C3 June 2013 (R) Q4
4. The functions f and g are defined by\[\begin{aligned}&\mathrm{f}:x\mapsto 2|x|+3, && x\in\mathbb{R},\\ &\mathrm{g}:x\mapsto 3-4x, && x\in\mathbb{R}\end{aligned}\]
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x)\geqslant 3\) | M1A1 |
| (2) |
Notes
M1 Attempt at calculating f at \(x=0\). Sight of 3 is sufficient. Accept \(\mathrm{f}(x)>3\) and \(x>3\) for M1,
A1 \(\mathrm{f}(x)\geqslant 3\). Accept \(y\geqslant 3\), \(\text{range}\geqslant 3\), \([3,\infty)\)
Do not accept \(\mathrm{f}(x)>3\), \(x\geqslant 3\)
The correct answer is sufficient for both marks.
| Scheme | Marks |
|---|---|
| An attempt to find \(2|3-4x|+3\) when \(x=1\) | M1 |
| Correct answer \(\mathrm{fg}(1)=5\) | A1 |
| (2) |
Notes
M1 A full method of finding fg(1). The order of substituting into the expressions must be correct and \(2|x|+3\) must be used as opposed to \(2x+3\)
Accept an attempt to calculate \(2|x|+3\) when \(x=-1\).
Accept an attempt to put \(x=1\) into \(3-4x\) and then substituting their answer to \(3-4x\big|_{x=1}\) into \(2|x|+3\)
Do not accept the substitution of \(x=1\) into \(2|x|+3\), followed by their result into ‘3-4\(x\)’
This is evidence of incorrect order.
A1 fg(1)=5.
Watch for \(1\xrightarrow{3-4x}1\xrightarrow{2|x|+3}5\) which is M1A0
| Scheme | Marks |
|---|---|
| \(y=3-4x\Rightarrow 4x=3-y\Rightarrow x=\dfrac{3-y}{4}\) | M1 |
| \(\mathrm{g}^{-1}(x)=\dfrac{3-x}{4}\) | A1 |
| (2) |
Notes
M1 Award for an attempt to make \(x\) or a swapped \(y\) the subject of the formula. It must be a full method and cannot finish \(4x=..\)
You can condone at most one ‘arithmetic’ error for this method mark.
\(y=3-4x\Rightarrow 4x=3+y\Rightarrow x=\dfrac{3+y}{4}\) is fine for the M1 as there is only one error
\(y=3-4x\Rightarrow 4x=3-y\Rightarrow x=\dfrac{3}{4}-y\) is fine for the M1 as there is only one error
\(y=3-4x\Rightarrow 4x=3+y\Rightarrow x=\dfrac{3}{4}+y\) is M0 as there are two arithmetic errors
A1 Obtaining a correct expression \(\mathrm{g}^{-1}(x)=\dfrac{3-x}{4}\) oe such as \(\mathrm{g}^{-1}(x)=\dfrac{x-3}{-4},\ \mathrm{g}^{-1}(x)=\dfrac{3}{4}-\dfrac{x}{4}\)
It must be in terms of x, but could be expressed ‘y=’ or \(g^{-1}(x)\to\)
| Scheme | Marks |
|---|---|
| \(\left[g(x)\right]^2=(3-4x)^2\) | B1 |
| \(\mathrm{gg}(x)=3-4(3-4x)\) | M1 |
| \(gg(x)+\left[g(x)\right]^2=0\Rightarrow -9+16x+9-24x+16x^2=0\) | |
| \(16x^2-8x=0\) | A1 |
| \(8x(2x-1)=0\Rightarrow x=0,\ 0.5\) oe | M1A1 |
| (5) | |
| (11 marks) |
Notes
B1 Sight of \(\left[g(x)\right]^2=(3-4x)^2\). If only the expanded version appears it must be correct
M1 A full attempt to find \(\mathrm{gg}(x)=3-4(3-4x)\)
Condone invisible brackets. Note that it may appear in an equation
A1 \(16x^2-8x=0\) Accept other alternatives such as \(2x^2=x\)
M1 For factorising their quadratic or cancelling their \(Ax^2=Bx\) by \(x\) to get \(\geqslant 1\) value of x
If they have a 3TQ then usual methods are applicable.
A1 Both values correct \(x=0,\ 0.5\) oe