| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Even/odd function verification |
| Difficulty | Easy -1.2 This question tests basic definitions of even/odd functions using straightforward algebraic verification. Part (i) requires showing |(-x)³| = |x³| (trivial absolute value property), and part (ii) only needs a single counterexample or showing g(-x) ≠ -g(x). Both parts are definitional recall with minimal calculation, significantly easier than average A-level questions which typically require multi-step problem-solving. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(-x) = \ | (-x)^3\ | = \ |
| Correct completion | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(g(-x) = (-x-1)^3\) | M1 | |
| \(g(1) \neq g(-1)\) (say), so the function is not odd | A1 | |
| Total: 2 |
## Question 2:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(-x) = \|(-x)^3\| = \|-x^3\| = \|x^3\| = f(x)$ | M1 | Using $-x$ |
| Correct completion | A1 | |
| | **Total: 2** | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $g(-x) = (-x-1)^3$ | M1 | |
| $g(1) \neq g(-1)$ (say), so the function is not odd | A1 | |
| | **Total: 2** | |
---2 (i) Show that $\mathrm { f } ( x ) = \left| x ^ { 3 } \right|$ is an even function.\\
(ii) It is suggested that the function $\mathrm { g } ( x ) = ( x - 1 ) ^ { 3 }$ is odd. Prove that this is false.
\hfill \mbox{\textit{OCR MEI C3 Q2 [4]}}