| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Find inverse function after simplification |
| Difficulty | Standard +0.3 This question requires partial fractions manipulation and finding an inverse function, but the algebraic simplification in part (a) is guided ('show that'), making it straightforward. Parts (b)-(d) involve standard C3 techniques: inverting a simple rational function, identifying domain restrictions, and solving a composite function equation with logarithms. While multi-part, each step uses routine methods without requiring novel insight or complex problem-solving. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.02y Partial fractions: decompose rational functions1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2x^2+7x-4 = (2x-1)(x+4)\) | B1 | May not appear on first line |
| \(\dfrac{3(x+1)}{(2x-1)(x+4)} - \dfrac{1}{(x+4)} = \dfrac{3(x+1)-(2x-1)}{(2x-1)(x+4)}\) | M1 | Combines to single fraction with common denominator; allow slips on numerator |
| \(= \dfrac{x+4}{(2x-1)(x+4)}\) | M1 | Simplifies to linear numerator over factorised quadratic denominator |
| \(= \dfrac{1}{2x-1}\) | A1* | Given/cso; all bracketing and algebra must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \dfrac{1}{2x-1} \Rightarrow y(2x-1)=1 \Rightarrow 2xy - y = 1\) | M1 | Attempt to make \(x\) subject; minimum: multiply by \((2x-1)\) and finish with \(x=\) |
| \(2xy = 1+y \Rightarrow x = \dfrac{1+y}{2y}\) | M1 | Correct order of operations; allow max one slip |
| \(f^{-1}(x) = \dfrac{1+x}{2x}\) | A1 | Must be in terms of \(x\); accept \(\frac{x^{-1}+1}{2}\), \(\frac{1}{2x}+\frac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x > 0\) | B1 | Accept \((0,\infty)\); do not accept \(x\geq 0\), \(y>0\), \([0,\infty]\), \(f^{-1}(x)>0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{1}{2\ln(x+1)-1} = \dfrac{1}{7}\) | M1 | Attempt \(fg(x)\) set equal to \(\frac{1}{7}\); accept incorrect/missing brackets |
| \(\ln(x+1) = 4\) | A1 | Accept also \(\ln(x+1)^2 = 8\) |
| \(x = e^4 - 1\) | M1 A1 | M1 for correct ln work \(\ln(x\pm A)=c \Rightarrow x=\cdots\); A1 accept \(e^4 - e^0\) |
## Question 7:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x^2+7x-4 = (2x-1)(x+4)$ | B1 | May not appear on first line |
| $\dfrac{3(x+1)}{(2x-1)(x+4)} - \dfrac{1}{(x+4)} = \dfrac{3(x+1)-(2x-1)}{(2x-1)(x+4)}$ | M1 | Combines to single fraction with common denominator; allow slips on numerator |
| $= \dfrac{x+4}{(2x-1)(x+4)}$ | M1 | Simplifies to linear numerator over factorised quadratic denominator |
| $= \dfrac{1}{2x-1}$ | A1* | Given/cso; all bracketing and algebra must be correct |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \dfrac{1}{2x-1} \Rightarrow y(2x-1)=1 \Rightarrow 2xy - y = 1$ | M1 | Attempt to make $x$ subject; minimum: multiply by $(2x-1)$ and finish with $x=$ |
| $2xy = 1+y \Rightarrow x = \dfrac{1+y}{2y}$ | M1 | Correct order of operations; allow max one slip |
| $f^{-1}(x) = \dfrac{1+x}{2x}$ | A1 | Must be in terms of $x$; accept $\frac{x^{-1}+1}{2}$, $\frac{1}{2x}+\frac{1}{2}$ |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x > 0$ | B1 | Accept $(0,\infty)$; do not accept $x\geq 0$, $y>0$, $[0,\infty]$, $f^{-1}(x)>0$ |
**Part (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{1}{2\ln(x+1)-1} = \dfrac{1}{7}$ | M1 | Attempt $fg(x)$ set equal to $\frac{1}{7}$; accept incorrect/missing brackets |
| $\ln(x+1) = 4$ | A1 | Accept also $\ln(x+1)^2 = 8$ |
| $x = e^4 - 1$ | M1 A1 | M1 for correct ln work $\ln(x\pm A)=c \Rightarrow x=\cdots$; A1 accept $e^4 - e^0$ |
---\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$\mathrm { f } : x \mapsto \frac { 3 ( x + 1 ) } { 2 x ^ { 2 } + 7 x - 4 } - \frac { 1 } { x + 4 } , \quad x \in \mathbb { R } , x > \frac { 1 } { 2 }$$
(a) Show that $\mathrm { f } ( x ) = \frac { 1 } { 2 x - 1 }$\\
(b) Find $\mathrm { f } ^ { - 1 } ( x )$\\
(c) Find the domain of $\mathrm { f } ^ { - 1 }$
$$\mathrm { g } ( x ) = \ln ( x + 1 )$$
(d) Find the solution of $\mathrm { fg } ( x ) = \frac { 1 } { 7 }$, giving your answer in terms of e .
\hfill \mbox{\textit{Edexcel C3 2012 Q7 [12]}}