| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Complete the square |
| Difficulty | Standard +0.3 This is a multi-part question covering completing the square, domain restrictions for composite functions, and finding inverse functions. Part (a) is routine completing the square. Parts (b) and (c) require understanding that gf exists when range of f ⊆ domain of g, which is standard but requires careful thinking. Part (d) is a standard inverse function calculation with domain restriction. Overall slightly easier than average due to straightforward techniques and clear structure, though part (b) requires some conceptual understanding of composite function domains. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2x-3)^2+4\) | B1 B1 | Or \(a=-3\), \(b=4\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| their \((2x-3)^2+4[\lt]8\) OR \(4x^2-12x+13[\lt]8\) | *M1 | Linking quadratic with 8 |
| \((2x-3)^2\lt4\) leading to \(-2\lt2x-3\lt2\) OR \(4x^2-12x+5\lt0\) leading to \((2x-1)(2x-5)\lt0\) | DM1 | Simplify to 3-term quadratic and solve. Condone no method shown |
| \(\frac{1}{2}\lt x\lt2\frac{1}{2}\) leading to [LEAST] \(p=\frac{1}{2}\), [GREATEST] \(q=2\frac{1}{2}\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{gf}(x)=12x^2-36x+40\) | B1 | OE \(\text{gf}(x)=3(2x-3)^2+13\) |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y=(2x-3)^2+4\) leading to \((2x-3)^2=y-4\) leading to \(2x-3=[\pm]\sqrt{y-4}\) | *M1 | |
| \(2x=3[\pm]\sqrt{y-4}\) leading to \(x=\frac{3}{2}[\pm]\frac{\sqrt{y-4}}{2}\) | DM1 | |
| \(h^{-1}(x)=\frac{3}{2}-\frac{\sqrt{x-4}}{2}\) | A1 | |
| Total | 3 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2x-3)^2+4$ | B1 B1 | Or $a=-3$, $b=4$ |
| **Total** | **2** | |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| their $(2x-3)^2+4[\lt]8$ OR $4x^2-12x+13[\lt]8$ | *M1 | Linking quadratic with 8 |
| $(2x-3)^2\lt4$ leading to $-2\lt2x-3\lt2$ OR $4x^2-12x+5\lt0$ leading to $(2x-1)(2x-5)\lt0$ | DM1 | Simplify to 3-term quadratic and solve. Condone no method shown |
| $\frac{1}{2}\lt x\lt2\frac{1}{2}$ leading to [LEAST] $p=\frac{1}{2}$, [GREATEST] $q=2\frac{1}{2}$ | A1 | |
| **Total** | **3** | |
---
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{gf}(x)=12x^2-36x+40$ | B1 | OE $\text{gf}(x)=3(2x-3)^2+13$ |
| **Total** | **1** | |
---
## Question 9(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=(2x-3)^2+4$ leading to $(2x-3)^2=y-4$ leading to $2x-3=[\pm]\sqrt{y-4}$ | *M1 | |
| $2x=3[\pm]\sqrt{y-4}$ leading to $x=\frac{3}{2}[\pm]\frac{\sqrt{y-4}}{2}$ | DM1 | |
| $h^{-1}(x)=\frac{3}{2}-\frac{\sqrt{x-4}}{2}$ | A1 | |
| **Total** | **3** | |
---9
\begin{enumerate}[label=(\alph*)]
\item Express $4 x ^ { 2 } - 12 x + 13$ in the form $( 2 x + a ) ^ { 2 } + b$, where $a$ and $b$ are constants.\\
The function f is defined by $\mathrm { f } ( x ) = 4 x ^ { 2 } - 12 x + 13$ for $p < x < q$, where $p$ and $q$ are constants. The function g is defined by $\mathrm { g } ( x ) = 3 x + 1$ for $x < 8$.
\item Given that it is possible to form the composite function gf , find the least possible value of $p$ and the greatest possible value of $q$.
\item Find an expression for $\operatorname { gf } ( x )$.\\
The function h is defined by $\mathrm { h } ( x ) = 4 x ^ { 2 } - 12 x + 13$ for $x < 0$.
\item Find an expression for $\mathrm { h } ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q9 [9]}}