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Functions, MET1 2023 VCE SM-Bank 2

Consider the functions \(f\) and \(g\), where

\begin{aligned}
& f: R \rightarrow R, f(x)=x^2-9 \\
& g:[0, \infty) \rightarrow R, g(x)=\sqrt{x}
\end{aligned}

  1. State the range of \(f\).  (1 mark)

  2. Determine the rule for the equation and state the domain of the function \(f \circ g\).  (2 marks)

  3. Let \(h\) be the function \(h: D \rightarrow R, h(x)=x^2-9\).
  4. Determine the maximal domain, \(D\), such that \(g \circ h\) exists.  (1 marks)

Show Answers Only

a.    \([-9, \infty)\)

b.    \(f\circ g(x)=x-9, \text{Domain}\ [0, \infty)\)

c.    \((-\infty, -3)\cap (3, \infty)\)

Show Worked Solution

a.    \(\text{Range}\ \rightarrow\  [-9, \infty)\)

b.     \(f\circ g(x)\) \(=(g(x))^2-9\)
    \(=(\sqrt{x})^2-9\)
    \(=x-9\)

\(g(x)=\sqrt{x} \ \rightarrow x\ \text{must be }\geq 0\)

\(\therefore\ \text{Domain}\ f\circ g(x) \text{ is }[0, \infty)\)

c.     \(g\circ h(x)\) \(=\sqrt{h(x)}\)
    \(=\sqrt{x^2-9}\)

\(\text{For }g\circ h(x)\ \text{to exist}\ h(x)\geq 0\)

\(x\text{-intercepts for }h(x)\ \text{are } x=-3, 3\)

\(\text{and }h(x)\ \text{is positive for } x\leq -3\ \text{and }x\geq 3\)

\(\therefore\ \text{Maximal domain} = (-\infty, -3)\cap (3, \infty)\)

Calculus, MET2 2023 VCAA 1

Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.

  1. State the coordinates of all axial intercepts of \(f\).   (1 mark)
  2. Find the coordinates of the stationary points of \(f\).   (2 marks)
    1. Let \(g:R\rightarrow R, g(x)=x-2\).
    2. Find the values of \(x\) for which \(f(x)=g(x)\).   (1 mark)
    1. Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\).  (2 marks)
    2. Hence, find the total area of the regions bound by \(f\) and \(g\), correct to two decimal places.   (1 mark)
  1. Let \(h:R\rightarrow R, h(x)=(x-a)(x-b)^2\), where \(h(x)=f(x)+k\) and \(a, b, k \in R\).
  2. Find the possible values of \(a\) and \(b\).   (4 marks)
Show Answers Only

a.    \((-1, 0), (0, 0), (2, 0)\)

b.    \(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)

c.i.  \(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

c.ii. \(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

 \(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)

c.iii. \(5.95\)

d.   \(\text{1st case }\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case }\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

b.    \(\text{Using CAS solve for}\ x:\)

\(\dfrac{d}{dx}(x(x-2)(x+1))=0\)

\(\therefore\ x=\dfrac{1-\sqrt{7}}{3}\ \text{and }x=\dfrac{1+\sqrt{7}}{3}\)

\(\text{Substitute }x\ \text{values into }f(x)\ \text{using CAS to get}\ y\ \text{values}\)

\(\text{The stationary points of }f\ \text{are}:\)

\(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
  

ci    \(\text{Given }f(x)=g(x)\)

\(x(x-2)(x+1)\) \(=x-2\)
\(x(x-2)(x+1)(x-2)\) \(=0\)
\((x-2)(x(x+1)-1)\) \(=0\)
\((x-2)(x^2+x-1)\) \(=0\)

  
\(\therefore\ \text{Using CAS: } \)

\(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

cii  \(\text{Area of bounded region:}\)

\(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
  

ciii  \(\text{Solve the integral in c.ii above using CAS:}\)
  \(\text{Total area}=5.946045..\approx 5.95\)

  

d.   \(\text{Method 1 – Equating coefficients}\)

\((x-a)(x-b)^2=x(x-2)(x+1)+k\)

\(x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k\)

\((x^3-(a+2b)x^2+(2ab+b^2)x-ab^2=x^3-x^2-2x+k\)

\(\therefore\ -(a+2b)=-1\ \to\ a=1-2b …(1)\)

\(2ab+b^2=-2\ \ …(2)\)

\(\text{Substitute (1) into (2) and solve for }b.\)

\(2b(1-2b)+b^2\) \(=-2\)
\(3b^2-2b-2\) \(=0\)
\(b\) \(=\dfrac{1\pm \sqrt{7}}{3}\)
\(\text{When }b\) \(=\dfrac{1+\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1+\sqrt{7}}{3}\Bigg)=\dfrac{-2\sqrt{7}+1}{3}\)
\(\text{When }b\) \(=\dfrac{1-\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1-\sqrt{7}}{3}\Bigg)=\dfrac{2\sqrt{7}+1}{3}\)

  

\(\text{Method 2 – Using transformations}\)

\(\text{The squared factor in }(x-a)(x-b)^2=x(x-2)(x+1)+k,\)

\(\text{shows that the turning point is on the }x\ \text{axis}.\)

\(\therefore\ \text{Lowering }f(x)\ \text{by }\dfrac{2(7\sqrt{7}-10)}{27}\ \text{and raising }f(x)\ \text{by }\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\text{will give the 2 possible sets of values for }a\ \text{and}\ b.\)

\(\text{1st case – lowering using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)-\dfrac{2(7\sqrt{7}-10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case – raising using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)+\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

 

Functions, MET2 2023 VCAA 20 MC

Let \(f(x)=\log_{e}\Bigg(x+\dfrac{1}{\sqrt{2}}\Bigg)\).

Let \(g(x)=\sin(x)\) where \(x\in (-\infty, 5)\).

The largest interval of \(x\) values for which \((f\circ g)(x)\) and \((g\circ f)(x)\) both exist is
 

  1. \(\Bigg(-\dfrac{1}{\sqrt{2}},\dfrac{5\pi}{4}\Bigg)\)
  2. \(\Bigg[-\dfrac{1}{\sqrt{2}},\dfrac{5\pi}{4}\Bigg)\)
  3. \(\Bigg(-\dfrac{\pi}{4},\dfrac{5\pi}{4}\Bigg)\)
  4. \(\Bigg[-\dfrac{\pi}{4},\dfrac{5\pi}{4}\Bigg]\)
  5. \(\Bigg[-\dfrac{\pi}{4},-\dfrac{1}{\sqrt{2}}\Bigg]\)
Show Answers Only

\(A\)

Show Worked Solution

\(f(x)=\log_e\Bigg(x+\dfrac{1}{\sqrt{2}}\Bigg)\ \text{and }g(x)=\sin(x)\ \text{for}\ x\in (-\infty, 5)\)

\(1.\  \ (f \circ g)(x)=\log_e\Bigg(\sin(x)+\dfrac{1}{\sqrt{2}}\Bigg)\)

\(\to\ \) \(\sin(x)+\dfrac{1}{\sqrt{2}}\) \(>0\)
  \(\sin(x)\) \(>-\dfrac{1}{\sqrt{2}}\)
  \(\therefore\ x\) \(\in\Bigg(-\dfrac{\pi}{4}, \dfrac{5\pi}{4}\Bigg)\cup \Bigg(-\dfrac{9\pi}{4}, -\dfrac{9\pi}{4}\Bigg)\cup\dots\ = \Bigg(-\dfrac{\pi}{4}+2\pi k, \dfrac{5\pi}{4}+2\pi k\Bigg)\)

  
\(2.\  \ (g \circ f)(x)=\sin\Bigg(\log_e\Bigg(x+\dfrac{1}{\sqrt{2}}\Bigg)\Bigg)\)

\(\to\ \) \(\log_e\Bigg(x+\dfrac{1}{\sqrt{2}}\Bigg)\) \(<5\)
  \(x\) \(=e^5-\dfrac{1}{\sqrt{2}}\)
  \(\therefore\ x\) \(\in \bigg(-\dfrac{1}{\sqrt{2}},e^5-\dfrac{1}{\sqrt{2}}\Bigg)\) 

  
\(\text{Largest interval of }x\ \text{for which both }(f \circ g)(x)\ \text{and }(g \circ f)(x)\text{ exist is:}\)

\(\Bigg(-\dfrac{\pi}{4}+2\pi k, \dfrac{5\pi}{4}+2\pi k\Bigg)\cap \bigg(-\dfrac{1}{\sqrt{2}},e^5-\dfrac{1}{\sqrt{2}}\Bigg)\) 

\(=\bigg(-\dfrac{1}{\sqrt{2}}, \dfrac{5\pi}{4}\Bigg)\) 

 

\(\Rightarrow A\)


♦♦ Mean mark 30%.
MARKER’S COMMENT: 26% incorrectly chose C.

Functions, MET2 2021 VCAA 9 MC

Let  `g(x) = x + 2`  and  `f(x) = x^2 - 4`

If `h` is the composite function given by  `h : [–5, –1) to R, h(x) = f(g(x))`, then the range of `h` is

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

`E`

Show Worked Solution
`h(x)` `= (x + 2)^2 – 4`
  `= x^2 + 4x`

 
`text{By CAS, graph} \ \ y = x^2 + 4x , \ x∈ [–5, –1)`

`text{Min at} \ (–2, –4)`

`text{Max at} \ (–5, 5)`

`:. \ text{Range of} \ h(x) ∈ [–4, 5]`

`=> E`

Functions, MET1-NHT 2018 VCAA 2

Let  `f(x) = -x^2 + x + 4`  and  `g(x) = x^2 - 2`.

  1. Find  `g(f(3))`.  (2 marks)
  2. Express the rule for  `f(g(x))`  in the form  `ax^4 + bx^2 + c`, where  `a`, `b`  and  `c`  are non-zero integers.  (2 marks)
Show Answers Only
  1. `2`
  2. `-x^4 + 5x^2 – 2`
Show Worked Solution
a.    `f(3)` `= -3^2 + 3 + 4`
  `= -2`

 

`g(f(3))` `= g(-2)`
  `= (-2)^2 – 2`
  `= 2`

 

b.    `f(g(x))` `= -(x^2 – 2)^2 + (x^2 – 2) + 4`
  `= -(x^4 – 4x^2 + 4) + x^2 + 2`
  `= -x^4 + 5x^2 – 2`

Algebra, MET2 2018 VCAA 6 MC

Let  `f` and `g` be two functions such that  `f(x) = 2x`  and  `g(x + 2) = 3x + 1`.

The function  `f (g(x))`  is

  1. `6x - 5`
  2. `6x + 1`
  3. `6x^2 + 1`
  4. `6x - 10`
  5. `6x + 2`
Show Answers Only

`D`

Show Worked Solution
`g(x + 2)` `= 3x + 1`
`g((x – 2) + 2)` `= 3(x – 2) + 1`
`g(x)` `= 3x – 5`

 

`f(x)` `= 2x`
`f(g(x))` `= 2(3x – 5)`
  `= 6x – 10`

 
`=>   D`

Functions, MET1 2017 VCAA 7

Let  `f: [0, oo) -> R,\ f(x) = sqrt(x + 1)`.

  1.  State the range of  `f`.  (1 mark)
  2.  Let  `g: (-oo, c] -> R,\ \ g(x) = x^2 + 4x + 3`.
  3.  i. Find the largest possible value of `c` such that the range of `g` is a subset of the domain of  `f`.  (2 marks)
  4. ii. For the value of `c` found in part b.i., state the range of  `f(g(x))`.  (1 mark) 
  5. Let  `h: R -> R,\ \ h(x) = x^2 + 3`.
  6. State the range of  `f(h(x))`.  (1 mark)
Show Answers Only
  1. `[1, oo)`
    1. `-3`
    2. `[1, oo)`
  3. `[2, oo)`
Show Worked Solution

a.  `text(Sketch of)\ \ f(x):`
 

`:.\ text(Range)\ \ (f) = [1, oo)`

 

b.i.  `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`

♦ Mean mark 38%.
 


 

`text(Domain of)\ \ f(x)=[0,oo)`

`text(Find domain of)\ \ g(x)\ \ text(such that Range)\ (g) = [0, oo)`

`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`

`=> x ∈ (–oo, –3] ∪ [–1,oo)`

`:. c = -3`

 

b.ii.  `text(Range)\ g(x) = [0, oo) = text(Domain)\ \ f(x)`

♦♦♦ Mean mark 20%.

`:.\ text(Range)\ \ f(g(x)) = [1, oo)`

 

c.  `text(Range)\ h(x) = [3, oo)`

♦♦ Mean mark 30%.
`f(3)` `= sqrt (3 + 1)`
  `= sqrt 4`
  `= 2`

 

`:.\ text(Range)\ \ f(h(x)) = [2, oo)`

Algebra, MET2 2007 VCAA 2 MC

Let  `g(x) = x^2 + 2x - 3 and f(x) = e^(2x + 3).`

Then  `f(g(x))`  is given by 

  1. `e^(4x + 6) + 2 e^(2x + 3) - 3`
  2. `2x^2 + 4x - 6`
  3. `e^(2x^2 + 4x + 9)`
  4. `e^(2x^2 + 4x - 3)`
  5. `e^(2x^2 + 4x - 6)`
Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

`text(Define)\ \ f(x) and g(x)\ \ text(on CAS)`

`f(g(x)) = e^(2x^2 + 4x – 3)`

`=>   D`
 

`text(Solution 2)`

`f(g(x))` `=e^(2 xx (x^2 + 2x – 3)+3)`
  `= e^(2x^2 + 4x – 3)`

`=>D`

Graphs, MET1 2016 VCAA 5

Let  `f : (0, ∞) → R`, where  `f(x) = log_e(x)`  and  `g: R → R`, where  `g (x) = x^2 + 1`.

  1.   i. Find the rule for `h`, where  `h(x) = f (g(x))`.  (1 mark)
  2.  ii. State the domain and range of `h`.  (2 marks)
  3. iii. Show that  `h(x) + h(−x) = f ((g(x))^2 )`.  (2 marks)
  4. iv. Find the coordinates of the stationary point of `h` and state its nature.  (2 marks)
     
  5. Let  `k: (– ∞, 0] → R`  where  `k (x) = log_e(x^2 + 1)`.
  6.  i. Find the rule for  `k^(−1)`.  (2 marks)
  7. ii. State the domain and range of  `k^(– 1)`.  (2 marks)
Show Answers Only
  1.   i. `log_e(x^2 + 1)`
  2.  ii. `[0,∞)`
  3. iii. `text(See Worked Solutions)`
  4. iv. `(0, 0)`
  5.  i. `−sqrt(e^x – 1)`
  6. ii. `text(Domain)\ (k) = (−∞,0]`
  7.    `text(Range)\ (k) = [0,∞)`
Show Worked Solution
a.i.    `h(x)` `= f(x^2 + 1)`
    `= log_e(x^2 + 1)`

 

a.ii.   `text(Domain)\ (h) =\ text(Domain)\ (g) = R`

♦♦ Mean mark part (a)(ii) 30%.
  `text(For)\ x ∈ R` `-> x^2 + 1 >= 1`
  `-> log_e(x^2 + 1) >= 0`

`:.\ text(Range)\ (h) = [0,∞)`

 

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.
a.iii.   `text(LHS)` `= h(x) + h(−x)`
    `= log_e(x^2 _ 1) + log_e((−x)^2 + 1)`
    `= log_e(x^2 + 1) + log_e(x^2 + 1)`
    `= 2log_e(x^2 + 1)`

 

`text(RHS)` `= f((x^2 + 1)^2)`
  `= 2log_e(x^2 + 1)`

 

`:. h(x) + h(−x) = f((g(x))^2)\ \ text(… as required)`

 

a.iv.   `text(Stationary points when)\ \ h′(x) = 0`

♦ Mean mark part (a)(iv) 41%.
MARKER’S COMMENT: Solving a fraction is zero

   `text(Using Chain Rule:)`

`h′(x)` `= (2x)/(x^2 + 1)`

`:.\ text(S.P. when)\ \ x=0`

 

`text(Find nature using 1st derivative test:)`

`:.\ text{Minimum stationary point at (0, 0)}.`

 

b.i.   `text(Let)\ \ y = k(x)`

♦ Mean mark (b)(i) 49%.
MARKER’s COMMENT: Many students failed to consider the restrictions on the domain in `k(x)` and only select the negative root.

  `text(Inverse: swap)\ x ↔ y`

`x` `= log_e(y^2 + 1)`
`e^x` `= y^2 + 1`
`y^2` `= e^x – 1`
`y` `= ±sqrt(e^x – 1)`

 

`text(But range)\ \ (k^(−1)) =\ text(domain)\ (k)`

`:.k^(−1)(x) = −sqrt(e^x – 1)`

♦ Mean mark part (b)(ii) 44%.

 

b.ii.   `text(Range)\ (k^(−1)) =\ text(Domain)\ (k) = (−∞,0]`

  `text(Domain)\ (k^(−1)) =\ text(Range)\ (k) = [0,∞)`

Algebra, MET2 2010 VCAA 4 MC

If  `f(x) = 1/2e^(3x)  and  g(x) = log_e(2x) + 3`  then  `g (f(x))` is equal to
 

  1. `2x^3 + 3`
  2. `e^(3x) + 3`
  3. `e^(8x + 9)`
  4. `3(x + 1)`
  5. `log_e (3x) + 3`
Show Answers Only

`D`

Show Worked Solution

`text(Define)\ \ f(x)= 1/2e^(3x), \ g(x)= log_e(2x) + 3`

`g(f(x))` `= log_e(2 xx 1/2e^(3x)) + 3`
  `=log_e e^(3x) + 3`
  `=3x + 3`
  `= 3 (x + 1)`

 
`=>   D`

Functions, MET1 2008 VCAA 10

Let  `f: R -> R,\ \ f(x) = e^(2x)-1`.

  1. Find the rule and domain of the inverse function  `f^(−1)`.  (2 marks)
  2. On the axes provided, sketch the graph of  `y = f(f^(−1)(x))`  for its maximal domain.  (1 mark)
     

     

    met1-2008-vcaa-q2
     

  3. Find  `f(−f^(−1)(2x))`  in the form  `(ax)/(bx + c)`  where `a`, `b` and `c` are real constants.  (2 marks)
Show Answers Only
  1. `f^(−1)(x) = 1/2log_e(x + 1), x ∈ (−1,∞)`
  2.  
    met1-2008-vcaa-q10-answer
  3. `f(−f^(−1)(2x)) = (−2x)/(2x + 1)`
Show Worked Solution

a.   `text(Let)\ \ y = f(x)`

`text(For Inverse, swap)\ x ↔ y`

`x` `= e^(2y)-1`
`x + 1` `= e^(2y)`
`2y` `= log_e(x + 1)`
`y` `= 1/2 log_e(x + 1)`
`text(Domain)(f^(−1))` `=\ text(Range)\ (f)`
  `= (−1,∞)`

 

`:. f^(−1)(x) = 1/2log_e(x + 1),quadx ∈ (−1,∞)`

 

b.   `f(f^(−1)(x)) = x`

♦ Mean mark part (b) 19%.
MARKER’S COMMENT: Few students were aware that a function of its own inverse function is the line  `y=x`  over the appropriate domain.

`text(Domain is)\ \ (-1, oo)`

met1-2008-vcaa-q10-answer

 

♦ Mean mark 34%.
c.    `−f^(−1)(2x)` `= −1/2 ln(2x + 1)`
  `:. f(−f^(−1)(2x))` `= e^(-log_e(2x + 1))-1`
    `=(2x+1)^-1 -1`
    `= 1/(2x + 1)-1`
    `= (−2x)/(2x + 1)`

Functions, MET1 2011 VCAA 4

If the function `f` has the rule  `f(x) = sqrt (x^2 - 9)`  and the function `g` has the rule  `g(x) = x + 5`

  1.  find integers `c` and `d` such that  `f(g(x)) = sqrt {(x + c) (x + d)}`(2 marks)
  2.  state the maximal domain for which  `f(g(x))`  is defined.  (2 marks)
Show Answers Only
  1. `c = 2, d = 8 or c = 8, d = 2`
  2. `x in (– oo, – 8] uu [– 2, oo)`
Show Worked Solution
a.   `f(g(x))` `= sqrt {(x + 5)^2 – 9}`
    `= sqrt (x^2 + 10x + 16)`
    `= sqrt {(x + 2) (x + 8)}`

 
 `:. c = 2, d = 8 or c = 8, d = 2`

 

b.   `text(Find)\ x\ text(such that:)`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: “Very poorly answered” with a common response of `[-3,3)` that ignored the information from part (a).

`(x+2)(x+8) >= 0`
 

 vcaa-2011-meth-4ii

`(x + 2) (x + 8) >= 0\ \ text(when)`

`x <= -8 or x >= -2`

`:.\ text(Maximal domain is:)`

`x in (– oo, – 8] uu [– 2, oo)`