| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.8 This is a straightforward composite and inverse functions question requiring routine algebraic manipulation. Part (i) involves basic function composition, part (ii) requires finding an inverse by swapping and rearranging (standard technique), and part (iii) involves solving a quadratic equation. All steps are mechanical applications of well-practiced techniques with no novel insight required, making it easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{gf}(x) = 3(2x^2+3)+2 = 6x^2+11\) | B1 | AG |
| \(\text{fg}(x) = 2(3x+2)^2+3\). Allow \(18x^2+24x+11\) | B1 | ISW if simplified incorrectly. Not retrospectively from (ii) |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 2(3x+2)^2+3 \Rightarrow 3x+2 = (\pm)\sqrt{(y-3)/2}\) oe | M1 | Subtract 3; divide by 2; square root. Or \(x/y\) interchanged. Allow \(\dfrac{\sqrt{y-3}}{2}\) for 1st M |
| \(\Rightarrow x = (\pm)\dfrac{1}{3}\sqrt{(y-3)/2} - \dfrac{2}{3}\) oe | M1 | Subtract 2; divide by 3; indep. of 1st M1. Or \(x/y\) interchanged |
| \(\Rightarrow (\text{fg})^{-1}(x) = \dfrac{1}{3}\sqrt{(x-3)/2} - \dfrac{2}{3}\) oe | A1 | Must be a function of \(x\). Allow alt. method \(g^{-1}f^{-1}(x)\). OR \(18\left(x+\dfrac{2}{3}\right)^2+3 \Rightarrow (\text{fg})^{-1}(x) = \sqrt{\dfrac{x-3}{18}}-\dfrac{2}{3}\) |
| Solve their \((\text{fg})^{-1}(x) \geqslant 0\) or attempt range of fg | M1 | Allow range \(\geqslant 3\) for M only. Can be implied by correct answer or \(x > 11\) |
| Domain is \(x \geqslant 11\) | A1 | |
| Total: | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6(2x)^2+11 = 2(3x+2)^2+3\) | M1 | Replace \(x\) with \(2x\) in gf and equate to their \(\text{fg}(x)\) from (i). Allow \(12x^2+11=\) |
| \(6x^2 - 24x = 0\) oe | A1 | Collect terms to obtain correct quadratic expression |
| \(x = 0,\ 4\) | A1 | Both required |
| Total: | 3 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{gf}(x) = 3(2x^2+3)+2 = 6x^2+11$ | B1 | AG |
| $\text{fg}(x) = 2(3x+2)^2+3$. Allow $18x^2+24x+11$ | B1 | ISW if simplified incorrectly. Not retrospectively from **(ii)** |
| **Total:** | **2** | |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 2(3x+2)^2+3 \Rightarrow 3x+2 = (\pm)\sqrt{(y-3)/2}$ oe | M1 | Subtract 3; divide by 2; square root. Or $x/y$ interchanged. Allow $\dfrac{\sqrt{y-3}}{2}$ for 1st M |
| $\Rightarrow x = (\pm)\dfrac{1}{3}\sqrt{(y-3)/2} - \dfrac{2}{3}$ oe | M1 | Subtract 2; divide by 3; indep. of 1st M1. Or $x/y$ interchanged |
| $\Rightarrow (\text{fg})^{-1}(x) = \dfrac{1}{3}\sqrt{(x-3)/2} - \dfrac{2}{3}$ oe | A1 | Must be a function of $x$. Allow alt. method $g^{-1}f^{-1}(x)$. OR $18\left(x+\dfrac{2}{3}\right)^2+3 \Rightarrow (\text{fg})^{-1}(x) = \sqrt{\dfrac{x-3}{18}}-\dfrac{2}{3}$ |
| Solve their $(\text{fg})^{-1}(x) \geqslant 0$ or attempt range of fg | M1 | Allow range $\geqslant 3$ for M only. Can be implied by correct answer or $x > 11$ |
| Domain is $x \geqslant 11$ | A1 | |
| **Total:** | **5** | |
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6(2x)^2+11 = 2(3x+2)^2+3$ | M1 | Replace $x$ with $2x$ in gf and equate to their $\text{fg}(x)$ from (i). Allow $12x^2+11=$ |
| $6x^2 - 24x = 0$ oe | A1 | Collect terms to obtain correct quadratic expression |
| $x = 0,\ 4$ | A1 | Both required |
| **Total:** | **3** | |8 The functions f and g are defined for $x \geqslant 0$ by
$$\begin{aligned}
& \mathrm { f } : x \mapsto 2 x ^ { 2 } + 3 \\
& \mathrm {~g} : x \mapsto 3 x + 2
\end{aligned}$$
(i) Show that $\operatorname { gf } ( x ) = 6 x ^ { 2 } + 11$ and obtain an unsimplified expression for $\operatorname { fg } ( x )$.\\
(ii) Find an expression for $( \mathrm { fg } ) ^ { - 1 } ( x )$ and determine the domain of $( \mathrm { fg } ) ^ { - 1 }$.\\
(iii) Solve the equation $\mathrm { gf } ( 2 x ) = \mathrm { fg } ( x )$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q8 [10]}}