| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on standard C3 topics. Part (a) requires completing the square to find range; part (b) is routine inverse function finding; part (c)(i) is direct composition; part (c)(ii) involves solving a quadratic after substitution. All techniques are standard textbook exercises 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 | Mark | Guidance |
| \(f(x) = (x-3)^2 - 4\), minimum at \(x=3\) giving \(f(3)=-4\) | M1 | Complete the square or use calculus |
| Range: \(f(x) \geq -4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let \(y = (x-3)^2 - 4\), rearrange: \(x - 3 = \sqrt{y+4}\) | M1 | Attempt to rearrange (positive root only since \(x \geq 3\)) |
| \(x = 3 + \sqrt{y+4}\) | A1 | |
| \(f^{-1}(x) = 3 + \sqrt{x+4}\) | A1 | |
| Domain: \(x \geq -4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(gf(x) = | x^2 - 6x + 5 - 6 | = |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \( | x^2 - 6x - 1 | = 6\) |
| \(x^2 - 6x - 1 = 6 \Rightarrow x^2 - 6x - 7 = 0 \Rightarrow (x-7)(x+1) = 0\) | M1 | Solving quadratic |
| \(x = 7\) (accept, since \(x \geq 3\)), \(x = -1\) rejected | A1 | |
| \(x^2 - 6x - 1 = -6 \Rightarrow x^2 - 6x + 5 = 0 \Rightarrow (x-5)(x-1)=0\) | M1 | |
| \(x = 5\) (accept), \(x = 1\) rejected | A1 | Both valid solutions \(x=5, x=7\) |
# Question 5:
## Part (a): Range of f
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = (x-3)^2 - 4$, minimum at $x=3$ giving $f(3)=-4$ | M1 | Complete the square or use calculus |
| Range: $f(x) \geq -4$ | A1 | |
## Part (b): Find $f^{-1}(x)$
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $y = (x-3)^2 - 4$, rearrange: $x - 3 = \sqrt{y+4}$ | M1 | Attempt to rearrange (positive root only since $x \geq 3$) |
| $x = 3 + \sqrt{y+4}$ | A1 | |
| $f^{-1}(x) = 3 + \sqrt{x+4}$ | A1 | |
| Domain: $x \geq -4$ | A1 | |
## Part (c)(i): Find $gf(x)$
| Answer | Mark | Guidance |
|--------|------|----------|
| $gf(x) = |x^2 - 6x + 5 - 6| = |x^2 - 6x - 1|$ | B1 | |
## Part (c)(ii): Solve $gf(x) = 6$
| Answer | Mark | Guidance |
|--------|------|----------|
| $|x^2 - 6x - 1| = 6$ | M1 | Setting up two equations |
| $x^2 - 6x - 1 = 6 \Rightarrow x^2 - 6x - 7 = 0 \Rightarrow (x-7)(x+1) = 0$ | M1 | Solving quadratic |
| $x = 7$ (accept, since $x \geq 3$), $x = -1$ rejected | A1 | |
| $x^2 - 6x - 1 = -6 \Rightarrow x^2 - 6x + 5 = 0 \Rightarrow (x-5)(x-1)=0$ | M1 | |
| $x = 5$ (accept), $x = 1$ rejected | A1 | Both valid solutions $x=5, x=7$ |5 The functions $f$ and $g$ are defined with their respective domains by
$$\begin{array} { l l }
\mathrm { f } ( x ) = x ^ { 2 } - 6 x + 5 , & \text { for } x \geqslant 3 \\
\mathrm {~g} ( x ) = | x - 6 | , & \text { for all real values of } x
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the range of f.
\item The inverse of f is $\mathrm { f } ^ { - 1 }$.
Find $\mathrm { f } ^ { - 1 } ( x )$. Give your answer in its simplest form.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { gf } ( x )$.
\item Solve the equation $\operatorname { gf } ( x ) = 6$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2014 Q5 [11]}}