| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 Finding the inverse requires algebraic manipulation with exponentials (cross-multiplying, rearranging, taking logarithms) which is slightly above routine but follows standard techniques. The range question is straightforward once the inverse is found. This is a typical C3/FP1-level inverse function question with moderate algebraic complexity. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06c Logarithm definition: log_a(x) as inverse of a^x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{e^x}{1-e^x}\) | ||
| \(y(1-e^x) = e^x\) | M1 | AO 1.1a |
| \([y = e^x(1+y)]\ \ e^x = \frac{y}{1+y}\) | A1 | AO 1.1 |
| \(f^{-1}(x) = \ln\left(\frac{x}{1+x}\right)\) | A1 [3] | AO 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f^{-1}(x) \neq 0\) | B1 [1] | AO 1.2 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{e^x}{1-e^x}$ | | | |
| $y(1-e^x) = e^x$ | M1 | AO 1.1a | Clearing fractions; $x$ and $y$ may be interchanged at any stage |
| $[y = e^x(1+y)]\ \ e^x = \frac{y}{1+y}$ | A1 | AO 1.1 | Expression for $e^x$ |
| $f^{-1}(x) = \ln\left(\frac{x}{1+x}\right)$ | A1 [3] | AO 2.1 | Condone '$y=$'; condone no brackets or mod |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f^{-1}(x) \neq 0$ | B1 [1] | AO 1.2 | Allow $y \neq 0$ but not $x \neq 0$ |9 The function $\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { 1 - \mathrm { e } ^ { x } }$ is defined on the domain $x \in \mathbb { R } , x \neq 0$.\\
(i) Find $\mathrm { f } ^ { - 1 } ( x )$.\\
(ii) Write down the range of $\mathrm { f } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{OCR MEI Paper 3 2018 Q9 [4]}}