| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Multiple separate transformations (sketch-based, modulus involved) |
| Difficulty | Standard +0.2 This is a straightforward C3 transformations question requiring standard techniques: sketching inverse and composite transformations with axis intercepts, finding intercepts by substitution, and finding an inverse function of a simple square root function. All steps are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x)1.02x Combinations of transformations: multiple transformations |
| Answer | Marks |
|---|---|
| (i) Graph with curve through \((b, 0)\) and \((0, a)\), shape ∩ | M1 A2 |
| (ii) Graph with curve through \((\frac{1}{a}, 0)\) and \((0, 2b)\), shape ∪ | M1 A2 |
| Answer | Marks |
|---|---|
| \(x = 0 ⟹ y = -1\) ∴ \(b = -1\) | B1 |
| \(y = 0 ⟹ 2 - \sqrt{x+9} = 0\) | |
| \(x = 2^2 - 9 = -5\) ∴ \(a = -5\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(y = 2 - \sqrt{x+9}, \sqrt{x+9} = 2 - y\) | M1 |
| \(x + 9 = (2-y)^2\) | |
| \(x = (2-y)^2 - 9\) | M1 A1 |
| \(f^{-1}(x) = (2-x)^2 - 9\) | |
| \(f(-9) = 2\) ∴ domain of \(f^{-1}(x)\) is \(x \in \mathbb{R}, x \leq 2\) | M1 A1 |
| (14) |
**Part (a)**
(i) Graph with curve through $(b, 0)$ and $(0, a)$, shape ∩ | M1 A2 |
(ii) Graph with curve through $(\frac{1}{a}, 0)$ and $(0, 2b)$, shape ∪ | M1 A2 |
**Part (b)**
$x = 0 ⟹ y = -1$ ∴ $b = -1$ | B1 |
$y = 0 ⟹ 2 - \sqrt{x+9} = 0$ |
$x = 2^2 - 9 = -5$ ∴ $a = -5$ | M1 A1 |
**Part (c)**
$y = 2 - \sqrt{x+9}, \sqrt{x+9} = 2 - y$ | M1 |
$x + 9 = (2-y)^2$ |
$x = (2-y)^2 - 9$ | M1 A1 |
$f^{-1}(x) = (2-x)^2 - 9$ |
$f(-9) = 2$ ∴ domain of $f^{-1}(x)$ is $x \in \mathbb{R}, x \leq 2$ | M1 A1 |
| (14) |
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**Total: (75)**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a36989df-555f-4727-b6c6-e66362380011-4_481_808_248_424}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the graph of $y = \mathrm { f } ( x )$ which meets the coordinate axes at the points $( a , 0 )$ and $( 0 , b )$, where $a$ and $b$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Showing, in terms of $a$ and $b$, the coordinates of any points of intersection with the axes, sketch on separate diagrams the graphs of
\begin{enumerate}[label=(\roman*)]
\item $\quad y = \mathrm { f } ^ { - 1 } ( x )$,
\item $y = 2 \mathrm { f } ( 3 x )$.
Given that
$$\mathrm { f } ( x ) = 2 - \sqrt { x + 9 } , \quad x \in \mathbb { R } , \quad x \geq - 9 ,$$
\end{enumerate}\item find the values of $a$ and $b$,
\item find an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state its domain.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q7 [14]}}