| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | State domain or range |
| Difficulty | Moderate -0.3 This question tests understanding of inverse function domains and ranges, requiring students to recognize that the domain of f^(-1) equals the range of f, and vice versa. While conceptually important, it only requires evaluating f at the endpoints of its domain (f(-1)=-5, f(2)=4) and applying the domain/range swap principle—a standard technique with no complex calculations or novel problem-solving required. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks |
|---|---|
| 4 | y = x3 – 4 x (cid:108) y |
| Answer | Marks |
|---|---|
| range is −1 ≤ y ≤ 2 | M1 |
| Answer | Marks |
|---|---|
| [5] | m |
| Answer | Marks |
|---|---|
| 1.1 | e |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | 5 | 0 |
Question 4:
4 | y = x3 – 4 x (cid:108) y
x = y3 – 4
(cid:159) x + 4 = y3
(cid:159) y = 3 x(cid:14)4 so f−1(x) = 3 x(cid:14)4
range of f is −5 ≤ y ≤ 4
so domain of f−1 is −5 ≤ x ≤ 4 e
range is −1 ≤ y ≤ 2 | M1
A1
c
M1
A1
B1
[5] | m
1.1
1.1
i
1.1
1.2
1.1 | e
attempt to invert
accept y(cid:32)3 x(cid:14)4 but not
x(cid:32)3 y(cid:14)4
May be implied
or [−5, 4]
or −1 ≤ f−1(x) ≤ 2 or [−1, 2]
4 | 5 | 0 | 0 | 0 | 5 | 0The function f(x) is defined by $\text{f}(x) = x^3 - 4$ for $-1 \leq x \leq 2$.
For $\text{f}^{-1}(x)$, determine
\begin{itemize}
\item the domain
\item the range.
\end{itemize} [5]
\hfill \mbox{\textit{OCR MEI Paper 2 Q4 [5]}}