| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.8 This question requires forming the composite function fg(x), setting it equal to x to create a rational equation, clearing denominators to get a quadratic, then applying the discriminant condition for equal roots to find k, followed by solving the resulting equation. It involves multiple algebraic techniques (composition, rational equations, discriminant) and careful algebraic manipulation across two connected parts, making it moderately challenging but still within standard P1 scope. |
| 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 | Guidance |
|---|---|---|
| (i) \(fg(x) = \frac{36}{2 - x} - 2k = x\) | M1 | Knowing to put \(g\) into \(f\) (not \(gf\)) |
| \(x^2 + 2kx - 2x + 36 - 4k\) | A1 | Correct quadratic. |
| \((2k - 2)^2 = 4(36 - 4k)\) | M1 | Any use of \(b^2 - 4ac\) on quadratic = 0 |
| \(k = 5\) or \(-7\) | A1 | Both correct. [4] |
| (ii) \(x^2 + 8x + 16 = 0, \quad x^2 - 16x + 64 = 0\) | M1 | Substituting one of the values of \(k\). |
| \(x = -4\) or \(x = 8\). | A1 A1 | [3] |
$f : x \mapsto 4x - 2k, \quad g : x \mapsto \frac{9}{2 - x}$
**(i)** $fg(x) = \frac{36}{2 - x} - 2k = x$ | M1 | Knowing to put $g$ into $f$ (not $gf$)
$x^2 + 2kx - 2x + 36 - 4k$ | A1 | Correct quadratic.
$(2k - 2)^2 = 4(36 - 4k)$ | M1 | Any use of $b^2 - 4ac$ on quadratic = 0
$k = 5$ or $-7$ | A1 | Both correct. [4]
**(ii)** $x^2 + 8x + 16 = 0, \quad x^2 - 16x + 64 = 0$ | M1 | Substituting one of the values of $k$.
$x = -4$ or $x = 8$. | A1 A1 | [3]8 Functions f and g are defined by
$$\begin{array} { l l }
\mathrm { f } : x \mapsto 4 x - 2 k & \text { for } x \in \mathbb { R } , \text { where } k \text { is a constant, } \\
\mathrm { g } : x \mapsto \frac { 9 } { 2 - x } & \text { for } x \in \mathbb { R } , x \neq 2 .
\end{array}$$
(i) Find the values of $k$ for which the equation $\mathrm { fg } ( x ) = x$ has two equal roots.\\
(ii) Determine the roots of the equation $\operatorname { fg } ( x ) = x$ for the values of $k$ found in part (i).
\hfill \mbox{\textit{CAIE P1 2008 Q8 [7]}}