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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%.

Functions, MET2 2024 VCAA 1

Consider the function  \( f: R \rightarrow R, f(x)=(x+1)(x+a)(x-2)(x-2 a) \text { where } a \in R \text {. } \)

  1. State, in terms of \(a\) where required, the values of \(x\) for which \(f(x)=0\).  (1 mark
  1. Find the values of \(a\) for which the graph of \(y=f(x)\) has
      
     i. exactly three \(x\)-intercepts.   (2 marks)

    ii. exactly four \(x\)-intercepts.   (1 mark)

  1. Let \(g\) be the function \(g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2\), which is the function \(f\) where \(a=1\).
      
      i. Find \(g^{\prime}(x)\)   (1 mark)

     ii. Find the coordinates of the local maximum of \(g\).   (1 mark)

    iii. Find the values of \(x\) for which \(g^{\prime}(x)>0\).   (1 mark)

     iv. Consider the two tangent lines to the graph of \(y=g(x)\) at the points where
    \(x=\dfrac{-\sqrt{3}+1}{2}\) and \(x=\dfrac{\sqrt{3}+1}{2}\). Determine the coordinates of the point of intersection of these two tangent lines.   (2 marks)

  1. Let \(g\) remain as the function \(g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2\), which is the function \(f\) where \(a=1\).

    Let \(h\) be the function \(h: R \rightarrow R, h(x)=(x+1)(x-1)(x+2)(x-2)\), which is the function \(f\) where \(a=-1\).
      
     i. Using translations only, describe a sequence of transformations of \(h\), for which its image would have a local maximum at the same coordinates as that of \(g\).   (1 mark)

    ii. Using a dilation and translations, describe a different sequence of transformations of \(h\), for which its image would have both local minimums at the same coordinates as that of \(g\).   (2 marks)

Show Answers Only

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

bii. \(R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}\)

ci.  \(g^{\prime}(x)=2(x-2)(x+1)(2x-1)\)

cii. \(\left(\dfrac{1}{2} , \dfrac{81}{16}\right)\)

ciii. \(x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)\)

civ. \(\left(\dfrac{1}{2}, \dfrac{27}{4}\right)\)

di.   \(\text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}\)

dii.  \(\text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }\)

\(\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units}\)

Show Worked Solution

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

bii. \(\text{The solution must be all }R\ \text{except those that give 3 or less solutions.}\)

\(\therefore\ R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}\)

ci.    \(g^{\prime}(x)\) \(=2(2-x)(x+1)^2+(x-2)^22(x+1)\)
    \(=2(x-2)(x+1)(x+1+x+2)\)
    \(=2(x-2)(x+1)(2x-1)\)

  
cii.  \(\text{When  }g^{\prime}(x)=0, x=2, -1, \dfrac{1}{2}\)

\(\text{From graph local maximum occurs when }x=\dfrac{1}{2}\)

\(g\left(\dfrac{1}{2}\right)\) \(=\left(\dfrac{1}{2}+1\right)^2\left(\dfrac{1}{2}-2\right)^2\)
  \(=\dfrac{9}{4}\times \dfrac{9}{4}=\dfrac{81}{16}\)

  
\(\therefore\ \text{Local maximum at}\ \left(\dfrac{1}{2} , \dfrac{81}{16}\right)\)

ciii. \(\text{From graph}\ g^{\prime}(x)>0\ \text{when }x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)\)

civ.  \(\text{Use CAS to find tangent lines and solve to find intersection.}\)

\(\text{Point of intersection of tangent lines}\ \left(\dfrac{1}{2}, \dfrac{27}{4}\right)\)

di.  \(\text{Local maximum of }g(x)\ \rightarrow\left(\dfrac{1}{2}, \dfrac{81}{16}\right)\)

\(\text{From CAS local maximum of }h(x)\ \rightarrow \left(0, 4\right)\)

\(\therefore\ \text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}\)

dii. \(\text{Using CAS to solve }g^{\prime}(x)=0\ \text{and }h^{\prime}(x)=0\)

\(\text{Local Minimums for }g(x)\ \text{at }(-1, 0)\ \text{and }(2, 0)\ \text{which are 3 apart.}\)

\(\text{Local minimums for at }h(x)\ \text{at }\left(-\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)\ \text{and }\left(\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)\)

\(\therefore\ \text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }\)

\(\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units.}\)

 

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%.
 

\(\Rightarrow A\)

Functions, MET2 2024 VCAA 12 MC

The graph of \(y=f(x)\) is shown below.

Which of the following options best represents the graph of \(y=f(2 x+1)\) ?
 

Show Answers Only

\(A\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{Graph has been dilated by a factor of}\ \dfrac{1}{2}\ \text{from}\ y\text{axis.}\)

→ \(\text{Eliminate C and D.}\)

\(\text{Graph is then translated}\ \dfrac{1}{2}\ \text{unit to the left.}\)

\(\text{Consider the turning point}\ (2, 1)\ \text{after translation:}\)

\(\left(2, 1\right)\ \rightarrow \ \left(2\times \dfrac{1}{2}-\dfrac{1}{2}, 1\right)=\left(\dfrac{1}{2}, 1\right)\)

\(\therefore\ \text{Option A is the only possible solution.}\)

\(\Rightarrow A\)

♦ Mean mark 47%.

Graphs, MET2 2020 VCAA 13 MC

The transformation  `T:R^(2)rarrR^(2)`  that maps the graph of  `y=cos(x)`  onto the graph of  `y=cos(2x+4)`  is

  1. `T([[x],[y]])=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-4],[0]])`
  2. `T([[x],[y]])=[[(1)/(2),0],[0,1]][[x],[y]]+[[-4],[0]]`
  3. `T([[x],[y]])=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-2],[0]])`
  4. `T([[x],[y]])=[[2,0],[0,1]]([[x],[y]]+[[2],[0]])`
  5. `T([[x],[y]])=[[2,0],[0,1]][[x],[y]]+[[2],[0]]`
Show Answers Only

`A`

Show Worked Solution

`y=cos(2x+4)=cos(2(x+2))`

♦♦♦ Mean mark 26%.

`text{Dilation of factor} \ 1/2 \ text{from} \ y text{-axis}.`

`text{Translation of 2 units to the left.}`
 

`T([[x],[y]])=[[(1)/(2),0],[0,1]][[x],[y]]+[[-2],[0]]=[[(1)/(2),0],[0,1]]([[x],[y]]+[[-4],[0]])`
 

`=>  A`

Functions, MET2 2021 VCAA 5 MC

Consider the following four functional relations.

  `f(x) = f(-x)\ \ \ \ \ -f(x) = f(-x)\ \ \ \ \ f(x) = -f(x)\ \ \ \ \ (f(x))^2 = f(x^2)`

The number of these functional relations that are satisfied by the function  `f : R -> R, \ f(x) = x`  is

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

`C`

Show Worked Solution

`text{Consider each relation}`

`f(x) = x \ ,` `f(–x) = –x \ \ ✘`
`–f(x) = –x  \ ,` ` f(-x) = –x \ \ ✓`
`f(x) = x \ ,` `–f(x) = –x \ \ ✘`
`(f(x))^2 = x^2 \ ,` `f(x^2) = x^2 \ \ ✓`

 

`=> C`

Algebra, MET1 2021 VCAA 5

Let  `f:R -> R, \ f(x) = x^2 - 4`  and  `g:R -> R, \ g(x) = 4(x - 1)^2 - 4`.

  1. The graphs of `f` and `g` have a common horizontal axis intercept at `(2, 0)`.
  2. Find the coordinates of the other horizontal axis intercept of the graph of `g`.   (2 marks)

  3. Let the graph of `h` be a transformation of the graph of `f` where the transformations have been applied in the following order:
    • dilation by a factor of  `1/2`  from the vertical axis (parallel to the horizontal axis)
    • translation by two units to the right (in the direction of the positive horizontal axis
  4. State the rule of `h` and the coordinates of the horizontal axis intercepts of the graph of `h`.   (2 marks)

Show Answers Only
  1. `(0,0)`
  2. `(1,0) and (3,0)`
Show Worked Solution

a.   `xtext(-axis intercept of)\ g(x):`

`4(x-1)^2-4` `=0`  
`(x-1)^2` `=1`  
`x-1` `=+-1`  

 
`text(When)\ \ x-1=-1\ \ =>\ \ x=0`

`:.\ text{Other horizontal intercept occurs at (0, 0)}`
 

b.   `text(1st transformation)`

♦♦ Mean mark part (b) 25%.

`text(Dilation by a factor of)\ 1/2\ text(from the)\ ytext(-axis:)`

`x^2 – 4 \ → \ (x/(1/2))^2 -4 = 4x^2-4`
 

`text(2nd transformation)`

`text(Translation by 2 units to the right:)`

`4x^2-4 \ → \ h(x) = 4(x-2)^2 – 4`
 

`xtext(-axis intercept of)\ h(x):`

`4(x-2)^2-4` `=0`  
`(x-2)^2` `=1`  
`x-2` `=+-1`  

 
`x-2=1 \ => \ x=3`

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

`:.\ text(Horizontal axis intercepts occur at)\ (1,0) and (3,0).`

Functions, MET2-NHT 2019 VCAA 17 MC

The graph of the function `g` is obtained from the graph of the function `f` with rule  `f(x) = cos(x) - (3)/(8)`  by a dilation of factor  `(4)/(pi)`  from the `y`-axis, a dilation of factor  `(4)/(3)`  from the `x`-axis, a reflection in the  `y`-axis and a translation of  `(3)/(2)`  units in the positive `y` direction, in that order.

The range and period of `g` are respectively

  1. `[–(1)/(3) , (7)/(3)] \ text(and) \ 2`
  2. `[–(1)/(3) , (7)/(3)] \ text(and) \ 8`
  3. `[–(7)/(3) , (1)/(3)] \ text(and) \ 2`
  4. `[–(7)/(3) , (1)/(3)] \ text(and) \ 8`
  5. `[–(4)/(3) , (4)/(3)] \ text(and) \ (pi^2)/(2)`
Show Answers Only

`B`

Show Worked Solution

`text(Dilation of) \ \ (4)/(pi) \ \ text(from) \ y text(-axis:)`

`cos x -(3)/(8) \ => \ cos ((pi x)/(4)) – (3)/(8)`
 
`text(Dilation of) \ \ (4)/(3) \ \ text(from) \ x text(-axis:)`

`cos((pi x)/(4)) – (3)/(8) \ => \ (4)/(3) (cos({pi x}/{4}) -(3)/(8)) = (4)/(3) \ cos((pi x)/(4)) – (1)/(2)`
 
`text(Reflection in) \ y text(-axis and translation up) \ (3)/(2) :`

`(4)/(3) cos((pi x)/(4)) – (1)/(2) \ => \ (4)/(3) \ cos((–pi x)/(4)) + 1`

`:. y = (4)/(3) \ cos ((–pi x)/(4)) + 1`
 
`text(Range:) \ [1 – (4)/(3) , 1 + (4)/(3)] = [–(1)/(3) , (7)/(3)]`
 
`text(Period:) \ (2 pi)/(n) = (pi)/(4) \ => \ n = 8`

`=> \ B`

Graphs, MET2 2019 VCAA 13 MC

The graph of the function  `f`  passes through the point  `(-2, 7)`.

If  `h(x) = f(x/2) + 5`, then the graph of the function  `h`  must pass through the point

  1. `(-1, -12)`
  2. `(-1, 19)`
  3. `(-4, 12)`
  4. `(-4, -14)`
  5. `(3, 3.5)`
Show Answers Only

`C`

Show Worked Solution

`h(x) = f(x/2) + 5`

`text(Dilate by a factor 2 from)\ y text(-axis)`

`(-2, 7) -> (-2 xx 2, 7) -= (-4, 7)`

`text(Translate up 5 vertically):`

`(-4, 7) -> (-4, 7 + 5) -= (-4, 12)`
 

`=>   C`

Graphs, MET2 2018 VCAA 4 MC

The point  `A (3, 2)`  lies on the graph of the function  `f`. A transformation maps the graph of  `f` to the graph of `g`,

where  `g(x) = 1/2 f(x - 1)`. The same transformation maps the point `A` to the point `P`.

The coordinates of the point `P` are

  1. `(2, 1)`
  2. `(2, 4)`
  3. `(4, 1)`
  4. `(4, 2)`
  5. `(4, 4)`
Show Answers Only

`C`

Show Worked Solution

`g(x) = 1/2 f(x – 1),\ A(3, 2)`

♦ Mean mark 48%.

`text(Dilate by a factor of)\ 1/2\ text(from)\ x text(-axis:)`

`A(3, 2) -> A′(3, 1)`
 

`text(Translate 1 unit to right:)`

`A′(3, 1) -> P(4, 1)`
 

`=>   C`

Graphs, MET2 2007 VCAA 15 MC

The graph of the function  `f: [0, oo) -> R`  where  `f(x) = 3x^(5/2)`  is reflected in the `x`-axis and then translated 3 units to the right and 4 units down.

The equation of the new graph is

`y = 3(x - 3)^(5/2) + 4`

`y = -3 (x - 3)^(5/2) - 4`

`y = -3 (x + 3)^(5/2) - 1`

`y = -3 (x - 4)^(5/2) + 3`

`y = 3(x - 4)^(5/2) + 3`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ y= 3x^(5/2)`

`text(Reflect in the)\ x text(-axis:)`

`y= – 3x^(5/2)`

 

`text(Translate 3 units to the right:)`

`y=- 3(x-3)^(5/2)`

 

`text(Translate 4 units down:)`

`y=- 3(x-3)^(5/2) – 4`

`=>   B`

Graphs, MET2 2008 VCAA 18 MC

Let  `f: [0, pi/2] -> R,\ f(x) = sin(4x) + 1.` The graph of  `f`  is transformed by a reflection in the `x`-axis followed by a dilation of factor 4 from the `y`-axis.

The resulting graph is defined by

  1. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = -1 - 4 sin (4x)`
  2. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (16x)`
  3. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = 1 - sin (x)`
  4. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = 1 - sin (4x)`
  5. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (x)`
Show Answers Only

`E`

Show Worked Solution

`f(x) = sin(4x)+1`

♦♦ Mean mark 36%.

`text(Reflecting in the)\ x text(-axis,)`

`=> h(x) = – sin (4x) – 1`

 

`text(Dilation by a factor of 4 from the)\ \ y text(-axis,)`

`=> g(x) = -sin(x)-1`

`text(Dilation factor: adjust the domain from)\ \ [0, pi/2]`

`text(to)\ \ [0xx4, pi/2 xx 4]=[0,2pi].`

`=>   E`

Graphs, MET2 2009 VCAA 19 MC

The graph of a function  `f`, with domain `R`, is as shown.

VCAA 2009 19mc

The graph which best represents  `1 - f (2x)`  is

VCAA 2009 19mci

VCAA 2009 19mcii

Show Answers Only

`E`

Show Worked Solution

`text(Determine transformations)`

`text(from)\ \ f -> – f (2x) + 1`

`text(- Dilate by factor)\ 1/2\ text(from)\ y text(-axis.)`

`text(- Reflect in)\ x text(-axis.)`

`text(- Shift up 1 unit.)`

`=>   E`

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`

Graphs, MET2 2014 VCAA 1 MC

The point `P\ text{(4, −3)}` lies on the graph of a function  `f`. The graph of  `f` is translated four units vertically up and then reflected in the `y`-axis.

The coordinates of the final image of `P` are

  1. `text{(−4, 1)}`
  2. `text{(−4, 3)}`
  3. `text{(0, −3)}`
  4. `text{(4, −6)}`
  5. `text{(−4, −1)}`
Show Answers Only

`A`

Show Worked Solution

`text(Using mapping notation:)`

`P(4,−3)\ {:(y + 4),(vec(quad(1)quad)):}\ (4,1)\ {:(−x),(vec(\ (2)\ )):}\ pprime(−4,1)`

`=>   A`