| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a straightforward inverse function question requiring algebraic manipulation to find f(x) from f^(-1)(x), determining the range of f^(-1) as the domain of f, and composing two given functions. All steps are routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempt to find \((f^{-1})^{-1}\) | M1 | |
| \(2xy = 1 - 5x\) or \(\frac{1}{2x} = y + \frac{5}{2}\) — Allow 1 sign error | A1 | Or with \(x/y\) transposed. |
| \(x = \frac{1}{2y+5}\) oe — Allow 1 sign error (total) | A1 | Or with \(x/y\) transposed. Allow \(x = -\frac{\frac{1}{2}}{y+\frac{5}{2}}\) |
| \((f(x)) = \frac{1}{2x+5}\) for \(x \geq -\frac{9}{4}\) | A1 B1 | Allow \(-\frac{\frac{1}{2}}{x+\frac{5}{2}}\). Condone \(x > -\frac{9}{4}\), \(\left(-\frac{9}{4}, \infty\right)\) |
| \(\left(\text{Allow } -\frac{9}{4} \leq x \leq \infty\right)\) | [5] | (etc.) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f^{-1}\!\left(\frac{1}{x}\right) = \frac{1 - \frac{5}{x}}{\frac{2}{x}}\) | M1 | Reasonable attempt to find \(f^{-1}\!\left(\frac{1}{x}\right)\) |
| \(\frac{x-5}{2}\) or \(\frac{1}{2}x - \frac{5}{2}\) | A1 [2] |
## Question 6(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to find $(f^{-1})^{-1}$ | M1 | |
| $2xy = 1 - 5x$ or $\frac{1}{2x} = y + \frac{5}{2}$ — Allow 1 sign error | A1 | Or with $x/y$ transposed. |
| $x = \frac{1}{2y+5}$ oe — Allow 1 sign error (total) | A1 | Or with $x/y$ transposed. Allow $x = -\frac{\frac{1}{2}}{y+\frac{5}{2}}$ |
| $(f(x)) = \frac{1}{2x+5}$ for $x \geq -\frac{9}{4}$ | A1 B1 | Allow $-\frac{\frac{1}{2}}{x+\frac{5}{2}}$. Condone $x > -\frac{9}{4}$, $\left(-\frac{9}{4}, \infty\right)$ |
| $\left(\text{Allow } -\frac{9}{4} \leq x \leq \infty\right)$ | [5] | (etc.) |
## Question 6(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f^{-1}\!\left(\frac{1}{x}\right) = \frac{1 - \frac{5}{x}}{\frac{2}{x}}$ | M1 | Reasonable attempt to find $f^{-1}\!\left(\frac{1}{x}\right)$ |
| $\frac{x-5}{2}$ or $\frac{1}{2}x - \frac{5}{2}$ | A1 [2] | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{8da9e73a-3126-471b-b904-25e3c156f6bf-2_519_670_1640_735}
The diagram shows the graph of $y = \mathrm { f } ^ { - 1 } ( x )$, where $\mathrm { f } ^ { - 1 }$ is defined by $\mathrm { f } ^ { - 1 } ( x ) = \frac { 1 - 5 x } { 2 x }$ for $0 < x \leqslant 2$.\\
(i) Find an expression for $\mathrm { f } ( x )$ and state the domain of f .\\
(ii) The function g is defined by $\mathrm { g } ( x ) = \frac { 1 } { x }$ for $x \geqslant 1$. Find an expression for $\mathrm { f } ^ { - 1 } \mathrm {~g} ( x )$, giving your answer in the form $a x + b$, where $a$ and $b$ are constants to be found.
\hfill \mbox{\textit{CAIE P1 2015 Q6 [7]}}