Edexcel C3 2011 January — Question 6 13 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2011
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyModerate -0.3 This is a standard C3 composite and inverse functions question. Part (a) requires routine algebraic manipulation to find the inverse of a rational function (3 marks). The remaining parts involve reading values from a piecewise linear graph and basic function composition. While multi-part, each component is straightforward textbook material requiring no novel insight, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.02z Models in context: use functions in modelling
  1. The function \(f\) is defined by
$$\mathrm { f } : x \mapsto \frac { 3 - 2 x } { x - 5 } , \quad x \in \mathbb { R } , x \neq 5$$
  1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
    (3) \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-10_901_1091_593_429} \captionsetup{labelformat=empty} \caption{Figure 2}
    \end{figure} The function g has domain \(- 1 \leqslant x \leqslant 8\), and is linear from \(( - 1 , - 9 )\) to \(( 2,0 )\) and from \(( 2,0 )\) to \(( 8,4 )\). Figure 2 shows a sketch of the graph of \(y = \mathrm { g } ( x )\).
  2. Write down the range of g.
  3. Find \(\operatorname { gg } ( 2 )\).
  4. Find \(\mathrm { fg } ( 8 )\).
  5. On separate diagrams, sketch the graph with equation
    1. \(y = | \mathrm { g } ( x ) |\),
    2. \(y = \mathrm { g } ^ { - 1 } ( x )\). Show on each sketch the coordinates of each point at which the graph meets or cuts the axes.
  6. State the domain of the inverse function \(\mathrm { g } ^ { - 1 }\).
Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(y = \frac{3-2x}{x-5} \Rightarrow y(x-5) = 3-2x\)M1 Attempt to make \(x\) (or swapped \(y\)) the subject
\(xy - 5y = 3-2x \Rightarrow xy + 2x = 3+5y \Rightarrow x(y+2) = 3+5y\)M1 Collect \(x\) terms together and factorise
\(x = \frac{3+5y}{y+2}\ \therefore f^{-1}(x) = \frac{3+5x}{x+2}\)A1 oe
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Range of \(g\) is \(-9 \leq g(x) \leq 4\) or \(-9 \leq y \leq 4\)B1 Correct range
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Deduces \(g(2) = 0\)M1 Seen or implied
\(g\,g(2) = g(0) = -6\)A1 \(-6\)
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(fg(8) = f(4)\)M1 Correct order: \(g\) followed by \(f\)
\(= \frac{3-4(2)}{4-5} = \frac{-5}{-1} = 5\)A1 \(5\)
Part (e)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Correct shape of \(g^{-1}\)B1
Graph goes through \((\{0\}, 2)\) and \((-6, \{0\})\), both markedB1
Part (f):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Domain of \(g^{-1}\) is \(-9 \leq x \leq 4\)B1\(\checkmark\) Either correct answer or follow through from part (b) 
## Question 6:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{3-2x}{x-5} \Rightarrow y(x-5) = 3-2x$ | M1 | Attempt to make $x$ (or swapped $y$) the subject |
| $xy - 5y = 3-2x \Rightarrow xy + 2x = 3+5y \Rightarrow x(y+2) = 3+5y$ | M1 | Collect $x$ terms together and factorise |
| $x = \frac{3+5y}{y+2}\ \therefore f^{-1}(x) = \frac{3+5x}{x+2}$ | A1 oe | |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Range of $g$ is $-9 \leq g(x) \leq 4$ or $-9 \leq y \leq 4$ | B1 | Correct range |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $g(2) = 0$ | M1 | Seen or implied |
| $g\,g(2) = g(0) = -6$ | A1 | $-6$ |

### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(8) = f(4)$ | M1 | Correct order: $g$ followed by $f$ |
| $= \frac{3-4(2)}{4-5} = \frac{-5}{-1} = 5$ | A1 | $5$ |

### Part (e)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape of $g^{-1}$ | B1 | |
| Graph goes through $(\{0\}, 2)$ and $(-6, \{0\})$, both marked | B1 | |

### Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Domain of $g^{-1}$ is $-9 \leq x \leq 4$ | B1$\checkmark$ | Either correct answer or follow through from part (b) |
\begin{enumerate}
  \item The function $f$ is defined by
\end{enumerate}

$$\mathrm { f } : x \mapsto \frac { 3 - 2 x } { x - 5 } , \quad x \in \mathbb { R } , x \neq 5$$

(a) Find $\mathrm { f } ^ { - 1 } ( x )$.\\
(3)

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-10_901_1091_593_429}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

The function g has domain $- 1 \leqslant x \leqslant 8$, and is linear from $( - 1 , - 9 )$ to $( 2,0 )$ and from $( 2,0 )$ to $( 8,4 )$. Figure 2 shows a sketch of the graph of $y = \mathrm { g } ( x )$.\\
(b) Write down the range of g.\\
(c) Find $\operatorname { gg } ( 2 )$.\\
(d) Find $\mathrm { fg } ( 8 )$.\\
(e) On separate diagrams, sketch the graph with equation\\
(i) $y = | \mathrm { g } ( x ) |$,\\
(ii) $y = \mathrm { g } ^ { - 1 } ( x )$.

Show on each sketch the coordinates of each point at which the graph meets or cuts the axes.\\
(f) State the domain of the inverse function $\mathrm { g } ^ { - 1 }$.

\hfill \mbox{\textit{Edexcel C3 2011 Q6 [13]}}
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