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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}

\(\text{Calculations:}\)

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

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

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

Calculus, 2ADV C3 2023 HSC 30

Let  \(f(x)=e^{-x} \sin x\).

  1. Find the coordinates of the stationary points of \(f(x)\) for  \(0\leq x\leq 2\pi\). You do NOT need to check the nature of the stationary points.  (3 marks)

  2. Without using any further calculus, sketch the graph of  \(y=f(x)\), for  \(0\leq x\leq 2\pi\), showing stationary points and intercepts.  (2 marks)
     


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a.    \( \Big(\dfrac{\pi}{4},\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}}\Big)\ \text{and}\  \Big(\dfrac{5\pi}{4},\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}}\Big)\)

b.    
         

Show Worked Solution

a.    \(f(x)=e^{-x} \sin x\)

\(f^{′}(x)=e^{-x} \cos x-e^{-x} \sin x = e^{-x}( \cos x-\sin x) \)

\(\text{SPs when}\ f^{′}(x)=0: \)

\(e^{-x}=0\ \ \rightarrow \ \text{no solution} \)

\(\cos x-\sin x\) \(=0\)  
\(1-\tan x\) \(=0\)  
\(\tan\) \(=1\)  
Mean mark (a) 54%.

\(x=\dfrac{\pi}{4}, \dfrac{5\pi}{4} \)
 

\(f(\dfrac{\pi}{4})=e^{-\frac{\pi}{4}}\sin \frac{\pi}{4}=\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}} \)

\(f(\dfrac{5\pi}{4})=e^{-\frac{5\pi}{4}}\sin \frac{5\pi}{4}=\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}} \)
 

\(\therefore\ \text{SPs at}\ \Big(\dfrac{\pi}{4},\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}}\Big)\ \text{and}\  \Big(\dfrac{5\pi}{4},\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}}\Big)\)

 
b.    \(\ x\text{-intercepts at}\ \ x=0, \pi,\ 2\pi \)
 

 

♦♦ Mean mark (b) 34%.

Calculus, 2ADV C3 2022 HSC 27

Let  `f(x)=xe^(-2x)`.

It is given that  `f^(′)(x)=e^(-2x)-2xe^(-2x)`.

  1. Show that  `f^(″)(x)=4(x-1)e^(-2x)`.  (2 marks)

  2. Find any stationary points of  `f(x)`  and determine their nature.  (2 marks)

  3. Sketch the curve  `y=x e^{-2 x}`, showing any stationary points, points of inflection and intercepts with the axes.  (3 marks)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Max S.P. at}\ \ (1/2, 1/(2e))`
  3.  
Show Worked Solution

a.   `f^(′)(x)=e^(-2x)-2xe^(-2x)`

`f^(″)(x)` `=-2e^(-2x)-[2x(-2)e^(-2x)+2e^(-2x)]`  
  `=-2e^(-2x)+4xe^(-2x)-2e^(-2x)`  
  `=4xe^(-2x)-4e^(-2x)`  
  `=4(x-1)e^(-2x)\ \ text{… as required}`  

 

b.   `text{S.P.’s occur when}\ \ f^(′)(x)=0`

`e^(-2x)-2xe^(-2x)` `=0`  
`e^(-2x)(1-2x)` `=0`  

 
`e^(-2x)=0\ \ =>\ \ text{No solution}`

`1-2x=0\ \ =>\ \ x=1/2`

`f^(″)(1/2)` `=4(1/2-1)e^(-2xx1/2)`  
  `=-2e^(-1)<0\ \ =>\ text{MAX}`  

 
`f(1/2)=1/2e^(-1)=1/(2e)`
 

c.   `text{POI occurs when}\ \ f^(″)(x)=0`

`f^(″)(x)=4(x-1)e^(-2x)=0\ \ =>\ \ x=1`

`text{Test for change in concavity:}`

`f^(″)(0)=4(0-1)e^0=-4`

`f^(″)(2)=4(2-1)e^(-4)>0\ \ =>\ text{Concavity changes}`

`:.\ text{POI exists at}\ \ (1,1/e^2)`

`text{As}\ \ x->oo\ \ =>\ \ y->0^+`
 


Mean mark part (c) 54%.

L&E, 2ADV E1 2021 HSC 5 MC

Which of the following best represents the graph of  `y = 10 (0.8)^x`?
 

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`A`

Show Worked Solution

`\text{By elimination:}`

`\text{When} \ x = 0 \ , \ y = 10(0.8) ^0 = 10`

`-> \ text{Eliminate B and D}`

`text(As)\ \ x→oo, \ y→0`

`-> \ text{Eliminate C}`

`=> A`

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)

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  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.

Algebra, STD2 A4 2016 HSC 29b

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)

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  1. `1.5\ text(kg)`
  2. `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 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`

Algebra, STD2 A4 2004 HSC 26a

  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.   (4 marks)

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

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  1. `text(14%)`
  2. `714`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `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)`

`text(approximately 8.4 hours.)`

Algebra, STD2 A4 2012 HSC 30c

In 2010, the city of Thagoras modelled the predicted population of the city using the equation

`P = A(1.04)^n`.

That year, the city introduced a policy to slow its population growth. The new predicted population was modelled using the equation

`P = A(b)^n`.

In both equations, `P` is the predicted population and `n` is the number of years after 2010.  

The graph shows the two predicted populations.
 

  1. Use the graph to find the predicted population of Thagoras in 2030 if the population policy had NOT been introduced.   (1 mark)

  2. In each of the two equations given, the value of `A` is 3 000 000.

     

    What does `A` represent?   (1 mark)

  3. The guess-and-check method is to be used to find the value of `b`, in  `P = A(b)^n`.

     

    (1) Explain, with or without calculations, why 1.05 is not a suitable first estimate for `b`.  (1 mark)

     

    (2) With  `n = 20`  and  `P = 4\ 460\ 000`, use the guess-and-check method and the equation  `P = A(b)^n`  to estimate the value of `b` to two decimal places. Show at least TWO estimate values for `b`, including calculations and conclusions.  (2 marks)

  4. The city of Thagoras was aiming to have a population under 7 000 000 in 2050. Does the model indicate that the city will achieve this aim?

     

    Justify your answer with suitable calculations.  (2 marks)

Show Answers Only
  1.  `6\ 600\ 000`
  2.  `text(The population in 2010.)`
  3.  `text{(1)  See Worked Solution}`

     

    `(2)\ \ b = 1.03, 1.02`

  4. `text(See Worked Solution)`
Show Worked Solution

i.    `text(2030 occurs at)\ \ n = 20\ \ text(on the)\ x text(-axis.)`

`text(Expected population (no policy) ) = 6\ 600\ 000`

COMMENT: Common ADV/STD2 content in new syllabus.

 

ii.   `A\ text(represents the population when)\ \ n=0` 

`text(which is the population in 2010.)`
 

♦ Mean mark part (ii) 48%

iii. (1)  `P = A(1.05)^n\ text(would be steeper and lie above)`

    `P = A(1.04)^n\ text(since)\ 1.05 > 1.04`
 

iii. (2)  `text(Let)\ \ b = 1.03`

`P` `= 3\ 000\ 000 xx 1.03^20`
  `= 5\ 418\ 000`

 
`text(Let)\ \ b = 1.02`

`P` `= 3\ 000\ 000 xx 1.02^20`
  `= 4\ 457\ 800`

 
`:. b = 1.02`
 

iv.  `text(In 2050,)\ n = 40`

`P` `= 3\ 000\ 000 xx 1.02^40`
  `= 6\ 624\ 119\ \ (text(nearest whole))`

 
`text(S)text(ince the population is below 7 million,)`

`text(the model will achieve the aim.)`