| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Transformations of functions |
| Difficulty | Moderate -0.3 This is a straightforward C3 question on function transformations. Part (i) requires finding where the square root equals zero (routine algebra), part (ii) is reading the range from the graph, and part (iii) applies standard horizontal stretch and vertical translation rules. All parts are textbook-standard with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1-9a^2=0\) | M1 | or \(1-9x^2=0\); \(\sqrt{(1-9a^2)}=1-3a\) is M0 |
| \(\Rightarrow a^2=1/9 \Rightarrow a=1/3\) | A1 [2] | or \(0.33\) or better; \(\sqrt{(1/9)}\) is A0; not \(a=\pm1/3\) nor \(x=1/3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Range \(0\leq y\leq 1\) | B1 [1] | or \(0\leq f(x)\leq 1\) or \(0\leq f\leq 1\), not \(0\leq x\leq 1\); \(0\leq y\leq\sqrt{1}\) is B0; allow \([0,1]\), or 0 to 1 inclusive, but not 0 to 1 or \((0,1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curve goes from \(x=-3a\) to \(x=3a\) (or \(-1\) to \(1\)) | M1 | must have evidence of using s.f. 3; allow also if s.f.3 stated and stretch is reasonably to scale |
| Vertex at origin | M1 | |
| Curve, 'centre' \((0,-1)\), from \((-1,-1)\) to \((1,-1)\); (\(y\)-coords of \(-1\) can be inferred from vertex at O and correct scaling) | A1 [3] | allow from \((-3a,-1)\) to \((3a,-1)\) provided \(a=1/3\) or \(x=[\pm]1/3\) in (i); A0 for badly inconsistent scale(s) |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1-9a^2=0$ | M1 | or $1-9x^2=0$; $\sqrt{(1-9a^2)}=1-3a$ is M0 |
| $\Rightarrow a^2=1/9 \Rightarrow a=1/3$ | A1 **[2]** | or $0.33$ or better; $\sqrt{(1/9)}$ is A0; not $a=\pm1/3$ nor $x=1/3$ |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Range $0\leq y\leq 1$ | B1 **[1]** | or $0\leq f(x)\leq 1$ or $0\leq f\leq 1$, not $0\leq x\leq 1$; $0\leq y\leq\sqrt{1}$ is B0; allow $[0,1]$, or 0 to 1 inclusive, but not 0 to 1 or $(0,1)$ |
---
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve goes from $x=-3a$ to $x=3a$ (or $-1$ to $1$) | M1 | must have evidence of using s.f. 3; allow also if s.f.3 stated and stretch is reasonably to scale |
| Vertex at origin | M1 | |
| Curve, 'centre' $(0,-1)$, from $(-1,-1)$ to $(1,-1)$; ($y$-coords of $-1$ can be inferred from vertex at O and correct scaling) | A1 **[3]** | allow from $(-3a,-1)$ to $(3a,-1)$ provided $a=1/3$ or $x=[\pm]1/3$ in (i); A0 for badly inconsistent scale(s) |
---4 Fig. 4 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \sqrt { 1 - 9 x ^ { 2 } } , - a \leqslant x \leqslant a$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7b77c646-2bc5-4166-b22e-3c1229abd722-2_476_572_861_751}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Find the value of $a$.\\
(ii) Write down the range of $\mathrm { f } ( x )$.\\
(iii) Sketch the curve $y = \mathrm { f } \left( \frac { 1 } { 3 } x \right) - 1$.
\hfill \mbox{\textit{OCR MEI C3 2012 Q4 [6]}}