Edexcel P3 2022 January — Question 6 11 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2022
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyStandard +0.3 This is a slightly-above-average routine question on rational functions. Part (a) requires standard differentiation (quotient rule) and sign analysis. Part (b) is a textbook inverse function calculation. Part (c) involves composition of rational functions and determining range from the domain restriction—all standard techniques with no novel insight required.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.07h Differentiation from first principles: for sin(x) and cos(x)1.07n Stationary points: find maxima, minima using derivatives

6. The function f is defined by $$f ( x ) = \frac { 5 x - 3 } { x - 4 } \quad x > 4$$
  1. Show, by using calculus, that f is a decreasing function.
  2. Find \(\mathrm { f } ^ { - 1 }\)
    1. Show that \(\mathrm { ff } ( x ) = \frac { a x + b } { x + c }\) where \(a , b\) and \(c\) are constants to be found.
    2. Deduce the range of ff.

Question 6:
Part (a):
AnswerMarks Guidance
AnswerMarks Guidance
\(f(x)=\frac{5x-3}{x-4}\Rightarrow f'(x)=\frac{5(x-4)-(5x-3)}{(x-4)^2}=\frac{k}{(x-4)^2}\)M1 dM1 M1: Quotient rule to form \(\frac{p(x-4)-q(5x-3)}{(x-4)^2}\); dM1: proceeds to \(f'(x)=\frac{k}{(x-4)^2}\)
\(f'(x)=\frac{-17}{(x-4)^2}<0\), hence decreasingA1* Requires correct \(f'(x)\), correct statement \(f'(x)<0\), and minimal conclusion
Part (b):
AnswerMarks Guidance
AnswerMarks Guidance
\(y=\frac{5x-3}{x-4}\Rightarrow xy-4y=5x-3\Rightarrow x=\frac{4y-3}{y-5}\)M1 Cross multiply, collect \(x\) terms on one side
\(f^{-1}(x)=\frac{4x-3}{x-5}\)A1 Condone lhs as \(f^{-1}\) or \(y\)
Domain \(x>5\)B1
Part (c)(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(ff(x)=\dfrac{5\cdot\frac{5x-3}{x-4}-3}{\frac{5x-3}{x-4}-4}\)M1 Substitutes \(\frac{5x-3}{x-4}\) into \(f\); form must be correct
\(ff(x)=\frac{5(5x-3)-3(x-4)}{5x-3-4(x-4)}=\frac{22x-3}{x+13}\)dM1 A1 dM1: Multiplies numerator and denominator by \((x-4)\); A1: \(\frac{22x-3}{x+13}\)
Part (c)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(5B1 B1 First B1: achieves one correct "end" value (not just the number \(5\)); Second B1: fully correct; accept \((5,22)\); condone non-strict inequalities \(5\leq ff<21\) for B1 B0
## Question 6:

### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x)=\frac{5x-3}{x-4}\Rightarrow f'(x)=\frac{5(x-4)-(5x-3)}{(x-4)^2}=\frac{k}{(x-4)^2}$ | M1 dM1 | M1: Quotient rule to form $\frac{p(x-4)-q(5x-3)}{(x-4)^2}$; dM1: proceeds to $f'(x)=\frac{k}{(x-4)^2}$ |
| $f'(x)=\frac{-17}{(x-4)^2}<0$, hence decreasing | A1* | Requires correct $f'(x)$, correct statement $f'(x)<0$, and minimal conclusion |

### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=\frac{5x-3}{x-4}\Rightarrow xy-4y=5x-3\Rightarrow x=\frac{4y-3}{y-5}$ | M1 | Cross multiply, collect $x$ terms on one side |
| $f^{-1}(x)=\frac{4x-3}{x-5}$ | A1 | Condone lhs as $f^{-1}$ or $y$ |
| Domain $x>5$ | B1 | |

### Part (c)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ff(x)=\dfrac{5\cdot\frac{5x-3}{x-4}-3}{\frac{5x-3}{x-4}-4}$ | M1 | Substitutes $\frac{5x-3}{x-4}$ into $f$; form must be correct |
| $ff(x)=\frac{5(5x-3)-3(x-4)}{5x-3-4(x-4)}=\frac{22x-3}{x+13}$ | dM1 A1 | dM1: Multiplies numerator and denominator by $(x-4)$; A1: $\frac{22x-3}{x+13}$ |

### Part (c)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5<ff(x)<22$ | B1 B1 | First B1: achieves one correct "end" value (not just the number $5$); Second B1: fully correct; accept $(5,22)$; condone non-strict inequalities $5\leq ff<21$ for B1 B0 |
6. The function f is defined by

$$f ( x ) = \frac { 5 x - 3 } { x - 4 } \quad x > 4$$
\begin{enumerate}[label=(\alph*)]
\item Show, by using calculus, that f is a decreasing function.
\item Find $\mathrm { f } ^ { - 1 }$
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\mathrm { ff } ( x ) = \frac { a x + b } { x + c }$ where $a , b$ and $c$ are constants to be found.
\item Deduce the range of ff.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel P3 2022 Q6 [11]}}