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Calculus, MET2 2024 VCAA 3

The points shown on the chart below represent monthly online sales in Australia.

The variable \(y\) represents sales in millions of dollars.

The variable \(t\) represents the month when the sales were made, where \(t=1\) corresponds to January 2021, \(t=2\) corresponds to February 2021 and so on.

  1. A cubic polynomial \(p ;(0,12] \rightarrow R, p(t)=a t^3+b t^2+c t+d\) can be used to model monthly online sales in 2021.

    The graph of \(y=p(f)\) is shown as a dashed curve on the set of axes above.

    It has a local minimum at (2,2500) and a local maximum at (11,4400).

    1. Find, correct to two decimal places, the values of \(a, b, c\) and \(d\).  (3 marks)
    1. Let \(q:(12,24] \rightarrow R, q(t)=p(t-h)+k\) be a cubic function obtained by translating \(p\), which can be used to model monthly online sales in 2022.

      Find the values of \(h\) and \(k\) such that the graph of \(y=q(t)\) has a local maximum at \((23,4750)\).  (2 marks)

  1. Another function \(f\) can be used to model monthly online sales, where

    \(f:(0,36] \rightarrow R, f(t)=3000+30 t+700 \cos \left(\dfrac{\pi t}{6}\right)+400 \cos \left(\dfrac{\pi t}{3}\right)\)

    Part of the graph of \(f\) is shown on the axes below.

    1. Complete the graph of \(f\) on the set of axes above until December 2023, that is, for \(t \in(24,36]\).Label the endpoint at \(t=36\) with its coordinates.  (2 marks)
    2. The function \(f\) predicts that every 12 months, monthly online sales increase by \(n\) million dollars.

      Find the value of \(n\).  (1 mark)

    1. Find the derivative \(f^{\prime}(t)\)  (1 mark)
    1. Hence, find the maximum instantaneous rate of change for the function \(f\), correct to the nearest million dollars per month, and the values of \(t\) in the interval \((0,36]\) when this maximum rate occurs, correct to one decimal place.  (2 marks)
Show Answers Only

ai.   \(a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18\)

aii.  \(h=12, k=350\)

bi.  

bii.  \( n=360\)

biii. \(f^{\prime}(t)=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)

biv. \(\text{Maximum rate occurs at }t=10.2, 22.2, 34.2\)

\(\text{Maximum rate}\ \approx 725\ \text{million/month}\)

Show Worked Solution
ai.   \(\text{Given}\ p(2)=2500, p(11)=4400, p^{\prime}(2)=0\ \text{and}\ p^{\prime}(11)=0\) 
  
\(p(t)=at^3+bt^2+ct+d\ \Rightarrow\ p^{\prime}(t)=3at^2+2bt+c\)
  
\(\text{Using CAS:}\)
    
\(\text{Solve}
\begin{cases}
8a+4b+2c+d=2500 \\
1331a+121b+11c+d=4400 \\
12a+4b+c=0  \\
363a+22b+c=0
\end{cases}\)
  

\(a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18\)

aii. \(\text{Local maximim }p(t)\ \text{is}\ (11, 4400)\)

\(\therefore\ h\) \(=23-11=12\)
\(k\) \(=4750-4400=350\)

 

bi.   \(\text{Plotting points from CAS:}\)

\((24,4820), (26, 3930), (28, 3290), (30, 3600), (32, 3410), (34, 4170), (36, 5180)\)

bii.  \(\text{Using CAS: }\)

\(f(12)-f(0)\) \(=4460-4100=360\)
\(f(24)-f(12)\) \(=4820-4460=360\)
\(f(36)-f(24)\) \(=5180-4820=360\) 

\(\therefore\ n=360\)

biii.  \(f(t)\) \(=3000+30t+700\cos\left(\dfrac{\pi t}{6}\right)+400\cos\left(\dfrac{\pi t}{3}\right)\)
  \(f^{\prime}(t)\) \(=30-\dfrac{700\pi}{6}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)
    \(=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)

  
biv.  \(\text{Max instantaneous rate of change occurs when }f^{\prime\prime}(t)=0\)

\(\text{Maximum rate occurs at }t=10.2, 22.2, 34.2\)

\(\text{Maximum rate using CAS:}\)

\(f{\prime}(10.2)=f{\prime}(22.2)=f{\prime}(34.2)=725.396\approx 725\ \text{million/month}\)

Algebra, MET2 2023 VCE SM-Bank 1 MC

\begin{aligned}
& x-2 y=3 \\
& 2 y-z=4
\end{aligned}

Which one of the following correctly describes the general solution to the system of linear equations given above?

  1. \(x=k, \quad y=\dfrac{1}{2}(k+3), \ z=k-1\), for all \(k \in R\)
  2. \(x=k, \quad y=\dfrac{1}{2}(k+3), \ z=k+1\), for all \(k \in R\)
  3. \(x=k, \quad y=\dfrac{1}{2}(k-3), \ z=k+7\), for all \(k \in R\)
  4. \(x=k, \quad y=\dfrac{1}{2}(k-3), \ z=k-7\), for all \(k \in R\)
  5. \(x=k, \quad y=\dfrac{1}{2}(k+3), \ z=k-7\), for all \(k \in R\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{When }x=k\quad (1)\)

\(2y\) \(=k-3\)
\(y\) \(=\dfrac{1}{2}(k-3)\quad (2)\)
\(\text{and}\ z\) \(=2y-4\)
\(z\) \(=2\times\dfrac{1}{2}(k-3)-4\)
  \(=k-7\quad (3)\)

  
\(\text{Using CAS:}\)

\(\Rightarrow D\)

Algebra, MET2 2007 VCAA 5 MC

The simultaneous linear equations

`mx + 12y = 24`

`3x + my = m`

have a unique solution only for

  1. `m = 6 or m = – 6`
  2. `m = 12 or m = 3`
  3. `m in R\ text(\){– 6, 6}`
  4. `m = 2 or m = 1`
  5. `m in R\ text(\){– 12, – 3}`
Show Answers Only

`C`

Show Worked Solution
`mx + 12y` `=24`
`y` `=-m/12 x +2\ \ …\ (1)`
`3x + my` `=m`
`y` `= -3/m x+1\ \ …\ (2)`

 

`text(Unique solution occurs when)\ \ m_1!= m_2,`

♦♦ Mean mark 36%.
`-m/12` `!=-3/m`
`m^2` `!= 36`
`:. m` `!= +- 6`

`=>   C`

Algebra, MET1 SM-Bank 25

Solve these simultaneous equations to find the values of  `x`  and  `y`.    (3 marks)

`y = 2x + 1`

`x − 2y − 4 = 0`

Show Answers Only

`x = -2,\ y = -3`

Show Worked Solution

`text(Solution 1 – Substitution)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`

 

`text(Substitute)\ \ y = 2x + 1\ \ text(into)\ text{(ii)}`

`x\ – 2(2x + 1)\ – 4` `= 0`
`x\ – 4x\ – 2\ – 4` `= 0`
`-3x\ – 6` `= 0`
`3x` `= -6`
`x` `= -2`

 
`text(Substitute)\ \ x = –2\ \ text(into)\ text{(i)}`

`y = 2(–2) + 1 = -3`
 

`:.\ text(Solution is)\ x = -2,\ y = -3`
 

`text(Alternative Solution – Elimination)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`
 

`text(Multiply)\ text{(i)} xx 2`

`2y` `= 4x + 2`
`-4x + 2y\ – 2` `= 0\ \ \ \ \ …\ text{(iii)}`

 

`text{Add  (ii) + (iii)}`

`-3x\ – 6` `= 0`
`x` `= -2`
`y` `= -3\ \ text{(see Solution 1)}`

Algebra, MET2 2009 VCAA 1 MC

The simultaneous linear equations

`kx - 3y = 0`

`5x - (k + 2)y = 0`

where `k` is a real constant, have a unique solution provided

  1. `k in {– 5, 3}`
  2. `k in R\ text(\){– 5, 3}`
  3. `k in {– 3, 5}`
  4. `k in R\ text(\){– 3, 5}`
  5. `k in R\ text(\){0}`
Show Answers Only

`B`

Show Worked Solution

`text(Unique solution occurs when:)`

♦ Mean mark 49%.
`m_1` `!= m_2`
`k/5` `!= (– 3)/(– (k + 2))`
`k^2+2k-15` `!=0`
`(k+5)(k-3)` `!=0`
`:. k` `!= – 5, 3`

`=>   B`

Algebra, MET1 2011 VCAA 6

Consider the simultaneous linear equations

`kx - 3y` `= k + 3`
`4x + (k + 7) y` `= 1`

where `k` is a real constant.

  1. Find the value of `k` for which there are infinitely many solutions.  (3 marks)
  2. Find the values of `k` for which there is a unique solution.  (1 mark)
Show Answers Only
  1. `– 4`
  2. `k in R\ text(\)\ {– 4, – 3}`
Show Worked Solution

a.   `text(Infinite solutions if gradients and)`

♦ Mean mark 39%.
MARKER’S COMMENT: A number of solutions are possible here, including using the determinant.

`y text(-intercepts are the same.)`

`kx-3y` `=k+3`
`3y` `=kx-k-3`
`y` `=k/3 x – (k+3)/3`
`:.m_1=k/3 and c_1= -((k+3)/3)`

 

`4x + (k + 7) y` `=1`
`y` `=((-4)/(k+7)) x + 1/(k+7)`
`:. m_2=(-4)/(k+7) and c_2=1/(k+7)`
   

 

`text(Equating gradients and)\ y text(-intercepts:)`

`m_1` `= m_2` `\ \ \ and` `\ \ \ c_1` `=c_2`
`k/3` `= (– 4)/(k + 7)`   `-((k+3)/3)` `= 1/(k+7)`
`k^2 + 7k` `= – 12`   `-(k+3)(k+7)` `=3`
`k^2 + 7k + 12` `= 0`   `k^2+10k+24` `=0`
`(k + 3) (k + 4)` `= 0`   `(k+6)(k+4)` `= 0`
`k` `= – 3, – 4`   `k`  `= – 6, – 4`

 

`:. k = – 4\ \ \ text{(satisfies both)}`

 

b.   `text(Unique solution if:)`

♦♦ Mean mark 33%.

`m_1 != m_2`

`:. k in R\ text(\)\ {– 4, – 3}`

Algebra, MET1 SM-Bank 28

Consider the simultaneous linear equations below.

 `4x - 2y = 18`

`3x + ky = 10`   (3 marks)

where  `k`  is a real constant.

  1. What are the values of  `k`  where no solutions exist?   (3 marks)
  2. What values of  `k`  do the simultaneous equations have a unique solution?   (1 mark)
Show Answers Only
  1. `k=-3/2`
  2. `k\ in R text(\) {-3/2}` 
Show Worked Solution
a.    `4x – 2y` `=18`
  `y` `=2x-9\ \ …\ (1)`

`=> m_1 = 2,\ \ c_1=-9`

 

`3x +ky` `=10`
`y` `=-3/k x +10/k\ \ …\ (2)`

`=> m_2 = – 3/k,\ \ c_2=10/k`

 

`text(No solution if)\ \ m_1=m_2,  and  c_1!=c_2.`

`-3/k` `=2`
`k` `=- 3/2`

`text(When)\ \ k=-3/2,  c_1!=c_2.`

`:.\ text(No solution when)\ \ k=-3/2.`

 

b.   `text(A unique solution exists when)\ \ m_1 != m_2,`

`k in R\ text(\) {-3/2}`