| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a straightforward composite and inverse functions question requiring standard techniques: finding inverses of linear and reciprocal functions (routine), sketching f and f^{-1} with reflection in y=x (standard), and solving fg(x) = 5-kx to find when no solutions exist (requires some algebraic manipulation but follows a clear method). All parts are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.02m Graphs of functions: difference between plotting and sketching1.02q Use intersection points: of graphs to solve equations1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \(f : x \mapsto 2x + 5\) \(g : x \mapsto \frac{8}{x - 3}\) | ||
| (i) \(f^{-1} = \frac{1}{2}(x - 5)\) | B1 M1 A1 | co Attempt at x the subject. co but (f(x)) Allow if a linear denominator. [4] |
| \(g^{-1} = \frac{8}{x} + 3, x \neq 0\) | B1♦ | +ve gradient, +ve y intercept +ve gradient, +ve y intercept States or shows the line y = x as a line of symmetry. |
| (ii) | B1 B1 B1 | [3] |
| (iii) fg(x) = \(\frac{16}{x - 3} + 5\) | B1 | co |
| Forms eqn \(\frac{16}{x - 3} + 5 = 5 - kx\) | M1 | Must lead to a quadratic |
| \(kx^2 - 3kx + 16 = 0\) | ||
| Uses \(b^2 = 4ac \to k = \frac{64}{9}\) or 0 | M1 A1 | Use of \(b^2 - 4ac\) even if <0, >0. co Condone < |
| Set of values \(0 < k < \frac{64}{9}\) | A1 | co [5] |
$f : x \mapsto 2x + 5$ $g : x \mapsto \frac{8}{x - 3}$ | |
(i) $f^{-1} = \frac{1}{2}(x - 5)$ | B1 M1 A1 | co Attempt at x the subject. co but (f(x)) Allow if a linear denominator. [4]
$g^{-1} = \frac{8}{x} + 3, x \neq 0$ | B1♦ | +ve gradient, +ve y intercept +ve gradient, +ve y intercept States or shows the line y = x as a line of symmetry.
(ii) | B1 B1 B1 | [3]
(iii) fg(x) = $\frac{16}{x - 3} + 5$ | B1 | co
Forms eqn $\frac{16}{x - 3} + 5 = 5 - kx$ | M1 | Must lead to a quadratic
$kx^2 - 3kx + 16 = 0$ | |
Uses $b^2 = 4ac \to k = \frac{64}{9}$ or 0 | M1 A1 | Use of $b^2 - 4ac$ even if <0, >0. co Condone <
Set of values $0 < k < \frac{64}{9}$ | A1 | co [5]Functions $f$ and $g$ are defined by
$$f : x \mapsto 2x + 5 \quad \text{for } x \in \mathbb{R},$$
$$g : x \mapsto \frac{8}{x - 3} \quad \text{for } x \in \mathbb{R}, x \neq 3.$$
\begin{enumerate}[label=(\roman*)]
\item Obtain expressions, in terms of $x$, for $f^{-1}(x)$ and $g^{-1}(x)$, stating the value of $x$ for which $g^{-1}(x)$ is not defined. [4]
\item Sketch the graphs of $y = f(x)$ and $y = f^{-1}(x)$ on the same diagram, making clear the relationship between the two graphs. [3]
\item Given that the equation $fg(x) = 5 - kx$, where $k$ is a constant, has no solutions, find the set of possible values of $k$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2012 Q10 [12]}}