| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a standard inverse function question with routine steps: finding range from a restricted quadratic, inverting by swapping and rearranging (straightforward since domain restriction ensures one-to-one), and solving a composite function equation. Part (c) requires careful algebra with two compositions but follows predictable patterns. Slightly above average due to the multi-part nature and algebraic manipulation in part (c), but all techniques are standard P1 material with no novel insight required. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| \([y] \leqslant -1\) | B1 | Accept f or \(f(x) \leqslant -1\), \(-\infty < y \leqslant -1\), \((-\infty, -1]\). Do not accept \(x \leqslant -1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -3x^2 + 2\) rearranged to \(3x^2 = 2 - y\), leading to \(x^2 = \frac{2-y}{3}\) or \(y^2 = \frac{2-x}{3}\) | M1 | |
| \(x = [\pm]\sqrt{\frac{2-y}{3}} \rightarrow [f^{-1}(x)] = \{-\}\left\{\sqrt{\frac{2-x}{3}}\right\}\) | A1 A1 | A1 for minus, A1 for \(\sqrt{\frac{2-x}{3}}\), allow \(-\sqrt{\frac{x-2}{-3}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{fg}(x) = -3(-x^2-1)^2 + 2\) | M1 | SOI expect \(-3x^4 - 6x^2 - 1\) |
| \(\text{gf}(x) = -(-3x^2+2)^2 - 1\) | M1 | SOI expect \(-9x^4 + 12x^2 - 5\) |
| \(\text{fg}(x) - \text{gf}(x) + 8 = 0\) leading to \(6x^4 - 18x^2 + 12\ [=0]\) | A1 | OE |
| \([6](x^2-1)(x^2-2)[=0]\) or formula or completion of the square | M1 | Solving a 3-term quadratic equation in \(x^2\) must be seen |
| \(x = -1\ ,\ -\sqrt{2}\) only these two solutions | A1 | Allow \(-\sqrt{1}\), \(-1.41[4]\). Answers only SC B1 |
## Question 9:
**Part 9(a):**
| $[y] \leqslant -1$ | B1 | Accept f or $f(x) \leqslant -1$, $-\infty < y \leqslant -1$, $(-\infty, -1]$. Do not accept $x \leqslant -1$ |
**Part 9(b):**
| $y = -3x^2 + 2$ rearranged to $3x^2 = 2 - y$, leading to $x^2 = \frac{2-y}{3}$ or $y^2 = \frac{2-x}{3}$ | M1 | |
| $x = [\pm]\sqrt{\frac{2-y}{3}} \rightarrow [f^{-1}(x)] = \{-\}\left\{\sqrt{\frac{2-x}{3}}\right\}$ | A1 A1 | A1 for minus, A1 for $\sqrt{\frac{2-x}{3}}$, allow $-\sqrt{\frac{x-2}{-3}}$ |
**Part 9(c):**
| $\text{fg}(x) = -3(-x^2-1)^2 + 2$ | M1 | SOI expect $-3x^4 - 6x^2 - 1$ |
| $\text{gf}(x) = -(-3x^2+2)^2 - 1$ | M1 | SOI expect $-9x^4 + 12x^2 - 5$ |
| $\text{fg}(x) - \text{gf}(x) + 8 = 0$ leading to $6x^4 - 18x^2 + 12\ [=0]$ | A1 | OE |
| $[6](x^2-1)(x^2-2)[=0]$ or formula or completion of the square | M1 | Solving a 3-term quadratic equation in $x^2$ must be seen |
| $x = -1\ ,\ -\sqrt{2}$ only these **two** solutions | A1 | Allow $-\sqrt{1}$, $-1.41[4]$. Answers only SC B1 |
---9 The function f is defined by $\mathrm { f } ( x ) = - 3 x ^ { 2 } + 2$ for $x \leqslant - 1$.
\begin{enumerate}[label=(\alph*)]
\item State the range of f.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
The function g is defined by $\mathrm { g } ( x ) = - x ^ { 2 } - 1$ for $x \leqslant - 1$.
\item Solve the equation $\mathrm { fg } ( x ) - \mathrm { gf } ( x ) + 8 = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q9 [9]}}