| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question on basic function operations. Part (i) requires sketching linear functions and their inverses (reflection in y=x), part (ii) involves simple composition and completing the square to find a maximum, parts (iii-iv) are standard completing the square and finding an inverse with a restricted domain. All techniques are routine P1/AS-level material requiring only direct application of learned procedures with no problem-solving insight needed. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| [Graph description: Graph of \(y = 3x - 2\) with evidence of mirror image in \(y = x\) or graph of \(\frac{1}{3}(x + 2)\). Whichever way, there must be symmetry shown or quoted or implied by same intercepts.] | B1, B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(gf(x) = (3x-2) - (3x-2)^2\) | M1 | Must be \(gf\), not \(fg\) |
| \(= -9x^2 + 30x - 16\) | A1 | Co |
| \(\frac{d}{dx} = -18x + 30\) | M1 | Differentiates or completes square |
| \(= 0 \text{ when } x = \frac{5}{3}\) | DM1 | Sets to 0, solves and attempts to find \(y\) |
| \(\rightarrow \text{Max of } 9\) | A1 | All ok – answer was given |
| Answer | Marks | Guidance |
|---|---|---|
| \(6x - x^2 = 9 - (x-3)^2\) | B1, B1 | Does not need \(a\) or \(b\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 9 - (3-x)^2\) | M1 | Order of operations in making \(x\) subject |
| \(3 - x = \pm\sqrt{9-y}\) | DM1 | Interchanging \(x\) and \(y\) |
| \(\rightarrow h^{-1}(x) = 3 + \sqrt{9-x}\) | A1 | Allow if \(\pm\) given. (Special case \(\rightarrow\) if correct with \(y\) instead of \(x\), give 2 out of 3) |
$f : x \mapsto 3x - 2$
**(i)**
[Graph description: Graph of $y = 3x - 2$ with evidence of mirror image in $y = x$ or graph of $\frac{1}{3}(x + 2)$. Whichever way, there must be symmetry shown or quoted or implied by same intercepts.] | B1, B1 | [2]
**(ii)**
$gf(x) = (3x-2) - (3x-2)^2$ | M1 | Must be $gf$, not $fg$
$= -9x^2 + 30x - 16$ | A1 | Co
$\frac{d}{dx} = -18x + 30$ | M1 | Differentiates or completes square
$= 0 \text{ when } x = \frac{5}{3}$ | DM1 | Sets to 0, solves and attempts to find $y$
$\rightarrow \text{Max of } 9$ | A1 | All ok – answer was given
$(gf(x) = 9 - (3x-5)^2 \rightarrow \text{Max } 9)$
**(iii)**
$6x - x^2 = 9 - (x-3)^2$ | B1, B1 | Does not need $a$ or $b$.
**(iv)**
$y = 9 - (3-x)^2$ | M1 | Order of operations in making $x$ subject
$3 - x = \pm\sqrt{9-y}$ | DM1 | Interchanging $x$ and $y$
$\rightarrow h^{-1}(x) = 3 + \sqrt{9-x}$ | A1 | Allow if $\pm$ given. (Special case $\rightarrow$ if correct with $y$ instead of $x$, give 2 out of 3)10 The function f is defined by
$$\mathrm { f } : x \mapsto 3 x - 2 \text { for } x \in \mathbb { R } .$$
(i) 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.
The function g is defined by
$$\mathrm { g } : x \mapsto 6 x - x ^ { 2 } \text { for } x \in \mathbb { R }$$
(ii) Express $\operatorname { gf } ( x )$ in terms of $x$, and hence show that the maximum value of $\operatorname { gf } ( x )$ is 9 .
The function h is defined by
$$\mathrm { h } : x \mapsto 6 x - x ^ { 2 } \text { for } x \geqslant 3$$
(iii) Express $6 x - x ^ { 2 }$ in the form $a - ( x - b ) ^ { 2 }$, where $a$ and $b$ are positive constants.\\
(iv) Express $\mathrm { h } ^ { - 1 } ( x )$ in terms of $x$.
\hfill \mbox{\textit{CAIE P1 2008 Q10 [12]}}