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Functions, 2ADV 2026 HSC 10 MC

It is given that  \(h(x)=\sin^2 x-\sin x\)  is a composite function of \(f(x)\) and \(g(x)\) where  \(f(x)=x^2-x\)  and  \(g(x)=\sin x\)  for all real values of \(x\).

For what values of \(k\) does the equation  \(\sin^2 x-\sin x=k\)  have real solutions?

  1. \(-1 \le k \le 1\)
  2. \(-1 \le k \le 2\)
  3. \(-\dfrac{1}{4} \le k \le 1\)
  4. \(-\dfrac{1}{4} \le k \le 2\)
Show Answers Only

\(D\)

Show Worked Solution

\(f(x)=x^2-x,\ \ g(x)=\sin x\ \ \Rightarrow\ \ f(g(x))=\sin^2 x-\sin x\)

\(h(x)=f(g(x))\)

\(\text{Since}\ \ g(x)=\sin x\ \ \Rightarrow\ \ -1 \leq g(x) \leq 1\)
 

\(\text{Consider}\ \ f(u)=u^2-u\ \ \text{for}\ \ -1 \le u \le 1:\)

\(\text{Vertex at}\ \ u=\dfrac{1}{2}:\ \ f\left(\dfrac{1}{2}\right)=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\)

\(\text{Endpoints:}\ \ f(-1)=2, \ \ f(1)=0\)
 

\(\therefore\ \text{Real solutions exist for}\ \ -\dfrac{1}{4} \le k \le 2\)

\(\Rightarrow D\)

Functions, 2ADV F1 2025 HSC 18

Find the range of \(g(f(x))\),  given  \(f(x)=\dfrac{3}{x-1}\)  and  \(g(x)=x+5\).   (2 marks)

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\(\text{Range} \ g(f(x)): \text{All} \  y, \  y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Show Worked Solution

\(f(x)=\dfrac{3}{x-1}, \ \ g(x)=x+5\)

\(g(f(x))=\dfrac{3}{x-1}+5 \Rightarrow \ \text{vertical translation +5 of} \ f(x)\)

\(\text{Range} \ f(x):\ \text{All} \ y, \ y \neq 0\)

\(\text{Range} \ g(f(x)): \text{All} \  y, \  y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Functions, 2ADV F1 EQ-Bank 23

Given that  \(f(x)=x^2+1\)  and  \(g(x)=x+2\),  determine \(f(g(x))\) and its range.   (2 marks)

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\(f(g(x))=(x+2)^2+1\)

\(\ \text{Range} \  f(g(x)): \ y \geqslant 1\)

Show Worked Solution

\(f(x)=x^2+1, \quad g(x)=x+2\)

\(f(g(x))=(x+2)^2+1\)

\(\Rightarrow f(g(x)) \ \text{is a concave up parabola, vertex}\ (-2,1)\)

\(\therefore \ \text{Range} \  f(g(x)): \ y \geqslant 1\)

Functions, 2ADV F1 EQ-Bank 18

If  \(f(x)=x^2-3\)  and  \(g(x)=\sqrt{x-2}\),

  1. Find  \(f(g(x))\).   (1 mark)
  2. What is the domain of \(f(g(x))\) ?   (1 mark)

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a.  \(f(g(x))=x-5\)

b.   \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\)

\(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\)

Show Worked Solution

a.    \(f(g(x))\) \(=(\sqrt{x-2})^2-3\)
    \(=x-2-3\)
    \(=x-5\)

 

b.   \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\)

\(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\)

Functions, 2ADV F1 EQ-Bank 17

Given  `f(x) = sqrtx`  and  `g(x) = 25-x^2`

  1. Find `g(f(x))`.   (1 mark)

  2. Find the domain and range of `f(g(x))`.   (2 marks)

Show Answers Only

a.    `25-x`

b.    `text(Domain:)\ −5<= x <= 5`

`text(Range:)\ 0<=y<= 5`

Show Worked Solution

a.    `g(f(x))= 25-(f(x))^2= 25-(sqrtx)^2= 25-x`
 

b.    `f(g(x))= sqrt(g(x))= sqrt(25-x^2)`

`:.\ text(Domain:)\ −5<= x <= 5`

`:.\ text(Range:)\ 0<=y<= 5`

Functions, 2ADV F1 EQ-Bank 30

Given  `f(x) = sqrt (x^2 - 9)`  and  `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 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 <=x<=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(Domain:)\ \ x<=-8\ \ and\ \  x>=-2`

Functions, 2ADV F1 EQ-Bank 26

Let  `f(x) = log_e(x)`  for  `x>0,` and  `g (x) = x^2 + 1`  for  all `x`.

  1. Find `h(x)`, where  `h(x) = f (g(x))`.   (1 mark)

  2. State the domain and range of `h(x)`.   (2 marks)

  3. Show that  `h(x) + h(−x) = f ((g(x))^2 )`.   (2 marks)
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a.    `log_e(x^2 + 1)`

b.    text(Domain)\ (h):\ text(all)\ x`

`text(Range)\ h(x):\ \  h>=0`

c.    `text(See Worked Solutions)`

Show Worked Solution
a.    `h(x)` `= f(x^2 + 1)`
    `= log_e(x^2 + 1)`

 

b.   `text(Domain)\ (h) =\ text(Domain)\ (g):\ text(all)\ x`

♦♦ Mean mark part (a)(ii) 30%.
`=> x^2 + 1 >= 1`
`=> log_e(x^2 + 1) >= 0`

 
`:.\ text(Range)\ h(x):\ \  h>=0`

 

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.
c.   `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)`

Functions, 2ADV F1 EQ-Bank 25

Let  `f(x) = sqrt(x + 1)`   for   `x>=0`

  1. State the range of  `f(x)`.   (1 mark)

  2. Let  `g(x)=x^2+4x+3`,  where  `x<=c`  and  `c<=0`.
  3. Find the largest possible value of `c` such that the range of `g(x)` is a subset of the domain of `f(x)`.   (2 marks)

Show Answers Only

a.    `y>= 1`

 b.    `-3`

Show Worked Solution

a.  `text(Sketch of)\ \ f(x):`
 

`:.\ text(Range:)\ \ y>= 1`
 

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

`text(Domain of)\ \ f(x):\ \ x>=0`

`text(Find domain of)\ g(x)\ text(such that range)\ g(x):\ y>=0`

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

`x <= –3\ \ text(and)\ \ x>=-1`
 

`:. c = -3`