| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation with inverses |
| Difficulty | Moderate -0.3 This is a straightforward composite and inverse functions question requiring standard techniques: evaluating a composition, sketching a linear function with its inverse (reflection in y=x), finding inverses algebraically, and solving a simple equation. All parts are routine P1 exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02m Graphs of functions: difference between plotting and sketching1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| solution of this \(x = 7.5\) or \(7\frac{1}{2}\) | M1, DM1, A1 | Puts \(g\) into \(f\) – order correct (or \(f = 3 - x - 3\frac{1}{2}\)). Correct method of solution (or \(f = 3 \Rightarrow x = 7\frac{1}{2}\)). Correct only |
| (ii) Graph: | B1, B1, B1 | Graph of \(f(x)\) – needs \(m>1\), \(+ve\) y intercept. Graph of \(f^{-1}(x)\) – needs \(m<1\), \(+ve\) x-intercept. Some idea of reflection in \(y = x\) – stated ok |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 6 + (3x+2)\) makes x the subject and swops x and y \(\Rightarrow \frac{1}{2}(6/x - 3)\) | B1, M1, A1 | Correct only. Any valid method. Correct only – any form |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 2\) or \(x = -4.5\) | M1, A1 | Complete method of solution. Correct only |
$f: x \mapsto 3x + 2$
$g: x \mapsto -6 + (2x + 3)$
**(i)** $fg(x) = 3$
$18 - (2x+3) + 2 = 3$
solution of this $x = 7.5$ or $7\frac{1}{2}$ | M1, DM1, A1 | Puts $g$ into $f$ – order correct (or $f = 3 - x - 3\frac{1}{2}$). Correct method of solution (or $f = 3 \Rightarrow x = 7\frac{1}{2}$). Correct only
**(ii)** Graph: | B1, B1, B1 | Graph of $f(x)$ – needs $m>1$, $+ve$ y intercept. Graph of $f^{-1}(x)$ – needs $m<1$, $+ve$ x-intercept. Some idea of reflection in $y = x$ – stated ok
**(iii)** $f^{-1}(x) = \frac{1}{3}(x-2)$
$y = 6 + (3x+2)$ makes x the subject and swops x and y $\Rightarrow \frac{1}{2}(6/x - 3)$ | B1, M1, A1 | Correct only. Any valid method. Correct only – any form
$\frac{1}{3}(x-2) = \frac{1}{2}(6/x - 3) \Rightarrow 2x^2 + 5x = 18$
$x = 2$ or $x = -4.5$ | M1, A1 | Complete method of solution. Correct only10 The functions $f$ and $g$ are defined by
$$\begin{array} { l l }
\mathrm { f } : x \mapsto 3 x + 2 , & x \in \mathbb { R } , \\
\mathrm {~g} : x \mapsto \frac { 6 } { 2 x + 3 } , & x \in \mathbb { R } , x \neq - 1.5 .
\end{array}$$
(i) Find the value of $x$ for which $\operatorname { fg } ( x ) = 3$.\\
(ii) Sketch, in a single diagram, the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$, making clear the relationship between the two graphs.\\
(iii) Express each of $\mathrm { f } ^ { - 1 } ( x )$ and $\mathrm { g } ^ { - 1 } ( x )$ in terms of $x$, and solve the equation $\mathrm { f } ^ { - 1 } ( x ) = \mathrm { g } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{CAIE P1 2002 Q10 [11]}}