| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This question requires finding an inverse function and solving a composite function equation, both standard P1 techniques. Part (i) involves algebraic manipulation to find f^(-1)(x), and part (ii) requires composing two functions then solving a straightforward equation. The restricted domains add minor complexity but the overall problem follows routine procedures without requiring novel insight. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y=\frac{2}{x^2-1} \Rightarrow x^2=\frac{2}{y}+1\) OE | M1 | |
| \(x=(\pm)\sqrt{\frac{2}{y}+1}\) OE | A1 | With or without \(x/y\) interchanged |
| \(f^{-1}(x)=-\sqrt{\frac{2}{x}+1}\) OE | A1 | Minus sign obligatory. Must be a function of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{2}{x^2-1}\right)^2 + 1 = 5\) | B1 | |
| \(\frac{2}{x^2-1} = (\pm)2\) OE OR \(x^4 - 2x^2 = 0\) OE | B1 | Condone \(x^2 = 0\) as an additional solution |
| \(x^2 - 1 = (\pm)1 \Rightarrow x^2 = 2\) (or 0) | ||
| \(x = -\sqrt{2}\) or \(-1.41\) only | ||
| Total | 4 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=\frac{2}{x^2-1} \Rightarrow x^2=\frac{2}{y}+1$ OE | M1 | |
| $x=(\pm)\sqrt{\frac{2}{y}+1}$ OE | A1 | With or without $x/y$ interchanged |
| $f^{-1}(x)=-\sqrt{\frac{2}{x}+1}$ OE | A1 | Minus sign obligatory. Must be a function of $x$ |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{2}{x^2-1}\right)^2 + 1 = 5$ | B1 | |
| $\frac{2}{x^2-1} = (\pm)2$ OE OR $x^4 - 2x^2 = 0$ OE | B1 | Condone $x^2 = 0$ as an additional solution |
| $x^2 - 1 = (\pm)1 \Rightarrow x^2 = 2$ (or 0) | | |
| $x = -\sqrt{2}$ or $-1.41$ only | | |
| **Total** | **4** | |
---6 The functions f and g are defined by
$$\begin{aligned}
& \mathrm { f } ( x ) = \frac { 2 } { x ^ { 2 } - 1 } \text { for } x < - 1 \\
& \mathrm {~g} ( x ) = x ^ { 2 } + 1 \text { for } x > 0
\end{aligned}$$
(i) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
(ii) Solve the equation $\operatorname { gf } ( x ) = 5$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q6 [7]}}