C3 June 2013 Q4
4. \[\mathrm{f}(x)=25x^2\mathrm{e}^{2x}-16,\qquad x\in\mathbb{R}\]
The equation \(\mathrm{f}(x)=0\) has a root \(\alpha\), where \(\alpha=0.5\) to 1 decimal place.
| Scheme | Marks |
|---|---|
| \(\mathrm{f}^{\prime}(x)=50x^2\mathrm{e}^{2x}+50x\mathrm{e}^{2x}\) oe. | M1A1 |
| Puts \(\mathrm{f}^{\prime}(x)=0\) to give \(x=-1\) and \(x=0\) or one coordinate | dM1A1 |
| Obtains \((0,-16)\) and \((-1,\ 25\mathrm{e}^{-2}-16)\) CSO | A1 |
| (5) |
Notes
No marks can be scored in part (a) unless you see differentiation as required by the question.M1 Uses \(vu'+uv'\). If the rule is quoted it must be correct.
It can be implied by their \(u=..,v=...,u'=...,v'=...\) followed by their \(vu'+uv'\)
If the rule is not quoted nor implied only accept answers of the form \(Ax^2\mathrm{e}^{2x}+Bx\mathrm{e}^{2x}\)
A1 \(\mathrm{f}^{\prime}(x)=50x^2\mathrm{e}^{2x}+50x\mathrm{e}^{2x}\).
Allow un simplified forms such as \(\mathrm{f}^{\prime}(x)=25x^2\times 2\mathrm{e}^{2x}+50x\times\mathrm{e}^{2x}\)
dM1 Sets \(\mathrm{f}^{\prime}(x)=0\), factorises out/ or cancels the \(\mathrm{e}^{2x}\) leading to at least one solution of \(x\)
This is dependent upon the first M1 being scored.
A1 Both \(x=-1\) and \(x=0\) or one complete coordinate. Accept \((0,-16)\) and \((-1,\ 25\mathrm{e}^{-2}-16)\) or \((-1,\ \text{awrt }-12.6)\)
A1 CSO. Obtains both solutions from differentiation. Coordinates can be given in any way.
\(x=-1,0\quad y=\dfrac{25}{\mathrm{e}^2}-16,\ -16\) or linked together by coordinate pairs \((0,-16)\) and \((-1,\ 25\mathrm{e}^{-2}-16)\) but the ‘pairs’ must be correct and exact.
| Scheme | Marks |
|---|---|
| Puts \(25x^2\mathrm{e}^{2x}-16=0\Rightarrow x^2=\dfrac{16}{25}\mathrm{e}^{-2x}\Rightarrow x=\pm\dfrac{4}{5}\mathrm{e}^{-x}\) | B1* |
| (1) |
Notes
B1 This is a show that question and all elements must be seen
Candidates must 1) State that \(\mathrm{f}(x)=0\) or writes \(25x^2\mathrm{e}^{2x}-16=0\) or \(25x^2\mathrm{e}^{2x}=16\)
2) Show at least one intermediate (correct) line with either \(x^2\) or \(x\) the subject. Eg \(x^2=\dfrac{16}{25}e^{-2x},\quad x=\sqrt{\dfrac{16}{25}e^{-2x}}\) oe
or square rooting \(25x^2\mathrm{e}^{2x}=16\Rightarrow 5x\mathrm{e}^{x}=\pm 4\)
or factorising by DOTS to give \((5x\mathrm{e}^{x}+4)(5x\mathrm{e}^{x}-4)=0\)
3) Show the given answer \(x=\pm\dfrac{4}{5}\mathrm{e}^{-x}\).
Condone the minus sign just appearing on the final line.
A ‘reverse’ proof is acceptable as long as there is a statement that \(\mathrm{f}(x)=0\)
| Scheme | Marks |
|---|---|
| Subs \(x_0=0.5\) into \(x=\dfrac{4}{5}\mathrm{e}^{-x}\Rightarrow x_1=\text{awrt }0.485\) | M1A1 |
| \(\Rightarrow x_2=\text{awrt }0.492,\ x_3=\text{awrt }0.489\) | A1 |
| (3) |
Notes
M1 Substitutes \(x_0=0.5\) into \(x=\dfrac{4}{5}\mathrm{e}^{-x}\Rightarrow x_1=\ldots.\)
This can be implied by \(x_1=\dfrac{4}{5}\mathrm{e}^{-0.5}\), or awrt 0.49
A1 \(x_1=\text{awrt }0.485\) 3dp. Mark as the first value given. Don’t be concerned by the subscript.
A1 \(x_2=\text{awrt }0.492,\ x_3=\text{awrt }0.489\) 3dp. Mark as the second and third values given.
| Scheme | Marks |
|---|---|
| \(\alpha=0.49\) | B1 |
| \(\mathrm{f}(0.485)=-0.487,\ \mathrm{f}(0.495)=(+)0.485\), sign change and deduction | B1 |
| (2) | |
| (11 marks) |
Notes
B1 States \(\alpha=0.49\)
B1 Justifies by
either calculating correctly \(\mathrm{f}(0.485)\) and \(\mathrm{f}(0.495)\) to awrt 1sf or 1dp,
\(\mathrm{f}(0.485)=-0.5,\ \mathrm{f}(0.495)=(+)0.5\) rounded
\(\mathrm{f}(0.485)=-0.4,\ \mathrm{f}(0.495)=(+)0.4\) truncated
giving a reason – accept change of sign, >0 <0 or \(\mathrm{f}(0.485)\times\mathrm{f}(0.495)<0\)
and giving a minimal conclusion. Eg. Accept hence root or \(\alpha=0.49\)
A smaller interval containing the root may be used, eg \(\mathrm{f}(0.49)\) and \(\mathrm{f}(0.495)\). Root = 0.49007
or by stating that the iteration is oscillating
or by calculating by continued iteration to at least the value of \(x_4=\text{awrt }0.491\) and stating (or seeing each value round to) 0.49