| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This is a straightforward composite function question requiring substitution of g(x) into f(x), solving a simple equation involving a fraction squared, then finding an inverse by reversing operations. All steps are routine for P1 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{fg}(x) = \frac{1}{(2x+1)^2} - 1\) | B1 | SOI |
| \(\frac{1}{(2x+1)^2} - 1 = 3\) leading to \(4(2x+1)^2 = 1\) or \(\frac{1}{(2x+1)} = [\pm]2\) or \(16x^2 + 16x + 3 = 0\) | M1 | Setting \(\text{fg}(x) = 3\) and reaching a stage before \(2x+1 = \pm\frac{1}{2}\) or reaching a 3 term quadratic in \(x\) |
| \(2x+1 = \pm\frac{1}{2}\) or \(2x+1 = -\frac{1}{2}\) or \((4x+1)(4x+3)[=0]\) | A1 | Or formula or completing square on quadratic |
| \(x = -\frac{3}{4}\) only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 - 1 = 3\) | M1 | |
| \(g(x) = -2\) | A1 | |
| \(\frac{1}{(2x+1)} = -2\) | M1 | |
| \(x = -\frac{3}{4}\) only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{1}{(2x+1)^2} - 1\) leading to \((2x+1)^2 = \frac{1}{y+1}\) leading to \(2x+1 = [\pm]\frac{1}{\sqrt{y+1}}\) | *M1 | Obtain \(2x+1\) or \(2y+1\) as the subject |
| \(x = [\pm]\frac{1}{2\sqrt{y+1}} - \frac{1}{2}\) | DM1 | Make \(x\) (or \(y\)) the subject |
| \(-\frac{1}{2\sqrt{x+1}} - \frac{1}{2}\) | A1 | OE e.g. \(-\frac{\sqrt{x+1}}{2x+2} - \frac{1}{2}\), \(-\left(\sqrt{\frac{-x}{4x+4}} + \frac{1}{4} + \frac{1}{2}\right)\) |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{fg}(x) = \frac{1}{(2x+1)^2} - 1$ | B1 | SOI |
| $\frac{1}{(2x+1)^2} - 1 = 3$ leading to $4(2x+1)^2 = 1$ or $\frac{1}{(2x+1)} = [\pm]2$ or $16x^2 + 16x + 3 = 0$ | M1 | Setting $\text{fg}(x) = 3$ and reaching a stage before $2x+1 = \pm\frac{1}{2}$ or reaching a 3 term quadratic in $x$ |
| $2x+1 = \pm\frac{1}{2}$ or $2x+1 = -\frac{1}{2}$ or $(4x+1)(4x+3)[=0]$ | A1 | Or formula or completing square on quadratic |
| $x = -\frac{3}{4}$ only | A1 | |
**Alternative method for Question 8(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - 1 = 3$ | M1 | |
| $g(x) = -2$ | A1 | |
| $\frac{1}{(2x+1)} = -2$ | M1 | |
| $x = -\frac{3}{4}$ only | A1 | |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{1}{(2x+1)^2} - 1$ leading to $(2x+1)^2 = \frac{1}{y+1}$ leading to $2x+1 = [\pm]\frac{1}{\sqrt{y+1}}$ | *M1 | Obtain $2x+1$ or $2y+1$ as the subject |
| $x = [\pm]\frac{1}{2\sqrt{y+1}} - \frac{1}{2}$ | DM1 | Make $x$ (or $y$) the subject |
| $-\frac{1}{2\sqrt{x+1}} - \frac{1}{2}$ | A1 | OE e.g. $-\frac{\sqrt{x+1}}{2x+2} - \frac{1}{2}$, $-\left(\sqrt{\frac{-x}{4x+4}} + \frac{1}{4} + \frac{1}{2}\right)$ |
---8 Functions f and g are defined as follows:
$$\begin{aligned}
& \mathrm { f } : x \mapsto x ^ { 2 } - 1 \text { for } x < 0 \\
& \mathrm {~g} : x \mapsto \frac { 1 } { 2 x + 1 } \text { for } x < - \frac { 1 } { 2 }
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\operatorname { fg } ( x ) = 3$.
\item Find an expression for $( \mathrm { fg } ) ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q8 [7]}}