| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Verify composite identity |
| Difficulty | Moderate -0.3 This is a slightly below-average A-level question. Part (i) requires straightforward algebraic substitution to verify f(f(x))=x and recognizing that this means f is self-inverse. Part (ii) involves routine verification that g(-x)=g(x) and stating the symmetry property. Both parts are standard textbook exercises with no problem-solving insight required, though the algebra in part (i) needs care. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = f\left(\frac{1-x}{1+x}\right) = \frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}} = \frac{\frac{1+x-1+x}{1+x}}{\frac{1+x+1-x}{1+x}} = \frac{2x}{2} = x^*\) | M1 | substituting \((1-x)/(1+x)\) for \(x\) in \(f(x)\) |
| \(= \frac{1+x-1+x}{1+x+1-x} = \frac{2x}{2} = x^*\) | A1 | correctly simplified to \(x\) |
| \(f^{-1}(x) = f(x) = (1-x)/(1+x)\) | B1 | or just \(f^{-1}(x) = f(x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(g(-x) = \frac{1-(-x)^2}{1+(-x)^2} = \frac{1-x^2}{1+x^2} = g(x)\) | M1 | substituting \(-x\) for \(x\) in \(g(x)\) condone use of 'f' for g |
| \(= \frac{1-x^2}{1+x^2} = g(x)\) | A1 | must indicate that \(g(-x) = g(x)\) somewhere |
| Graph is symmetrical about the y-axis. | B1 | allow 'reflected', 'reflection' for symmetrical |
| Total: [3] | if brackets are omitted or misplaced allow M1A0; condone use of 'f' for g; must state axis (y-axis or x = 0) |
**(i)**
$f(x) = f\left(\frac{1-x}{1+x}\right) = \frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}} = \frac{\frac{1+x-1+x}{1+x}}{\frac{1+x+1-x}{1+x}} = \frac{2x}{2} = x^*$ | M1 | substituting $(1-x)/(1+x)$ for $x$ in $f(x)$
$= \frac{1+x-1+x}{1+x+1-x} = \frac{2x}{2} = x^*$ | A1 | correctly simplified to $x$ | NB AG
$f^{-1}(x) = f(x) = (1-x)/(1+x)$ | B1 | or just $f^{-1}(x) = f(x)$
**Total: [3]**
**(ii)**
$g(-x) = \frac{1-(-x)^2}{1+(-x)^2} = \frac{1-x^2}{1+x^2} = g(x)$ | M1 | substituting $-x$ for $x$ in $g(x)$ condone use of 'f' for g
$= \frac{1-x^2}{1+x^2} = g(x)$ | A1 | must indicate that $g(-x) = g(x)$ somewhere
Graph is symmetrical about the y-axis. | B1 | allow 'reflected', 'reflection' for symmetrical
**Total: [3]** | | if brackets are omitted or misplaced allow M1A0; condone use of 'f' for g; must state axis (y-axis or x = 0)
---7 (i) The function $\mathrm { f } ( x )$ is defined by
$$f ( x ) = \frac { 1 - x } { 1 + x } , x \neq - 1$$
Show that $\mathrm { f } ( \mathrm { f } ( x ) ) = x$.\\
Hence write down $\mathrm { f } ^ { - 1 } ( x )$.\\
(ii) The function $\mathrm { g } ( x )$ is defined for all real $x$ by
$$\mathrm { g } ( x ) = \frac { 1 - x ^ { 2 } } { 1 + x ^ { 2 } }$$
Prove that $\mathrm { g } ( x )$ is even. Interpret this result in terms of the graph of $y = \mathrm { g } ( x )$.
\hfill \mbox{\textit{OCR MEI C3 2015 Q7 [6]}}