| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Evaluate composite at point |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic function operations. Part (b) specifically asks to evaluate fg(1), which requires computing g(1) = -1, then f(-1) = 2|-1| + 3 = 5. This is routine substitution with no problem-solving required, easier than the average A-level question. |
| 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/Working | Mark | Guidance |
| \(f(x) \geqslant 3\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to find \(2 | 3-4x | +3\) when \(x=1\) |
| \(fg(1) = 5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 3-4x \Rightarrow 4x = 3-y \Rightarrow x = \frac{3-y}{4}\) | M1 | |
| \(g^{-1}(x) = \frac{3-x}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([g(x)]^2 = (3-4x)^2\) | B1 | |
| \(gg(x) = 3 - 4(3-4x)\) | M1 | |
| \(gg(x) + [g(x)]^2 = 0 \Rightarrow -9+16x+9-24x+16x^2 = 0\) | ||
| \(16x^2 - 8x = 0\) | A1 | |
| \(8x(2x-1) = 0 \Rightarrow x = 0, \ 0.5\) | M1, A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) \geqslant 3\), range \(\geqslant 3\), \([3, \infty)\) | B1, A1 | Attempt at calculating f at \(x=0\). Sight of 3 sufficient. Accept \(f(x)>3\) and \(x>3\) for M1. Do not accept \(f(x)>3\), \(x \geqslant 3\). Correct answer sufficient for both marks. |
| Answer | Marks | Guidance |
|---|---|---|
| Full method of finding \(fg(1)\) using \(2 | x | +3\) (not \(2x+3\)) |
| \(fg(1)=5\) | A1 | Watch for \(1 \xrightarrow{3-4x} 1 \xrightarrow{2 |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to make \(x\) or swapped \(y\) the subject; cannot finish at \(4x=\ldots\) | M1 | Condone at most one arithmetic error; e.g. \(y=3-4x \Rightarrow 4x=3+y \Rightarrow x=\frac{3+y}{4}\) fine (one error); \(y=3-4x \Rightarrow 4x=3+y \Rightarrow x=\frac{3}{4}+y\) is M0 (two errors) |
| \(g^{-1}(x)=\frac{3-x}{4}\), e.g. \(g^{-1}(x)=\frac{x-3}{-4}\), \(g^{-1}(x)=\frac{3}{4}-\frac{x}{4}\) | A1 | Must be in terms of \(x\); could be expressed as \(y=\) or \(g^{-1}(x)\to\) |
| Answer | Marks | Guidance |
|---|---|---|
| Sight of \([g(x)]^2=(3-4x)^2\) | B1 | If only expanded version appears it must be correct |
| Full attempt to find \(gg(x)=3-4(3-4x)\) | M1 | Condone invisible brackets |
| \(16x^2-8x=0\) | A1 | Accept alternatives such as \(2x^2=x\) |
| Factorising/cancelling to get \(\geq 1\) value of \(x\) | M1 | If 3TQ then usual methods apply |
| Both values correct: \(x=0,\ 0.5\) | A1 |
## Question 4:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) \geqslant 3$ | M1, A1 | |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to find $2|3-4x|+3$ when $x=1$ | M1 | |
| $fg(1) = 5$ | A1 | |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 3-4x \Rightarrow 4x = 3-y \Rightarrow x = \frac{3-y}{4}$ | M1 | |
| $g^{-1}(x) = \frac{3-x}{4}$ | A1 | |
**Part (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[g(x)]^2 = (3-4x)^2$ | B1 | |
| $gg(x) = 3 - 4(3-4x)$ | M1 | |
| $gg(x) + [g(x)]^2 = 0 \Rightarrow -9+16x+9-24x+16x^2 = 0$ | | |
| $16x^2 - 8x = 0$ | A1 | |
| $8x(2x-1) = 0 \Rightarrow x = 0, \ 0.5$ | M1, A1 | oe |
## Question 4:
### Part (a):
| $f(x) \geqslant 3$, range $\geqslant 3$, $[3, \infty)$ | B1, A1 | Attempt at calculating f at $x=0$. Sight of 3 sufficient. Accept $f(x)>3$ and $x>3$ for M1. Do not accept $f(x)>3$, $x \geqslant 3$. Correct answer sufficient for both marks. |
### Part (b):
| Full method of finding $fg(1)$ using $2|x|+3$ (not $2x+3$) | M1 | Order of substitution must be correct; accept attempt to calculate $2|x|+3$ when $x=-1$; accept putting $x=1$ into $3-4x$ then substituting into $2|x|+3$ |
| $fg(1)=5$ | A1 | Watch for $1 \xrightarrow{3-4x} 1 \xrightarrow{2|x|+3} 5$ which is **M1A0** |
### Part (c):
| Attempt to make $x$ or swapped $y$ the subject; cannot finish at $4x=\ldots$ | M1 | Condone at most one arithmetic error; e.g. $y=3-4x \Rightarrow 4x=3+y \Rightarrow x=\frac{3+y}{4}$ fine (one error); $y=3-4x \Rightarrow 4x=3+y \Rightarrow x=\frac{3}{4}+y$ is M0 (two errors) |
| $g^{-1}(x)=\frac{3-x}{4}$, e.g. $g^{-1}(x)=\frac{x-3}{-4}$, $g^{-1}(x)=\frac{3}{4}-\frac{x}{4}$ | A1 | Must be in terms of $x$; could be expressed as $y=$ or $g^{-1}(x)\to$ |
### Part (d):
| Sight of $[g(x)]^2=(3-4x)^2$ | B1 | If only expanded version appears it must be correct |
| Full attempt to find $gg(x)=3-4(3-4x)$ | M1 | Condone invisible brackets |
| $16x^2-8x=0$ | A1 | Accept alternatives such as $2x^2=x$ |
| Factorising/cancelling to get $\geq 1$ value of $x$ | M1 | If 3TQ then usual methods apply |
| Both values correct: $x=0,\ 0.5$ | A1 | |
---\begin{enumerate}
\item The functions f and g are defined by
\end{enumerate}
$$\begin{array} { l l }
\mathrm { f } : x \mapsto 2 | x | + 3 , & x \in \mathbb { R } , \\
\mathrm {~g} : x \mapsto 3 - 4 x , & x \in \mathbb { R }
\end{array}$$
(a) State the range of f.\\
(b) Find $\mathrm { fg } ( 1 )$.\\
(c) Find $\mathrm { g } ^ { - 1 }$, the inverse function of g .\\
(d) Solve the equation
$$\operatorname { gg } ( x ) + [ \mathrm { g } ( x ) ] ^ { 2 } = 0$$
\hfill \mbox{\textit{Edexcel C3 2013 Q4 [11]}}