Edexcel P3 2022 June — Question 2 9 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2022
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeEvaluate composite at point
DifficultyModerate -0.8 Part (a) is straightforward function composition requiring simple substitution. Part (b) is routine inverse function finding for a rational function. Part (c) requires solving an equation involving both functions but follows standard algebraic manipulation. All parts are textbook-standard exercises with no novel insight required, making this easier than average.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence

2. The functions f and g are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = \frac { 5 - x } { 3 x + 2 } & x \in \mathbb { R } , x \neq - \frac { 2 } { 3 } \\ \mathrm {~g} ( x ) = 2 x - 7 & x \in \mathbb { R } \end{array}$$
  1. Find the value of \(\mathrm { fg } ( 5 )\)
  2. Find \(\mathrm { f } ^ { - 1 }\)
  3. Solve the equation $$\mathrm { f } \left( \frac { 1 } { a } \right) = \mathrm { g } ( a + 3 )$$

Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(fg(5) = f(3) = \frac{5-\text{"3"}}{3\times\text{"3"}+2} = \frac{2}{11}\)M1, A1 Correct order: g before f on 5; allow two-step approach \(g(5)=A \rightarrow f(A)=\ldots\); condone arithmetic slips; A1 for \(\frac{2}{11}\) or exact equivalent, accept \(0.1\dot{8}\) but not \(0.18\)
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(f^{-1}(x) = \frac{5-2x}{3x+1}\)M1A1 M1 for changing subject of \(y=\frac{5-x}{3x+2}\) to get \(x=\frac{\ldots\pm2y}{\pm3y\pm\ldots}\); A1 for \(\frac{5-2x}{3x+1}\) or equivalent e.g. \(\frac{2x-5}{-3x-1}\); accept mapping notation; do not allow just \(y=\ldots\)
\(x \in \mathbb{R},\ x \neq -\frac{1}{3}\)B1 Omission of \(x \in \mathbb{R}\) condoned
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{5-\frac{1}{a}}{3\times\frac{1}{a}+2} = 2(a+3)-7\)B1ft Follow through on their \(f^{-1}(x)\); or equivalent \(\frac{1}{a} = f^{-1}(2a-1)\)
\(\frac{5-\frac{1}{a}}{3\times\frac{1}{a}+2} = 2(a+3)-7 \Rightarrow 4a^2-a-2=0\)M1A1 Must start with allowable equation form; rearranges to 3-term quadratic; A1 for \(4a^2-a-2=0\) or equivalent
\(a = \frac{1\pm\sqrt{33}}{8}\)A1 Exact equivalent only; do not accept rounded decimals
## Question 2:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(5) = f(3) = \frac{5-\text{"3"}}{3\times\text{"3"}+2} = \frac{2}{11}$ | M1, A1 | Correct order: g before f on 5; allow two-step approach $g(5)=A \rightarrow f(A)=\ldots$; condone arithmetic slips; A1 for $\frac{2}{11}$ or exact equivalent, accept $0.1\dot{8}$ but not $0.18$ |

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f^{-1}(x) = \frac{5-2x}{3x+1}$ | M1A1 | M1 for changing subject of $y=\frac{5-x}{3x+2}$ to get $x=\frac{\ldots\pm2y}{\pm3y\pm\ldots}$; A1 for $\frac{5-2x}{3x+1}$ or equivalent e.g. $\frac{2x-5}{-3x-1}$; accept mapping notation; do not allow just $y=\ldots$ |
| $x \in \mathbb{R},\ x \neq -\frac{1}{3}$ | B1 | Omission of $x \in \mathbb{R}$ condoned |

### Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5-\frac{1}{a}}{3\times\frac{1}{a}+2} = 2(a+3)-7$ | B1ft | Follow through on their $f^{-1}(x)$; or equivalent $\frac{1}{a} = f^{-1}(2a-1)$ |
| $\frac{5-\frac{1}{a}}{3\times\frac{1}{a}+2} = 2(a+3)-7 \Rightarrow 4a^2-a-2=0$ | M1A1 | Must start with allowable equation form; rearranges to 3-term quadratic; A1 for $4a^2-a-2=0$ or equivalent |
| $a = \frac{1\pm\sqrt{33}}{8}$ | A1 | Exact equivalent only; do not accept rounded decimals |

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2. The functions f and g are defined by

$$\begin{array} { l l } 
\mathrm { f } ( x ) = \frac { 5 - x } { 3 x + 2 } & x \in \mathbb { R } , x \neq - \frac { 2 } { 3 } \\
\mathrm {~g} ( x ) = 2 x - 7 & x \in \mathbb { R }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\mathrm { fg } ( 5 )$
\item Find $\mathrm { f } ^ { - 1 }$
\item Solve the equation

$$\mathrm { f } \left( \frac { 1 } { a } \right) = \mathrm { g } ( a + 3 )$$
\end{enumerate}

\hfill \mbox{\textit{Edexcel P3 2022 Q2 [9]}}