| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Evaluate composite at point |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic function operations: identifying range from a simple quadratic, evaluating a composite function at a given point (substitute once, then once more), and finding an inverse of a simple rational function. All parts are routine procedures requiring no problem-solving or insight, making it easier than the average A-level question. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) \leq 9\) | B1 | Correct range. Allow equivalent notation e.g. \(y \leq 9\), \(f \leq 9\), \(y \in (-\infty, 9]\) but not \(x \leq 9\). Condone just "\(\leq 9\)" and "less than or equal to 9" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(fg(1.5) = f\left(\frac{3}{2\times1.5+1}\right) = 9-\left(\frac{3}{2\times1.5+1}\right)^2\) | M1 | Full attempt to find \(fg(1.5)\) condoning slips. Implied by correct answer or 8.44. Requires attempt to apply \(g(1.5)\) first and then \(f\) to their \(g(1.5)\). Also allow substituting \(x=1.5\) into \(9-\left(\frac{3}{2x+1}\right)^2\) |
| \(= \frac{135}{16}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(g(x) = \frac{3}{2x+1} \Rightarrow g^{-1}(x) = \frac{3-x}{2x}\) | M1 A1 | M1: Changes subject of \(y = \frac{3}{2x+1}\) and obtains \(x = \frac{3\pm y}{2y}\) or \(x = \frac{3}{2y} \pm \frac{1}{2}\) or equivalent. A1: \(g^{-1}(x) = \frac{3-x}{2x}\), or \(g^{-1}(x) = \frac{1}{2}\left(\frac{3-x}{x}\right)\), or \(g^{-1}(x) = \frac{3}{2x}-\frac{1}{2}\). Condone \(y = \frac{3-x}{2x}\). NOT \(f^{-1} = \frac{3-x}{2x}\). Don't allow ambiguous fractions like \(g^{-1}(x) = \frac{\frac{3}{x}-1}{2}\) |
| \(0 < x \leq 3\) | B1 | Correct domain. Allow \(x \in (0,3]\) but not just \((0,3]\) |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) \leq 9$ | B1 | Correct range. Allow equivalent notation e.g. $y \leq 9$, $f \leq 9$, $y \in (-\infty, 9]$ but not $x \leq 9$. Condone just "$\leq 9$" and "less than or equal to 9" |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $fg(1.5) = f\left(\frac{3}{2\times1.5+1}\right) = 9-\left(\frac{3}{2\times1.5+1}\right)^2$ | M1 | Full attempt to find $fg(1.5)$ condoning slips. Implied by correct answer or 8.44. Requires attempt to apply $g(1.5)$ first and then $f$ to their $g(1.5)$. Also allow substituting $x=1.5$ into $9-\left(\frac{3}{2x+1}\right)^2$ |
| $= \frac{135}{16}$ | A1 | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $g(x) = \frac{3}{2x+1} \Rightarrow g^{-1}(x) = \frac{3-x}{2x}$ | M1 A1 | M1: Changes subject of $y = \frac{3}{2x+1}$ and obtains $x = \frac{3\pm y}{2y}$ or $x = \frac{3}{2y} \pm \frac{1}{2}$ or equivalent. A1: $g^{-1}(x) = \frac{3-x}{2x}$, or $g^{-1}(x) = \frac{1}{2}\left(\frac{3-x}{x}\right)$, or $g^{-1}(x) = \frac{3}{2x}-\frac{1}{2}$. Condone $y = \frac{3-x}{2x}$. NOT $f^{-1} = \frac{3-x}{2x}$. Don't allow ambiguous fractions like $g^{-1}(x) = \frac{\frac{3}{x}-1}{2}$ |
| $0 < x \leq 3$ | B1 | Correct domain. Allow $x \in (0,3]$ but not just $(0,3]$ |
---\begin{enumerate}
\item The functions f and g are defined by
\end{enumerate}
$$\begin{array} { l l l }
\mathrm { f } ( x ) = 9 - x ^ { 2 } & x \in \mathbb { R } & x \geqslant 0 \\
\mathrm {~g} ( x ) = \frac { 3 } { 2 x + 1 } & x \in \mathbb { R } & x \geqslant 0
\end{array}$$
(a) Write down the range of f\\
(b) Find the value of fg(1.5)\\
(c) Find $\mathrm { g } ^ { - 1 }$
\hfill \mbox{\textit{Edexcel P3 2023 Q1 [6]}}