| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This question tests standard A-level content on quadratic functions and inverses. Part (a) requires completing the square to find the vertex (routine). Part (b)(i) tests understanding that parabolas aren't one-to-one (basic concept). Part (b)(ii) involves restricting the domain and finding an inverse by completing the square and rearranging—all standard textbook procedures with no novel problem-solving required. The 8 total marks reflect length rather than conceptual difficulty. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02n Sketch curves: simple equations including polynomials1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch of U-shaped curve with vertex at \((-3,4)\) | G1, G1 | (Shape) (Stationary point) |
| Answer | Marks |
|---|---|
| A correct statement, e.g. \(f^{-1}\) doesn't exist because \(f\) is not a one-one function | E1 |
| Answer | Marks |
|---|---|
| Any appropriate domain, e.g. There are many possible appropriate domains. It is essential that any domain must be contained in one branch of the curve shown. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(y = x^2 + 6x + 13 = (x+3)^2 + 4\) | M1 | (Attempt to find \(x\) in terms of \(y\)) |
| \(x + 3 = \pm\sqrt{y - 4}\) | A1 | |
| So that \(x = -3 \pm \sqrt{y - 4}\) | A1 | |
| Since \(x > -3\), the positive sign is appropriate | A1 | |
| \(\therefore x = -3 + \sqrt{y - 4}\) | A1 | |
| And \(f^{-1}(x) = -3 + \sqrt{x - 4}\) | A1 |
### Part (a):
Sketch of U-shaped curve with vertex at $(-3,4)$ | G1, G1 | (Shape) (Stationary point)
### Part (b)(i):
A correct statement, e.g. $f^{-1}$ doesn't exist because $f$ is not a one-one function | E1 |
### Part (b)(ii):
Any appropriate domain, e.g. There are many possible appropriate domains. It is essential that any domain must be contained in one branch of the curve shown. | B1 |
Here we consider $(-3, \infty)$.
Let $y = x^2 + 6x + 13 = (x+3)^2 + 4$ | M1 | (Attempt to find $x$ in terms of $y$)
$x + 3 = \pm\sqrt{y - 4}$ | A1 |
So that $x = -3 \pm \sqrt{y - 4}$ | A1 |
Since $x > -3$, the positive sign is appropriate | A1 |
$\therefore x = -3 + \sqrt{y - 4}$ | A1 |
And $f^{-1}(x) = -3 + \sqrt{x - 4}$ | A1 |
**Total: [8]**\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = x^2 + 6x + 13$, identifying the stationary point. [2]
\item The function $f$ is defined by $f(x) = x^2 + 6x + 13$ with domain $(a,b)$.
\begin{enumerate}[label=(\roman*)]
\item Explain why $f^{-1}$ does not exist when $a = -10$ and $b = 10$. [1]
\item Write down a value of $a$ and a value of $b$ for which the inverse of $f$ does exist and derive an expression for $f^{-1}(x)$. [5]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q3 [8]}}