| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.8 This is a straightforward composite function question requiring basic substitution and solving simple linear equations. Finding constants from given conditions (ff(2)=10 and g^(-1)(2)=3) involves routine algebraic manipulation, and finding fg(x) is direct function composition. Below average difficulty as it's purely procedural with no conceptual challenges. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks |
|---|---|
| \(f^2(2) = 18 + 4a = 10 \to a = -2\) | M1 A1 |
| \(g^{-1}(x) = \frac{b - x}{2} \to \frac{b - 2}{2} = 3 \to b = 8\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(fg(x) = 3(b - 2x) + a = 22 - 6x\) | B1 |
**Question 2:**
$f: x \mapsto 3x + a$, $g: x \mapsto b - 2x$
**(i)**
$f^2(x) = 3(3x + a) + a$
$f^2(2) = 18 + 4a = 10 \to a = -2$ | M1 A1
$g^{-1}(x) = \frac{b - x}{2} \to \frac{b - 2}{2} = 3 \to b = 8$ | M1 A1
or $g(3) = 2 \to b - 6 = 2 \to b = 8$
**[4]**
**(ii)**
$fg(x) = 3(b - 2x) + a = 22 - 6x$ | B1
**[2]**
**Guidance:**
Must be correct – unsimplified ok
Correct method leading to a value for $b$
Must be $fg$ not $gf$
$\checkmark$ on $a$ and $b$; $(3b + a - 6x)$ must be two-term answer2 The functions f and g are defined for $x \in \mathbb { R }$ by
$$\begin{aligned}
& \mathrm { f } : x \mapsto 3 x + a \\
& \mathrm {~g} : x \mapsto b - 2 x
\end{aligned}$$
where $a$ and $b$ are constants. Given that $\mathrm { ff } ( 2 ) = 10$ and $\mathrm { g } ^ { - 1 } ( 2 ) = 3$, find\\
(i) the values of $a$ and $b$,\\
(ii) an expression for $\mathrm { fg } ( x )$.
\hfill \mbox{\textit{CAIE P1 2011 Q2 [6]}}