| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard P3/C3 techniques: finding a specific inverse value (straightforward algebra with logarithms), proving monotonicity via differentiation (routine quotient rule), finding an inverse function of a rational function (standard algebraic manipulation), and finding a range of a composite function. All parts are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f^{-1}(22) \Rightarrow 2 + 5\ln x = 22 \Rightarrow \ln x = 4 \Rightarrow x = e^4\) | M1 A1 | M1: sets \(2+5\ln x = 22\) and rearranges to \(\ln x = \ldots\); A1: must be simplified and exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g(x) = \frac{6x-2}{2x+1} \Rightarrow g'(x) = \frac{6(2x+1)-2(6x-2)}{(2x+1)^2}\) | M1 A1 | M1: quotient/product/chain rule attempt; A1: correct unsimplified form |
| States \(g'(x) = \frac{10}{(2x+1)^2} > 0\) hence increasing | A1* | Requires reason why always positive AND minimal conclusion e.g. "hence increasing" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{6x-2}{2x+1} \Rightarrow 2xy + y = 6x - 2 \Rightarrow 2xy - 6x = -y - 2\) | M1 | Attempts to change subject; achieves form \(\pm2xy \pm 6x = \pm y \pm 2\) |
| \(x = \frac{-y-2}{2y-6}\), so \(g^{-1}(x) = \frac{-x-2}{2x-6}\) | A1 | Equivalent forms accepted e.g. \(\frac{x+2}{6-2x}\) |
| Domain \(0 < x < 3\) | B1 | e.g. \(0 < x \cap x < 3\); not \(0 < y < 3\) or \(0 < g^{-1}(x) < 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Range of \(fg\) is \(fg < 2 + 5\ln 3\) | M1, A1 | M1: \(2+5\ln 3\) found (allow awrt 7.5); A1: fully correct, \((-\infty <)\, fg < 2+5\ln 3\) |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f^{-1}(22) \Rightarrow 2 + 5\ln x = 22 \Rightarrow \ln x = 4 \Rightarrow x = e^4$ | M1 A1 | M1: sets $2+5\ln x = 22$ and rearranges to $\ln x = \ldots$; A1: must be simplified and exact |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g(x) = \frac{6x-2}{2x+1} \Rightarrow g'(x) = \frac{6(2x+1)-2(6x-2)}{(2x+1)^2}$ | M1 A1 | M1: quotient/product/chain rule attempt; A1: correct unsimplified form |
| States $g'(x) = \frac{10}{(2x+1)^2} > 0$ hence increasing | A1* | Requires reason why always positive AND minimal conclusion e.g. "hence increasing" |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{6x-2}{2x+1} \Rightarrow 2xy + y = 6x - 2 \Rightarrow 2xy - 6x = -y - 2$ | M1 | Attempts to change subject; achieves form $\pm2xy \pm 6x = \pm y \pm 2$ |
| $x = \frac{-y-2}{2y-6}$, so $g^{-1}(x) = \frac{-x-2}{2x-6}$ | A1 | Equivalent forms accepted e.g. $\frac{x+2}{6-2x}$ |
| Domain $0 < x < 3$ | B1 | e.g. $0 < x \cap x < 3$; not $0 < y < 3$ or $0 < g^{-1}(x) < 3$ |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Range of $fg$ is $fg < 2 + 5\ln 3$ | M1, A1 | M1: $2+5\ln 3$ found (allow awrt 7.5); A1: fully correct, $(-\infty <)\, fg < 2+5\ln 3$ |
---\begin{enumerate}
\item The functions $f$ and $g$ are defined by
\end{enumerate}
$$\begin{aligned}
& \mathrm { f } ( x ) = 2 + 5 \ln x \quad x > 0 \\
& \mathrm {~g} ( x ) = \frac { 6 x - 2 } { 2 x + 1 } \quad x > \frac { 1 } { 3 }
\end{aligned}$$
(a) Find $\mathrm { f } ^ { - 1 } ( 22 )$\\
(b) Use differentiation to prove that g is an increasing function.\\
(c) Find $\mathrm { g } ^ { - 1 }$\\
(d) Find the range of fg
\hfill \mbox{\textit{Edexcel P3 2024 Q5 [10]}}