| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | November |
| Marks | 12 |
| 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 basic function manipulation. Part (i) requires simple substitution and solving a linear equation. Part (ii) involves setting a linear function equal to a quadratic and using the discriminant condition—a standard technique. Parts (iii) and (iv) are routine completing the square and finding an inverse function with restricted domain. All parts are textbook exercises requiring only direct application of standard methods with no problem-solving insight needed. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \(ff(x)=11\): \(2(2x-3)-3=11\) | M1, DM1 | Everything completed to give answer; if \(-3\) omitted \(\rightarrow 4\frac{1}{4}\), allow M1 only; \(2(2x-3)=ff(x)\) gets M1 not DM1 |
| \(\rightarrow x = 5\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x - a = x^2 - 6x \rightarrow x^2 - 8x + a = 0\) | M1 | Setting up 3-term quadratic in \(x\) |
| Use of \(b^2 - 4ac = 0\); \(\rightarrow a = 16\) (or inspection) | M1, A1 [3] | Using \(b^2-4ac\) on quadratic \(=0\) or \(\geq 0\); co; can be stated from \((-8x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 - 6x = (x-3)^2 - 9\); \(\rightarrow p=3\), \(q=9\) | B1 B1 [2] | Allow if \((x-3)^2-9\) without \(p\) or \(q\) stated |
| Answer | Marks | Guidance |
|---|---|---|
| \(y=(x-3)^2-9\); \(x = \pm\sqrt{(y+9)}+3\) | M1 | Attempt to make \(x\) the subject, but only from completing the square expression |
| \(y = h^{-1}(x) = \sqrt{(x+9)}+3\) | DM1A1 | Replace \(y\) by \(x\) — sign lost for A; special case "ans \(= \sqrt{(y+9)}+3\)" allow 2/3 |
| Domain of \(h^{-1} = \{x: x \geq -9\}\) | B1 [4] | Co (allow \(\geq -9\) or \(y \geq -9\) etc.) |
# Question 9:
## Part (i)
| $ff(x)=11$: $2(2x-3)-3=11$ | M1, DM1 | Everything completed to give answer; if $-3$ omitted $\rightarrow 4\frac{1}{4}$, allow M1 only; $2(2x-3)=ff(x)$ gets M1 not DM1 |
| $\rightarrow x = 5$ | A1 [3] | |
## Part (ii)
| $2x - a = x^2 - 6x \rightarrow x^2 - 8x + a = 0$ | M1 | Setting up 3-term quadratic in $x$ |
| Use of $b^2 - 4ac = 0$; $\rightarrow a = 16$ (or inspection) | M1, A1 [3] | Using $b^2-4ac$ on quadratic $=0$ or $\geq 0$; co; can be stated from $(-8x)$ |
## Part (iii)
| $x^2 - 6x = (x-3)^2 - 9$; $\rightarrow p=3$, $q=9$ | B1 B1 [2] | Allow if $(x-3)^2-9$ without $p$ or $q$ stated |
## Part (iv)
| $y=(x-3)^2-9$; $x = \pm\sqrt{(y+9)}+3$ | M1 | Attempt to make $x$ the subject, but only from completing the square expression |
| $y = h^{-1}(x) = \sqrt{(x+9)}+3$ | DM1A1 | Replace $y$ by $x$ — sign lost for A; special case "ans $= \sqrt{(y+9)}+3$" allow 2/3 |
| Domain of $h^{-1} = \{x: x \geq -9\}$ | B1 [4] | Co (allow $\geq -9$ or $y \geq -9$ etc.) |
---9 The function f : $x \mapsto 2 x - a$, where $a$ is a constant, is defined for all real $x$.\\
(i) In the case where $a = 3$, solve the equation $\mathrm { ff } ( x ) = 11$.
The function $\mathrm { g } : x \mapsto x ^ { 2 } - 6 x$ is defined for all real $x$.\\
(ii) Find the value of $a$ for which the equation $\mathrm { f } ( x ) = \mathrm { g } ( x )$ has exactly one real solution.
The function $\mathrm { h } : x \mapsto x ^ { 2 } - 6 x$ is defined for the domain $x \geqslant 3$.\\
(iii) Express $x ^ { 2 } - 6 x$ in the form $( x - p ) ^ { 2 } - q$, where $p$ and $q$ are constants.\\
(iv) Find an expression for $\mathrm { h } ^ { - 1 } ( x )$ and state the domain of $\mathrm { h } ^ { - 1 }$.
\hfill \mbox{\textit{CAIE P1 2004 Q9 [12]}}