| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.8 This is a straightforward question on linear function inverses requiring only routine algebraic manipulation to find f^(-1)(x) = (x+2)/3, sketching two straight lines with reflection symmetry, and solving a simple linear inequality 3x-2 > (x+2)/3. All three parts are standard textbook exercises with no conceptual challenges beyond basic technique. |
| Spec | 1.02m Graphs of functions: difference between plotting and sketching1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 3x - 2\) so \(x = \frac{y+2}{3}\) | M1 | 1.1a |
| \(f^{-1}(x) = \frac{x+2}{3}\) | A1 | 1.1 — Condone '\(y=\)' but e.g. '\(f(x)=\)' does not score |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph of \(y = 3x - 2\) going through \((0, -2)\), or \(y = \frac{x+2}{3}\) going through \((0, \frac{2}{3})\) | B1 | 1.1 |
| \(y = f^{-1}(x)\) is a reflection of \(y = f(x)\) in \(y = x\); both graphs linear to get B2 | B1 | 1.1 — Line of symmetry roughly at 45° or implied by coords |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3x - 2 = x\) when \(x = 1\); attempt to find when \(f(x) = x\), or attempt to solve \(3x - 2 = \frac{x+2}{3}\) or \(3x - 2 > \frac{x+2}{3}\), so \(x = 1\) | M1 | 3.1a — Using *their* inverse |
| \(x > 1\) | A1 | 2.2a |
| [2] |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 3x - 2$ so $x = \frac{y+2}{3}$ | M1 | 1.1a |
| $f^{-1}(x) = \frac{x+2}{3}$ | A1 | 1.1 — Condone '$y=$' but e.g. '$f(x)=$' does not score |
| **[2]** | | |
---
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph of $y = 3x - 2$ going through $(0, -2)$, or $y = \frac{x+2}{3}$ going through $(0, \frac{2}{3})$ | B1 | 1.1 |
| $y = f^{-1}(x)$ is a reflection of $y = f(x)$ in $y = x$; both graphs linear to get B2 | B1 | 1.1 — Line of symmetry roughly at 45° or implied by coords |
| **[2]** | | |
---
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3x - 2 = x$ when $x = 1$; attempt to find when $f(x) = x$, or attempt to solve $3x - 2 = \frac{x+2}{3}$ or $3x - 2 > \frac{x+2}{3}$, so $x = 1$ | M1 | 3.1a — Using *their* inverse |
| $x > 1$ | A1 | 2.2a |
| **[2]** | | |1 The function $\mathrm { f } ( x )$ is defined for all real $x$ by\\
$f ( x ) = 3 x - 2$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\item Sketch the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$ on the same diagram.
\item Find the set of values of $x$ for which $\mathrm { f } ( x ) > \mathrm { f } ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q1 [6]}}