| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Find inverse function after simplification |
| Difficulty | Standard +0.8 This question requires factorizing a quadratic, combining algebraic fractions with partial fractions in reverse, finding the range of a rational function with domain restriction, and solving g(a) = g^{-1}(a) which involves finding the inverse function and solving a resulting equation. The multi-step nature, the need to work with inverse functions, and the domain restriction make this moderately challenging, though each individual technique is standard C3 material. |
| Spec | 1.02k 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 existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2+x-6=(x+3)(x-2)\) | B1 | |
| \(\frac{x}{x+3}+\frac{3(2x+1)}{(x+3)(x-2)} = \frac{x(x-2)+3(2x+1)}{(x+3)(x-2)}\) | M1 | |
| \(=\frac{x^2+4x+3}{(x+3)(x-2)}\) | A1 | |
| \(=\frac{(x+3)(x+1)}{(x+3)(x-2)} = \frac{(x+1)}{(x-2)}\) | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| One end: \((y)>1,(y)\geq 1\) or \((y)<4,(y)\leq 4\) | B1 | |
| \(1 < y < 4\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to set \(g(x)=x\) or \(g(x)=g^{-1}(x)\) or \(g^{-1}(x)=x\) or \(g^2(x)=x\) | M1 | |
| \(\frac{(x+1)}{(x-2)}=x \Rightarrow x^2-3x-1=0\) | A1, dM1 | |
| \(a = \frac{3+\sqrt{13}}{2}\) oe \(\left(1.5+\sqrt{3.25}\right)\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 + x - 6 = (x+3)(x-2)\) | B1 | Can occur anywhere in solution |
| Combining two fractions with common denominator, e.g. \(\frac{x}{x+3} + \frac{3(2x+1)}{x^2+x-6} = \frac{x(x^2+x-6)+3(2x+1)(x+3)}{(x+3)(x^2+x-6)}\) | M1 | Denominator must be correct, at least one numerator adapted. Condone missing brackets |
| Correct intermediate form e.g. \(\frac{x^2+4x+3}{(x+3)(x-2)}\) or \(\frac{x^2+4x+3}{x^2+x-6}\) | A1 | Simplified quadratic over simplified quadratic |
| \(\frac{(x+1)(x+3)(x+3)}{(x+3)(x^2+x-6)} \rightarrow \frac{(x+1)}{(x-2)}\) | A1* | Must factorise, cancel correctly. All bracketing correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States either end of range: \(y < 4\), \(y \leq 4\) or \(y > 1\), \(y \geq 1\) | B1 | Accept with or without \(y\)'s |
| Correct range \(1 < y < 4\) | B1 | Accept \(1 < g < 4\), \(y > 1\) and \(y < 4\), \((1,4)\). Do not accept \(1 < x < 4\), \(1 < y \leq 4\), \([1,4)\). Special case B1B0 for \(1 < x < 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to set \(g(x) = x\), \(g^{-1}(x) = x\) or \(g(x) = g^{-1}(x)\) or \(g^2(x) = x\) | M1 | If \(g^{-1}(x)\) used, full attempt to make \(x\) subject required |
| Result in form \(g^{-1}(x) = \frac{\pm 2x \pm 1}{\pm x \pm 1}\), e.g. \(\frac{(x+1)}{(x-2)} = x\) | A1 | Accept evidence such as \(\frac{x+1}{x-2} = \frac{\pm 2x\pm 1}{\pm x\pm 1}\) |
| \(x^2 - 3x - 1 = 0\) | A1 | The \(=0\) may be implied by subsequent work |
| Solving a 3TQ \(= 0\) | dM1 | Dependent on first M1. Do not accept factors unless clearly factorises. Allow awrt 3.30 |
| \(x = \frac{3+\sqrt{13}}{2}\) | A1 | Ignore reference to \(\frac{3-\sqrt{13}}{2}\). Withhold if additional values given for \(x > 3\) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2+x-6=(x+3)(x-2)$ | B1 | |
| $\frac{x}{x+3}+\frac{3(2x+1)}{(x+3)(x-2)} = \frac{x(x-2)+3(2x+1)}{(x+3)(x-2)}$ | M1 | |
| $=\frac{x^2+4x+3}{(x+3)(x-2)}$ | A1 | |
| $=\frac{(x+3)(x+1)}{(x+3)(x-2)} = \frac{(x+1)}{(x-2)}$ | A1* | cso |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| One end: $(y)>1,(y)\geq 1$ or $(y)<4,(y)\leq 4$ | B1 | |
| $1 < y < 4$ | B1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to set $g(x)=x$ or $g(x)=g^{-1}(x)$ or $g^{-1}(x)=x$ or $g^2(x)=x$ | M1 | |
| $\frac{(x+1)}{(x-2)}=x \Rightarrow x^2-3x-1=0$ | A1, dM1 | |
| $a = \frac{3+\sqrt{13}}{2}$ oe $\left(1.5+\sqrt{3.25}\right)$ | A1 | cso |
# Question (a) - Algebraic Fractions
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 + x - 6 = (x+3)(x-2)$ | B1 | Can occur anywhere in solution |
| Combining two fractions with common denominator, e.g. $\frac{x}{x+3} + \frac{3(2x+1)}{x^2+x-6} = \frac{x(x^2+x-6)+3(2x+1)(x+3)}{(x+3)(x^2+x-6)}$ | M1 | Denominator must be correct, at least one numerator adapted. Condone missing brackets |
| Correct intermediate form e.g. $\frac{x^2+4x+3}{(x+3)(x-2)}$ or $\frac{x^2+4x+3}{x^2+x-6}$ | A1 | Simplified quadratic over simplified quadratic |
| $\frac{(x+1)(x+3)(x+3)}{(x+3)(x^2+x-6)} \rightarrow \frac{(x+1)}{(x-2)}$ | A1* | Must factorise, cancel correctly. All bracketing correct |
# Question (b) - Range
| Answer/Working | Mark | Guidance |
|---|---|---|
| States either end of range: $y < 4$, $y \leq 4$ or $y > 1$, $y \geq 1$ | B1 | Accept with or without $y$'s |
| Correct range $1 < y < 4$ | B1 | Accept $1 < g < 4$, $y > 1$ and $y < 4$, $(1,4)$. Do not accept $1 < x < 4$, $1 < y \leq 4$, $[1,4)$. Special case B1B0 for $1 < x < 4$ |
# Question (c) - Inverse Function and Equation
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to set $g(x) = x$, $g^{-1}(x) = x$ or $g(x) = g^{-1}(x)$ or $g^2(x) = x$ | M1 | If $g^{-1}(x)$ used, full attempt to make $x$ subject required |
| Result in form $g^{-1}(x) = \frac{\pm 2x \pm 1}{\pm x \pm 1}$, e.g. $\frac{(x+1)}{(x-2)} = x$ | A1 | Accept evidence such as $\frac{x+1}{x-2} = \frac{\pm 2x\pm 1}{\pm x\pm 1}$ |
| $x^2 - 3x - 1 = 0$ | A1 | The $=0$ may be implied by subsequent work |
| Solving a 3TQ $= 0$ | dM1 | Dependent on first M1. Do not accept factors unless clearly factorises. Allow awrt 3.30 |
| $x = \frac{3+\sqrt{13}}{2}$ | A1 | Ignore reference to $\frac{3-\sqrt{13}}{2}$. Withhold if additional values given for $x > 3$ |
---5.
$$\mathrm { g } ( x ) = \frac { x } { x + 3 } + \frac { 3 ( 2 x + 1 ) } { x ^ { 2 } + x - 6 } , \quad x > 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { g } ( x ) = \frac { x + 1 } { x - 2 } , \quad x > 3$
\item Find the range of g.
\item Find the exact value of $a$ for which $\mathrm { g } ( a ) = \mathrm { g } ^ { - 1 } ( a )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q5 [10]}}