| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Complete the square |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing routine techniques: completing the square (standard manipulation), reading the vertex from completed square form, solving quadratic equations, and finding an inverse function with a restricted domain. All parts follow predictable patterns with no novel problem-solving required, making it easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02u Functions: definition and vocabulary (domain, range, mapping) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x-1)^2 - 16\) | B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-16\) | B1✓ | Ft from (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9 \leq (x-1)^2 - 16 \leq 65\) OR \(x^2 - 2x - 15 = 9 \rightarrow 6, -4\) | M1 | OR \(x^2-2x-24 \geq 0,\ x^2-2x-80 \leq 0\) |
| \(25 \leq (x-1)^2 \leq 81\) and \(x^2-2x-15=65 \rightarrow 10,-8\) | M1 | \((x-6)(x+4) \geq 0\ \ (x-10)(x+8) \leq 0\) |
| \(5 \leq x-1 \leq 9\) and \(p = 6\) | A1 | \(x \geq 6\); \(x \leq 10\) |
| \(6 \leq x \leq 10\) and \(q = 10\) | A1 | SC B2, B2 for trial/improvement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = (y-1)^2 - 16\) [interchange \(x/y\)] | M1 | OR \((x-1)^2 = y + 16\) |
| \(y - 1 = (\pm)\sqrt{x+16}\) | M1 | \(x = 1 + (\pm)\sqrt{y+16}\) |
| \(f^{-1}(x) = 1 + \sqrt{x+16}\) | A1 | \(f^{-1}(x) = 1 + \sqrt{x+16}\) |
## Question 10:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x-1)^2 - 16$ | **B1B1** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-16$ | **B1**✓ | Ft from **(i)** |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9 \leq (x-1)^2 - 16 \leq 65$ **OR** $x^2 - 2x - 15 = 9 \rightarrow 6, -4$ | **M1** | **OR** $x^2-2x-24 \geq 0,\ x^2-2x-80 \leq 0$ |
| $25 \leq (x-1)^2 \leq 81$ and $x^2-2x-15=65 \rightarrow 10,-8$ | **M1** | $(x-6)(x+4) \geq 0\ \ (x-10)(x+8) \leq 0$ |
| $5 \leq x-1 \leq 9$ and $p = 6$ | **A1** | $x \geq 6$; $x \leq 10$ |
| $6 \leq x \leq 10$ and $q = 10$ | **A1** | **SC B2, B2** for trial/improvement |
### Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = (y-1)^2 - 16$ [interchange $x/y$] | **M1** | **OR** $(x-1)^2 = y + 16$ |
| $y - 1 = (\pm)\sqrt{x+16}$ | **M1** | $x = 1 + (\pm)\sqrt{y+16}$ |
| $f^{-1}(x) = 1 + \sqrt{x+16}$ | **A1** | $f^{-1}(x) = 1 + \sqrt{x+16}$ |
---10 (i) Express $x ^ { 2 } - 2 x - 15$ in the form $( x + a ) ^ { 2 } + b$.
The function f is defined for $p \leqslant x \leqslant q$, where $p$ and $q$ are positive constants, by
$$f : x \mapsto x ^ { 2 } - 2 x - 15$$
The range of f is given by $c \leqslant \mathrm { f } ( x ) \leqslant d$, where $c$ and $d$ are constants.\\
(ii) State the smallest possible value of $c$.
For the case where $c = 9$ and $d = 65$,\\
(iii) find $p$ and $q$,\\
(iv) find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{CAIE P1 2014 Q10 [10]}}