AQA C3 2007 June — Question 5 9 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2007
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind composite function expression
DifficultyModerate -0.3 This is a straightforward C3 composite and inverse functions question requiring standard techniques: stating range from a square root function, finding fg(x) by substitution, solving a simple equation, and finding an inverse by swapping and rearranging. All parts are routine textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.08h Integration by substitution

5 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{aligned} & \mathrm { f } ( x ) = \sqrt { x - 2 } \text { for } x \geqslant 2 \\ & \mathrm {~g} ( x ) = \frac { 1 } { x } \quad \text { for real values of } x , x \neq 0 \end{aligned}$$
  1. State the range of f .
    1. Find fg(x).
    2. Solve the equation \(\operatorname { fg } ( x ) = 1\).
  2. The inverse of f is \(\mathrm { f } ^ { - 1 }\). Find \(\mathrm { f } ^ { - 1 } ( x )\).

5 The functions $f$ and $g$ are defined with their respective domains by

$$\begin{aligned}
& \mathrm { f } ( x ) = \sqrt { x - 2 } \text { for } x \geqslant 2 \\
& \mathrm {~g} ( x ) = \frac { 1 } { x } \quad \text { for real values of } x , x \neq 0
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item State the range of f .
\item \begin{enumerate}[label=(\roman*)]
\item Find fg(x).
\item Solve the equation $\operatorname { fg } ( x ) = 1$.
\end{enumerate}\item The inverse of f is $\mathrm { f } ^ { - 1 }$. Find $\mathrm { f } ^ { - 1 } ( x )$.
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2007 Q5 [9]}}