| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on composite and inverse functions. Part (a) requires substituting g(x) into f(x) and solving a quadratic equation. Part (b) is routine inverse function finding using algebraic manipulation. Part (c) involves sketching a quadratic and its reflection in y=x with clearly defined domains. All techniques are standard P3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02m Graphs of functions: difference between plotting and sketching1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{fg}(x) = \frac{4(5-2x^2)+6}{5-2x^2-5}\) | M1 | M1: Attempts \(\text{fg}(x)\) in correct order. |
| \(\text{fg}(x) = 3 \Rightarrow \frac{26-8x^2}{-2x^2} = 3 \Rightarrow x = -\sqrt{13}\) | dM1 A1 A1 | dM1: Proceeds to \(x^2=...\). Depends on first mark. A1: \(x^2=13\) (may be implied by \(x=\sqrt{13}\)). A1: \(x=-\sqrt{13}\) only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{4x+6}{x-5} \Rightarrow yx-5y = 4x+6 \Rightarrow yx-4x = 5y+6\) | M1 | M1: Attempts to change subject. Both terms in \(x\) isolated on one side. |
| \(\Rightarrow x = \frac{5y+6}{y-4}\) | A1 | For \(x=\frac{5y+6}{y-4}\) or equivalent such as \(x=5+\frac{26}{y-4}\). |
| \(\Rightarrow \text{f}^{-1}(x) = \frac{5x+6}{x-4},\quad x\in\mathbb{R},\ x\neq 4\) | A1 | Requires both expression in \(x\) and domain. Condone \(\text{f}^{-1}=...\). Do not allow \(y=...\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Shape and position for \(y=g(x)\): parabola starting on \(y\)-axis extending down into quadrant 2, to at least the \(x\)-axis. | B1 | Condone pen slips e.g. curve heads slightly back toward \(y\)-axis in Q3. |
| Shape for \(y=g(x)\) crosses \(x\)-axis and \(y=g^{-1}(x)\) is a parabola starting on \(x\)-axis, with graphs crossing in quadrant 3. | B1 | Condone if \(g^{-1}(x)\) heads slightly back toward \(x\)-axis in Q3. |
| Correct intercepts: \(\left(\frac{\sqrt{5}}{\sqrt{2}}, 0\right)\), \(\left(0, 5\right)\), \(\left(0, -\frac{\sqrt{5}}{\sqrt{2}}\right)\), \(\left(-\frac{\sqrt{5}}{\sqrt{2}}, 0\right)\) as shown. Negative intercepts must be exact. | B1 | Allow values or coordinates (wrong-way round condoned e.g. \((0,5)\) for \((5,0)\)) as long as in correct positions. Graphs must cross axes at these points. |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{fg}(x) = \frac{4(5-2x^2)+6}{5-2x^2-5}$ | M1 | M1: Attempts $\text{fg}(x)$ in correct order. |
| $\text{fg}(x) = 3 \Rightarrow \frac{26-8x^2}{-2x^2} = 3 \Rightarrow x = -\sqrt{13}$ | dM1 A1 A1 | dM1: Proceeds to $x^2=...$. **Depends on first mark.** A1: $x^2=13$ (may be implied by $x=\sqrt{13}$). A1: $x=-\sqrt{13}$ **only**. |
**(4 marks)**
---
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{4x+6}{x-5} \Rightarrow yx-5y = 4x+6 \Rightarrow yx-4x = 5y+6$ | M1 | M1: Attempts to change subject. Both terms in $x$ isolated on one side. |
| $\Rightarrow x = \frac{5y+6}{y-4}$ | A1 | For $x=\frac{5y+6}{y-4}$ or equivalent such as $x=5+\frac{26}{y-4}$. |
| $\Rightarrow \text{f}^{-1}(x) = \frac{5x+6}{x-4},\quad x\in\mathbb{R},\ x\neq 4$ | A1 | Requires both expression in $x$ **and** domain. Condone $\text{f}^{-1}=...$. Do **not** allow $y=...$. |
**(3 marks)**
---
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Shape and position for $y=g(x)$: parabola starting on $y$-axis extending down into quadrant 2, to at least the $x$-axis. | B1 | Condone pen slips e.g. curve heads slightly back toward $y$-axis in Q3. |
| Shape for $y=g(x)$ crosses $x$-axis **and** $y=g^{-1}(x)$ is a parabola starting on $x$-axis, with graphs crossing in quadrant 3. | B1 | Condone if $g^{-1}(x)$ heads slightly back toward $x$-axis in Q3. |
| Correct intercepts: $\left(\frac{\sqrt{5}}{\sqrt{2}}, 0\right)$, $\left(0, 5\right)$, $\left(0, -\frac{\sqrt{5}}{\sqrt{2}}\right)$, $\left(-\frac{\sqrt{5}}{\sqrt{2}}, 0\right)$ as shown. Negative intercepts must be exact. | B1 | Allow values or coordinates (wrong-way round condoned e.g. $(0,5)$ for $(5,0)$) as long as in correct positions. Graphs must cross axes at these points. |
**(3 marks) — Total: 10 marks**4. The functions f and g are defined by
$$\begin{array} { l l }
\mathrm { f } ( x ) = \frac { 4 x + 6 } { x - 5 } & x \in \mathbb { R } , x \neq 5 \\
\mathrm {~g} ( x ) = 5 - 2 x ^ { 2 } & x \in \mathbb { R } , x \leqslant 0
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Solve the equation
$$\operatorname { fg } ( x ) = 3$$
\item Find $\mathrm { f } ^ { - 1 }$
\item Sketch and label, on the same axes, the curve with equation $y = \mathrm { g } ( x )$ and the curve with equation $y = \mathrm { g } ^ { - 1 } ( x )$. Show on your sketch the coordinates of the points where each curve meets or cuts the coordinate axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q4 [10]}}