| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Determine if inverse exists |
| Difficulty | Moderate -0.3 This is a standard composite and inverse functions question covering routine techniques: finding inverse of a rational function (algebraic manipulation), computing composite functions, evaluating at a point, finding range, and explaining why an inverse doesn't exist (non-injective function). All parts are textbook exercises requiring no novel insight, though part (b) requires careful algebra. Slightly easier than average due to straightforward application of standard methods. |
| 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/Working | Mark | Guidance |
| \(y(x+1)=3x-5 \Rightarrow xy+y=3x-5 \Rightarrow y+5=3x-xy\) | M1 | Cross-multiply; attempt to collect \(x\)-terms on one side |
| \(y+5=x(3-y) \Rightarrow \frac{y+5}{3-y}=x\) | M1 | Fully correct method to find inverse |
| \(f^{-1}(x)=\frac{x+5}{3-x},\quad x\in\mathbb{R},\ x\neq 3\) | A1 | Correct \(f^{-1}(x)\) with domain, fully in function notation |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(ff(x)=\frac{3\left(\frac{3x-5}{x+1}\right)-5}{\left(\frac{3x-5}{x+1}\right)+1}\) | M1 | Substitutes \(f(x)=\frac{3x-5}{x+1}\) into \(\frac{3f(x)-5}{f(x)+1}\) |
| \(=\frac{\frac{3(3x-5)-5(x+1)}{x+1}}{\frac{(3x-5)+(x+1)}{x+1}}\) | M1 | Rationalises denominator for both numerator and denominator |
| \(=\frac{3(3x-5)-5(x+1)}{(3x-5)+(x+1)}\) | A1 | Correct unsimplified form |
| \(=\frac{9x-15-5x-5}{3x-5+x+1}=\frac{4x-20}{4x-4}=\frac{x-5}{x-1}\) (note \(a=-5\)) | A1 | Shows \(ff(x)=\frac{x+a}{x-1}\) where \(a=-5\), no errors |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(fg(2)=f(4-6)=f(-2)=\frac{3(-2)-5}{-2+1}\) | M1 | Substitutes result of \(g(2)\) into \(f\) |
| \(= 11\) | A1 | Correct answer |
| Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g(x)=x^2-3x=(x-1.5)^2-2.25\), hence \(g_{\min}=-2.25\) | M1 | Full method to establish minimum of \(g\) |
| Either \(g_{\min}=-2.25\) or \(g(x)\geq -2.25\) or \(g(5)=25-15=10\) | B1 | Finding correct minimum value or stating \(g(5)=10\) |
| \(-2.25\leq g(x)\leq 10\) or \(-2.25\leq y\leq 10\) | A1 | States correct range |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. the function \(g\) is many-one / not one-one / the inverse is one-many / \(g(0)=g(3)=0\) | B1 | Valid reason |
| Total: (1) |
## Question 10:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y(x+1)=3x-5 \Rightarrow xy+y=3x-5 \Rightarrow y+5=3x-xy$ | M1 | Cross-multiply; attempt to collect $x$-terms on one side |
| $y+5=x(3-y) \Rightarrow \frac{y+5}{3-y}=x$ | M1 | Fully correct method to find inverse |
| $f^{-1}(x)=\frac{x+5}{3-x},\quad x\in\mathbb{R},\ x\neq 3$ | A1 | Correct $f^{-1}(x)$ with domain, fully in function notation |
| **Total: (3)** | | |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $ff(x)=\frac{3\left(\frac{3x-5}{x+1}\right)-5}{\left(\frac{3x-5}{x+1}\right)+1}$ | M1 | Substitutes $f(x)=\frac{3x-5}{x+1}$ into $\frac{3f(x)-5}{f(x)+1}$ |
| $=\frac{\frac{3(3x-5)-5(x+1)}{x+1}}{\frac{(3x-5)+(x+1)}{x+1}}$ | M1 | Rationalises denominator for both numerator and denominator |
| $=\frac{3(3x-5)-5(x+1)}{(3x-5)+(x+1)}$ | A1 | Correct unsimplified form |
| $=\frac{9x-15-5x-5}{3x-5+x+1}=\frac{4x-20}{4x-4}=\frac{x-5}{x-1}$ (note $a=-5$) | A1 | Shows $ff(x)=\frac{x+a}{x-1}$ where $a=-5$, no errors |
| **Total: (4)** | | |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(2)=f(4-6)=f(-2)=\frac{3(-2)-5}{-2+1}$ | M1 | Substitutes result of $g(2)$ into $f$ |
| $= 11$ | A1 | Correct answer |
| **Total: (2)** | | |
**Part (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g(x)=x^2-3x=(x-1.5)^2-2.25$, hence $g_{\min}=-2.25$ | M1 | Full method to establish minimum of $g$ |
| Either $g_{\min}=-2.25$ or $g(x)\geq -2.25$ or $g(5)=25-15=10$ | B1 | Finding correct minimum value or stating $g(5)=10$ |
| $-2.25\leq g(x)\leq 10$ or $-2.25\leq y\leq 10$ | A1 | States correct range |
| **Total: (3)** | | |
**Part (e):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. the function $g$ is many-one / not one-one / the inverse is one-many / $g(0)=g(3)=0$ | B1 | Valid reason |
| **Total: (1)** | | |
---\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$f : x \mapsto \frac { 3 x - 5 } { x + 1 } , \quad x \in \mathbb { R } , \quad x \neq - 1$$
(a) Find $\mathrm { f } ^ { - 1 } ( x )$.\\
(b) Show that
$$\mathrm { ff } ( x ) = \frac { x + a } { x - 1 } , \quad x \in \mathbb { R } , \quad x \neq \pm 1$$
where $a$ is an integer to be found.
The function g is defined by
$$\mathrm { g } : x \mapsto x ^ { 2 } - 3 x , \quad x \in \mathbb { R } , 0 \leqslant x \leqslant 5$$
(c) Find the value of $\mathrm { fg } ( 2 )$.\\
(d) Find the range of g.\\
(e) Explain why the function $g$ does not have an inverse.
\hfill \mbox{\textit{Edexcel Paper 1 Q10 [13]}}