C3 June 2014 Q5
5. \[\mathrm{g}(x)=\frac{x}{x+3}+\frac{3(2x+1)}{x^2+x-6},\qquad x>3\]
| Scheme | Marks |
|---|---|
| \(x^2+x-6=(x+3)(x-2)\) | B1 |
| \(\dfrac{x}{x+3}+\dfrac{3(2x+1)}{(x+3)(x-2)}=\dfrac{x(x-2)+3(2x+1)}{(x+3)(x-2)}\) | M1 |
| \(=\dfrac{x^2+4x+3}{(x+3)(x-2)}\) | A1 |
| \(=\dfrac{\cancel{(x+3)}(x+1)}{\cancel{(x+3)}(x-2)}\) | |
| \(=\dfrac{(x+1)}{(x-2)}\) cso | A1* |
| (4) |
Notes
B1 \(x^2+x-6=(x+3)(x-2)\) This can occur anywhere in the solution.
M1 For combining the two fractions with a common denominator. The denominator must be correct for their fractions and at least one numerator must have been adapted. Accept as separate fractions. Condone missing brackets.
Accept \(\dfrac{x}{x+3}+\dfrac{3(2x+1)}{x^2+x-6}=\dfrac{x(x^2+x-6)+3(2x+1)(x+3)}{(x+3)(x^2+x-6)}\)
Condone \(\dfrac{x}{x+3}+\dfrac{3(2x+1)}{(x+3)(x-2)}=\dfrac{x\times x-2}{(x+3)(x-2)}+\dfrac{3(2x+1)}{(x+3)(x-2)}\)
A1 A correct intermediate form of \(\dfrac{\text{simplified quadratic}}{\text{simplified quadratic}}\)
Accept \(\dfrac{x^2+4x+3}{(x+3)(x-2)},\ \dfrac{x^2+4x+3}{x^2+x-6}\), OR \(\dfrac{x^3+7x^2+15x+9}{(x+3)(x^2+x-6)}\ \to\ \dfrac{(x+1)(x+3)\cancel{(x+3)}}{\cancel{(x+3)}(x^2+x-6)}\)
As in question one they can score this mark having 'invisible' brackets on line 1.
A1* Further factorises and cancels (which may be implied) to complete the proof to reach the given answer \(=\dfrac{(x+1)}{(x-2)}\). All aspects including bracketing must be correct. If a cubic is formed then it needs to be correct.
| Scheme | Marks |
|---|---|
| One end either \((y)>1,\ (y)\geqslant 1\) or \((y)<4,\ (y)\leqslant 4\) | B1 |
| \(1<y<4\) | B1 |
| (2) |
Notes
B1 States either end of the range. Accept either \(y<4,\ y\leqslant 4\) or \(y>1,\ y\geqslant 1\) with or without the \(y\)'s.
B1 Correct range. Accept \(1<y<4,\ 1<\mathrm{g}<4,\ y>1\) and \(y<4,\ (1,4),\ 1<\text{Range}<4\), even \(1<\mathrm{f}<4\),
Do not accept \(1<x<4,\ 1<y\leqslant 4,\ [1,4)\) etc.
Special case, allow B1B0 for \(1<x<4\)
| Scheme | Marks |
|---|---|
| Attempt to set Either \(g(x)=x\) or \(g(x)=g^{-1}(x)\) or \(g^{-1}(x)=x\) or \(g^2(x)=x\) \(\dfrac{(x+1)}{(x-2)}=x\qquad\dfrac{x+1}{x-2}=\dfrac{2x+1}{x-1}\qquad\dfrac{2x+1}{x-1}=x\qquad\dfrac{\frac{x+1}{x-2}+1}{\frac{x+1}{x-2}-2}=x\) | M1 |
| \(x^2-3x-1=0\Rightarrow x=\ldots\) | A1, dM1 |
| \(a=\dfrac{3+\sqrt{13}}{2}\) oe \(\left(1.5+\sqrt{3.25}\right)\) cso | A1 |
| (4) | |
| (10 marks) |
Notes
M1 Attempting to set \(g(x)=x,\ g^{-1}(x)=x\) or \(g(x)=g^{-1}(x)\) or \(g^2(x)=x\).
If \(g^{-1}(x)\) has been used then a full attempt must have been made to make \(x\) the subject of the formula.
A full attempt would involve cross multiplying, collecting terms, factorising and ending with division.
As a result, it must be in the form \(g^{-1}(x)=\dfrac{\pm 2x\pm 1}{\pm x\pm 1}\)
Accept as evidence \(\dfrac{(x+1)}{(x-2)}=x\) OR \(\dfrac{x+1}{x-2}=\dfrac{\pm 2x\pm 1}{\pm x\pm 1}\) OR \(\dfrac{\pm 2x\pm 1}{\pm x\pm 1}=x\) OR \(\dfrac{\frac{x+1}{x-2}+1}{\frac{x+1}{x-2}-2}=x\)
A1 \(x^2-3x-1=0\) or exact equivalent. The =0 may be implied by subsequent work.
dM1 For solving a 3TQ=0. It is dependent upon the first M being scored. Do not accept a method using factors unless it clearly factorises. Allow the answer written down awrt 3.30 (from a graphical calculator).
A1 \(a\) or \(x=\dfrac{3+\sqrt{13}}{2}\). Ignore any reference to \(\dfrac{3-\sqrt{13}}{2}\)
Withhold this mark if additional values are given for \(x,\ x>3\)