| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a standard C3 inverse functions question with routine parts: finding range/domain of inverse (straightforward from domain/range swap), algebraic manipulation to find f^(-1)(x) (simple quadratic rearrangement), reading x-intercepts from a graph, determining if a function has an inverse (one-to-one test), and composing functions. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02u 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: \(f(x) \leq -\frac{4}{3}\) | B1 B1 | \(x \leq 0\) gives \(f(0) = -\frac{4}{3}\) as maximum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Domain of \(f^{-1}\): \(x \leq -\frac{4}{3}\) | B1 | Domain of inverse = range of f |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{x^2-4}{3} \Rightarrow 3y = x^2 - 4 \Rightarrow x^2 = 3y+4\) | M1 | Rearrange correctly |
| \(x = -\sqrt{3x+4}\) | A1 | Negative root since \(x \leq 0\) |
| \(f^{-1}(x) = -\sqrt{3x+4}\) | A1 | Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln | 3x-1 | = 0 \Rightarrow 3x - 1 = 1 \Rightarrow x = \frac{2}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| g does not have an inverse because it is not one-to-one (many-to-one function) | B1 | Correct reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{gf}(x) = g\left(\frac{x^2-4}{3}\right) = \ln\left | 3\cdot\frac{x^2-4}{3}-1\right | \) |
| \(= \ln | x^2 - 4 - 1 | = \ln |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln | x^2 - 5 | = 0 \Rightarrow |
| \(x^2 - 5 = 1 \Rightarrow x^2 = 6 \Rightarrow x = \pm\sqrt{6}\) | A1 | |
| \(x^2 - 5 = -1 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\) | A1 | But check domain (\(x \leq 0\)) |
| All four values (or restricted by domain): \(x = -\sqrt{6},\ x = -2,\ x = 2,\ x = \sqrt{6}\) | A1 | All correct solutions |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Range: $f(x) \leq -\frac{4}{3}$ | B1 B1 | $x \leq 0$ gives $f(0) = -\frac{4}{3}$ as maximum |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Domain of $f^{-1}$: $x \leq -\frac{4}{3}$ | B1 | Domain of inverse = range of f |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{x^2-4}{3} \Rightarrow 3y = x^2 - 4 \Rightarrow x^2 = 3y+4$ | M1 | Rearrange correctly |
| $x = -\sqrt{3x+4}$ | A1 | Negative root since $x \leq 0$ |
| $f^{-1}(x) = -\sqrt{3x+4}$ | A1 | Correct expression |
## Part (c)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln|3x-1| = 0 \Rightarrow 3x - 1 = 1 \Rightarrow x = \frac{2}{3}$ | M1 A1 | $P = \left(\frac{2}{3}, 0\right)$ |
## Part (c)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| g does not have an inverse because it is not one-to-one (many-to-one function) | B1 | Correct reason |
## Part (c)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{gf}(x) = g\left(\frac{x^2-4}{3}\right) = \ln\left|3\cdot\frac{x^2-4}{3}-1\right|$ | M1 | Correct composition |
| $= \ln|x^2 - 4 - 1| = \ln|x^2 - 5|$, so $k = 5$ | A1 | |
## Part (c)(iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln|x^2 - 5| = 0 \Rightarrow |x^2 - 5| = 1$ | M1 | |
| $x^2 - 5 = 1 \Rightarrow x^2 = 6 \Rightarrow x = \pm\sqrt{6}$ | A1 | |
| $x^2 - 5 = -1 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$ | A1 | But check domain ($x \leq 0$) |
| All four values (or restricted by domain): $x = -\sqrt{6},\ x = -2,\ x = 2,\ x = \sqrt{6}$ | A1 | All correct solutions |5 The function f is defined by
$$\mathrm { f } ( x ) = \frac { x ^ { 2 } - 4 } { 3 } , \text { for real values of } x , \text { where } \boldsymbol { x } \leqslant \mathbf { 0 }$$
\begin{enumerate}[label=(\alph*)]
\item State the range of f.
\item The inverse of f is $\mathrm { f } ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Write down the domain of $\mathrm { f } ^ { - 1 }$.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.
\end{enumerate}\item The function g is defined by
$$\mathrm { g } ( x ) = \ln | 3 x - 1 | , \quad \text { for real values of } x , \text { where } x \neq \frac { 1 } { 3 }$$
The curve with equation $y = \mathrm { g } ( x )$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{b8614dd6-2197-40c3-a673-5bef3e3653a5-6_469_819_1254_612}
\begin{enumerate}[label=(\roman*)]
\item The curve $y = \mathrm { g } ( x )$ intersects the $x$-axis at the origin and at the point $P$.
Find the $x$-coordinate of $P$.
\item State whether the function $g$ has an inverse. Give a reason for your answer.
\item Show that $\operatorname { gf } ( x ) = \ln \left| x ^ { 2 } - k \right|$, stating the value of the constant $k$.
\item Solve the equation $\mathrm { gf } ( x ) = 0$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2013 Q5 [15]}}