| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a standard C3 inverse function question requiring routine techniques: sketching transformations (reflection in y=x, vertical stretch and horizontal stretch), finding axis intercepts by substitution, and finding an inverse by swapping x and y then rearranging. All steps are textbook procedures 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) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Graph (i): correct shape with \((b, 0)\) and \((0, a)\) marked | M1 A1 | |
| Graph (ii): correct shape with \((\frac{1}{3}a, 0)\) and \((0, 2b)\) marked | M1 A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x = 0 \Rightarrow y = -1\) \(\therefore b = -1\) | B1 | |
| \(y = 0 \Rightarrow 2 - \sqrt{x+9} = 0\) | ||
| \(x = 2^2 - 9 = -5\) \(\therefore a = -5\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(y = 2 - \sqrt{x+9}\), \(\sqrt{x+9} = 2 - y\), \(x + 9 = (2-y)^2\) | M1 | |
| \(\therefore f^{-1}(x) = (2-x)^2 - 9\) | A1 | |
| \(f(-9) = 2\) \(\therefore\) domain of \(f^{-1}(x)\) is \(x \in \mathbb{R}\), \(x \leq 2\) | M1 A1 | (12) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| Graph (i): correct shape with $(b, 0)$ and $(0, a)$ marked | M1 A1 | |
| Graph (ii): correct shape with $(\frac{1}{3}a, 0)$ and $(0, 2b)$ marked | M1 A2 | |
## Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x = 0 \Rightarrow y = -1$ $\therefore b = -1$ | B1 | |
| $y = 0 \Rightarrow 2 - \sqrt{x+9} = 0$ | | |
| $x = 2^2 - 9 = -5$ $\therefore a = -5$ | M1 A1 | |
## Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $y = 2 - \sqrt{x+9}$, $\sqrt{x+9} = 2 - y$, $x + 9 = (2-y)^2$ | M1 | |
| $\therefore f^{-1}(x) = (2-x)^2 - 9$ | A1 | |
| $f(-9) = 2$ $\therefore$ domain of $f^{-1}(x)$ is $x \in \mathbb{R}$, $x \leq 2$ | M1 A1 | **(12)** |
---7.\\
\includegraphics[max width=\textwidth, alt={}, center]{039ebdba-4ad5-4974-9345-d66712fa0a08-3_401_712_228_479}
The diagram 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 $y = \mathrm { f } ^ { - 1 } ( x )$,
\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{OCR C3 Q7 [12]}}