| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Verify composite identity |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on composite functions requiring routine algebraic manipulation. Part (a) involves solving a quadratic in √x using standard techniques, and part (b) requires expanding gh(x) and comparing coefficients—both are standard textbook exercises with no novel insight required. Slightly easier than average due to the mechanical nature of the tasks. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02u Functions: definition and vocabulary (domain, range, mapping)1.03f Circle properties: angles, chords, tangents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^{\frac{1}{2}} = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}\) | M1 A1 | OE. Answer must come from formula or completing square. If M0A0 scored then SC B1 for \(2 \pm \sqrt{3}\) only. |
| \(x = (2 \pm \sqrt{3})^2\) | M1 | Attempt to square *their* \(2 \pm \sqrt{3}\) |
| \(7 + 4\sqrt{3}\), \(7 - 4\sqrt{3}\) | A1 | Accept \(7 \pm 4\sqrt{3}\) or \(a=7, b=\pm4, c=3\). SC B1 instead of second M1A1 for correct final answer only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-4x^{\frac{1}{2}} + 1 = 0\) leading to \((x+1)^2 = 16x\) leading to \(x^2 - 14x + 1 = 0\) | *M1 A1 | OE |
| \(x = \frac{14 \pm \sqrt{196-4}}{2}\) | DM1 | Attempt to solve for \(x\) |
| \(7 + 4\sqrt{3}\), \(7 - 4\sqrt{3}\) | A1 | SC B1 instead of second M1A1 for correct final answer only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(gh(x) = m\left(x^{\frac{1}{2}} - 2\right)^2 + n\) | M1 | SOI |
| \(gh(x) = m\left(x - 4x^{\frac{1}{2}} + 4\right) + n \equiv x - 4x^{\frac{1}{2}} + 1\) | A1 | SOI |
| \(m = 1\), \(n = -3\) | A1 A1 | WWW |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^{\frac{1}{2}} = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$ | M1 A1 | OE. Answer must come from formula or completing square. If M0A0 scored then SC B1 for $2 \pm \sqrt{3}$ only. |
| $x = (2 \pm \sqrt{3})^2$ | M1 | Attempt to square *their* $2 \pm \sqrt{3}$ |
| $7 + 4\sqrt{3}$, $7 - 4\sqrt{3}$ | A1 | Accept $7 \pm 4\sqrt{3}$ or $a=7, b=\pm4, c=3$. SC B1 instead of second M1A1 for correct final answer only. |
**Alternative method for 9(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-4x^{\frac{1}{2}} + 1 = 0$ leading to $(x+1)^2 = 16x$ leading to $x^2 - 14x + 1 = 0$ | *M1 A1 | OE |
| $x = \frac{14 \pm \sqrt{196-4}}{2}$ | DM1 | Attempt to solve for $x$ |
| $7 + 4\sqrt{3}$, $7 - 4\sqrt{3}$ | A1 | SC B1 instead of second M1A1 for correct final answer only. |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $gh(x) = m\left(x^{\frac{1}{2}} - 2\right)^2 + n$ | M1 | SOI |
| $gh(x) = m\left(x - 4x^{\frac{1}{2}} + 4\right) + n \equiv x - 4x^{\frac{1}{2}} + 1$ | A1 | SOI |
| $m = 1$, $n = -3$ | A1 A1 | WWW |
---9 Functions f, g and h are defined as follows:
$$\begin{aligned}
& \mathrm { f } : x \mapsto x - 4 x ^ { \frac { 1 } { 2 } } + 1 \quad \text { for } x \geqslant 0 \\
& \mathrm {~g} : x \mapsto m x ^ { 2 } + n \quad \text { for } x \geqslant - 2 , \text { where } m \text { and } n \text { are constants, } \\
& \mathrm { h } : x \mapsto x ^ { \frac { 1 } { 2 } } - 2 \quad \text { for } x \geqslant 0 .
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\mathrm { f } ( x ) = 0$, giving your solutions in the form $x = a + b \sqrt { c }$, where $a , b$ and $c$ are integers.
\item Given that $\mathrm { f } ( x ) \equiv \mathrm { gh } ( x )$, find the values of $m$ and $n$.\\
\includegraphics[max width=\textwidth, alt={}, center]{05e75fa2-81ae-44b1-b073-4100f5d911e0-16_652_1045_255_550}
The diagram shows a circle with centre $A$ of radius 5 cm and a circle with centre $B$ of radius 8 cm . The circles touch at the point $C$ so that $A C B$ is a straight line. The tangent at the point $D$ on the smaller circle intersects the larger circle at $E$ and passes through $B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q9 [8]}}