| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation with inverses |
| Difficulty | Standard +0.3 This is a standard composite/inverse functions question requiring routine algebraic manipulation. Part (i) is textbook inverse finding, part (ii) is straightforward substitution, and part (iii) involves setting up a quadratic with equal roots (discriminant = 0). While multi-part, each step uses familiar techniques without requiring novel insight, 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 |
| \(y = \frac{2}{3-2x} \rightarrow y(3-2x) = 2 \rightarrow 3 - 2x = \frac{2}{y}\) | M1 | Correct first 2 steps |
| \(\rightarrow 2x = 3 - \frac{2}{y} \rightarrow f^{-1}(x) = \frac{3}{2} - \frac{1}{x}\) | M1 A1 | Correct order of operations, any correct form with \(f(x)\) or \(y =\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(gf(-1) = 3\), \(f(-1) = \frac{2}{5}\) | M1 | Correct first step |
| \(\frac{8}{5} + a = 3 \rightarrow a = \frac{7}{5}\) | M1 A1 | Forms an equation in \(a\) and finds \(a\), OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(g^{-1}(x) = \frac{x-a}{4} = f^{-1}(x)\) | M1 | Finding \(g^{-1}(x)\) and equating to their \(f^{-1}(x)\) even if \(a = 7/5\) |
| \(\rightarrow x^2 - x(a+6) + 4\ (= 0)\) | M1 | Use of \(b^2 - 4ac\) on a quadratic with \(a\) in a coefficient |
| Solving \((a+6)^2 = 16\) or \(a^2 + 12a + 20\ (= 0)\) | M1 | Solution of a 3 term quadratic |
| \(\rightarrow a = -2\) or \(-10\) | A1 |
## Question 9:
**Part 9(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{2}{3-2x} \rightarrow y(3-2x) = 2 \rightarrow 3 - 2x = \frac{2}{y}$ | M1 | Correct first 2 steps |
| $\rightarrow 2x = 3 - \frac{2}{y} \rightarrow f^{-1}(x) = \frac{3}{2} - \frac{1}{x}$ | M1 A1 | Correct order of operations, any correct form with $f(x)$ or $y =$ |
**Part 9(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $gf(-1) = 3$, $f(-1) = \frac{2}{5}$ | M1 | Correct first step |
| $\frac{8}{5} + a = 3 \rightarrow a = \frac{7}{5}$ | M1 A1 | Forms an equation in $a$ and finds $a$, OE |
**Part 9(iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $g^{-1}(x) = \frac{x-a}{4} = f^{-1}(x)$ | M1 | Finding $g^{-1}(x)$ and equating to their $f^{-1}(x)$ even if $a = 7/5$ |
| $\rightarrow x^2 - x(a+6) + 4\ (= 0)$ | M1 | Use of $b^2 - 4ac$ on a quadratic with $a$ in a coefficient |
| Solving $(a+6)^2 = 16$ or $a^2 + 12a + 20\ (= 0)$ | M1 | Solution of a 3 term quadratic |
| $\rightarrow a = -2$ or $-10$ | A1 | |9 The function f is defined by $\mathrm { f } : x \mapsto \frac { 2 } { 3 - 2 x }$ for $x \in \mathbb { R } , x \neq \frac { 3 } { 2 }$.\\
(i) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
The function g is defined by $\mathrm { g } : x \mapsto 4 x + a$ for $x \in \mathbb { R }$, where $a$ is a constant.\\
(ii) Find the value of $a$ for which $\operatorname { gf } ( - 1 ) = 3$.\\
(iii) Find the possible values of $a$ given that the equation $\mathrm { f } ^ { - 1 } ( x ) = \mathrm { g } ^ { - 1 } ( x )$ has two equal roots.\\
\hfill \mbox{\textit{CAIE P1 2017 Q9 [10]}}