SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, MET2 2022 VCAA 2

On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.

The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.

One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.

The graph has been drawn to scale.
 

The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.

  1.   i. State the initial population of rabbits.   (1 mark)

  2.  ii. State the minimum and maximum population of rabbits.   (1 mark)

  3. iii. State the number of weeks between maximum populations of rabbits.   (1 mark)

The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.

  1. Show that `a=900` and `b=\frac{\pi}{80}`.   (2 marks)

  2. Find the maximum combined population of foxes and rabbits. Give your answer correct to the nearest whole number.   (1 mark)

  3. What is the number of weeks between the periods when the combined population of foxes and rabbits is a maximum?   (1 mark)

The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
 

  1. Find the average population during the first 300 weeks for the combined population of foxes and rabbits, where the population of foxes is modelled by the transformation of `y=\sin(t)` under the transformation `Q`. Give your answer correct to the nearest whole number.   (4 marks)

Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.

The population of rabbits over a longer period of time can be modelled by the rule

`s(t)=1700cdote^(-0.003t)cdot sin((pit)/80)+2500,\qquad text(for all)\ t>=0`

  1. Find the average rate of change between the first two times when the population of rabbits is at a maximum. Give your answer correct to one decimal place.   (2 marks)

  2. Find the time, where `t>40`, in weeks, when the rate of change of the rabbit population is at its greatest positive value. Give your answer correct to the nearest whole number.   (2 marks)

  3. Over time, the rabbit population approaches a particular value.
  4. State this value.   (1 mark)

Show Answers Only

ai.   `r(0)=2500`

aii.  Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

aiii. `160`  weeks

b.    See worked solution.

c.    `~~ 5339` (nearest whole number)

d.    Weeks between the periods is 160

e.    `~~ 4142` (nearest whole number)

f.    Average rate of change `=-3.6` rabbits/week (1 d.p.)

g.    `t = 156` weeks (nearest whole number)  

h.    ` s → 2500`

Show Worked Solution

ai.  Initial population of rabbits

From graph when `t=0, \ r(0) = 2500`

Using formula when `t=0`

`r(t)` `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500`  
`r(0)` `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500`  rabbits  


aii. From graph,

Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

OR

Using formula

Minimum is when `t = 120`

`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`

Maximum is when `t = 40`

`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`

 
aiii. Number of weeks between maximum populations of rabbits `= 200-40 = 160`  weeks

 
b.  Period of foxes = period of rabbits = 160:

`frac{\2pi}{b} = 160`

`:.\  b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.

Using the point `(100 , 2500)`

Amplitude when `b = frac{\pi}{80}`: 

`f(t)` `=a \ sin (pi/80(t-60))+1600`  
`f(100)` `= 2500`  
`2500` `= a \ sin (pi/80(100-60))+1600`  
`2500` `= a \ sin (pi/2)+1600`  
`a` `= 2500-1600 = 900`  

 
`:.\  f(t)= 900 \ sin (pi/80)(t-60) + 1600`

  

c.    Using CAS find `h(t) = f(t) + r(t)`:

`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`

  
`text{fMax}(h(t),t)|0 <= t <= 160`      `t = 53.7306….`
  

`h(53.7306…)=5339.46`

  
Maximum combined population `~~ 5339` (nearest whole number)


♦♦ Mean mark (c) 40%.
MARKER’S COMMENT: Many rounding errors with a common error being 5340. Many students incorrectly added the max value of rabbits to the max value of foxes, however, these points occurred at different times.

d.    Using CAS, check by changing domain to 0 to 320.

`text{fMax}(h(t),t)|0 <= t <= 320`     `t = 213.7305…`
 
`h(213.7305…)=5339.4568….`
 
Therefore, the number of weeks between the periods is 160.
  

e.    Fox population:

`t^{\prime} = frac{90}{pi}t + 60`   →   `t = frac{pi}{90}(t^{\prime}-60)`

`y^{\prime} = 900y+1600`   →   `y = frac{1}{900}(y^{\prime}-1600)`

`frac{y^{\prime}-1600}{900} = sin(frac{pi(t^{\prime}-60)}{90})`

`:.\  f(t) = 900\ sin\frac{pi}{90}(t-60) + 1600`

  
Average combined population  [Using CAS]   
  
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
  

`= 4142.2646…..  ~~ 4142` (nearest whole number)


♦ Mean mark (e) 40%.
MARKER’S COMMENT: Common incorrect answer 1600. Some incorrectly subtracted `r(t)`. Others used average rate of change instead of average value.

f.   Using CAS

`s(t):= 1700e^(-0.003t) dot\sin\frac{pit}{80} + 2500`
 

`text{fMax}(s(t),t)|0<=t<=320`        `x = 38.0584….`
 

`s(38.0584….)=4012.1666….`
 

`text{fMax}(s(t),t)|160<=t<=320`     `x = 198.0584….`
 

`s(198.0584….)=3435.7035….`
 

Av rate of change between the points

`(38.058 , 4012.167)`  and  `(198.058 , 3435.704)`

`= frac{4012.1666….-3435.7035….}{38.0584….-198.0584….} =-3.60289….`
 

`:.` Average rate of change `=-3.6` rabbits/week (1 d.p.)


♦ Mean mark (f ) 45%.
MARKER’S COMMENT: Some students rounded too early.
`frac{s(200)-s(40)}{200-40}` was commonly seen.
Some found average rate of change between max and min populations.

g.   Using CAS

`s^(primeprime)(t) = 0` , `t = 80(n-0.049) \ \forall n \in Z`

After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2` 

`t` `= 80(n-0.049)`  
  `= 80(2-0.049)`  
  `= 156.08`  

 
`:. \ t = 156` weeks (nearest whole number)


♦♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: Many students solved `frac{ds}{dt}=0`.
A common answer was 41.8, as `s(156.11…)=41.79`.
Another common incorrect answer was 76 weeks.

h.   As `t → ∞`, `e^(-0.003t) → 0`

`:.\ s → 2500`

Graphs, MET2 2023 VCAA 18 MC

Consider the function \(f:[-a\pi, a\pi] \rightarrow\ R, f(x)=\sin(ax)\), where \(a\) is a positive integer.

The number of local minima in the graph of \(y=f(x)\) is always equal to

  1. \(2\)
  2. \(4\)
  3. \(a\)
  4. \(2a\)
  5. \(a^2\)
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Check graphs for different values of }a\)

\(a=1\to \text{1 local minimum}\)
 

     

\(a=2\to \text{4 local minimums}\)
 

     

\(a=3\to \text{9 local minimums}\)
 

     
  

\(\therefore\ \text{Number of local minimums is always equal to }a^2\)
 

\(\Rightarrow E\)


♦♦ Mean mark 29%.
MARKER’S COMMENT: 22% incorrectly chose C and 26% incorrectly chose D.

Functions, MET2 2023 VCAA 9 MC

The function \(f\) is given by
 

\(f(x) = \begin {cases}
\tan\Bigg(\dfrac{x}{2}\Bigg)         &\ \ 4 \leq x \leq 2\pi \\
\sin(ax) &\ \ \ 2\pi\leq x\leq 8
\end{cases}\)

 
The value of \(a\) for which \(f\) is continuous and smooth at  \( x\) = \(2\pi\)  is

  1. \(-2\)
  2. \(-\dfrac{\pi}{2}\)
  3. \(-\dfrac{1}{2}\)
  4. \(\dfrac{1}{2}\)
  5. \(2\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Solve for}\ a\ \text{given}\ \ x=2\pi:\)

\(\tan\Bigg(\dfrac{2\pi}{2}\Bigg)=\sin(a2\pi)=0\)

\(a=\pm \dfrac{1}{2}\)
 

\(\text{For smoothness, solve for}\ a\ \text{given}\ \ x=2\pi:\)

\(\dfrac{d}{dx}\tan\Bigg(\dfrac{x}{2}\Bigg)=\dfrac{d}{dx}\sin(ax)\)

\(\therefore a=-\dfrac{1}{2}\)

 
\(\Rightarrow C\)


♦ Mean mark 46%.

Graphs, MET2 2023 VCAA 1 MC

The amplitude, \(A\), and the period, \(P\), of the function \(f(x)=-\dfrac{1}{2}\sin(3x+2\pi)\) are

  1. \(A=-\dfrac{1}{2},\ P=\dfrac{\pi}{3}\)
  2. \(A=-\dfrac{1}{2},\ P=\dfrac{2\pi}{3}\)
  3. \(A=-\dfrac{1}{2},\ P=\dfrac{3\pi}{2}\)
  4. \(A=\dfrac{1}{2},\ P=\dfrac{\pi}{3}\)
  5. \(A=\dfrac{1}{2},\ P=\dfrac{2\pi}{3}\)
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Period:}\ \ P= \dfrac{2\pi}{n}= \dfrac{2\pi}{3}\)  

\(\text{Amplitude}:\ \ A=\bigg|-\dfrac{1}{2}\bigg|=\dfrac{1}{2}\)

\(\Rightarrow E\)

Calculus, MET1 2022 VCAA 7

A tilemaker wants to make square tiles of size 20 cm × 20 cm.

The front surface of the tiles is to be painted with two different colours that meet the following conditions:

  • Condition 1 - Each colour covers half the front surface of a tile.
  • Condition 2 - The tiles can be lined up in a single horizontal row so that the colours form a continuous pattern.

An example is shown below.
 

There are two types of tiles: Type A and Type B.

For Type A, the colours on the tiles are divided using the rule `f(x)=4 \sin \left(\frac{\pi x}{10}\right)+a`, where `a \in R`.

The corners of each tile have the coordinates (0,0), (20,0), (20,20) and (0,20), as shown below.
 

  1.  i. Find the area of the front surface of each tile.   (1 mark)

    ii. Find the value of `a` so that a Type A tile meets Condition 1.   (1 mark)


Type B tiles, an example of which is shown below, are divided using the rule `g(x)=-\frac{1}{100} x^3+\frac{3}{10} x^2-2 x+10`.
 

  1. Show that a Type B tile meets Condition 1.   (3 marks)

  2. Determine the endpoints of `f(x)` and `g(x)` on each tile. Hence, use these values to confirm that Type A and Type B tiles can be placed in any order to produce a continuous pattern in order to meet Condition 2.   (2 marks)

Show Answers Only

a.i.    `400` cm²

a.ii.   `a = 10`

b.     `200` cm²

c.     See worked solution

Show Worked Solution
a.i   Area `= 20 xx 20`  
  `= 400` cm²  

  
a.ii  `a = 10`
 

b.   Area `=\int_0^{20} \frac{-x^3}{100}+\frac{3 x^2}{10}-2 x+10\ d x`  
  `=\left[\frac{-x^4}{400}+\frac{x^3}{10}-x^2+10 x\right]_0^{20}`  
  `=\left[-\frac{20^4}{400}+\frac{20^3}{10}-20^2+10 xx 20\right]-\left[0\right]`  
  `= – 400 +800 -400 +200`  
  `= 200` cm²  

 
`:.` The area of the coloured section of the Type B tile is 200 cm² which is half the 400 cm² area of the tile.
 

c.   
`f(0)` `=4 \sin \left(\frac{\pi(0)}{10}\right)+10 = 10`  
  `f(20)` `=4 \sin (2 \pi)+10 = 10`  
  `g(0)` `=\frac{-0}{100}+\frac{3(0)}{10}-2(0)+10=10`  
  `g(20)` `=-\frac{8000}{100}+\frac{200}{10}-2(20)+10 = 10`  

 
→`\ f(0) = f(20) = g(0) = g(20) =10`
 

`:.`    The endpoints for `f(x)` are `(0,10)` and `(20,10)` and for `g(x)` are also `(0,10)` and `(20,10)`.

So the tiles can be placed in any order to make the continuous pattern.


♦♦ Mean mark (c) 30%.
MARKER’S COMMENT: Students often only found `f(20)` and `g(20)`, however, `f(0)` and `g(0)` also needed to be found to verify the pattern match.

Graphs, MET1 2022 VCAA 6

The graph of `y=f(x)`, where `f:[0,2 \pi] \rightarrow R, f(x)=2 \sin(2x)-1`, is shown below.
 

  1. On the axes above, draw the graph of `y=g(x)`, where `g(x)` is the reflection of `f(x)` in the horizontal axis.   (2 marks)
  2. Find all values of `k` such that `f(k)=0` and `k \in[0,2 \pi]`.   (3 marks)
  3. Let `h: D \rightarrow R, h(x)=2 \sin(2x)-1`, where `h(x)` has the same rule as `f(x)` with a different domain.
  4. The graph of `y=h(x)` is translated `a` units in the positive horizontal direction and `b` units in the positive vertical direction so that it is mapped onto the graph of `y=g(x)`, where `a, b \in(0, \infty)`.
    1. Find the value for `b`.   (1 mark)

    2. Find the smallest positive value for `a`.   (1 mark)

    3. Hence, or otherwise, state the domain, `D`, of `h(x)`.   (1 mark)

Show Answers Only

a.    Graph `y=g(x)`

b.    `\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`

c.i.    `b=2`

cii.    `=\frac{\pi}{2}`

ciii.   `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

Show Worked Solution

a.

b.  `2 \sin (2 k)-1` `=0`       `0<=k<=2\pi`  
`sin (2 k)` `=1/2`  
`2k` `=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}`  
`k` `=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`  

 

c.i   ` 2 \sin 2(x-a)-1+b` `=-(2 \sin 2 x-1)`  
`:.\ -1+b` `=1`  
`b` `=2`  

 

c.ii  `2 \sin 2(x-a)` `=-(2 \sin 2 x-1)`  
`\sin (2 x-2 a)` `=-\sin 2 x`  
`\therefore 2 a` `=\pi`  
`a` `=\frac{\pi}{2}`  

♦ Mean mark (c.ii) 50%.
MARKER’S COMMENT: Students confused vertical and horizontal translations. Common error `a=\frac{\pi}{4}.`

c.iii The domain for `f(x)` is `[0,2pi]`

`:. \ D` is `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

 
♦♦♦ Mean mark (c.iii) 10%.
MARKER’S COMMENT: Common error was translating in the wrong direction. A common incorrect answer was `\left[-\frac{\pi}{2}, \frac{5 \pi}{2}\right].`

Functions, MET1 2021 VCAA 3

Consider the function  `g: R -> R, \ g(x) = 2sin(2x).`

  1. State the range of `g`.   (1 mark)

  2. State the period of `g`.   (1 mark)

  3. Solve  `2 sin(2x) = sqrt3`  for  `x ∈ R`.   (3 marks)

Show Answers Only
  1. `[-2,2]`
  2. `pi`
  3. `x= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`
Show Worked Solution

a.   `text(S)text(ince)  -1<sin(2x)<1,`

`text(Range)\  g(x) = [-2,2]`
 

b.   `text(Period) = (2pi)/n = (2pi)/2 = pi`
 

c.    `2sin(2x)` `=sqrt3`
  `sin(2x)` `=sqrt3/2`
  `2x` `=pi/3, (2pi)/3, pi/3 + 2pi, (2pi)/3 + 2pi, …`
  `x` `=pi/6, pi/3, pi/6+pi, pi/3+pi, …`

 
`:.\ text(General solution)`

`= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`

Calculus, MET1-NHT 2018 VCAA 7

Let  `f : [ 0, (pi)/(2)] → R, \ f(x) = 4 cos(x)`  and  `g : [0, (pi)/(2)]  → R, \ g(x) = 3 sin(x)`.

  1. Sketch the graph of `f` and the graph of `g` on the axes provided below.   (2 marks)

     
     
    `qquad qquad `
     

  2. Let `c` be such that  `f(c) = g(c)`,  where  `c∈[0, (pi)/(2)]`

     
    Find the value of  `sin(c)`  and the value of  `cos(c)`.   (3 marks)

     

  3.  Let `A` be the region enclosed by the horizontal axis, the graph of `f` and the graph of `g`.
    1. Shade the region `A` on the axes provided in part a. and also label the position of `c` on the horizontal axis.   (1 mark)

    2. Calculate the area of the region `A`.   (3 marks)

Show Answers Only
  1.  

     

  2. `sin(c) = (4)/(5), \ cos(c) = (3)/(5)`

     

  3. i.

     
    ii. `2 \ u^2`
Show Worked Solution

a.   

 

b.   `text(At intersection:)`

`4cos(c)` `= 3sin(c)`
`tan(c)` `= (4)/(3)`

`sin(c) = (4)/(5)`

`cos(c) = (3)/(5)`

 

c. i.

 

   ii.       `A` `= int_0^c g(x)\ dx + int_c^((pi)/(2)) f(x)\ dx`
  `= int_0^c 3sin x \ dx + int_c^((pi)/(2)) 4cos \ x \ dx`
  `= 3[-cos x]_0^c + 4[sin x]_c^((pi)/(2))`
  `= 3(-cos(c) + cos \ 0) + 4(sin \ (pi)/(2)-sin(c))`
  `= 3(-(3)/(5) + 1) + 4(1-(4)/(5))`
  `= (6)/(5) + (4)/(5)`
  `= 2 \ \ text(u²)`

Trigonometry, MET2-NHT 2019 VCAA 2

The wind speed at a weather monitoring station varies according to the function

`v(t) = 20 + 16sin ((pi t)/(14))`

where `v` is the speed of the wind, in kilometres per hour (km/h), and  `t`  is the time, in minutes, after 9 am.

  1. What is the amplitude and the period of  `v(t)`?   (2 marks)

  2. What are the maximum and minimum wind speeds at the weather monitoring station?   (1 mark)

  3. Find  `v(60)`, correct to four decimal places.   (1 mark)

  4. Find the average value of  `v(t)`  for the first 60 minutes, correct to two decimal places.   (2 marks)

A sudden wind change occurs at 10 am. From that point in time, the wind speed varies according to the new function

                    `v_1(t) = 28 + 18 sin((pi(t-k))/(7))`

where  `v_1`  is the speed of the wind, in kilometres per hour, `t` is the time, in minutes, after 9 am and  `k ∈ R^+`. The wind speed after 9 am is shown in the diagram below.
 

  1. Find the smallest value of `k`, correct to four decimal places, such that  `v(t)`  and  `v_1(t)`  are equal and are both increasing at 10 am.   (2 marks)

  2. Another possible value of `k` was found to be 31.4358

     

    Using this value of `k`, the weather monitoring station sends a signal when the wind speed is greater than 38 km/h.

     

    i.  Find the value of `t` at which a signal is first sent, correct to two decimal places.   (1 mark)

     

    ii. Find the proportion of one cycle, to the nearest whole percent, for which  `v_1 > 38`.   (2 marks)

  3. Let  `f(x) = 20 + 16 sin ((pi x)/(14))`  and  `g(x) = 28 + 18 sin ((pi(x-k))/(7))`.
     
    The transformation  `T([(x),(y)]) = [(a \ \ \ \ 0),(0 \ \ \ \ b)][(x),(y)] + [(c),(d)]`  maps the graph of  `f`  onto the graph of  `g`.

     

    State the values of  `a`, `b`, `c` and `d`, in terms of `k` where appropriate.   (3 marks)

Show Answers Only
  1. `28`
  2. `v_text(max) \ = 36 \ text(km/h)`

     

    `v_text(min) \ = 4 \ text(km/h)`

  3. `32.5093 \ text(km/h)`
  4. `20.45 \ text(km/h)`
  5. `3.4358`
  6. i. `60.75 \ text(m)`

     

    ii. `31text(%)`

  7. `a = (1)/(2) \ , \ b = (9)/(8) \ , \ c = k \ , \ d = (11)/(2)`
Show Worked Solution

a.    `text(Amplitude) = 16`

`text{Find Period (n):}`

`(2 pi)/(n)` `= (pi)/(14)`
`n` `= 28`

 

b.    `v_text(max) = 20 + 16 = 36 \ text(km/h)`

`v_text(min) = 20-16 = 4 \ text(km/h)`

 

c.    `v(60)` `= 20 + 16 sin ((60 pi)/(14))`
  `= 32.5093 \ \ text(km/h)`

 

d.    `v(t)\ \ text(is always positive.)`

`s(t) = int_0^60 v(t) \ dt`

`v(t)_(avg)` `= (1)/(60) int_0^60 20 + 16 sin ((pi t)/(14))\ dt`
  `= 20.447`
  `= 20.45 \ text(km/h) \ \ text((to 2 d.p.))`

 

e.    `text(S) text{olve (for}\ k text{):} \ \ v(60) = v_1(60)`

`k = 3.4358 \ \ text((to 4 d.p.))`

 

f.i.  `text(S) text(olve for) \ t , \ text(given) \ \ v_1(t) = 38 \ \ text(and) \ \ k = 31.4358`

`=> t = 60.75 \ text(minutes)`
 

f.ii.  `text(S) text(olving for) \ \ v_1(t) = 38 \ , \ k = 31.4358`

`t_1 = 60.75 \ text{(part i)}, \ t_2 = 65.123`

`text(Period of) \ \ v_1 = (2 pi)/(n) = (pi)/(7)\ \ => \ n = 14`

`:. \ text(Proportion of cycle)` `= (65.123-60.75)/(14)`
  `= 0.312`
  `= 31 text{%  (nearest %)}`

 

g.    `f(x) → g(x)`

`y^{prime} = 28 + 18 sin ({pi(x^{prime}-k)}/{7})`

`x^{prime} = ax + c` `\ \ \ \ \ \ y^{prime} = by + d`

 

`text(Using) \ \ y^{prime} = by + d`

`28 + 18 sin ({pi(x^{prime}-k)}/{7}) = b (20 + 16 sin ({pi x}/{14})) + d`
 

`text(Equating coefficients of) \ \ sin theta :`

`16b = 18 \ \ \ => \ b = (9)/(8)`
 

`text(Equating constants:)`

`20 xx (9)/(8) + d = 28 \ \ \ => \ \ d = (11)/(2)`

`(x^{prime}-k)/(7)` `= (x)/(14)`
`x^{prime}` `= (x)/(2) + k \ \ => \ \ a = (1)/(2) \ , \ c = k`

 

`a = (1)/(2) \ , \ b = (9)/(8) \ , \ c = k \ , \ d = (11)/(2)`

Calculus, MET2 2019 VCAA 3

During a telephone call, a phone uses a dual-tone frequency electrical signal to communicate with the telephone exchange.

The strength, `f`, of a simple dual-tone frequency signal is given by the function  `f(t) = sin((pi t)/3) + sin ((pi t)/6)`, where  `t`  is a measure of time and  `t >= 0`.

Part of the graph of `y = f(t)`  is shown below

  1. State the period of the function.   (1 mark)

  2. Find the values of  `t`  where  `f(t) = 0`  for the interval  `t in [0, 6]`.   (1 mark)

  3. Find the maximum strength of the dual-tone frequency signal, correct to two decimal places.   (1 mark)

  4. Find the area between the graph of  `f`  and the horizontal axis for  `t in [0, 6]`.   (2 marks)

Let  `g`  be the function obtained by applying the transformation  `T`  to the function  `f`, where
 

`T([(x), (y)]) = [(a, 0), (0, b)] [(x), (y)] + [(c), (d)]`
 

and `a, b, c` and `d` are real numbers.

  1. Find the values of `a, b, c` and `d` given that  `int_2^0 g(t)\ dt + int_2^6 g(t)\ dt`  has the same area calculated in part d.   (2 marks)

  2. The rectangle bounded by the line  `y = k, \ k in R^+`, the horizontal axis, and the lines  `x = 0`  and  `x = 12`  has the same area as the area between the graph of  `f`  and the horizontal axis for one period of the dual-tone frequency signal.

     

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

Show Answers Only
  1. `12`
  2. `0, 4, 6`
  3. `1.760`
  4. `15/pi\ text(u²)`
  5. `a = 1,\ b =-1,\ c =-6,\ d = 0`
  6. `5/(2pi)`
Show Worked Solution

a.   `text(Period) = 12`
  

b.   `t = 0, 4, 6`
  

c.   `f(t) = sin ((pi t)/3) + sin ((pi t)/6)`

`f(t)_max ~~ 1.76\ \ text{(by CAS)}`

 

d.    `text(Area)` `= int_0^4 sin((pi t)/3) + sin ((pi t)/6) dt-int_4^6 sin ((pi t)/3) + sin ((pi t)/6) dt`
    `= 15/pi\ text(u²)`

 

e.   `text(Same area) => f(t)\ text(is reflected in the)\ x text(-axis and)`

`text(translated 6 units to the left.)`

`x′=ax+c`

`y′=by+d`

`text(Reflection in)\ xtext(-axis) \ => \ b=-1, \ d=0`

`text(Translate 6 units to the left) \ => \ a=1, \ c=-6`

`:. a = 1,\ b = -1,\ c = -6,\ d = 0`
  

f.    `text(Area of rectangle)` `= 2 xx text(Area between)\ f(t) and x text(-axis)\ \ t in [0, 6]`
  `12k` `= 2 xx 15/pi`
  `:. k` `=5/(2pi)`

Graphs, MET2 2019 VCAA 10 MC

Which one of the following statements is true for  `f: R -> R, \ f(x) = x + sin(x)`?

  1. The graph of  `f` has a horizontal asymptote
  2. There are infinitely many solutions to  `f(x) = 4`
  3. `f` has a period of  `2 pi`
  4. `f^{\prime}(x) >= 0`  for  `x in R`
  5. `f^{\prime}(x) = cos(x)`
Show Answers Only

`D`

Show Worked Solution

`text(By CAS, sketch)\ \ f(x) = x + sin(x):`

`text(By inspection, eliminate A, B, C)`

`text(By CAS, sketch)\ \ d/(dx)\ f(x):`

` text(Graph range) >= 0\ \ text(for)\ \ x in R`
 

`=>   D`

Graphs, MET2 2019 VCAA 1 MC

Let  `f: R -> R,\ \ f(x) = 3 sin ((2x)/5) - 2`.

The period and range of  `f`  are respectively

  1. `5 pi`  and  `[-3, 3]`
  2. `5 pi`  and  `[-5, 1]`
  3. `5 pi`  and  `[-1, 5]`
  4. `(5 pi)/2`  and  `[-5, 1]`
  5. `(5 pi)/2`  and  `[-3, 3]`
Show Answers Only

`B`

Show Worked Solution
`text(Period)` `= (2pi)/n`
  `= (2 pi)/(2/5)`
  `= 5 pi`
   
`text(Range)` `= [-2 -3, -2 + 3]`
  `= [-5, 1]`

 
`=>   B`

Graphs, MET2 2017 VCAA 1 MC

Let  `f : R → R, \ f (x) = 5sin(2x) - 1`.

The period and range of this function are respectively

  1. `π\ text(and)\ [−1, 4]`
  2. `2π\ text(and)\ [−1, 5]`
  3. `π\ text(and)\ [−6, 4]`
  4. `2π\ text(and)\ [−6, 4]`
  5. `4π\ text(and)\ [−6, 4]`
Show Answers Only

`C`

Show Worked Solution

`text(Period) = (2pi)/2 = pi`

`text(Range)` `= [−1 – 5, −1 + 5]`
  `= [−6 ,4]`

`=> C`

Graphs, MET1 SM-Bank 27

The graph shown is  `y = A sin bx`.

  1. Write down the value of  `A`.   (1 mark)

  2. Find the value of  `b`.   (1 mark)

  3. Copy or trace the graph into your writing booklet.

     

    On the same set of axes, draw the graph  `y = 3 sin x + 1`  for  `0 <= x <= pi`.   (2 marks)

Show Answers Only
  1. `A = 4`
  2. `b = 2`
  3. `text(See Worked Solutions for sketch)`
Show Worked Solution

a.   `A = 4`

b.  `text(S)text(ince the graph passes through)\ \ (pi/4, 4)`

`text(Substituting into)\ \ y = 4 sin bx`

`4 sin (b xx pi/4)` `=4`
`sin (b xx pi/4)` `= 1`
`b xx pi/4` `= pi/2`
`:. b` `= 2`

  

 MARKER’S COMMENT: Graphs are consistently drawn too small by many students. Aim to make your diagrams 1/3 to 1/2 of a page. 
c.

Graphs, MET2 2016 VCAA 6 MC

Consider the graph of the function defined by  `f: [0, 2 pi] -> R,\ f(x) = sin (2x).`

The square of the length of the line segment joining the points on the graph for which  `x = pi/4 and x = (3 pi)/4` is

  1. `(pi^2 + 16)/4`
  2. `pi + 4`
  3. `4`
  4. `(3 pi^2 + 16 pi)/4`
  5. `(10 pi^2)/16`
Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=pi/4,\ \ f(x) = sin(pi/2)=1`

`text(When)\ \ x=(3pi)/4,\ \ f(x) = sin((3pi)/2)=-1`

`text(Let)\ \ z` `= text(distance between)\ (pi/4, 1) and ((3pi)/4, −1)`
`z^2` `= ((3pi)/4 – pi/4)^2 + (−1 −1)^2`
   `=pi^2/4 + 4`
  `= (pi^2 + 16)/4`

`=>   A`

Functions, MET1 2012 VCAA 6

The graphs of  `y = cos (x) and y = a sin (x)`,  where `a` is a real constant, have a point of intersection at  `x = pi/3.`

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

  2. If  `x in [0, 2 pi]`, find the `x`-coordinate of the other point of intersection of the two graphs.  (1 mark)

Show Answers Only
  1. `1/sqrt 3`
  2. `(4 pi)/3`
Show Worked Solution

a.   `text(Intersection occurs when)\ \ x=pi/3,`

`a sin(pi/3)` `= cos (pi/3)`
`tan(pi/3)` `= 1/a`
`sqrt 3` `=1/a`
`:. a` `=1/sqrt3`

 

b.   `tan (x)` `= sqrt 3`
  `x` `= pi/3, (4 pi)/3, 2pi+ pi/3, …\ text(but)\ x in [0, 2 pi]`
  `:. x` `= (4 pi)/3`

Graphs, MET2 2015 VCAA 1 MC

Let `f: R -> R,\ f(x) = 2sin(3x) - 3.`

The period and range of this function are respectively

  1. `text(period) = (2 pi)/3 and text(range) = text{[−5, −1]}`
  2. `text(period) = (2 pi)/3 and text(range) = text{[−2, 2]}`
  3. `text(period) = pi/3 and text(range) = text{[−1, 5]}`
  4. `text(period) = 3 pi and text(range) = text{[−1, 5]}`
  5. `text(period) = 3 pi and text(range) = text{[−2, 2]}`
Show Answers Only

`A`

Show Worked Solution

`text(Range:)\ [−3 – 2, −3 + 2]`

`= [−5,−1]`

`text(Period) = (2pi)/n = (2pi)/3`

`=>   A`