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Functions, MET1 2024 VCAA 5

The function  \(h:[0, \infty) \rightarrow R, \ h(t)=\dfrac{3000}{t+1}\)  models the population of a town after \(t\) years.

  1. Use the model \(h(t)\) to predict the population of the town after four years.   (1 mark)

  2. A new function, \(h_1\), models a population where  \(h_1(0)=h(0)\) but \(h_1\) decreases at half the rate of \(h\) at any point in time.
  3. State a sequence of two transformations that maps \(h\) to this new model \(h_1\).   (2 marks)

  4. In the town, 100 people were randomly selected and surveyed, with 60 people indicating that they were unhappy with the roads.
    1. Determine an approximate 95% confidence interval for the proportion of people in the town who are unhappy with the roads.
    2. Use  \(z=2\)  for this confidence interval.   (2 marks)
    3. A new sample of \(n\) people results in the same sample proportion.
    4. Find the smallest value of \(n\) to achieve a standard deviation of  \(\dfrac{\sqrt{2}}{50}\)  for the sample proportion.   (1 mark)

Show Answers Only

a.    \(600\)

b.   \(\text{Transformations:}\)

\(\text{1- Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\)

\(\text{2- Translation of 1500 units upwards}\)

ci.   \(\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\ \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)

cii.  \(300\)

Show Worked Solution

a.  \(h(4)=\dfrac{3000}{4+1}=600\)

b.     \(h(t)\) \(=3000(t+1)^{-1}\)
  \(h^{\prime}(t)\) \(=-\dfrac{3000}{(t+1)^2}\)
  \(h_1^{\prime}(t)\) \(=\dfrac{1}{2}h^{\prime}(t)=-\dfrac{1500}{(t+1)^2}\)
  \(h_1(t)\) \(=\dfrac{1500}{t+1}+C\)

♦♦♦ Mean mark (a) 17%.

\(\text{Given}\ \ h(0)=h_1(0):\)

\(\dfrac{1500}{0+1} +C= 3000\ \ \Rightarrow\ \ C=1500\)

\(h_1(t)=\dfrac{1500}{t+1}+1500\)
 

\(\text{Transformations:}\)

\(\text{1- Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\)

\(\text{2- Translation of 1500 units upwards}\)
 

ci.    \(\hat{p}=\dfrac{60}{100}=\dfrac{3}{5},\quad 1-\hat{p}=\dfrac{2}{5},\quad z=2\)

\(\text{Approx CI}\) \(=\left(\dfrac{3}{5}-2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}},\ \dfrac{3}{5}+2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}}\right)\)
  \(=\left(\dfrac{3}{5}-\dfrac{2\sqrt{6}}{50},\quad \dfrac{3}{5}+\dfrac{2\sqrt{6}}{50}\right)\)
  \(=\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\quad \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)

  

cii.   \(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\dfrac{6}{25n}\) \(=\dfrac{2}{2500}\)
  \(\dfrac{25n}{6}\) \(=1250\)
  \(n\) \(=\dfrac{6}{25}\times 1250\)
    \(=300\)
♦♦ Mean mark (c.ii.) 33%.

Calculus, MET2 2024 VCAA 13 MC

The function  \(f:(0, \infty) \rightarrow R, f(x)=\dfrac{x}{2}+\dfrac{2}{x}\)  is mapped to the function \(g\) with the following sequence of transformations:

  1. dilation by a factor of 3 from the \(y\)-axis
  2. translation by 1 unit in the negative direction of the \(y\)-axis.

The function \(g\) has a local minimum at the point with the coordinates

  1. \((6,1)\)
  2. \(\left(\dfrac{2}{3}, 1\right)\)
  3. \((2,5)\)
  4. \(\left(2,-\dfrac{1}{3}\right)\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Dilate by a factor of 3 from the}\ y\text{-axis:}\)

\(f(x) \rightarrow f_1(x)=\dfrac{\frac{x}{3}}{2}+\dfrac{2}{\frac{x}{3}}=\dfrac{x}{6}+\dfrac{6}{x}\)

\(\text{Translate 1 unit down:}\)

\(f_1(x) \rightarrow g(x)=\dfrac{x}{6}+\dfrac{6}{x}-1\)

\(g'(x)=\dfrac{1}{6}-6x^{-2}\)

♦ Mean mark 45%.
 

\(\Rightarrow A\)

Graphs, MET2 2016 VCAA 12 MC

The graph of a function  `f` is obtained from the graph of the function `g` with rule  `g(x) = sqrt (2x - 5)`  by a reflection in the `x`-axis followed by a dilation from the `y`-axis by a factor of  `1/2`.

Which one of the following is the rule for the function  `f`?

  1. `f(x) = sqrt (5 - 4x)`
  2. `f(x) = - sqrt (x - 5)`
  3. `f(x) = sqrt (x + 5)`
  4. `f(x) = −sqrt (4x - 5)`
  5. `f(x) = −sqrt (4x - 10)`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ y=sqrt(2x-5)`

`text(1st transformation:)`

`y = – sqrt(2x-5)`
 

`text(2nd transformation:)`

`y` `=-sqrt(2(2x)-5)`
  `=- sqrt(4x-5)`
`:. f(x)` `= −sqrt(4x – 5)`

 
`=>   D`

Calculus, MET2 2012 VCAA 2

Let  `f: R text(\{2}) -> R,\ f(x) = 1/(2x-4) + 3.`

  1. Sketch the graph of  `y = f(x)` on the set of axes below. Label the axes intercepts with their coordinates and label each of the asymptotes with its equation.   (3 marks)

           VCAA 2012 2a
     

  2.   i. Find `f^{′}(x)`.   (1 mark)

  3.  ii. State the range of  `f ^{′}`.   (1 mark)

  4. iii. Using the result of part ii. explain why `f` has no stationary points.   (1 mark)

  5. If  `(p, q)`  is any point on the graph of  `y = f(x)`, show that the equation of the tangent to  `y = f(x)`  at this point can be written as  `(2p-4)^2 (y-3) = -2x + 4p-4.`   (2 marks)

  6. Find the coordinates of the points on the graph of  `y = f(x)`  such that the tangents to the graph at these points intersect at  `(-1, 7/2).`   (4 marks)

  7. A transformation  `T: R^2 -> R^2`  that maps the graph of  `f` to the graph of the function  `g: R text(\{0}) -> R,\ g(x) = 1/x`  has rule
  8.      `T([(x), (y)]) = [(a, 0), (0, 1)] [(x), (y)] + [(c), (d)]`, where `a`, `c` and `d` are non-zero real numbers.
  9. Find the values of `a, c` and `d`.   (2 marks)

Show Answers Only
  1. met2-2012-vcaa-sec2-answer
  2.   i. `f^{′}(x) = (−2)/((2x-4)^3)`
     ii. `text(Range) = (−∞,0)`
    iii. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(Coordinates:) (1,5/2)\ text(or)\ (5,19/6)`
  5. `a = 2, c = −4, d = −3`
Show Worked Solution

a.   `text(Asymptotes:)`

`x = 2`

`y = 3`

met2-2012-vcaa-sec2-answer

 

b.i.   `f^{′}(x) = (−2)/((2x-4)^2)`

 

b.ii.   `text(Range) = (−∞,0), or  R^-`

MARKER’S COMMENT: Incorrect notation in part (b)(ii) was common, including `{-oo,0}, -R, (0,-oo)`.

 

b.iii.    `text(As)\ \ ` `f^{′}(x) < 0quadtext(for)quadx ∈ R text(\{2})`
    `f^{′}(x) != 0`

 
`:. f\ text(has no stationary points.)`
 

c.   `text(Point of tangency) = P(p,1/(2p-4) + 3)`

♦♦ Mean mark 29%.
`m_text(tang)` `= f^{′}(p)`
  `= (-2)/((2p-4)^2)`

 
`text(Equation of tangent using:)`

`y-y_1` `= m(x-x_1)`
`y-(1/(2p-4) + 3)` `= (-2)/((2p-4)^2)(x-p)`
`y-3` `= (-2(x-p))/((2p-4)^2) + (2p-4)/((2p-4)^2)`
`(2p-4)^2(y-3)` `=-2x + 2p + 2p-4`
`:. (2p-4)^2(y-3)` `=-2x + 4p-4\ \ text(… as required)`

 

d.   `text(Substitute)\ \ (−1,7/2)\ text{into tangent (part c),}`

♦♦♦ Mean mark 19%.

`text(Solve)\ \ (2p-4)^2(7/2-3) = −2(-1) + 4p-4\ \ text(for)\ p:`

`:. p = 1,\ text(or)\ 5`

`text(Substitute)\ \ p = 1\ text(and)\ p = 5\ text(into)\ \ P(p,1/(2p-4) + 3)`

`:. text(Coordinates:)\ (1,5/2)\ text(or)\ (5,19/6)`
 

e.   `text(Determine transformations that that take)\ f -> g:`

`text(Dilate the graph of)\ \ f(x) = 1/(2x-4) + 3\ \ text(by a)`

`text(factor of 2 from the)\ \ ytext(-axis).`

`y = 1/(2(x/2)-4) + 3= 1/(x-4) + 3`

`text(Translate the graph 4 units to the left and 3)`

`text(units down to obtain)\ \ g(x).`
 

`text(Using the transformation matrix,)`

`x^{′}` `=ax+c`
`y^{′}` `=y+d`

 
`f -> g:\ \ 1/(2x-4) -> 1/(x^{′})`

`x^{′}=2x-4`

`=> a=2,\ \ c=-4`
 

`f -> g:\ \ y -> y^{′} + 3`

`y^{′}=y -3`

`=>\ \ d=-3`

Graphs, MET2 2014 VCAA 12 MC

The transformation  `T: R^2 -> R^2`  with rule

`T ([(x), (y)]) = [(-1, 0), (0, 2)] [(x), (y)] + [(1), (-2)]`

maps the line with equation  `x - 2y = 3`  onto the line with equation

  1. `x + y = 0`
  2. `x + 4y = 0`
  3. `-x - y = 4`
  4. `x + 4y = -6`
  5. `x - 2y = 1`
Show Answers Only

`C`

Show Worked Solution
`xprime` `= −x + 1`
`1 – xprime` `= x`

 

`yprime` `= 2y – 2`
`(yprime + 2)/2` `= y`

 

`text(Substitute)\ \ x,y\ \ text(into)\ \ x – 2y = 3,`

`(1 – xprime) – 2((yprime + 2)/2)` `= 3`
`−xprime – yprime` `= 4`

`=>   C`