| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation with inverses |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on composite and inverse functions requiring standard techniques: solving a quadratic with equal roots (discriminant = 0), evaluating a composite function and solving a linear equation, and finding an inverse by swapping and rearranging. All parts are routine P1/AS-level exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks |
|---|---|
| \(f: x \mapsto k - x\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(k - x = \frac{9}{x+2}\) | M1 | Forming a quadratic equation. |
| → \(x^2 + (2-k)x + 9 - 2k = 0\) | M1 | Use of \(b^2 - 4ac\) on quadratic = 0 |
| Use of \(b^2 - 4ac\) | DM1 | DM1 for solution. A1 both correct. |
| → \(a = 4\) or \(-8\) | M1 | Any valid method. |
| \(k = 4\), root is \(\frac{-b}{2a} = 1\) | A1 | Both correct. |
| \(k = -8\), root is \(-5\) | A1 | |
| (ii) \(fg(x) = 6 - \frac{9}{x+2}\) | M1 | Must be fg, not for gf. |
| Equates and solves with 5 | DM1 | Reasonable algebra. |
| \(x = 7\) | A1 | co. [\(g(x)=1\) M1 → DM1 \(x=7\) A1] |
| (iii) \(y = \frac{9}{x+2} \to x = \frac{9}{y} - 2\) | M1 | Virtually correct algebra. Allow + for −. |
| \(g^{-1}(x) = \frac{9}{x} - 2\) or \(\frac{9-2x}{x}\) | A1 | Correct and in terms of x. |
$f: x \mapsto k - x$ |
$g: x \mapsto \frac{9}{x+2}$
**(i)** $k - x = \frac{9}{x+2}$ | M1 | Forming a quadratic equation.
→ $x^2 + (2-k)x + 9 - 2k = 0$ | M1 | Use of $b^2 - 4ac$ on quadratic = 0
Use of $b^2 - 4ac$ | DM1 | DM1 for solution. A1 both correct.
→ $a = 4$ or $-8$ | M1 | Any valid method.
$k = 4$, root is $\frac{-b}{2a} = 1$ | A1 | Both correct.
$k = -8$, root is $-5$ | A1
**(ii)** $fg(x) = 6 - \frac{9}{x+2}$ | M1 | Must be fg, not for gf.
Equates and solves with 5 | DM1 | Reasonable algebra.
$x = 7$ | A1 | co. [$g(x)=1$ M1 → DM1 $x=7$ A1]
**(iii)** $y = \frac{9}{x+2} \to x = \frac{9}{y} - 2$ | M1 | Virtually correct algebra. Allow + for −.
$g^{-1}(x) = \frac{9}{x} - 2$ or $\frac{9-2x}{x}$ | A1 | Correct and in terms of x.
**Note:** DM1 for quadratic. Quadratic must be set to 0. Factors. Attempt at two brackets. Each bracket set to 0 and solved. Formula. Correct formula. Correct use, but allow for numerical slips in $b^2$ and $-4ac$.
**Total: [6] [3] [2]**11 Functions $f$ and $g$ are defined by
$$\begin{array} { l l }
\mathrm { f } : x \mapsto k - x & \text { for } x \in \mathbb { R } , \text { where } k \text { is a constant, } \\
\mathrm { g } : x \mapsto \frac { 9 } { x + 2 } & \text { for } x \in \mathbb { R } , x \neq - 2 .
\end{array}$$
(i) Find the values of $k$ for which the equation $\mathrm { f } ( x ) = \mathrm { g } ( x )$ has two equal roots and solve the equation $\mathrm { f } ( x ) = \mathrm { g } ( x )$ in these cases.\\
(ii) Solve the equation $\operatorname { fg } ( x ) = 5$ when $k = 6$.\\
(iii) Express $\mathrm { g } ^ { - 1 } ( x )$ in terms of $x$.
\hfill \mbox{\textit{CAIE P1 2006 Q11 [11]}}