| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Piecewise function inverses |
| Difficulty | Standard +0.3 This question requires finding an intersection point by solving a quadratic equation, then inverting a piecewise function by reversing each piece and swapping domains/ranges. While it involves multiple steps and careful attention to domains, the techniques are standard P1 material with no novel insight required—slightly easier than average A-level. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(11 - x^2 = 5 - x \Rightarrow x^2 - x - 6(= 0)\); \((x+2)(x-3)\); \(p = 3; q = 2\) | B1, B1, B1 | oe [3] |
| (ii) \(f^{-1}(x) = \begin{cases}\sqrt{11-x} \text{ for } 2 \leq x \leq 11 \\ 5-x \text{ for } x < 2\end{cases}\) | B1, B1, B1, B1, B1 | [5] |
(i) $11 - x^2 = 5 - x \Rightarrow x^2 - x - 6(= 0)$; $(x+2)(x-3)$; $p = 3; q = 2$ | B1, B1, B1 | oe [3]
(ii) $f^{-1}(x) = \begin{cases}\sqrt{11-x} \text{ for } 2 \leq x \leq 11 \\ 5-x \text{ for } x < 2\end{cases}$ | B1, B1, B1, B1, B1 | [5]7\\
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(i) The diagram shows part of the curve $y = 11 - x ^ { 2 }$ and part of the straight line $y = 5 - x$ meeting at the point $A ( p , q )$, where $p$ and $q$ are positive constants. Find the values of $p$ and $q$.\\
(ii) The function f is defined for the domain $x \geqslant 0$ by
$$f ( x ) = \begin{cases} 11 - x ^ { 2 } & \text { for } 0 \leqslant x \leqslant p \\ 5 - x & \text { for } x > p \end{cases}$$
Express $\mathrm { f } ^ { - 1 } ( x )$ in a similar way.
\hfill \mbox{\textit{CAIE P1 2012 Q7 [8]}}