| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This is a multi-part question covering standard composite and inverse function techniques. Parts (a)-(c) involve routine substitution, finding an inverse of a rational function, and composing functions—all textbook exercises. Part (d) requires using the discriminant to determine when a quadratic has no real solutions, which is a standard technique. While it has multiple parts (5 marks worth), each individual step is straightforward and requires no novel insight, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{fg}(2) = 4-3\!\left(\dfrac{5}{2(2)-9}\right)^{\!2} = \ldots\) | M1 | Correct method: attempts to find \(g(2)=\dfrac{5}{4-9}\) and substitutes into f, or attempts \(\text{fg}(x)=4-3\!\left(\dfrac{5}{2x-9}\right)^{\!2}\) and substitutes \(x=2\) |
| \(\text{fg}(2) = 1\) | A1 | Correct answer only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \dfrac{5}{2x-9} \Rightarrow 2xy - 9y = 5 \Rightarrow 2xy = 5+9y\) | M1 | Eliminates fraction and puts \(xy\) term (or \(x\) term) onto one side. Condone minor slips |
| \(x = \dfrac{5+9y}{2y}\) | A1 | Correct expression for inverse, \(x\) in terms of \(y\) (or \(y\) in terms of \(x\)) |
| \(g^{-1}(x) = \dfrac{5+9x}{2x},\ x\neq 0\ \{x\in\mathbb{R}\}\) | A1 | Fully correct notation including domain \(x\neq 0\). Condone \(x\neq 0\) without \(x\in\mathbb{R}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\text{gf}(x) =\}\ \dfrac{5}{2(4-3x^2)-9}\) | M1 | Correct method: substitutes f into g, condoning slips e.g. missing the 3 |
| \(= \dfrac{5}{-1-6x^2}\) or \(\dfrac{-5}{1+6x^2}\) | A1 | Correct simplified fraction. Ignore any reference to domains |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-5\), \(\text{gf}(x)<0\) | B1 | Deduces correct range. Allow e.g. \(-5\), \(y<0\), \(y\in[-5,0)\), \([-5,0)\). Do not allow e.g. \(f(x)<0\) or \(g(x)<0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x)=h(x) \Rightarrow 4-3x^2 = 2x^2-6x+k \Rightarrow 5x^2-6x+k-4=0\) | M1 | Sets \(f(x)=h(x)\) and collects terms to obtain a \(3TQ=0\) |
| \(b^2-4ac < 0 \Rightarrow 36-4(5)(k-4)<0 \Rightarrow k>\ldots\) | dM1 | Recognises need to use \(b^2-4ac\ldots 0\) on their \(3TQ\); establishes value/range for \(k\) in terms of \(k\) only |
| \(k > 5.8\) o.e. | A1 | Deduces correct range e.g. \(k>\dfrac{29}{5}\), \(k\in\!\left(\dfrac{29}{5},\infty\right)\) |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{fg}(2) = 4-3\!\left(\dfrac{5}{2(2)-9}\right)^{\!2} = \ldots$ | M1 | Correct method: attempts to find $g(2)=\dfrac{5}{4-9}$ and substitutes into f, or attempts $\text{fg}(x)=4-3\!\left(\dfrac{5}{2x-9}\right)^{\!2}$ and substitutes $x=2$ |
| $\text{fg}(2) = 1$ | A1 | Correct answer only |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \dfrac{5}{2x-9} \Rightarrow 2xy - 9y = 5 \Rightarrow 2xy = 5+9y$ | M1 | Eliminates fraction and puts $xy$ term (or $x$ term) onto one side. Condone minor slips |
| $x = \dfrac{5+9y}{2y}$ | A1 | Correct expression for inverse, $x$ in terms of $y$ (or $y$ in terms of $x$) |
| $g^{-1}(x) = \dfrac{5+9x}{2x},\ x\neq 0\ \{x\in\mathbb{R}\}$ | A1 | Fully correct notation including domain $x\neq 0$. Condone $x\neq 0$ without $x\in\mathbb{R}$ |
## Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{\text{gf}(x) =\}\ \dfrac{5}{2(4-3x^2)-9}$ | M1 | Correct method: substitutes f into g, condoning slips e.g. missing the 3 |
| $= \dfrac{5}{-1-6x^2}$ or $\dfrac{-5}{1+6x^2}$ | A1 | Correct simplified fraction. Ignore any reference to domains |
## Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-5$, $\text{gf}(x)<0$ | B1 | Deduces correct range. Allow e.g. $-5$, $y<0$, $y\in[-5,0)$, $[-5,0)$. Do not allow e.g. $f(x)<0$ or $g(x)<0$ |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x)=h(x) \Rightarrow 4-3x^2 = 2x^2-6x+k \Rightarrow 5x^2-6x+k-4=0$ | M1 | Sets $f(x)=h(x)$ and collects terms to obtain a $3TQ=0$ |
| $b^2-4ac < 0 \Rightarrow 36-4(5)(k-4)<0 \Rightarrow k>\ldots$ | dM1 | Recognises need to use $b^2-4ac\ldots 0$ on their $3TQ$; establishes value/range for $k$ in terms of $k$ only |
| $k > 5.8$ o.e. | A1 | Deduces correct range e.g. $k>\dfrac{29}{5}$, $k\in\!\left(\dfrac{29}{5},\infty\right)$ |
---\begin{enumerate}
\item The functions f and g are defined by
\end{enumerate}
$$\begin{array} { l l }
f ( x ) = 4 - 3 x ^ { 2 } & x \in \mathbb { R } \\
g ( x ) = \frac { 5 } { 2 x - 9 } & x \in \mathbb { R } , x \neq \frac { 9 } { 2 }
\end{array}$$
(a) Find fg(2)\\
(b) Find $\mathrm { g } ^ { - 1 }$\\
(c) (i) Find $\mathrm { gf } ( x )$, giving your answer as a simplified fraction.\\
(ii) Deduce the range of $\operatorname { gf } ( x )$.
The function h is defined by
$$h ( x ) = 2 x ^ { 2 } - 6 x + k \quad x \in \mathbb { R }$$
where $k$ is a constant.\\
(d) Find the range of values of $k$ for which the equation
$$\mathrm { f } ( x ) = \mathrm { h } ( x )$$
has no real solutions.
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q8 [11]}}