \ text{(should be below 2nd data point)}`

`text{Consider data point} \ (2.4, 0.34)`

`→ \ text{Eliminate} \ B \ text{(not on graph)}`

`=> \ C`

Trigonometry, EXT1 T3 2020 HSC 14b

Show that `sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`. (2 marks)

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By letting `x = 4sin theta` in the cubic equation `x^3-12x + 8 = 0`.

Show that `sin (3theta) = 1/2`. (2 marks)

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Prove that `sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `text(Prove:)\ \ sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`

`text(LHS)`	`= sin^3 theta-3/4 sin theta + 1/4 (sin 2thetacostheta + cos2thetasintheta)`
	`= sin^3 theta-3/4 sintheta + 1/4(2sinthetacos^2theta + sintheta(1 – 2sin^2theta))`
	`= sin^3theta-3/4 sintheta + 1/4(2sintheta(1-sin^2theta) + sintheta – 2sin^3theta)`
	`= sin^3theta-3/4 sintheta + 1/4(2sintheta-2sin^3theta + sintheta-2sin^3theta)`
	`= sin^3theta-3/4sintheta + 3/4sintheta-sin^3theta`
	`= 0`

ii. `text(Show)\ \ sin(3theta) = 1/2`

`text{Using part (i):}`

`(sin(3theta))/4`	`= 3/4 sintheta-sin^3 theta`
`sin(3theta)`	`= 3sintheta-4sin^3theta\ …\ (1)`

`x^3-12x + 8 = 0`

`text(Let)\ \ x = 4 sin theta`

`(4sintheta)^3-12(4sintheta) + 8`	`= 0`
`64sin^3theta-48sintheta`	`= 0`
`−16underbrace{(3sintheta-4sin^2theta)}_text{see (1) above}`	`= −8`
`−16 sin(3theta)`	`= −8`
`sin(3theta)`	`= 1/2`

♦♦♦ Mean mark (iii) 21%.

iii. `text(Prove:)\ \ sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`

`text(Solutions to)\ \ x^3-12x + 8 = 0\ \ text(are)`

`x = 4sintheta\ \ text(where)\ \ sin(3theta) = 1/2`

`text(When)\ \ sin3theta = 1/2,`

`3theta`	`= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, (25pi)/6, (29pi)/6, …`
`theta`	`= pi/18, (5pi)/18, (13pi)/18, (17pi)/18, (25pi)/18, (29pi)/18, …`

`:.\ text(Solutions)`

`x = 4sin\ pi/18 \ \ \ (= 4sin\ (17pi)/18)`

`x = 4sin\ (5pi)/18 \ \ \ (= 4sin\ (13pi)/18)`

`x = 4sin\ (25pi)/18 \ \ \ (= 4sin\ (29pi)/18)`

`text(If roots of)\ \ x^3-12x + 8 = 0\ \ text(are)\ \ α, β, γ:`

`α + β + γ = −b/a = 0`

`αβ + βγ + αγ = c/a = −12`

`(4sin\ pi/18)^2 + (4sin\ (5pi)/18)^2 + (4sin\ (25pi)/18)^2`	`= (α + β + γ)^2 – 2(αβ + βγ + αγ)`
`16(sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18)`	`= 0-2(−12)`
`:. sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18`	`= 24/16=3/2`

Statistics, STD1 S1 2020 HSC 2 MC

A random sample of students was taken from each of two universities, and their ages were recorded. The boxplots of their ages are shown.

For the given samples of students' ages, which of the following statements is FALSE?

The range for University A is smaller than the range for University B.
The median for University A is higher than the median for University B.
The interquartile range (IQR) for University A is larger than the IQR for University B.
The oldest student in the sample from University A is older than the oldest student in the sample from University B.

Show Answers Only

`A`

Show Worked Solution

`text{Consider} \ A :`

`text{Range of} \ A \ ≈ 40 – 17.5 ≈ 22.5`

`text{Range of} \ B \ ≈ 30 – 17.5 ≈ 12.5`

`therefore \ text{Range for} \ A > text{Range for}\ B`

`=> A`

Measurement, STD1 M3 2020 HSC 11

Consider the triangle shown.

Find the value of `theta`, correct to the nearest degree. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Find the value of `x`, correct to one decimal place. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

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`39^@`
`12.1 \ text{(to 1 d.p.)}`

Show Worked Solution

a.	`tan theta`	`= frac{8}{10}`
	`theta`	`= tan ^(-1) frac{8}{10}`
		`= 38.659…`
		`= 39^@ \ text{(nearest degree)}`

♦ Mean mark 44% and 39% for part (a) and (b) respectively.

b. `text{Using Pythagoras:}`

`x`	`= sqrt{8^2 + 10^2}`
	`= 12.806…`
	`= 12.8 \ \ text{(to 1 d.p.)}`

Algebra, STD1 A1 2020 HSC 7 MC

The distance between Bricktown and Koala Creek is 75 km. A person travels from Bricktown to Koala Creek at an average speed of 50 km/h.

How long does it take the person to complete the journey?

40 minutes
1 hour 25 minutes
1 hour 30 minutes
1 hour 50 minutes

Show Answers Only

`C`

Show Worked Solution

Mean mark 52%.

`text(Time)`	`= frac(text(Distance))(text(Speed))`
	`= frac(75)(50)`
	`=1.5 \ text(hours)`
	`= 1 \ text(hour) \ 30 \ text(minutes)`

`=> \ C`

Measurement, STD2 M6 2020 HSC 31

Mr Ali, Ms Brown and a group of students were camping at the site located at `P`. Mr Ali walked with some of the students on a bearing of 035° for 7 km to location `A`. Ms Brown, with the rest of the students, walked on a bearing of 100° for 9 km to location `B`.

Show that the angle `APB` is 65°. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the distance `AB`. (2 marks)

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Find the bearing of Ms Brown's group from Mr Ali's group. Give your answer correct to the nearest degree. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

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`text(See Worked Solutions)`
`8.76\ text{km (to 2 d.p.)}`
`146^@`

Show Worked Solution

a.	`angle APB`	`= 100 – 35`
		`= 65^@`

b. `text(Using cosine rule:)`

Mean mark 53%.

`AB^2`	`= AP^2 + PB^2 – 2 xx AP xx PB cos 65^@`
	`= 49 + 81 – 2 xx 7 xx 9 cos 65^@`
	`= 76.750…`
`:.AB`	`= 8.760…`
	`= 8.76\ text{km (to 2 d.p.)}`

`anglePAC = 35^@\ (text(alternate))`

♦♦ Mean mark 22%.

`text(Using cosine rule, find)\ anglePAB:`

`cos anglePAB`	`= (7^2 + 8.76 – 9^2)/(2 xx 7 xx 8.76)`
	`= 0.3647…`
`:. angle PAB`	`= 68.61…^@`
	`= 69^@\ \ (text(nearest degree))`

`:. text(Bearing of)\ B\ text(from)\ A\ (theta)`

`= 180 – (69 – 35)`

`= 146^@`

Calculus, EXT1 C3 2020 HSC 13b

The region `R` is bounded by the `y`-axis, the graph of `y = cos(2x)` and the graph of `y = sin x`, as shown in the diagram.

Find the volume of the solid of revolution formed when the region `R` is rotated about the `x`-axis. (4 marks)

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Show Answers Only

`(3sqrt3 pi)/16\ text(u)³`

Show Worked Solution

`text(Find intersection:)`

Mean mark 53%.

`sin x = cos 2x`

`sin x = 1 – 2sin^2 x`

`2sin^2 x + sinx – 1`	`= 0`
`(2 sinx – 1)(sinx + 1)`	`= 0`

`sin x`	`= 1/2`	`text(or)`	`sin x`	`= −1`
`x`	`= pi/6`		`x`	`= (3pi)/2`

`V`	`= pi int_0^(pi/6) (cos 2x)^2\ dx – pi int_0^(pi/6)(sin x)^2\ dx`
	`= pi int_0^(pi/6) cos^2 2x – sin^2 x\ dx`
	`= pi int_0^(pi/6) 1/2 (1 + cos 4x) – 1/2 (1 – cos 2x)\ dx`
	`= pi/2 int_0^(pi/6) cos 4x + cos 2x\ dx`
	`= pi/2 [1/4 sin 4x + 1/2 sin 2x]_0^(pi/6)`
	`= pi/8 [sin\ (2pi)/3 + 2sin\ pi/3]`
	`= pi/8 (sqrt3/2 + 2 xx sqrt3/2)`
	`= (3sqrt3 pi)/16\ text(u)³`

Calculus, EXT1 C2 2020 HSC 13a

Find `d/(d theta) (sin^3 theta)`. (1 mark)

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Use the substitution `x = tan theta` to evaluate `int_0^1 (x^2)/(1 + x^2)^(5/2)\ dx`. (4 marks)

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`3 cos theta sin^2 theta`
`sqrt2/12`

Show Worked Solution

i. `d/(d theta) (sin^3 theta) = 3 cos theta sin^2 theta`

ii. `text(Let)\ x = tan theta`

`(dx)/(d theta) = sec^2 theta \ => \ dx = sec^2 theta\ d theta`

`text(When)\ x = 1, \ theta = pi/4`

`text(When)\ x = 0, \ theta = 0`

`int_0^1 (x^2)/(1 + x^2)^(5/2) dx`	`= int_0^(pi/4) (tan^2 theta)/((1 + tan^2 theta)^(5/2)) xx sec^2 theta\ d theta`
	`= int_0^(pi/4) (tan^2 theta)/((sec^2 theta)^(5/2)) xx sec^2 theta\ d theta`
	`= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/((sec^2 theta)^(3/2))\ d theta`
	`= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/(sec^3 theta)\ d theta`
	`= int_0^(pi/4) sin^2 theta cos theta\ d theta`
	`= 1/3[sin^3 theta]_0^(pi/4)`
	`= 1/3(sin^3\ pi/4 – 0)`
	`= 1/3 (1/sqrt2)^3`
	`= 1/(6sqrt2)`
	`= sqrt2/12`

Calculus, EXT1 C2 2020 HSC 12d

Find `int_0^(pi/2) cos 5x\ sin 3x\ dx`. (3 marks)

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`−1/2`

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`int_0^(pi/2) cos 5x\ sin 3x\ dx`	`= 1/2 int_0^(pi/2) 2cos 5x\ sin 3x\ dx`
	`= 1/2 int_0^(pi/2) sin 8x-sin 2x\ dx`
	`= 1/2[−1/8 cos 8x + 1/2 cos 2x]_0^(pi/2)`
	`= 1/2[(−1/8 cos 4pi + 1/2 cospi)-(−1/8 cos0 + 1/2 cos0)]`
	`= 1/2(−1/8-1/2 + 1/8-1/2)`
	`= −1/2`

Combinatorics, EXT1 A1 2020 HSC 12c

To complete a course, a student must choose and pass exactly three topics.

There are eight topics from which to choose.

Last year 400 students completed the course.

Explain, using the pigeonhole principle, why at least eight students passed exactly the same three topics. (2 marks)

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`text(See Worked Solution)`

Show Worked Solution

`\ ^8C_3 = 56\ text(ways of choosing 3 topics)`

Mean mark 52%.

`text(400 students pass)`

`text(Pigeonholes)\ (k)= 56`

`text(Pigeons)\ (n) = 400`

`n/k`	`= 400/56`
	`= 7.14…`

`:.\ text(By PHP, at least 8 students passed the)`

`text(same 3 subjects.)`

Statistics, EXT1 S1 2020 HSC 12b

When a particular biased coin is tossed, the probability of obtaining a head is `3/5`.

This coin is tossed 100 times.

Let `X` be the random variable representing the number of heads obtained. This random variable will have a binomial distribution.

Find the expected value, `E(X)`. (1 mark)

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By finding the variance, `text(Var)(X)`, show that the standard deviation of `X` is approximately 5. (1 mark)

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By using a normal approximation, find the approximate probability that `X` is between 55 and 65. (1 mark)

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`60`
`text(See Worked Solutions)`
`68text(%)`

Show Worked Solution

i. `X = text(number of heads)`

`X\ ~\ text(Bin) (n, p)\ ~\ text(Bin) (100, 3/5)`

`E(X)`	`= np`
	`= 100 xx 3/5`
	`= 60`

ii.	`text(Var)(X)`	`= np(1 – p)`
		`= 60 xx 2/5`
		`= 24`

`sigma(x)`	`= sqrt24`
	`~~ 5`

iii.	`P(55 <= x <=65)`	`~~ P(−1 <= z <= 1)`
		`~~ 68text(%)`

Measurement, STD2 M1 2020 HSC 25

A composite solid consists of a triangular prism which fits exactly on top of a cube, as shown.

Find the surface area of the composite solid. (3 marks)

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`424 \ text{cm}^2`

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`text{S.A. of 1 face of cube} = 8 xx 8 = 64 \ text{cm}^2`

`text{Height of triangle} = 11 – 8 = 3 \ text{cm}`

`therefore \ text{S.A. (triangular prism)}`	`= 2 xx ( frac{1}{2} xx 8 xx 3 ) + 2 xx (5 xx 8)`
	`= 24 + 80`
	`= 104 \ text{cm}^2`

`therefore \ text{Total S.A.}`	`= 5 xx 64 + 104`
	`= 424 \ text{cm}^2`

\(z\)	\(=\dfrac{-1-i \sqrt{3}}{1+i} \times \dfrac{1-i}{1-i}\)
	\(=\dfrac{(-1-i \sqrt{3})(1-i)}{1-i^2}\)
	\(=\dfrac{-1+i-i \sqrt{3}+i^2 \sqrt{3}}{2}\)
	\(=\dfrac{-1-\sqrt{3}}{2}+i\left(\dfrac{1-\sqrt{3}}{2}\right)\)

\(\abs{z}\)	\(=\sqrt{\left(\dfrac{-1-\sqrt{3}}{2}\right)^2+\left(\dfrac{1-\sqrt{3}}{2}\right)^2}\)
	\(=\sqrt{\dfrac{1+2 \sqrt{3}+3+1-2 \sqrt{3}+3}{4}}\)
	\(=\sqrt{2}\)

\(\tan \theta\)	\(=\dfrac{\frac{1-\sqrt{3}}{2}}{\frac{-1-\sqrt{3}}{2}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
\(\theta\)	\(=15^{\circ}=\dfrac{\pi}{12}\)

\(z\)	\(=\sqrt{2} \operatorname{cis}\left(-\dfrac{11 \pi}{12}\right)\)
\(z^2\)	\(=(\sqrt{2})^2 \operatorname{cis}\left(-\dfrac{11 \pi}{12} \times 2+2 \pi\right)\)
	\(=2 \operatorname{cis}\left(\dfrac{\pi}{6}\right)\)
	\(=2 e^{\small{\dfrac{i \pi}{6}}}\)

\(\abs{z}\)	\(=\sqrt{3^2+(\sqrt{3})^2}\)
	\(=\sqrt{12}\)
	\(=2 \sqrt{3}\)