| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.2 This question involves routine composite function evaluation, solving a linear equation from ff(x), finding conditions on a discriminant for no real solutions, and finding an inverse of a restricted quadratic—all standard P1 techniques with straightforward algebraic manipulation and no novel problem-solving required. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2(2x+3)+3 = 25\) | M1 | \(ff(x)\) needs to be correctly formed |
| \(\rightarrow x = 4\) or \(\{f(11)=25,\ f(4)=11\}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - 6x + 8 = 2x + k\) | M1 | Eliminates \(y\) to form eqn in \(x\) |
| \(x^2 - 8x + 8 - k = 0\) | M1 | Uses the discriminant – even if \(= 0. > 0\) |
| Uses \(b^2 - 4ac < 0\) | A1 | |
| \(\rightarrow k < -8\) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - 6x + 8 = (x-3)^2 - 1\) | B1 B1 | For "−3" and "−1" |
| \(y = (x-3)^2 - 1\) | ||
| Makes \(x\) the subject \(\rightarrow \pm\sqrt{(x+1)}+3\); needs specifically to lose the "−" | M1 A1\(\sqrt{}\) [4] | Makes \(x\) the subject, in terms of \(x\) and without \(-\) or \(\pm\) |
## Question 10:
$f: x \mapsto 2x + k$, $g: x \mapsto x^2 - 6x + 8$
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2(2x+3)+3 = 25$ | M1 | $ff(x)$ needs to be correctly formed |
| $\rightarrow x = 4$ or $\{f(11)=25,\ f(4)=11\}$ | A1 [2] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - 6x + 8 = 2x + k$ | M1 | Eliminates $y$ to form eqn in $x$ |
| $x^2 - 8x + 8 - k = 0$ | M1 | Uses the discriminant – even if $= 0. > 0$ |
| Uses $b^2 - 4ac < 0$ | A1 | |
| $\rightarrow k < -8$ | [3] | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - 6x + 8 = (x-3)^2 - 1$ | B1 B1 | For "−3" and "−1" |
| $y = (x-3)^2 - 1$ | | |
| Makes $x$ the subject $\rightarrow \pm\sqrt{(x+1)}+3$; needs specifically to lose the "−" | M1 A1$\sqrt{}$ [4] | Makes $x$ the subject, in terms of $x$ and without $-$ or $\pm$ |10 The function f is defined by $\mathrm { f } : x \mapsto 2 x + k , x \in \mathbb { R }$, where $k$ is a constant.\\
(i) In the case where $k = 3$, solve the equation $\mathrm { ff } ( x ) = 25$.
The function g is defined by $\mathrm { g } : x \mapsto x ^ { 2 } - 6 x + 8 , x \in \mathbb { R }$.\\
(ii) Find the set of values of $k$ for which the equation $\mathrm { f } ( x ) = \mathrm { g } ( x )$ has no real solutions.
The function $h$ is defined by $h : x \mapsto x ^ { 2 } - 6 x + 8 , x > 3$.\\
(iii) Find an expression for $\mathrm { h } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{CAIE P1 2013 Q10 [9]}}