C3 June 2014 (R) Q5
5.
Find the complete set of values of \(x\) for which
| Scheme | Marks |
|---|---|
| V shaped graph | B1 |
| Touches \(x\) axis at \(\tfrac{3}{4}\) and cuts y axis at 3 | B1 |
| (2) |
Notes
B1 A ‘V’ shaped graph. The position is not important. Do not accept curves. See practice and qualification items for clarity. Accept a V shape with a ‘dotted’ extension of \(y=4x-3\) appearing under the \(x\) axis.
B1 The graph meets the x axis at \(x=\dfrac{3}{4}\) and crosses the y axis at \(y=3\). Do not allow multiple meets or crosses
If they have lost the previous B1 mark for an extra section of graph underneath the \(x\) axis allow for crossing the \(x\) axis at \(x=\dfrac{3}{4}\) and crosses the y axis at \(y=3\).
Accept marked elsewhere on the page with \(A\) and \(B\) marked on the graph and \(A=\left(\dfrac{3}{4},0\right)\) and \(B=(0,3)\)
Condone \(\left(0,\dfrac{3}{4}\right)\) and \((3,0)\) marked on the correct axis
| Scheme | Marks |
|---|---|
| Solves \(4x-3=2-2x\) or \(3-4x=2-2x\) to give either value of \(x\) | M1 |
| Both \(x=\dfrac{5}{6}\) and \(x=\dfrac{1}{2}\) or \(x>\dfrac{5}{6}\) or \(x<\dfrac{1}{2}\) | A1 |
| \(x<\dfrac{1}{2}\) or \(x>\dfrac{5}{6}\) | dM1A1 |
| (4) |
Notes
M1 Attempts to solve \(\left|4x-3\right|\ldots 2-2x\) finding at least one solution. You may see \(\ldots\) replaced by either = or >
Accept as evidence \(\pm 4x\pm 3=2-2x\Rightarrow x=..\)
Accept as evidence \(\pm 4x\pm 3>2-2x\Rightarrow x>..,\) or \(x<..\)
A1 Both critical values \(x=\dfrac{5}{6}\) and \(x=\dfrac{1}{2}\), or one inequality, accept \(x>\dfrac{5}{6}\) or \(x<\dfrac{1}{2}\)
Accept \(x=0.83\) and \(x=0.5\) for the critical values
Accept both of these answers with no incorrect working for both marks
dM1 Dependent upon the previous M, this is scored for selecting the outside region of their two points.
Eg if M1 has been scored for \(4x-3=2-2x\Rightarrow x=0.83\) and \(-4x-3=2-2x\Rightarrow x=-2.5\)
A correct application of M1 would be \(x<-2.5,x>0.83\)
A1 Correct answer only \(x<\dfrac{1}{2}\) or \(x>\dfrac{5}{6}\).
Accept \(x<0.5,x>0.8\dot{3}\)
| Scheme | Marks |
|---|---|
| Draws graph Or solves \(\left|4x-3\right|=1\tfrac{1}{2}-2x\) to give one soln \(x=\tfrac{3}{4}\) | M1 |
| Accept for all values of \(x\) except \(x=\tfrac{3}{4}\) Or \((x\in\mathbb{R},)\ x\neq\tfrac{3}{4}\), or \(x<\tfrac{3}{4},\ x>\tfrac{3}{4}\) | A1 |
| (2) | |
| (8 marks) |
Notes
M1 Either sketch both lines showing a single intersection at the point \(x=\dfrac{3}{4}\)
Or solves \(\left|4x-3\right|=1\tfrac{1}{2}-2x\) using both \(4x-3=1\tfrac{1}{2}-2x\) and \(-4x+3=1\tfrac{1}{2}-2x\) giving one solution \(x=\dfrac{3}{4}\)
Accept \(\left|4x-3\right|>1\tfrac{1}{2}-2x\) using both \(4x-3>1\tfrac{1}{2}-2x\) and \(-4x+3>1\tfrac{1}{2}-2x\) giving one solution \(x\ldots\dfrac{3}{4}\)
If two values are obtained using either method it is M0A0
A1 States that the solution set is all values apart from \(x=\dfrac{3}{4}\). Do not isw in this question. Score their final statement. Accept versions of all values of \(x\) except \(x=\tfrac{3}{4}\) or \(x\in\mathbb{R},\ x\neq\tfrac{3}{4}\), or \(x<\tfrac{3}{4},\ x>\tfrac{3}{4}\)