| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a straightforward composite and inverse function question requiring standard techniques: composing two simple functions, finding domain restrictions, and verifying a given inverse formula. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(h(x) = \frac{1}{\sqrt{10-2x}}\) | B1 | ACF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x < 5\) | B1 | ACF; condone incorrect set notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Forms \(y = \frac{1}{\sqrt{10-2x}}\) and squares both sides correctly, OR rearranges to obtain expression for \(\sqrt{10-2x}\) | M1 | \(x\) and \(y\) can be switched at any point |
| \(10 - 2x = \frac{1}{y^2}\) or \(5 - x = \frac{1}{2y^2}\) | A1 | \(x\) and \(y\) can be switched at any point |
| \(h^{-1}(x) = 5 - \frac{1}{2x^2}\) | R1 | Completes reasoned argument with no incorrect steps; must use correct notation \(h^{-1}(x)\); consistent use of variables; AG |
## Question 7:
**Part 7(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $h(x) = \frac{1}{\sqrt{10-2x}}$ | B1 | ACF |
**Part 7(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x < 5$ | B1 | ACF; condone incorrect set notation |
**Part 7(c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Forms $y = \frac{1}{\sqrt{10-2x}}$ and squares both sides correctly, OR rearranges to obtain expression for $\sqrt{10-2x}$ | M1 | $x$ and $y$ can be switched at any point |
| $10 - 2x = \frac{1}{y^2}$ or $5 - x = \frac{1}{2y^2}$ | A1 | $x$ and $y$ can be switched at any point |
| $h^{-1}(x) = 5 - \frac{1}{2x^2}$ | R1 | Completes reasoned argument with no incorrect steps; must use correct notation $h^{-1}(x)$; consistent use of variables; AG |
---7 The functions f and g are defined by
$$\begin{aligned}
& \mathrm { f } ( x ) = \sqrt { 10 - 2 x } \text { for } \quad x \leq 5 \\
& \mathrm {~g} ( x ) = \frac { 1 } { x } \quad \text { for } \quad x \neq 0
\end{aligned}$$
The function $h$ has maximum possible domain and is defined by
$$\mathrm { h } ( x ) = \operatorname { gf } ( x )$$
7
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\mathrm { h } ( x )$\\
7
\item Find the domain of h\\
7
\item Show that $\mathrm { h } ^ { - 1 } ( x ) = 5 - \frac { 1 } { 2 x ^ { 2 } }$\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-11_2488_1716_219_153}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q7 [5]}}