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L&E, 2ADV EQ-Bank 5

The mass `M` kg of a baby pig at age `x` days is given by  `M = A(1.1)^x`  where `A` is a constant. The graph of this equation is shown.
 

2ug-2016-hsc-q29_1

  1. What is the value of `A`?   (1 mark)

  2. What is the daily growth rate of the pig’s mass? Write your answer as a percentage.   (1 mark)

Show Answers Only

i.    `1.5\ text(kg)`

ii.   `10text(%)`

Show Worked Solution

i.   `text(When)\ x = 0:`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!
`1.5` `= A(1.1)^0`
`:. A` `= 1.5\ text(kg)`

 
ii.   `text(Daily growth rate)\ = 0.1 = 10text(%)`

L&E, 2ADV EQ-Bank 4

  1. The number of bacteria in a culture grows from 100 to 114 in one hour.

     

    What is the percentage increase in the number of bacteria?   (1 mark)

  2. The bacteria continue to grow according to the formula  `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

     

    What is the number of bacteria after 15 hours?   (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\;  &  \;\; 5 \;\;  & \;\; 10 \;\;  & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\;  &  \;\; 193 \;\;  & \;\; 371 \;\;  & \;\; ? \;\; \\
\hline
\end{array}

  1. Use the values of `n` from  `t = 0`  to  `t = 15`  to draw a graph of  `n = 100(1.14)^t`.

     

    Use about half a page for your graph and mark a scale on each axis.   (2 marks)

  2. Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300.   (1 mark)

Show Answers Only

i.    `text(14%)`

ii.   `714`

iii.  `text(Proof)\ \ text{(See Worked Solutions)}`

iv.   `text(8.4 hours)`

Show Worked Solution

i.   `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`
 

ii.  `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n= 100(1.14)^15= 713.793\ …\ = 714\ \ \ text{(nearest whole)}`
 

iii. 

 

iv.  `text(Using the graph)`

`text(The number of bacteria reaches 300 after ~ 8.4 hours.)`

L&E, 2ADV E1 2024 HSC 13

The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .

Population \(W\) is modelled by the equation  \(y=A \times(1.055)^x\).

Population \(K\) is modelled by the equation  \(y=B \times(0.97)^x\).
 

Complete the table using the information provided.   (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

Show Answers Only

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

\(\text{Calculations:}\)

\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)

\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)

\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)

L&E, 2ADV E1 SM-Bank 14

The spread of a highly contagious virus can be modelled by the function

`f(x) = 8000/(1 + 1000e^(−0.12x))`

Where `x` is the number of days after the first case of sickness due to the virus is diagnosed and `f(x)` is the total number of people who are infected by the virus in the first `x` days.

  1. Calculate `f(0)`.   (1 mark)

  2. Find the value of `f(365)` and interpret it result.   (2 marks)

Show Answers Only
  1. `7.99…`
  2. `text(After 1 year, the model predicts the total number)`
    `text(of people infected by the virus is 8000.)`
Show Worked Solution
i.   `f(0)` `= 8000/(1 + 1000e^0)`
    `= 8000/1001`
    `= 7.99…`

 

ii.    `f(365)` `= 8000/(1 + 1000e^(−0.12 xx 365))`
    `= 8000/(1 + 1000e^(−43.8))`
    `~~ 8000`

 
`text(After 1 year, the model predicts the total number)`

`text(of people infected by the virus is 8000.)`

Functions, 2ADV F2 SM-Bank 13

 

  1. Show that the function  `y = (1-e^x)/(1 + e^x)`  is an odd function?   (1 mark) 

  2. Sketch  `y = (1-e^x)/(1 + e^x)`, labelling all intercepts and asymptotes.   (2 marks)

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  1. `{text(all real)\ x,  x!=1/4}`

 

 

 

 

 

 

Show Worked Solution

i.   `f(x) = (1-e^x)/(1 + e^x)`

`f(−x)` `= (1-e^(−x))/(1 + e^(−x)) xx (e^x)/(e^x)`
  `= (e^x-1)/(e^x + 1)`
  `= −(1-e^x)/(1 + e^x)`
  `= −f(x)`

 
`:. f(x)\ text(is ODD.)`

 

ii.   `y = (1-e^x)/(1 + e^x) xx (e^(−x))/(e^(−x)) = (e^(−x)-1)/(e^(−x) + 1) = 1-2/(e^(−x) + 1)`

`text(As)\ x -> ∞, \ 2/(e^(−x) + 1) -> 2, \ y -> −1`

`text(As)\ x ->-∞, \ 2/(e^(−x) + 1) -> 0, \ y -> 1`

`text(When)\ x = 0, \ y = 0`
 

L&E, 2ADV E1 SM-Bank 2

The population of Indian Myna birds in a suburb can be described by the exponential function

`N = 35e^(0.07t)`

where `t` is the time in months.

  1.  What will be the population after 2 years?  (1 mark)

  2.  Draw a graph of the population.  (2 marks)

Show Answers Only
  1. `188\ text(birds)`

  2.  

Show Worked Solution

i.   `N = 35e^(0.07t)`

`text(Find)\ N\ text(when)\ \ t = 24:`

`N` `= 35e^(0.07 xx 24)`
  `= 35e^(1.68)`
  `= 187.79…`
  `= 188\ text(birds)`

 

ii.

L&E, 2ADV E1 2015 HSC 8 MC

The diagram shows the graph of  `y = e^x (1 + x).`
 

How many solutions are there to the equation  `e^x (1 + x) = 1-x^2`?

  1. `0`
  2. `1`
  3. `2`
  4. `3`
Show Answers Only

`C`

Show Worked Solution

`text(The graphs intersect at 2 points)`

`:. e^x(1 + x) = 1-x^2\ text(has 2 solutions)`

`=> C`