| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a multi-part question on composite and inverse functions that requires finding the range of a quadratic, finding an inverse function (straightforward for a restricted quadratic), solving a quadratic equation, and working backwards through a composition. While part (d) requires careful algebraic manipulation through ggf^(-1), each individual step uses standard techniques. Slightly above average due to the multi-step nature and the final part requiring methodical composition work, but no novel insights needed. |
| 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 |
| Range of f is \(f(x) \geqslant -4\) | B1 | Allow \(y\), f or 'range' or \([-4, \infty)\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = (x-2)^2 - 4 \Rightarrow (x-2)^2 = y+4 \Rightarrow x-2 = +\sqrt{(y+4)}\) or \(\pm\sqrt{(y+4)}\) | M1 | May swap variables here |
| \(\left[f^{-1}(x)\right] = \sqrt{(x+4)} + 2\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-2)^2 - 4 = -\frac{5}{3}x + 2 \Rightarrow x^2 - 4x + 4 - 4 = -\frac{5}{3}x + 2\ [\Rightarrow x^2 - \frac{7}{3}x - 2 = 0]\) | M1 | Equating and simplifying to a 3-term quadratic |
| \((3x+2)(x-3)[=0]\) or \(\frac{7 \pm \sqrt{7^2 - 4(3)(-6)}}{6}\) OE | M1 | Solving quadratic |
| \(x = 3\) only | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f^{-1}(12) = 6\) | M1 | Substitute 12 into *their* \(f^{-1}(x)\) and evaluate |
| \(g(f^{-1}(12)) = 6a + 2\) | M1 | Substitute *their* '6' into \(g(x)\) |
| \(g(g(f^{-1}(12))) = a(6a+2) + 2 = 62\) | M1 | Substitute the result into \(g(x)\) and \(= 62\) |
| \(6a^2 + 2a - 60\ [=0]\) | M1 | Forming and solving a 3-term quadratic |
| \(a = -\frac{10}{3}\) or \(3\) | A1 | |
| Alternative method: | ||
| \(g(f^{-1}(x)) = a(\sqrt{x+4}+2)+2\) or \(gg(x) = a(ax+2)+2\) | M1 | Substitute *their* \(f^{-1}(x)\) or \(g(x)\) into \(g(x)\) |
| \(g(g(f^{-1}(x))) = a\Big(a\big(\sqrt{x+4}+2\big)+2\Big)+2\) | M1 | Substitute the result into \(g(x)\) |
| \(g(g(f^{-1}(12))) = a(6a+2)+2 = 62\) | M1 | Substitute 12 and \(= 62\) |
| \(6a^2 + 2a - 60\ [=0]\) | M1 | Forming and solving a 3-term quadratic |
| \(a = -\frac{10}{3}\) or \(3\) | A1 | |
| 5 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Range of f is $f(x) \geqslant -4$ | B1 | Allow $y$, f or 'range' or $[-4, \infty)$ |
| | **1** | |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = (x-2)^2 - 4 \Rightarrow (x-2)^2 = y+4 \Rightarrow x-2 = +\sqrt{(y+4)}$ or $\pm\sqrt{(y+4)}$ | M1 | May swap variables here |
| $\left[f^{-1}(x)\right] = \sqrt{(x+4)} + 2$ | A1 | |
| | **2** | |
---
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-2)^2 - 4 = -\frac{5}{3}x + 2 \Rightarrow x^2 - 4x + 4 - 4 = -\frac{5}{3}x + 2\ [\Rightarrow x^2 - \frac{7}{3}x - 2 = 0]$ | M1 | Equating and simplifying to a 3-term quadratic |
| $(3x+2)(x-3)[=0]$ or $\frac{7 \pm \sqrt{7^2 - 4(3)(-6)}}{6}$ OE | M1 | Solving quadratic |
| $x = 3$ only | A1 | |
| | **3** | |
---
## Question 9(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f^{-1}(12) = 6$ | M1 | Substitute 12 into *their* $f^{-1}(x)$ and evaluate |
| $g(f^{-1}(12)) = 6a + 2$ | M1 | Substitute *their* '6' into $g(x)$ |
| $g(g(f^{-1}(12))) = a(6a+2) + 2 = 62$ | M1 | Substitute the result into $g(x)$ and $= 62$ |
| $6a^2 + 2a - 60\ [=0]$ | M1 | Forming and solving a 3-term quadratic |
| $a = -\frac{10}{3}$ or $3$ | A1 | |
| **Alternative method:** | | |
| $g(f^{-1}(x)) = a(\sqrt{x+4}+2)+2$ or $gg(x) = a(ax+2)+2$ | M1 | Substitute *their* $f^{-1}(x)$ or $g(x)$ into $g(x)$ |
| $g(g(f^{-1}(x))) = a\Big(a\big(\sqrt{x+4}+2\big)+2\Big)+2$ | M1 | Substitute the result into $g(x)$ |
| $g(g(f^{-1}(12))) = a(6a+2)+2 = 62$ | M1 | Substitute 12 and $= 62$ |
| $6a^2 + 2a - 60\ [=0]$ | M1 | Forming and solving a 3-term quadratic |
| $a = -\frac{10}{3}$ or $3$ | A1 | |
| | **5** | |
---9 Functions f and g are defined as follows:
$$\begin{aligned}
& \mathrm { f } ( x ) = ( x - 2 ) ^ { 2 } - 4 \text { for } x \geqslant 2 , \\
& \mathrm {~g} ( x ) = a x + 2 \text { for } x \in \mathbb { R } ,
\end{aligned}$$
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item State the range of f.
\item Find $\mathrm { f } ^ { - 1 } ( x )$.
\item Given that $a = - \frac { 5 } { 3 }$, solve the equation $\mathrm { f } ( x ) = \mathrm { g } ( x )$.
\item Given instead that $\operatorname { ggf } ^ { - 1 } ( 12 ) = 62$, find the possible values of $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q9 [11]}}