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Functions, MET2 2024 VCAA 8 MC

Some values of the functions  \(f: R \rightarrow R\)  and  \(g: R \rightarrow R\)  are shown below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \ x \quad \rule[-1ex]{0pt}{0pt} & \quad \ 1 \quad \rule[-1ex]{0pt}{0pt} & \quad \ 2 \quad \rule[-1ex]{0pt}{0pt} & \quad \ 3 \quad \\
\hline
\rule{0pt}{2.5ex} \quad  f(x) \quad \rule[-1ex]{0pt}{0pt} & \quad \ 0 \quad \rule[-1ex]{0pt}{0pt} & \quad 4 \quad \rule[-1ex]{0pt}{0pt} & \quad 5 \quad \\
\hline
\rule{0pt}{2.5ex} \quad  g(x) \quad \rule[-1ex]{0pt}{0pt} & \quad 3 \quad \rule[-1ex]{0pt}{0pt} & \quad 4 \quad \rule[-1ex]{0pt}{0pt} & \quad -5 \quad \\
\hline
\end{array}

The graph of the function  \(h(x)=f(x)-g(x)\)  must have an \(x\)-intercept at

  1. \((2,0)\)
  2. \((3,0)\)
  3. \((4,0)\)
  4. \((5,0)\)
Show Answers Only

\(A\)

Show Worked Solution

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad  x \quad \rule[-1ex]{0pt}{0pt} & \quad  1 \quad \rule[-1ex]{0pt}{0pt} & \quad  2 \quad \rule[-1ex]{0pt}{0pt} & \quad  3 \quad \\
\hline
\rule{0pt}{2.5ex}   h(x)  \rule[-1ex]{0pt}{0pt} & \ \ -3 \ \ \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ \rule[-1ex]{0pt}{0pt} & \ \ 10 \ \ \\
\hline
\end{array}

\(\therefore\ x\text{-intercept}\ \rightarrow\ (2, 0)\)

\(\Rightarrow A\)

Functions, MET1 EQ-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.     \(g\circ h(x)\) \(=\sqrt{h(x)}\)
    \(=\sqrt{x^2-9}\)

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

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

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

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

Algebra, MET2 2023 VCAA 16 MC

Let \(f(x)=e^{x-1}\).

Given that the product function \(f(x)\times g(x)=e^{(x-1)^2}\), the rule for the function \(g\) is

  1. \(g(x)=e^{x-1}\)
  2. \(g(x)=e^{(x-2)(x-1)}\)
  3. \(g(x)=e^{(x+2)(x-1)}\)
  4. \(g(x)=e^{x(x-2)}\)
  5. \(g(x)=e^{x(x-3)}\)
Show Answers Only

\(B\)

Show Worked Solution
\(f(x)\times g(x)\) \(=e^{(x-1)^2}\)
\(e^{x-1}\times g(x)\) \(=e^{(x-1)^2}\)
\( g(x)\) \(=\dfrac{e^{(x-1)^2}}{e^{(x-1)}}\)
  \(=e^{(x-1)^2}\times e^{(x-1)^-1}\)
  \(=e^{x^2-3x+2}\)
  \(=e^{(x-2)(x-1)}\)

 
\(\Rightarrow B\)

Graphs, MET2 2023 VCAA 3 MC

Two function, \(p\) and \(q\), are continuous over their domains, which are \([-2, 3)\) and \((-1, 5]\), respectively.

The domain of the sum function  \(p+q\)  is

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

\(E\)

Show Worked Solution

\(\text{Domain of sum function = intersection of two domains.}\)

\([-2, 3)\cap(-1, 5]=(-1, 3)\)

\(\Rightarrow E\)


♦ Mean mark 47%.

Functions, MET2 2021 VCAA 10 MC

Consider the functions  `f(x) = sqrt{x+2}`  and  `g(x) = sqrt{1-2x}`, defined over their maximal domains.

The maximal domain of the function  `h = f + g`  is.

  1. `(–2, 1/2)`
  2. `[–2,∞)`
  3. `(–∞, –2) ∪ (1/2, ∞)`
  4. `[–2, 1/2]`
  5. `[–2, 1]`
Show Answers Only

`D`

Show Worked Solution

`f(x) \ = \ sqrt{x + 2} \ => \ text{domain} \ x ≥ -2`

`g(x) \ = \ sqrt{1-2x} \ => \ text{domain} \ x ≤ 1/2`

`text{Intersection of domains = domain} \ h(x)`

`:. \ h(x) ∈ [-2, 1/2]`

`=> D`

Algebra, MET1 2019 VCAA 8

The function  `f: R -> R, \ f(x)`  is a polynomial function of degree 4. Part of the graph of  `f`  is shown below.

The graph of  `f`  touches the `x`-axis at the origin.
 


 

  1. Find the rule of  `f`.   (1 mark) 

Let  `g`  be a function with the same rule as  `f`.

Let  `h: D -> R, \ h(x) = log_e (g(x))-log_e (x^3 + x^2)`, where  `D`  is the maximal domain of  `h`.

  1. State  `D`.   (1 mark)

  2. State the range of  `h`.   (2 marks)

Show Answers Only
  1. `f(x) = -4x^2(x^2-1)`
  2. `x in (-1, 1) text(\{0})`
  3. `h(x) in (-oo, 3 log_e 2) text(\ {)2 log_e 2 text(})`
Show Worked Solution
a.    `y` `= ax^2 (x-1)(x + 1)`
    `= ax^2 (x^2-1)\ …\ (1)`

 
`text(Substitute)\ (1/ sqrt 2, 1)\ text{into  (1):}`

`1 = a ⋅ (1/sqrt 2)^2 ((1/sqrt 2)^2-1)`

`1 = a(1/2)(-1/2)`

`a = -4`

`:. f(x) = -4x^2(x^2-1)`

 

b.    `g(x) > 0` `=> -4x^2 (x^2-1) > 0`
    `=> x in (-1, 1)\  text(\{0})`

`text(and)`

`x^3 + x^2 > 0 => text(true for)\ \ x in (-1 , 1)\ text(\{0})`

`:. D:\ x in (-1, 1)\ text(\{0})`

 

c.    `h(x)` `= log_e ((-4x^2(x^2-1))/(x^3 + x^2))`
    `= log_e ((-4x^2(x + 1)(x-1))/(x^2(x + 1)))`
    `= log_e (4(1-x))\ \ text(where)\ \ x in (-1, 1)\ text(\{0})`

  
`text(As)\ \ x -> -1,\ \ h(x) -> log_e 8 = 3 log_e 2`

`text(As)\ \ x -> 1,\ \ h(x) -> -oo`

`text(As)\ \ x -> 0,\ \ h(x) -> log_e 4 = 2 log_e 2`

`text{(}h(x)\ text(undefined when)\ \ x = 0 text{)}`
 

`:.\ text(Range)\ \ h(x) in (-oo, 3 log_e 2)\ text(\{) 2 log_e 2 text(})`

Algebra, MET2 2018 VCAA 10 MC

The function `f` has the property  `f (x + f (x)) = f (2x)`  for all non-zero real numbers `x`.

Which one of the following is a possible rule for the function?

  1. `f(x) = 1 - x`
  2. `f(x) = x - 1`
  3. `f(x) = x`
  4. `f(x) = x/2`
  5. `f(x) = (1 - x)/2`
Show Answers Only

`C`

Show Worked Solution

`text(By trial and error,)`

`text(Consider option C:)`

`x + f(x)` `= x + x = 2x`
`f(2x)` `=2x`
`:. f(x + f(x))` `= f(2x)`

 
`=>   C`

Algebra, MET2 2017 VCAA 13 MC

Let  `h:(−1,1) -> R`,  `h(x) = 1/(x - 1)`.

Which one of the following statements about `h` is not true?

  1. `h(x)h(–x) = –h(x^2)`
  2. `h(x) + h(–x) = 2h(x^2)`
  3. `h(x) - h(0) = xh(x)`
  4. `h(x) - h(–x) = 2xh(x^2)`
  5. `(h(x))^2 = h(x^2)`
Show Answers Only

`E`

Show Worked Solution

`text(By trial and error, consider option)\ E:`

♦ Mean mark 46%.

`h(x) = 1/(x – 1)`

`(h(x))^2 = 1/(x – 1)^2=1/(x^2-2x+1)`

`h(x^2)=1/(x^2-1) != (h(x))^2`

 

`=> E`

Algebra, MET2 2008 VCAA 12 MC

Let  `f: R -> R,\ f(x) = e^x + e^(–x).`

For all  `u in R,\ f(2u)`  is equal to

  1. `f(u) + f(-u)`
  2. `2 f(u)`
  3. `(f(u))^2 - 2`
  4. `(f(u))^2`
  5. `(f(u))^2 + 2`
Show Answers Only

`C`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 44%.

`text(Define)\ \ f(x) = e^x + e^-x`

`text(Enter each functional equation)`

`[text(i.e.)\ \ f(2u) = (f(u))^2 – 2]`

`text(until CAS output is “true”)`

`=>   C`

 

`text(Solution 2)`

`f(2u)` `=e^(2u) + e^(-2u)`
`(f(u))^2` `=(e^u + e^(-u))^2`
  `=e^(2u) + 2 + e^(-2u)`
   

`:. f(2u) = (f(u))^2-2`

`=>C`

Algebra, MET2 2009 VCAA 5 MC

Let  `f: R -> R,\ f (x) = x^2`

Which one of the following is not true?

  1. `f(xy) = f (x) f (y)`
  2. `f(x) - f(-x) = 0`
  3. `f (2x) = 4 f (x)`
  4. `f (x - y) = f(x) - f(y)`
  5. `f (x + y) + f (x - y) = 2 (f (x) + f(y))`
Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

`text(Consider option)\ D:`

`f(x-y)` `=(x-y)^2`
  `=x^2 -2xy+y^2`
`f(x)-f(y)` `= x^2-y^2`
`:.f(x-y)` `!=f(x)-f(y)`

 
`=>D`

 

`text(Solution 2)`

`text(Define)\ \ f(x) = x^2`

`text(Enter each functional equation on CAS)`

`text(until output does NOT read “true”.)`

`=>   D`

Algebra, MET2 2016 VCAA 11 MC

The function  `f` has the property  `f(x) - f(y) = (y - x)\ f(xy)`  for all non-zero real numbers `x` and `y`.

Which one of the following is a possible rule for the function?

  1. `f(x) = x^2`
  2. `f(x) = x^2 + x^4`
  3. `f(x) = x log_e (x)`
  4. `f(x) = 1/x`
  5. `f(x) = 1/x^2`
Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 47%.

`text(Consider option)\ D:`

`text(LHS)\ = 1/x – 1/y`

`text(RHS)\ =(y-x) xx 1/(xy) = 1/x – 1/y =\ text(LHS)`

`=>D`

 

`text{Solution 2 (using technology)}`

`text(Define each specific function on CAS)`

`[text(i.e.)\ \ f(x) = 1/x]`

`text(Enter functional equation)\ \ f(x) – f(y) = (y – x)\ f(xy)`

`text(unitl CAS output is “TRUE”)`

`=>   D`

Algebra, MET2 2013 VCAA 13 MC

If the equation  `f(2x) - 2f(x) = 0`  is true for all real values of `x`, then the rule for  `f` could be

  1. `x^2/2`
  2. `sqrt (2x)`
  3. `2x`
  4. `log_e (x/2)`
  5. `x - 2`
Show Answers Only

`C`

Show Worked Solution

`text(We need)\ \ f(2x)=2\ f(x),`

`text(Consider)\ C,`

`f(x)` `=2x,`
`f(2x)` `= 2(2x)`
  `= 2\ f(x)`

 

`text(CAS can be used to quickly prove other answers)`

`text(do not satisfy equation.)`

`=>   C`

Algebra, MET2 2013 VCAA 5 MC

If   `f: text{(−∞, 1)} -> R,\ \ f(x) = 2 log_e (1 - x)\ \ text(and)\ \ g: text{[−1, ∞)} -> R, g(x) = 3 sqrt (x + 1),`  then the maximal domain of the function   `f + g`  is

  1. `text{[−1, 1)}`
  2. `(1, oo)`
  3. `text{(−1, 1]}`
  4. `text{(−∞, −1]}`
  5. `R`
Show Answers Only

`A`

Show Worked Solution

`text(Consider)\ \ f(x) = 2 log_e (1 – x):`

`(1-x)` `>0`
`:. x` `<1`

 

`text(Consider)\ \ g(x) = 3 sqrt (x + 1):`

`(x+1)` `>=0`
`:. x` `>= -1`

 

`:.\ text(The maximal domain of)\ \ f + g\ \ text{is [−1, 1)}.`

`=>   A`