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Calculus, 2ADV C3 2020 HSC 25

A landscape gardener wants to build a garden in the shape of a rectangle attached to a quarter-circle. Let `x` and `y` be the dimensions of the rectangle in metres, as shown in the diagram.
 


 

The garden bed is required to have an area of 36 m² and to have a perimeter which is as small as possible. Let `P` metres be the perimeter of the garden bed.

  1. Show that  `P = 2x + 72/x`.  (3 marks)

  2. Find the smallest possible perimeter of the garden bed, showing why this is the minimum perimeter.  (4 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `24\ text(m)`
Show Worked Solution
a.    `text(Area)` `= xy + 1/4 pir^2`
  `36` `= xy + 1/4 pix^2`
  `xy` `= 36 – (pix^2)/4`
  `y` `= 36/x – (pix)/4`

♦ Mean mark part (a) 47%.

 

`:. P` `= 2x + 2y + 1/4(2pix)`
  `= 2x + 2(36/x – (pix)/4) + (pix)/2`
  `= 2x + 72/x – (pix)/2 + (pix)/2`
  `= 2x + 72/x`

 

b.   `P = 2x + 72/x`

`(dP)/(dx)` `= 2 – 72x^(−2)`
`(d^2P)/(dx^2)` `= 144x^(−3)`

 
`text(Max or min when)\ \ (dP)/(dx) = 0:`

`2 – 72/(x^2)` `= 0`
`2x^2` `= 72`
`x^2` `= 36`
`x` `= 6,\ \  (x > 0)`

 

`text(When)\ \ x = 6,`

`(d^2P)/(dx^2) = 144/(6^3) = 2/3 > 0`

 
`=>\  text(MIN at)\ x = 6`

`:. P_text(min)` `= 2 xx 6 + 72/6`
  `= 24\ text(m)`

Statistics, 2ADV S3 2020 HSC 23

A continuous random variable, `X`, has the following probability density function.
 

`f(x) = {(sin x, text(for)\ \ 0 <= x <= k),(0, text(for all other values of)\ x):}`
 

  1. Find the value of `k`.  (2 marks)

  2. Find  `P(X <= 1)`. Give your answer correct to four decimal places.  (2 marks)

Show Answers Only
  1. `pi/2`
  2. `0.4597\ \ (text(to 2 d.p.))`
Show Worked Solution
a.    `int_0^k sin x` `= 1`
  `[−cos x]_0^k` `= 1`
  `−cos k + cos 0` `= 1`
  `−cos k` `= 0`
  `cos k` `= 0`
  `k` `= pi/2`

 

♦ Mean mark part (b) 44%.
b.    `P(X <= 1)` `= int_0^1 sin x\ dx`
    `= [−cos x]_0^1`
    `= −cos1 + cos0`
    `= 1 – cos1`
    `= 0.45969…`
    `= 0.4597\ \ (text(to 4 d.p.))`

Calculus, 2ADV C4 2020 HSC 20

Kenzo is driving his car along a road while his friend records the velocity of the car, `v(t)`, in km/h every minute over a 5-minute period. The table gives the velocity  `v(t)`  at time  `t`  hours.
 

 

The distance covered by the car over the 5-minute period is given by

`int_0^(5/60) v(t)\ dt`.

Use the trapezoidal rule and the velocity at each of the six time values to find the approximate distance in kilometres the car has travelled in the 5-minute period. Give your answer correct to one decimal place.  (2 marks)

Show Answers Only

`5.4\ text(km)`

Show Worked Solution

`int_0^(5/60) v(t)\ dt` `~~ 1/2 xx 1/60 [60 + 2(55 + 65 + 68 + 70) + 67]`
  `~~ 1/120 (643)`
  `~~ 5.358…`
  `~~ 5.4\ text(km)`

Trigonometry, 2ADV T2 2020 HSC 19

Prove that  `sec theta-cos theta = sin theta\ tan theta.`   (2 marks)

Show Answers Only
`text(LHS)` `=1/cos theta-cos theta`  
  `=(1-cos^2 theta)/cos theta`  
  `=sin^2 theta/cos theta`  
  `=sin theta * sin theta/cos theta`  
  `=sin theta\ tan theta\ …\ text(as required)`  
Show Worked Solution
`text(LHS)` `=1/cos theta-cos theta`  
  `=(1-cos^2 theta)/cos theta`  
  `=sin^2 theta/cos theta`  
  `=sin theta * sin theta/cos theta`  
  `=sin theta\ tan theta\ …\ text(as required)`  

Calculus, 2ADV C3 2020 HSC 16

Sketch the graph of the curve  `y = −x^3 + 3x^2 - 1`, labelling the stationary points and point of inflection. Do NOT determine the `x`-intercepts of the curve.  (4 marks)

Show Answers Only

Show Worked Solution
`y` `= −x^3 + 3x^2 – 1`
`(dy)/(dx)` `= −3x^2 + 6x`
`(d^2y)/(dx^2)` `= −6x + 6`

 
`text(SP’s when)\ (dy)/(dx) = 0`

`−3x^2 + 6x` `= 0`
`−3x(x – 2)` `= 0`

`x = 0\ \ text(or)\ \ 2`

 
`text(When)\ \ x = 0,`

`y = −1`

`(d^2 y)/(dx^2) = 6 > 0`

 
`:. text(MIN at)\ \ (0, −1)`
 

`text(When)\ \ x = 2,`

`y` `= −8 + 12 – 1 = 3`
`(d^2y)/(dx^2)` `= −6 xx 2 + 6 = −6 < 0`

 
`:. text(MAX at)\ \ (2, 3)`
 

`(d^2y)/(dx^2) = 0\ text(when)`

`−6x + 6` `= 0`
`x` `= 1`

 
`text(Checking change of concavity)`

`text(Concavity changes either side of)\ x = 1`

`:. text(POI at)\ (1, 1)`
 

Trigonometry, 2ADV T1 2020 HSC 15

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`.
 


 

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

  2. Find the distance `AB`.  (2 marks)

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

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8.76\ text{km  (to 2 d.p.)}`
  3. `146^@`
Show Worked Solution
a.    `angle APB` `= 100-35`
    `= 65^@`

 

b.   `text(Using cosine rule:)`

`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.)}`

 
c.

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

`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^@`

Measurement, STD2 M7 2020 HSC 17

Ayla wishes to estimate the number of trees on a square block of land measuring 1000 m by 1000 m. She counts the number of trees on a 5 m by 5 m section of the block and finds there are 8 trees.

Based on this, estimate the number of tress on the entire square block of land.   (2 marks)

Show Answers Only

`320\ 000`

Show Worked Solution
`text{Area of block}` `= 1000 xx 1000`
  `= 1\ 000\ 000 \ text{m}^2`

 

`8\ text{trees per} \ 25 text{m}^2`

`therefore \ text{Total trees}` `= 8 xx frac{1\ 000\ 000}{25}`
  `= 320\ 000`

Statistics, 2ADV S3 2020 HSC 9 MC

Suppose the weight of melons is normally distributed with a mean of `mu` and a standard deviation of `sigma`.

A melon has a weight below the lower quartile of the distribution but NOT in the bottom 10% of the distribution.

Which of the following most accurately represents the region in which the weight of this melon lies?
 

A. B.
C. D.
Show Answers Only

`C`

Show Worked Solution

`text(Distributions using)\ mu and sigma:`
 

 
`text(Adjusting the above to identify the interval  10% < weight < 25%)`

  
`=>C`

Statistics, STD2 S4 2020 HSC 12 MC

For a set of bivariate data, Pearson's correlation coefficient is  –1.

Which graph could best represent this set of bivariate data?
 

 

 

  

 
Show Answers Only

`D`

Show Worked Solution

`text(Negative correlation coefficient:)`

`text(Line of Best Fit will go from top left to bottom right)`

`text(Correlation coefficient of magnitude 1:)`

`text(Data will be in a straight line.)`

`=> \ D`

Statistics, 2ADV S3 2020 HSC 3 MC

John recently did a class test in each of three subjects. The class scores on each test were normally distributed.

The table shows the subjects and John's scores as well as the mean and standard deviation of the class scores on each test.
 

 
Relative to the rest of class, which row of the table below shows John's strongest subject and his weakest subject?
 

Show Answers Only

`A`

Show Worked Solution

`text(Calculate the)\ ztext(-score of each subject:)`

`ztext{-score (French)} = frac(82 – 70)(8) = 1.5`

`ztext{-score (Commerce)} = frac(80 – 65)(5) = 3.0`

`ztext{-score (Music)} = frac(74 – 50)(12) = 2.0`

 
`therefore \ text{Commerce is strongest, French is weakest}`

`=> \ A`

Statistics, STD2 S5 2020 HSC 8 MC

John recently did a class test in each of three subjects. The class scores on each test were normally distributed.

The table shows the subjects and John's scores as well as the mean and standard deviation of the class scores on each test.

Relative to the rest of class, which row of the table below shows John's strongest subject and his weakest subject?

Show Answers Only

`A`

Show Worked Solution

`text(Calculate the)\ ztext(-score of each subject:)`

`ztext{-score (French)} = frac(82-70)(8) = 1.5`

`ztext{-score (Commerce)} = frac(80-65)(5) = 3.0`

`ztext{-score (Music)} = frac(74-50)(12) = 2.0`

 
`therefore \ text{Commerce is strongest, French is weakest}`

`=> \ A`

Algebra, STD2 A2 2020 HSC 6 MC

Suppose  `y = -1-2x`.

When the value of  `x`  increases by 5, the value of  `y`  decreases by

  1.  1.
  2.  2.
  3.  5.
  4.  10.
Show Answers Only

`D`

Show Worked Solution

`text(Strategy 1)`

`text(If) \ \ x\  \ text(increases by) \ 5`

`=> y \ \ text(decreases by) \ \ 2x = 2 xx 5 = 10`

 
`text(Strategy 2)`

`text(Test) \ 2 \ text(values:)`

`text(If) \ \ x = 0 , \ y = -1`

`text(If) \ \ x = 5 , \ y = -1-10 = -11`

`:.\ y \ text(decreases by) \ 10.`

`=>D`

Financial Maths, STD2 F4 2020 HSC 4 MC

Joan invests $200. She earns interest at 3% per annum, compounded monthly.

What is the future value of Joan's investment after 1.5 years?

  1. $209.07
  2. $209.19
  3. $279.51
  4. $311.93
Show Answers Only

`B`

Show Worked Solution

`text(Monthly interest rate) \ = frac(0.03)(12)`

`n \ = \ 1.5 xx 12 = 18`
  

`text(FV)` `= text(PV) \ (1 + r)^n`
  `= 200 (1 + frac(0.03)(12))^18`
  `= $209.19`

 
`=> \ B`

Functions, 2ADV F1 2020 HSC 1 MC

Which inequality gives the domain of  `y = sqrt(2x-3)`?

  1. `x < 3/2`
  2. `x > 3/2`
  3. `x <= 3/2`
  4. `x >= 3/2`
Show Answers Only

`D`

Show Worked Solution

`text(Domain exists when:)`

`2x-3` `>= 0`
`2x` `>= 3`
`x` `>= 3/2`

  
`=>D`

Mechanics, EXT2 M1 EQ-Bank 3

A car and its driver has a mass of 1200 kilograms and is travelling at 90 km/h.

Calculate the magnitude of the uniform breaking force, in Newtons, required to bring the car to a stop in 40 metres.   (3 marks)

Show Answers Only

`9375\ text(N)`

Show Worked Solution

`text(Let)\ \ k =\ text{Uniform breaking force (constant)}`

`text(Newton’s 2nd law:)`

`F = mddotx` `= −k`
`1200ddotx` `= −k`
`ddotx` `= −k/1200`
`v · (dv)/(dx)` `= −k/1200`
`(dv)/(dx)` `= −k/(1200v)`
`(dx)/(dv)` `= −1200/k v`
`x` `= −1200/k intv\ dv`
  `= −600/k v^2 + C`

 
`text(When)\ \ x = 0, v = (90\ 000)/(60 xx 60) = 25\ text(ms)^(−1):`

`0 = −600/k  · 25^2 + C`

`C = (375\ 000)/k`

`x = (375\ 000)/k – 600/k v^2`

 

`text(When)\ \ x = 40, v = 0:`

`40 = (375\ 000)/k`

`:. k = 9375\ text(N)`

Mechanics, EXT2 M1 EQ-Bank 2

A particle with mass `m` moves horizontally against a resistance force `F`, equal to  `mv(1 + v^2)`  where `v` is the particle's velocity.

Initially, the particle is travelling in a positive direction from the origin at velocity `T`.

  1. Show that the particle's displacement from the origin, `x`, can be expressed as
     
         `x = tan^(-1)((T - v)/(1 + Tv))`  (2 marks)

     

  2. Show that the time, `t`, when the particle is travelling at velocity `v`, is given by
     
         `t = 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`  (4 marks)

     

  3. Express `v^2` as a function of `t`, and hence find the limiting values of `x` and `v`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)“
  3. `v -> 0, \ x -> tan^(−1)T`
Show Worked Solution

i.   `text(Newton’s 2nd law:)`

`F = mddotx = −mv(1 + v^2)`

`v · (dv)/(dx)` `= −v(1 + v^2)`
`(dv)/(dx)` `= −(1 + v^2)`
`(dx)/(dv)` `= −1/(1 + v^2)`
`x` `= −int 1/(1 + v^2)\ dv`
  `= −tan^(−1) v + C`

 

`text(When)\ \ x = 0, v = T:`

`C = tan^(−1)T`

`x = tan^(−1)T – tan^(−1)v`
 

`text(Let)\ \ x = A – B`

`A = tan^(−1) T \ => \ T = tan A`

`B = tan^(−1)v \ => \ v = tan B`

`tan x` `= tan(A – B)`
  `= (tan A – tan B)/(1 + tan A tan B)`
  `= (T – v)/(1 + Tv)`
`:. x` `= tan^(−1)((T – v)/(1 + Tv))`

 

ii.   `text(Using)\ \ ddotx = (dv)/(dt):`

`(dv)/(dt) = −1/(v(1 + v^2))`

`t = −int 1/(v(1 + v^2)) dv`
 

`text(Using Partial Fractions):`

`1/(v(1 + v^2)) = A/v + (Bv + C)/(1 + v^2)`

`A(1 + v^2) + (Bv + C)v = 1`

`A = 1`

`(A + B)v^2` `= 0`   `=> `    `B` `= −1`
`Cv` `= 0`   `=> `    `C` `= 0`

 

`t` `= −int 1/v\ dv + int v/(1 + v^2)\ dv`
  `= −log_e v + 1/2 int(2v)/(1 + v^2)\ dv`
  `= −log_e v + 1/2 log_e (1 + v^2) + C`
  `= −1/2 log_e v^2 + 1/2 log_e (1 + v^2) + C`
  `= 1/2 log_e ((1 + v^2)/(v^2)) + C`

 

`text(When)\ \ t = 0, v = T:`

`C = −1/2 log_e ((1 + T^2)/(T^2))`
 

`:. t` `= 1/2 log_e ((1 + v^2)/(v^2)) – 1/2 log_e((1 + T^2)/(T^2))`
  `= 1/2 log_e (((1 + v^2)/(v^2))/((1 + T^2)/(T^2)))`
  `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`

 

iii. `t` `= 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))`
  `e^(2t)` `= (T^2(1 + v^2))/(v^2(1 + T^2))`
  `1 + v^2` `= (e^(2t)v^2(1 + T^2))/(T^2)`
  `1` `= v^2((e^(2t)(1 + T^2))/(T^2) – 1)`
  `1` `= v^2((e^(2t)(1 + T^2) – T^2)/(T^2))`
  `:. v^2` `= (T^2)/(e^(2t)(1 + T^2) – T^2)`

 
`text(As)\ \ t -> ∞:`

`v^2 -> 0, \ v -> 0`

`x = tan^(−1)T – tan^(−1)v`

`x -> tan^(−1)T`

Mechanics, EXT2 M1 EQ-Bank 1

A canon ball of mass 9 kilograms is dropped from the top of a castle at a height of `h` metres above the ground.

The canon ball experiences a resistance force due to air resistance equivalent to  `(v^2)/500`, where `v` is the speed of the canon ball in metres per second. Let  `g=9.8\ text(ms)^-2`  and the displacement, `x` metres at time `t` seconds, be measured in a downward direction.

  1. Show the equation of motion is given by
     
         `ddotx = g - (v^2)/4500`  (1 mark)

  2. Show, by integrating using partial fractions, that
     
         `v = 210((e^(7/75 t) - 1)/(e^(7/75 t) + 1))`  (5 marks)

     

  3. If the canon hits the ground after 4 seconds, calculate the height of the castle, to the nearest metre.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `78\ text(metres)`
Show Worked Solution

i.   `text(Newton’s 2nd Law:)`

`text(Net Force)` `= mddotx`
`mddotx` `= mg – (v^2)/500`
`9ddotx` `= 9g – (v^2)/500`
`ddotx` `= g – (v^2)/4500`

 

ii.    `(dv)/(dt)` `= g – (v^2)/4500`
    `= (44\ 100 – v^2)/4500`

 
`(dt)/(dv) = 4500/(44\ 100 – v^2)`
 

`text(Using Partial Fractions:)`

`1/(44\ 100 – v^2) = A/(210- v) + B/(210 – v)`

`A(210 – v) + B(210 + v) = 1`
 

`text(If)\ \ v = 210,`

`420B = 1 \ => \ B = 1/420`
 

`text(If)\ \ v = −210,`

`420A = 1 \ => \ A = 1/420`
 

`t` `= int 4500/(210^2 – v^2)\ dv`
  `= 4500/420 int 1/(210 + v) + 1/(210 – v)\ dv`
  `= 75/7 [ln(210 + v) – ln(210 – v)] + c`
  `= 75/7 ln((210 + v)/(210 – v)) + c`

 

`text(When)\ \ t = 0, v = 0:`

`0` `= 75/7 ln(210/210) + c`
`:. c` `= 0`

 

` t` `= 75/7 ln((210 + v)/(210 – v))`
`7/75 t` `= ln((210 + v)/(210 – v))`
`e^(7/75 t)` `= (210 + v)/(210 – v)`
`e^(7/75 t) (210 – v)` `= 210 + v`
`210e^(7/75 t) – 210` `= ve^(7/75 t) + v`
`210(e^(7/75 t) – 1)` `= v(e^(7/75 t) + 1)`
`:. v` `= 210((e^(7/75 t) – 1)/(e^(7/75 t) + 1))`

 

iii.    `v · (dv)/(dx)` `= (44\ 100 – v^2)/4500`
  `(dx)/(dv)` `= (4500v)/(44\ 100 – v^2)`
  `int (dx)/(dv)\ dv` `= −4500/2 int (−2v)/(44\ 100 – v^2)\ dv`
  `x` `= −2550 log_e(44\ 100 – v^2) + c`

 
`text(When)\ \ x = 0, v = 0:`

`0` `= −2250 log_e(44\ 100) + c`
`c` `= 2250 log_e(44\ 100)`
`x` `= −2250 log_e(44\ 100 – v^2) + 2250 log_e44\ 100`
  `= 2250 log_e((44\ 100)/(44\ 100 – v^2))`

 
`text(When)\ \ t = 4:`

`v` `= 210((e^(28/75) – 1)/(e^(28/75) + 1))`
  `= 38.7509…`

 

`:. h` `= 2250 log_e((44\ 100)/(44\ 100 – 38.751^2))`
  `= 77.94…`
  `= 78\ text(metres)`

Calculus, EXT2 C1 EQ-Bank 1

Using partial fractions, show that

`int_0^(1/2) 8/(1-x^4)\ dx = log_e 9-4 tan^(−1) (1/2)`  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Using partial fractions:)`

`8/(1-x^4) = A/(1-x^2) + B/(1 + x^2)`

`A(1 + x^2) + B(1-x^2)` `= 8`
`A + B + (A-B)x^2` `= 8`
`A + B` ` = 8\ \ …\ (1)`
`A-B` ` = 0\ \ …\ (2)`

 
`A = 4, \ B = 4`

 

`4/(1-x^2) = A/(1-x) + B/(1 + x)`

`A(1 + x) + B(1-x)` `= 4`
`A + B + (A-B)x` `= 4`
`A + B` `= 4\ \ …\ (1)`
`A-B` `= 0\ \ …\ (2)`

 
`A = 2, B = 2`
 

`int_0^(1/2) 8/(1-x^4)\ dx` `= int_0^(1/2) 2/(1-x) + 2/(1 + x) + 4/(1 + x^2)\ dx`
  `= [−2ln |1-x| + 2ln |1 + x| + 4tan^(−1)x]_0^(1/2)`
  `= [2ln |(1 + x)/(1-x)| + 4tan^(−1)x]_0^(1/2)`
  `= 2ln |(1 1/2)/(1/2)| + 4tan^(−1)(1/2)-(2ln1 + 4tan^(−1) 0)`
  `= 2ln3 + 4tan^(−1)(1/2)`
  `= ln9 + 4tan^(−1)(1/2)`

Trigonometry, EXT1 T1 EQ-Bank 4 MC

The graph of the function  `y = sin^(-1)(x-2)`  is transformed by being dilated by a factor of 3 from the `y`-axis and then translated to the right by 2.

What is the equation of the transformed graph?

  1. `y = sin^(-1)((x-8)/3)`
  2. `y = sin^(-1)((x-12)/3)`
  3. `y = sin^(-1)(3x-4)`
  4. `y = sin^(-1)(3x-8)`
Show Answers Only

`A`

Show Worked Solution

`y = sin^(-1)(x-2)`
 

`text(Dilate by factor 3 from the)\ ytext(-axis)`

`text(Swap:)\ \ x -> x/3`

`y_1` `= sin^(-1)(x/3-2)`
  `= sin^(-1)((x-6)/3)`

 

`text(Translate to the right by 2)`

`text(Swap:)\ \ x -> x-2`

`y_2` `= sin^(-1)(((x-2)-6)/3)`
  `= sin^(-1)((x-8)/3)`

 
`=>A`

Trigonometry, EXT1 T1 EQ-Bank 3 MC

The graph of the function  `y = arccos(x-3)`  is transformed by being dilated horizontally with a scale factor of `1/2` and then translated to the left by 1.

What is the equation of the transformed graph?

  1. `y = cos^(-1)((x-5)/2)`
  2. `y = cos^(-1)((x-8)/2)`
  3. `y = cos^(-1)(2x-1)`
  4. `y = cos^(-1)(2x-2)`
Show Answers Only

`C`

Show Worked Solution

`y = cos^(-1)(x-3)`

`text(Dilate horizontally with scale factor)\ 1/2`

`text(Swap:)\ \ x -> 2x`

`y_1 = cos^(-1)(2x-3)`

 

`text(Translate to the left by 1)`

`text(Swap:)\ \ x -> x + 1`

`y_2` `= cos^(-1)(2 (x + 1)-3)`
  `= cos^(-1)(2x + 2-3)`
  `= cos^(-1)(2x-1)`

 
`=>C`

Algebra, STD2 A2 SM-Bank 12 MC

 
Which of these equations represents the line in the graph?

  1.  `y = 11 - 2x`
  2.  `y = 11 + 5.5x`
  3.  `y = 11 - 5.5x`
  4.  `y = 11 + 2x`
Show Answers Only

`A`

Show Worked Solution

`text{Graph passes through (0, 11) and (5.5, 0)}`

`text(Gradient)` `=(y_2-y_1)/(x_2-x_1)`  
  `=(11-0)/(0-5.5)`  
  `=-2`  

`:.\ text(Equation is:)\ \ y = 11 – 2x`

`=> A`

Algebra, STD2 A2 SM-Bank 10 MC

Leo drew a straight line through the points (0, 5) and (3, -2) as shown in the diagram below.
 

 
What is the gradient of the line that Leo drew?

  1. `7/3`
  2. `3/7`
  3. `-3/7`
  4. `-7/3`
Show Answers Only

`D`

Show Worked Solution

`text{Line passes through (0, 5) and (3, – 2)}`

`text(Gradient)` `= (y_2-y_1)/(x_2-x_1)`
  `= (5-(-2))/(0-3)`
  `= -7/3`

 
`=>D`

Algebra, STD2 A2 SM-Bank 20

Peter uses matchsticks to make a pattern of shapes, as shown in the table below.

    

How many sticks (`S`) will be needed to make Shape Number (`N`) 13?  (2 marks)

Show Answers Only

`66`

Show Worked Solution

`text(Shape 1:)\ \ S=5 xx 1 +1 =6`

`text(Shape 2:)\ \ S=5 xx 2 +1 =11`

`vdots`

`text(Shape)\ N:\ \ S=5N +1 `
 

`:.\ text(Sticks required when)\ \ N=13`

`=13 xx 5 +1`

`=66`

Algebra, STD2 A2 SM-Bank 6 MC

Renee went bike riding on a holiday.

The hiring charges are  listed in the table below:

\begin{array} {|l|c|c|}
\hline \text{Hours hired} \ (h) & 1 & 2 & 3 & 4 & 5 \\
\hline \text{Cost} \ (C) & 18 & 24 & 30 & 36 & 42 \\
\hline \end{array}

Which linear equation shows the relationship between `C` and `h`?

  1. `C = 12 + 6h`
  2. `C = 6 + 12h`
  3. `C=18 + 12h`
  4. `C=12 + 18h`
Show Answers Only

`A`

Show Worked Solution

`text(Consider Option 1:)`

`12 + (6 xx 1) = 12+6=18`

`12 + (6 xx 2) = 12+12=24`

`12 + (6 xx 3) = 12+18=30\ \ \ \ text(etc …)`

`:.\ text(The linear equation is:)\ \ C = 12 + 6h`

`=>A`

Measurement, STD2 M7 SM-Bank 19

A laundromat can wash 12 loads of laundry in one hour at full capacity.

A standard load of laundry weighs 7 kilograms.

Here is some information about two different washing machines.
 


 

Working at full capacity, how many litres of water would the laundromat expect to save in one hour by using the front loader instead of the top loader?   (2 marks)

Show Answers Only

`text(315 L)`

Show Worked Solution

`text(Top loader water used (1 load))`

`= 7 xx 10.25`

`= 71.75\ text(L)`
 

`text(Front loader water used (1 load))`

`= 7 xx 6.5`

`= 45.5\ text(L)`
 

`:.\ text(Expected water saved)`

`= 12 xx (71.75 – 45.5)`

`= 315\ text(L)`

Measurement, STD2 M7 SM-Bank 18

A health food packet had the following information on its label.
 

 
What was the mass of the health food packet to the nearest gram?   (2 marks)

Show Answers Only

`578\ text(grams)`

Show Worked Solution

`text(Multiple of 100 grams in one packet:)`

`844 ÷ 146 = 5.78…`

`5.78 xx 100 = 578.08…`
 

`:.\ text{The mass of the packet = 578 grams  (nearest gram)}`

Algebra, STD2 A1 SM-Bank 2 MC

Fleur lives 15 kilometres from her work.

On Wednesday, she drove to work and averaged 60 kilometres per hour.

On Thursday, she took the bus which averaged 15 kilometres per hour.

What was the extra time of the bus journey, in minutes, compared to when she drove on Wednesday?

  1.  15
  2.  45
  3.  60
  4.  75
Show Answers Only

`B`

Show Worked Solution
`text(Time on Wednesday)` `= 15/60`
  `= 0.25\ text(hour)`
  `= 15\ text(minutes)`

 

`text(Time on Thursday)` `= 15/15`
  `= 1\ text(hour)`
  `= 60\ text(minutes)`

 

`:.\ text(The extra time taking the bus)\ =60-15 = 45\ text(minutes)`

`=>B`

Measurement, STD2 M7 SM-Bank 1 MC

A dog breeder owns 500 dogs.

150 are Ridgebacks and the rest are Pugs.

What is the ratio of Ridgebacks to Pugs?

  1. 3 : 5
  2. 3 : 7
  3. 3 : 9
  4. 3 : 10
Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince 150 are ridgebacks,)`

`=> 500-150=350\ text(pugs)`

`text(Ridgebacks)` `:\ text(Pugs)`
`150` `: 350`
`3` `: 7`

 
`=> B`

Measurement, STD2 M7 SM-Bank 17

There are 48 people in a meeting room.

The ratio of females to males in the room is 13 to 11.

How many females need to leave the room so that the ratio of females to males in the room is 1 to 1?   (2 marks)

Show Answers Only

`4\ text(females)`

Show Worked Solution

`text(Females : Males) = 13 : 11`

`text(S)text(ince there are 48 people in the room,)`

`26\ text(females : 22 males.)`

 

`:. 4\ text(females need to leave for the ratio to become 1 : 1)`

Measurement, STD2 M7 SM-Bank 16

A refreshment stand at the market sells its iced tea in four different sizes.
 

Showing calculations, find which serving size offers the cheapest iced tea.   (2 marks)

Show Answers Only

`text(375 mL)`

Show Worked Solution

`text(Consider the price per mL of each option)`

`text(150 mL :)\ \ 1.50/150 = 1\ text(cent/mL)`

`text(750 mL :)\ \ 6.10/750 = 0.81\ text(cent/mL)`

`text(375 mL :)\ \ 3.00/375 = 0.8\ text(cent/mL)`

`text(200 mL :)\ \ 1.80/200 = 0.9\ text(cent/mL)`

 

`:.\ text(375 mL serving size offers the cheapest price)`

Measurement, STD2 M7 SM-Bank 15 MC

A brand of pasta is sold in two different packets.
 

What is the difference in the price per kilogram of these two packets?

  1. $1.50
  2. $3.00
  3. $3.60
  4. $7.20
Show Answers Only

`B`

Show Worked Solution

`text(Convert to price per kg:)`

`text(Packet 1)`

`200\ text(g) = $1.80`

`1\ text(kg) = 5 xx 1.80 = $9`

 

`text(Packet 2)`

`750\ text(g) = $9`

`100\ text(g) = 9.00 ÷ 7.5 = $1.20`

`1\ text(kg) = 1.20 xx 10 = $12`

 

`:.\ text(Difference per kg)`

`= 12 – 9`

`= $3.00`
 

`=> B`

Measurement, STD2 M7 SM-Bank 14

Reece looks at the price of four sunscreens.
 

 
Which sunscreen is the cheapest per litre? Show your calculations.   (2 marks)

Show Answers Only

`text(Sport sunscreen)`

Show Worked Solution

`text(Sun)\ =1090/1200 = 0.91\ text(c/mL)`

STRATEGY: Finding the cost per mL is much more efficient in answering this question than finding the cost per litre.

`text(Sport)\ =540/600 = 0.9\ text(c/mL)`

`text(Block)\ =475/500 = 0.95\ text(c/mL)`

`text(Wet)\ =740/750 = 0.99\ text(c/mL)`

 

`:.\ text(Sport sunscreen is the cheapest.)`

Calculus, EXT1 C3 EQ-Bank 1

  1. Sketch the region bounded by the curve  `y = x^2`  and the lines  `y = 16`  and  `y = 9`.  (1 mark)

  2. Calculate the area of this region.  (3 marks)

Show Answers Only
  1.  
  2. `148/3 \ text(u²)`
Show Worked Solution
i.   

 

ii.   `text(Areas either side of)\ ytext(-axis are equal.)`

`y = x^2\ \ =>\ \ x = sqrty`

`:. A` `= 2 int_9^16 x\ dy`
  `= 2 int_9^16 sqrty\ dy`
  `= 2[2/3 y^(3/2)]_9^16`
  `= 4/3[(sqrt16)^3 – (sqrt9)^3]`
  `= 4/3[64 – 27]`
  `= 148/3 \ text(u²)`

Functions, 2ADV F2 EQ-Bank 13

The curve  `y = kx^2 + c`  is subject to the following transformations

    • Translated 2 units in the positive `x`-direction
    • Dilated in the positive `y`-direction by a factor of 4
    • Reflected in the `y`-axis

The final equation of the curve is  `y = 8x^2 + 32x - 8`.

  1.  Find the equation of the graph after the dilation.  (1 mark)

  2.  Find the values of  `k`  and  `c`.  (2 marks)

Show Answers Only
  1. `y = 4k(x – 2)^2 + 4c`
  2. `k = 2, c = −10`
Show Worked Solution

i.    `y = kx^2 + c`

`text(Translate 2 units in positive)\ xtext(-direction.)`

`y = kx^2 + c \ => \ y = k(x – 2)^2 + c`

`text(Dilate in the positive)\ ytext(-direction by a factor of 4.)`

`y = k(x – 2)^2 + c \ => \ y = 4k(x – 2)^2 + 4c`

 

ii.    `y` `= 4k(x^2 – 4x + 4) + 4c`
    `= 4kx^2 – 16kx + 16k + 4c`

 

 
`text(Reflect in the)\ ytext(-axis.)`

COMMENT: Using “swap” terminology for reflections in the y-axis is simpler and more intelligible for students in our view.

`=>\ text(Swap:)\ \ x →\ – x`

`y` `= 4k(−x)^2 – 16k(−x) + 16k + 4c`
  `= 4kx^2 + 16kx + 16k + 4c`

 

 
`text(Equating co-efficients:)`

`4k` `=8`  
`:. k` `=2`  

 

`16k + 4c` `= −8`
`4c` `= −40`
`:. c` `=-10`

Functions, 2ADV F2 EQ-Bank 14

List a set of transformations that, when applied in order, would transform  `y = x^2`  to the graph with equation  `y = 1 - 6x - x^2`.  (3 marks)

Show Answers Only

`text(T1: Translate 3 units in negative)\ xtext(-direction)`

`text(T2: Translate 10 units in negative)\ ytext(-direction)`

`text(T3: Reflect in the)\ xtext(-axis)`

Show Worked Solution

`y = x^2`

`text(Transformation 1:)`

`text(Translate 3 units in negative)\ xtext(-direction)`

`y = (x + 3)^2`

`y = x^2 + 6x + 9`
 

`text(Transformation 2:)`

`text(Translate 10 units in negative)\ ytext(-direction)`

`y = x^2 + 6x – 1`
 

`text(Transformation 3:)`

`text(Reflect in the)\ xtext(-axis)`

`y` `= −(x^2 + 6x – 1)`
  `= 1 – 6x – x^2`

Statistics, STD2 S4 EQ-Bank 4

Ten high school students have their height and the length of their right foot measured.

The results are recorded in the table below.
 


 

  1. Using technology, calculate Pearson's correlation coefficient for the data. Give your answer to 3 decimal places.  (1 mark)

  2. Describe the strength of the association between height and length of right foot for these students.  (1 mark)

  3. Using technology, determine the least squares regression line that allows height to be predicted from right foot length.  (1 mark)

Show Answers Only
  1. `0.941\ \ (text(to 3 d.p.))`
  2. `text(The association is positive and strong.)`
  3. `text(Height) =47.4 + 4.7 xx text(foot length)`
Show Worked Solution

i.   `text(By calculator,)`

COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 correlation” .

`r` `= 0.94095…`
  `= 0.941\ \ (text(to 3 d.p.))`

 

ii.   `text(The association is positive and strong.)`

 

iii.   `x\ text(value ⇒ foot length (independent variables))`

`y\ text(value ⇒ height.)`

`text(By calculator:)`

`text(Height) = 47.4 + 4.7 xx text(foot length)`

Functions, 2ADV F2 EQ-Bank 16

`y = -(x + 2)^4/3`  has been produced by three successive transformations: a translation, a dilation and then a reflection.

  1. Describe each transformation and state the equation of the graph after each transformation.  (2 marks)

  2. Sketch the graph.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2.  
Show Worked Solution

i.   `text(Transformation 1:)`

`text(Translate)\ \ y = x^4\ \ 2\ text(units to the left.)`

`y = x^4 \ => \ y = (x + 2)^4`
  

`text(Transformation 2:)`

`text(Dilate)\ \ y = (x + 2)^4\ \ text(by a factor of)\ 1/3\ text(from the)\ xtext(-axis)`

`y = (x + 2)^4 \ => \ y = ((x + 2)^4)/3`
 

`text(Transformation 3:)`

`text(Reflect)\ \ y = ((x + 2)^4)/3\ \ text(in the)\ xtext(-axis).`

`y = ((x + 2)^4)/3 \ => \ y = −(x + 2)^4/3`

 

ii.   

Functions, 2ADV F1 EQ-Bank 8

Jacques is a marine biologist and finds that the mass of a crab is directly proportional to the cube of the diameter of its shell.

If a crab with a shell diameter of 15 cm weighs 680 grams, what will be the diameter of a crab that weighs 1.1 kilograms? Give your answer to 1 decimal place.  (2 marks)

Show Answers Only

`17.6\ text(cm)`

Show Worked Solution
`M` `prop d^3`  
`M` `= kd^3`  

 
`text(When)\ \ M=680, \ d=15`

`680` `=k xx 15^3`  
`k` `=0.201481…`  

 
`text(Find)\ \ d\ \ text(when)\ \ M=1100:`

`1100` `=0.20148… xx d^3`  
`d` `=root3(1100/(0.20148…))`  
  `=17.608…`  
  `=17.6\ text{cm  (to 1 d.p.)}`  

Functions, 2ADV F1 EQ-Bank 7

The current of an electrical circuit, measured in amps `(A)`, varies inversely with its resistance, measured in ohms `(R)`.

When the resistance of a circuit is 28 ohms, the current is 3 amps.

What is the current when the resistance is 8 ohms?   (2 marks)

Show Answers Only

`10.5`

Show Worked Solution

`A \prop 1/R\ \ =>\ \ A=k/R`

`text(When)\ \ A=3, \ R=28:`

`3` `=k/28`  
`k` `=84`  

 
`text(Find)\ A\ \text{when}\ R=8:`

`A=84/8=10.5`

Statistics, 2ADV S3 EQ-Bank 2

Let  `X` denote a normal random variable with mean 0 and standard deviation 1.

The random variable `X` has the probability density function

`f(x) = 1/sqrt(2pi) e^((−x^2)/2)`   where  `x ∈ (−∞, ∞)`

The diagram shows the graph of  `y = f(x)`.
 

 
 

  1. Complete the table of values for the given function, correct to four decimal places.  (1 mark)
     

     
  2. Use the trapezoidal rule and 5 function values in the table in part i. to estimate

     

         `int_(−2)^2 f(x)\ dx`   (2 marks)

  3. The weights of Rhodesian ridgebacks are normally distributed with a mean of 48 kilograms and a standard deviation of 6 kilograms.

     

    Using the result from part ii., calculate the probability of a randomly selected Rhodesian ridgeback weighing less than 36 kilograms.  (2 marks)

Show Answers Only
i.   

ii.  `0.9369`

iii.  `3.155text(%)`

Show Worked Solution
i.   

 

ii.   

 

`int_(−2)^2 f(x)` `~~ 1/2[0.0540 + 2(0.2420 + 0.3989 + 0.2420) + 0.0540]`
  `~~ 1/2 xx 1.8738`
  `~~ 0.9369`

 

iii.   `mu = 48, sigma = 6`

`ztext(-score (36))` `= (x – mu)/sigma= (32 – 48)/6=-2`

`text(Shaded area = 93.69%)`

`text(By symmetry,)`

`P(X < 36\ text(kgs))` `= P(z < −2)`
  `= (100 – 93.69)/2`
  `= 3.155text(%)`

Statistics, 2ADV S2 EQ-Bank 3

The table below lists the average life span (in years) and average sleeping time (in hours/day) of 9 animal species.
 


 

  1. Using sleeping time as the independent variable, calculate the least squares regression line. (1 mark)

  2. A wallaby species sleeps for 4.5 hours, on average, each day.

     

    Use your equation from part i to predict its expected life span, to the nearest year.   (1 mark)

Show Answers Only
  1. `text(life span) = 42.89 – 2.85 xx text(sleeping time)`
  2. `30\ text(years)`
Show Worked Solution

i.    `text(By calculator:)`

COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 regression line” .

`text(life span) = 42.89 – 2.85 xx text(sleeping time)`
 

ii.   `text(Predicted life span of wallaby)`

`= 42.89 – 2.85 xx 4.5` 

`= 30.06…`

`= 30\ text(years)`

Calculus, 2ADV C4 EQ-Bank 5

The velocity of a particle moving along the `x`-axis at `v` metres per second at `t` seconds, is shown in the graph below.
 


 

Initially, the displacement `x` is equal to 12 metres.

  1. Write an equation that describes the displacement, `x`, at time `t` seconds.  (2 marks)

  2. Draw a graph that shows the displacement of the particle, `x`  metres from the origin, at a time `t` seconds between  `t= 0`  and  `t = 5`. Label the coordinates of the endpoints of your graph.  (2 marks)

Show Answers Only
  1. `x = 3t – (3)/(10) t ^2 + 12`
  2.  

Show Worked Solution
i.    `m_v` `= -(3)/(5)`
  `v` `= 3 – (3)/(5) t`

 

`x` `= int v \ dt`
  `= int 3 – (3)/(5) t \ dt`
  `= 3t – (3)/(10) t^2 + c`

 
`text(When) \ \ t = 0, x = 12  \ => \ c = 12`

`:. \ x = 3t – (3)/(10) t ^2 + 12`

 

ii.  `text(When) \ \ t = 5:`

`x = 15 – (3)/(10) xx 25 + 12 = 19.5`
 

Financial Maths, 2ADV M1 EQ-Bank 1

Ralph opens an annuity account and makes a contribution of $12 000 at the end of each year for 9 years.

For the first 8 years, the interest rate is 4% per annum, compounded annually.

For the 9th year, the interest rate decreases to 3% per annum, compounded annually.
 


 

Use the Future Value of an Annuity table to calculate the amount in the account immediately after the 9th contribution is made.  (3 marks)

Show Answers Only

`$125\ 885.04`

Show Worked Solution

`text(After 8 years:)`

`text(Total)` `= 9.214 xx $12\ 000`
  `= $110\ 568`

 
`text{After 9 years (9th contribution made year end):}`

`text(Total)` `= 110\ 568 xx 1.03 + 12\ 000`
  `= $125\ 885.04`

Statistics, 2ADV S2 EQ-Bank 2

The table below lists the average body weight (in kilograms) and average brain weight (in grams) of nine animal species.
 


 

A least squares regression line is fitted to the data using body weight as the independent variable.

  1. Calculate the equation of the least squares regression line. (1 mark)

  2. If dingos have an average body weight of 22.3 kilograms, calculate the predicted average brain weight of a dingo using your answer to part i.   (1 mark)

Show Answers Only
  1. `text(brain weight) = 49.4 + 2.68 xx text(body weight)`
  2. `109\ text(grams)`
Show Worked Solution

i.   `text(By calculator:)`

COMMENT: Know this critical calculator skill!.

`text(brain weight) = 49.4 + 2.68 xx text(body weight)`

 

ii.   `text(Predicted brain weight of a dingo)`

`= 49.4 + 2.68 xx 22.3` 

`=109.164`

`= 109\ text(grams)`

Statistics, 2ADV S2 EQ-Bank 1

The arm spans (in cm) and heights (in cm) for a group of 13 boys have been measured. The results are displayed in the table below.
 

CORE, FUR2 2008 VCAA 4

The aim is to find a linear equation that allows arm span to be predicted from height.

  1. What will be the independent variable in the equation?  (1 mark)

  2. Assuming a linear association, determine the equation of the least squares regression line that enables arm span to be predicted from height. Write this equation in terms of the variables arm span and height. Give the coefficients correct to two decimal places.  (2 marks)

  3. Using the equation that you have determined in part b., interpret the slope of the least squares regression line in terms of the variables height and arm span.  (1 mark)

Show Answers Only
  1. `text(Height)`
  2. `text(Arm span)\ = 1.09 xx text(height) – 15.63`
  3. `text(On average, arm span increases by 1.09 cm)`
    `text(for each 1 cm increase in height.)`
Show Worked Solution

a.   `text(Height)`

COMMENT: Calculator skills for finding the least squares regression line were required in NESA sample exam – know this critical skill well!

 

b.   `text(By calculator,)`

`text(Arm span)\ = 1.09 xx text(height) – 15.63`

 

c.   `text(On average, arm span increases by 1.09 cm)`

`text(for each 1 cm increase in height.)`

Functions, 2ADV F2 EQ-Bank 1

The function  `f(x) = |x|`  is transformed and the equation of the new function is  `y = kf(x + b) + c`.

The graph of the new function is shown below.
 


 

What are the values of  `k`, `b`  and  `c`.  (2 marks)

Show Answers Only

`k = −1/3, b = 3, c = 2`

Show Worked Solution

`y = |x|`

`text(Translate 3 units left) \ => \ y = |x + 3|`

`text(Reflect in the)\ xtext(-axis) \ => \ y = −|x + 3|`

`text(Dilate by)\ 1/3\ text(from the)\ x text(-axis)`

`=>\ text(Multiply by)\ 1/3 \ => \ y = −1/3|x + 3|`

`text(Translate 2 units up) \ => \ y = −1/3 |x + 3| + 2`
 

`:. k = −1/3, b = 3, c = 2`

Functions, 2ADV F1 EQ-Bank 6

The graph of `f(x)` is shown below. It has asymptotes at  `y = 3`  and  `x = 2`.
 


 

Using interval notation, state the domain and range of `f(x)`.   (2 marks)

Show Answers Only

`text(Domain:)\ [−4, 2) ∪ (2, ∞)`

`text(Range:)\ (−∞, ∞)`

Show Worked Solution

`text(Domain:)\ [−4, 2) ∪ (2, ∞)`

`text(Range:)\ (−∞, ∞)`

Statistics, 2ADV S3 EQ-Bank 1

A probability density function can be used to model the lifespan of a termite, `X`, in weeks, is given by
 

`f(x) = {(k(36 - x^2)),(0):}\ \ \ {:(3 <= x <= 6),(text(otherwise)):}`
 

  1. Show that the value of  `k`  is  `1/45`.  (2 marks)

  2. Find the cumulative distribution function.  (2 marks)

  3. Find the probability that a termite's lifespan is greater than 5 weeks.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `F(x) = {(0),(1/135 (108t – t^3  – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`
  3. `17/135`
Show Worked Solution
i.    `k int_3^6 36 – x^2\ dx` `= 1`
  `k[36x – (x^3)/3]_3^6` `= 1`
  `k[(216 – 72)-(108 – 9)]` `= 1`
  `45k` `= 1`
  `k` `= 1/45`

 

ii.    `F(t)` `= int_(-∞)^t f(x)\ dx`
    `= int_3^t f(x)\ dx`
    `= 1/45 int_3^t 36 – x^2\ dx`
    `= 1/45 [36x – (x^3)/3]_3^t`
    `= 1/135[108x – x^3]_3^t`
    `= 1/35[(108t – t^3) – (324 – 27)]`
    `= 1/135(108t – t^3 – 297)`

 
`:. F(x) = {(0),(1/135 (108t – t^3  – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`

 

iii.    `P(X > 5)` `= 1 – F(5)`
    `= 1 – 1/135(108 xx 5 – 5^3 – 297)`
    `= 1 – 118/135`
    `= 17/135`

Calculus, 2ADV C4 EQ-Bank 2

The population, `D`, of Tasmanian Devils in a sanctuary is given by  `D(t)`, where  `t`  is the time in years after the sanctuary was established.

The devil population changes at a rate modelled by the function  `(dD)/(dt) = 28 e^(0.35t)`.

Calculate the increase in the number of Tasmanian Devils at the end of the first 8 years. Give your answer correct to three significant figures.  (3 marks)

Show Answers Only

`1240 \ text((to 3 sig. fig.))`

Show Worked Solution
`int_0^8 28e^(0.35t)` `= [28 xx (1)/(0.35) e^(0.35t)]_0^8`
  `= 80(e^(0.35 xx 8) – e°)`
  `= 80(16.44 … – 1)`
  `= 1235.57 …`
  `= 1240 \ text((to 3 sig. fig.))`

Calculus, 2ADV C2 EQ-Bank 10

The function  `f(theta) = sin^3(2 theta)`.

If  `f^{′}(theta) = 6 cos(2 theta)-6 cos^n (2 theta)`, find the value of `n`.  (2 marks)

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`3`

Show Worked Solution

`f(theta)= sin^3(2 theta)= (sin(2theta))^3`

`f^{′}(theta)` `= 3 xx 2cos(2 theta) xx sin^2(2 theta)`
  `= 6 cos(2 theta)(1-cos^2(2 theta))`
  `= 6 cos (2 theta)-6 cos^3(2 theta)`

 
`:. \ n = 3`

Functions, 2ADV F1 EQ-Bank 11

Given the function  `f(x) = sqrt(3-x)`  and  `g(x) = x^2-2`, sketch  `y = g(f(x))`  over its natural domain.   (2 marks)

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Show Worked Solution

`g(x) = x^2-2,\ \ f(x) = sqrt(3-x)`

`g(f(x))` `= (sqrt(3-x))^2-2`
  `= 3-x-2`
  `= 1-x`

 
`text(S)text(ince)\ \ f(x) = sqrt(3-x),`

`=> text(Domain:)\ x <= 3`
 

Functions, 2ADV F1 EQ-Bank 12

Two archers play a game where each can aim for a large target or a small target.

If an arrow hits the large target it scores `L` points, and if it hits the small target, it scores `S` points.

The results of a game are shown in the table below.
 

By forming a pair of simultaneous equations, or otherwise, find the value of `L` and `S`.  (3 marks)

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`L = 3, S = 7`

Show Worked Solution

`5L + 8S = 71\ \ …\ (1)`

`12L + 5S = 71\ \ …\ (2)`
 

`text(Multiply  (1)) xx 5`

`25L + 40S = 355\ \ …\ (3)`
 

`text(Multiply  (2)) xx 8`

`96L + 40S = 568\ \ …\ (4)`

 
`text{Subtract  (4) − (3)}`

`71L` `= 213`
`:. L` `= 3`

 
`text(Substitute)\ \ L = 3\ \ text(into (1))`

`8S` `= 56`
`:. S` `= 7`

Probability, 2ADV S1 EQ-Bank 42

The discrete random variable `X` has the probability distribution shown in the table below.
 

     `X = x` `4` `5` `6` `7` `8`
  `P(x)` `0.3` `a` `0.1` `0.15` `0.2`

   
 Find the value of  `a`, and hence calculate the the expected value and variance of  `X`.  (3 marks)

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`2.31`

Show Worked Solution

`0.3 + a + 0.1 + 0.15 + 0.2 = 1`

`=> \ a = 0.25`
 

`E(X) = ∑ x P(x)`
 

  `qquad X = x` `4` `5` `6` `7` `8`
  `qquad P(x) qquad` `0.3` `0.25` `0.1` `0.15` `0.2`
  `qquad x xx P(x) qquad` `1.2` `1.25` `0.6` `1.05` `1.6`

 

`E(X)` `= 1.2 + 1.25 + 0.6 + 1.05 + 1.6`
  `= 5.7`

 

`text(Var)(X)` `= E(X^2) – [E(X)]^2`
  `= (4^2 xx 0.3) + (5^2 xx 0.25) + (6^2 xx 0.1) + (7^2 xx 0.15) + (8^2 xx 0.2) – 5.7^2`
  `= 2.31`

Calculus, 2ADV C2 EQ-Bank 4 MC

If  `f(x)=log_2(x^(2x))`, which expression is equal to  `f^(′)(x)`?

  1. `2/(x^(2x)ln2`
  2. `2/ln2 + 2log_2x`
  3. `log_2x+2/ln2`
  4. `2/ln2 xx log_2(x^(2x-1))`
Show Answers Only

`B`

Show Worked Solution
`f(x)` `=log_2(x^(2x))`  
  `=2x log_2x`  
  `=(2x lnx)/ln2`  

 

`f^(′)(x)` `=1/ln2 (2x*1/x + 2lnx)`  
  `=2/ln2 + (2lnx)/ln2`  
  `=2/ln2 + 2log_2x`  

 
`=>  B`

NETWORKS, FUR2-NHT 2019 VCAA 3

The zoo’s management requests quotes for parts of the new building works.

Four businesses each submit quotes for four different tasks.

Each business will be offered only one task.

The quoted cost, in $100 000, of providing the work is shown in Table 1 below.
  


 

The zoo’s management wants to complete the new building works at minimum cost.

The Hungarian algorithm is used to determine the allocation of tasks to businesses.

The first step of the Hungarian algorithm involves row reduction; that is, subtracting the smallest element in each row of Table 1 from each of the elements in that row.

The result of the first step is shown in Table 2 below.
 


 

The second step of the Hungarian algorithm involves column reduction; that is, subtracting the smallest element in each column of Table 2 from each of the elements in that column.

The results of the second step of the Hungarian algorithm are shown in Table 3 below. The values of Task 1 are given as `A, B, C` and `D`.
 


 

  1. Write down the values of `A, B, C` and `D`.   (1 mark)

  2. The next step of the Hungarian algorithm involves covering all the zero elements with horizontal or vertical lines. The minimum number of lines required to cover the zeros is three.

     

    Draw these three lines on Table 3 above.   (1 mark)

  3. An allocation for minimum cost is not yet possible.

     

    When all steps of the Hungarian algorithm are complete, a bipartite graph can show the allocation for minimum cost.

     

    Complete the bipartite graph below to show this allocation for minimum cost.   (1 mark)

     

  1. Business 4 has changed its quote for the construction of the pathways. The new cost is $1 000 000. The overall minimum cost of the building works is now reduced by reallocating the tasks.

     

    How much is this reduction?   (1 mark)

Show Answers Only
  1. `A = 2, B = 1, C = 1, D = 0`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
Show Worked Solution

a.   `A = 2, \ B = 1, \ C = 1, \ D = 0`

 

b.  

 

c.  `text(After next step:)`
 

`text(Allocation): `
 

 

d.  `text{Hungarian Algorithm table (complete):}`
 

`text(Allocation): B2 -> T3,\ B1 -> T4,\ B3 -> T2,\ B4 -> T1`
 

`text(C)text(ost) = 4 + 7 + 8 + 10 = 29`

`text(Previous cost) = 5 + 10 + 8 + 8 = 31`
 

`:.\ text(Reduction)` `= 2 xx 100\ 000`
  `= $200\ 000`