| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Verify composite identity |
| Difficulty | Moderate -0.3 This is a slightly below-average A-level question. Parts (a) and (b) involve routine algebraic manipulation of rational functions—finding an inverse and verifying a composite function identity are standard C3 techniques. Parts (c) and (d) are straightforward function composition and finding range of a quadratic on a restricted domain. No novel insight or complex problem-solving is required; it's methodical application of learned procedures. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Method 1: \(y(x+1) = 3x-5 \Rightarrow xy+y = 3x-5\), \(y+5 = 3x-xy \Rightarrow y+5 = x(3-y)\) | M1 | Brings \((x+1)\) to LHS and multiplies out by \(y\). Or if \(x\) and \(y\) swapped first, \((y+1)\) to LHS and multiplies out by \(x\) |
| \(\Rightarrow \frac{y+5}{3-y} = x\) | M1 | Full method to make \(x\) (or swapped \(y\)) the subject by collecting terms and factorising |
| \(f^{-1}(x) = \frac{x+5}{3-x}\) \((x \in \mathbb{R},\ x \neq 3)\) | A1 oe | Equivalent forms e.g. \(-\frac{x+5}{x-3}\) or \(\frac{-x-5}{x-3}\) or \(-1+\frac{8}{3-x}\) etc. Ignore LHS. Does not need to include domain for this mark |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Method 1: \(\text{ff}(x) = \dfrac{3\left(\frac{3x-5}{x+1}\right)-5}{\left(\frac{3x-5}{x+1}\right)+1}\) | M1 A1 | M1: Attempt to substitute f into itself. e.g. \(\text{ff}(x) = \frac{3f(x)-5}{f(x)+1}\). Squaring f\((x)\) is M0. Allow \(\text{ff}(x) = \frac{3f(x)-5}{x+1}\) or \(\frac{3x-5}{f(x)+1}\) for M1A0 |
| \(= \frac{3(3x-5)-5(x+1)}{(3x-5)+(x+1)} = \frac{9x-15-5x-5}{3x-5+x+1} = \frac{4x-20}{4x-4}\) | M1 | Attempt to combine numerator and denominator into single rational fraction with same common denominator |
| \(= \frac{x-5}{x-1}\) (note that \(a=-5\)) | A1 | Does not need to include domain or statement that \(x \in \mathbb{R}\), \(x \neq -1\), \(x \neq 1\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{fg}(2) = f(4-6) = f(-2) = \frac{3(-2)-``5"}{-2+1}\ {:=}\ 11\) or substitute 2 into \(\text{fg}(x) = \frac{3(x^2-3x)-5}{x^2-3x+1}\ {:=}\ 11\) | M1; A1 | M1: Full method of inserting \(g(2)\) (i.e. \(-2\)) into \(f(x)\). Or substitutes 2 into \(\text{fg}(x) = \frac{3(x^2-3x)-5}{x^2-3x+1}\) |
| [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. e.g. \((x \pm \alpha)^2 + \beta\) leading to \(g_{\min}=\beta\). Or derivative method, or symmetry of roots, or listing values. This mark may also be implied by \(-2.25\) |
| Either \(g_{\min} = -2.25\) or \(g(x) \geq -2.25\) or \(g(5) = 25-15 = 10\) | B1 | For finding either the correct minimum value of g (can be implied by \(g(x) \geq -2.25\)) or for stating \(g(5)=10\) or finding value 10 as maximum |
| \(-2.25 \leq g(x) \leq 10\) or \(-2.25 \leq y \leq 10\) | A1 | Note: \(-2.25 \leq x \leq 10\) (wrong variable) is A0; \(-2.25 < y < 10\) (wrong inequality) is A0; \(-2.25 \leq f \leq 10\) (wrong function) is A0; Accept \([-2.25, 10]\) for A1 but not \((-2.25, 10)\) |
| [3] Total: 12 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Method 1:** $y(x+1) = 3x-5 \Rightarrow xy+y = 3x-5$, $y+5 = 3x-xy \Rightarrow y+5 = x(3-y)$ | M1 | Brings $(x+1)$ to LHS and multiplies out by $y$. Or if $x$ and $y$ swapped first, $(y+1)$ to LHS and multiplies out by $x$ |
| $\Rightarrow \frac{y+5}{3-y} = x$ | M1 | Full method to make $x$ (or swapped $y$) the subject by collecting terms and factorising |
| $f^{-1}(x) = \frac{x+5}{3-x}$ $(x \in \mathbb{R},\ x \neq 3)$ | A1 oe | Equivalent forms e.g. $-\frac{x+5}{x-3}$ or $\frac{-x-5}{x-3}$ or $-1+\frac{8}{3-x}$ etc. Ignore LHS. Does not need to include domain for this mark |
| **[3]** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Method 1:** $\text{ff}(x) = \dfrac{3\left(\frac{3x-5}{x+1}\right)-5}{\left(\frac{3x-5}{x+1}\right)+1}$ | M1 A1 | M1: Attempt to substitute f into itself. e.g. $\text{ff}(x) = \frac{3f(x)-5}{f(x)+1}$. Squaring f$(x)$ is M0. Allow $\text{ff}(x) = \frac{3f(x)-5}{x+1}$ or $\frac{3x-5}{f(x)+1}$ for M1A0 |
| $= \frac{3(3x-5)-5(x+1)}{(3x-5)+(x+1)} = \frac{9x-15-5x-5}{3x-5+x+1} = \frac{4x-20}{4x-4}$ | M1 | Attempt to combine numerator and denominator into single rational fraction with same common denominator |
| $= \frac{x-5}{x-1}$ (note that $a=-5$) | A1 | Does not need to include domain or statement that $x \in \mathbb{R}$, $x \neq -1$, $x \neq 1$ |
| **[4]** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{fg}(2) = f(4-6) = f(-2) = \frac{3(-2)-``5"}{-2+1}\ {:=}\ 11$ or substitute 2 into $\text{fg}(x) = \frac{3(x^2-3x)-5}{x^2-3x+1}\ {:=}\ 11$ | M1; A1 | M1: Full method of inserting $g(2)$ (i.e. $-2$) into $f(x)$. Or substitutes 2 into $\text{fg}(x) = \frac{3(x^2-3x)-5}{x^2-3x+1}$ |
| **[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. e.g. $(x \pm \alpha)^2 + \beta$ leading to $g_{\min}=\beta$. Or derivative method, or symmetry of roots, or listing values. This mark may also be implied by $-2.25$ |
| Either $g_{\min} = -2.25$ or $g(x) \geq -2.25$ or $g(5) = 25-15 = 10$ | B1 | For finding either the correct minimum value of g (can be implied by $g(x) \geq -2.25$) or for stating $g(5)=10$ or finding value 10 as maximum |
| $-2.25 \leq g(x) \leq 10$ or $-2.25 \leq y \leq 10$ | A1 | Note: $-2.25 \leq x \leq 10$ (wrong variable) is A0; $-2.25 < y < 10$ (wrong inequality) is A0; $-2.25 \leq f \leq 10$ (wrong function) is A0; Accept $[-2.25, 10]$ for A1 but not $(-2.25, 10)$ |
| **[3] Total: 12** | | |
---7. The function f is defined by
$$\mathrm { f } : x \mapsto \frac { 3 x - 5 } { x + 1 } , \quad x \in \mathbb { R } , x \neq - 1$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$
\item Show that
$$\operatorname { ff } ( x ) = \frac { x + a } { x - 1 } , \quad x \in \mathbb { R } , x \neq - 1 , x \neq 1$$
where $a$ is an integer to be determined.
The function $g$ is defined by
$$\mathrm { g } : x \mapsto x ^ { 2 } - 3 x , \quad x \in \mathbb { R } , 0 \leqslant x \leqslant 5$$
\item Find the value of fg(2)
\item Find the range of g
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q7 [12]}}