| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Evaluate composite at point |
| Difficulty | Standard +0.3 This question involves reading values from a piecewise linear graph, evaluating a composite function at a point, finding an inverse of a rational function (standard technique), and solving an equation involving composition. All parts use routine C3 techniques with no novel insight required, though part (d) requires combining multiple steps. Slightly easier than average due to the graphical nature and straightforward algebraic manipulation. |
| 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 |
| \(0 \leqslant f(x) \leqslant 10\) | B1 | Allow \(0 \leqslant f \leqslant 10\), \(0 \leqslant y \leqslant 10\), \([0,10]\). Do NOT allow \(0 \leqslant x \leqslant 10\). Must include BOTH ends (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(ff(0) = f(5)\) | B1 | Correct one application at \(x=0\). Accept \(f(0)=5\), \(f(5)\), \(0 \to 5 \to \ldots\) |
| \(= 3\) | B1 | '3' can score both marks if no incorrect working seen (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \dfrac{4+3x}{5-x} \Rightarrow y(5-x) = 4+3x \Rightarrow 5y - 4 = xy + 3x\) | M1 | Attempt to make \(x\) (or replaced \(y\)) subject. Condone sign slips |
| \(5y - 4 = x(y+3) \Rightarrow x = \dfrac{5y-4}{y+3}\) | dM1 | Take out common factor of \(x\), divide. Only allow one sign error |
| \(g^{-1}(x) = \dfrac{5x-4}{3+x}\) | A1 | No need to state domain. Accept \(y = \dfrac{4-5x}{-3-x}\) and \(y = \dfrac{5 - \frac{4}{x}}{1 + \frac{3}{x}}\) (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(gf(x) = 16 \Rightarrow f(x) = g^{-1}(16) = 4\) | M1A1 | Stating/implying \(f(x) = g^{-1}(16)\). Accept \(\dfrac{4+3f(x)}{5-f(x)} = 16 \Rightarrow f(x) = \ldots\) |
| \(f(x) = 4 \Rightarrow x = 6\) | B1 | May be given if no working shown |
| \(f(x) = 4 \Rightarrow 5 - 2.5x = 4 \Rightarrow x = 0.4\) | M1A1 | Full method from line \(y = 5 - 2.5x\). Accept \(\dfrac{2}{5} = \dfrac{2-x}{4} \Rightarrow x = \ldots\) (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(gf(x) = 16 \Rightarrow \dfrac{4+3(ax+b)}{5-(ax+b)} = 16\) | M1 | Writes \(gf(x) = 16\) with linear \(f(x)\). Order must be correct. Condone invisible brackets |
| \(ax + b = x - 2\) or \(5 - 2.5x\) | A1 | |
| \(x = 6\) | B1 | May be given if no working |
| \(\dfrac{4+3(5-2.5x)}{5-(5-2.5x)} = 16 \Rightarrow x = \ldots\) | M1 | Bracketing must be correct, no more than one error |
| \(x = 0.4\) | A1 | (5) |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 \leqslant f(x) \leqslant 10$ | B1 | Allow $0 \leqslant f \leqslant 10$, $0 \leqslant y \leqslant 10$, $[0,10]$. Do NOT allow $0 \leqslant x \leqslant 10$. Must include BOTH ends **(1)** |
---
# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $ff(0) = f(5)$ | B1 | Correct one application at $x=0$. Accept $f(0)=5$, $f(5)$, $0 \to 5 \to \ldots$ |
| $= 3$ | B1 | '3' can score both marks if no incorrect working seen **(2)** |
---
# Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \dfrac{4+3x}{5-x} \Rightarrow y(5-x) = 4+3x \Rightarrow 5y - 4 = xy + 3x$ | M1 | Attempt to make $x$ (or replaced $y$) subject. Condone sign slips |
| $5y - 4 = x(y+3) \Rightarrow x = \dfrac{5y-4}{y+3}$ | dM1 | Take out common factor of $x$, divide. Only allow **one** sign error |
| $g^{-1}(x) = \dfrac{5x-4}{3+x}$ | A1 | No need to state domain. Accept $y = \dfrac{4-5x}{-3-x}$ and $y = \dfrac{5 - \frac{4}{x}}{1 + \frac{3}{x}}$ **(3)** |
---
# Question 7(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $gf(x) = 16 \Rightarrow f(x) = g^{-1}(16) = 4$ | M1A1 | Stating/implying $f(x) = g^{-1}(16)$. Accept $\dfrac{4+3f(x)}{5-f(x)} = 16 \Rightarrow f(x) = \ldots$ |
| $f(x) = 4 \Rightarrow x = 6$ | B1 | May be given if no working shown |
| $f(x) = 4 \Rightarrow 5 - 2.5x = 4 \Rightarrow x = 0.4$ | M1A1 | Full method from line $y = 5 - 2.5x$. Accept $\dfrac{2}{5} = \dfrac{2-x}{4} \Rightarrow x = \ldots$ **(5)** |
**Alt 1 to 7(d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $gf(x) = 16 \Rightarrow \dfrac{4+3(ax+b)}{5-(ax+b)} = 16$ | M1 | Writes $gf(x) = 16$ with linear $f(x)$. Order must be correct. Condone invisible brackets |
| $ax + b = x - 2$ or $5 - 2.5x$ | A1 | |
| $x = 6$ | B1 | May be given if no working |
| $\dfrac{4+3(5-2.5x)}{5-(5-2.5x)} = 16 \Rightarrow x = \ldots$ | M1 | Bracketing must be correct, no more than one error |
| $x = 0.4$ | A1 | **(5)** |
---7. The function $f$ has domain $- 2 \leqslant x \leqslant 6$ and is linear from $( - 2,10 )$ to $( 2,0 )$ and from $( 2,0 )$ to (6, 4). A sketch of the graph of $y = \mathrm { f } ( x )$ is shown in Figure 1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e29d66c-c3c6-4e4b-acfb-c73c60604d93-09_906_965_367_566}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Write down the range of f .
\item Find $\mathrm { ff } ( 0 )$.
The function $g$ is defined by
$$\mathrm { g } : x \rightarrow \frac { 4 + 3 x } { 5 - x } , \quad x \in \mathbb { R } , \quad x \neq 5$$
\item Find $\mathrm { g } ^ { - 1 } ( x )$
\item Solve the equation $\operatorname { gf } ( x ) = 16$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q7 [11]}}