C3 January 2013 Q2
2. \[\mathrm{g}(x) = \mathrm{e}^{x - 1} + x - 6\]
The root of \(\mathrm{g}(x) = 0\) is \(\alpha\).
The iterative formula
\[x_{n+1} = \ln(6 - x_n) + 1, \qquad x_0 = 2\]
is used to find an approximate value for \(\alpha\).
| Scheme | Marks |
|---|---|
| \(0 = e^{x - 1} + x - 6 \Rightarrow x = \ln(6 - x) + 1\) | M1A1* |
| (2) |
Notes
M1 Sets g(x)=0, and using correct ln work, makes the x of the \(e^{x - 1}\) term the subject of the formula.
Look for \(e^{x - 1} + x - 6 = 0 \Rightarrow e^{x - 1} = \pm 6 \pm x \Rightarrow x = \ln(\pm 6 \pm x) \pm 1\)
Do not accept \(e^{x - 1} = 6 - x\) without firstly seeing \(e^{x - 1} + x - 6 = 0\) or a statement that g(x)=0 \(\Rightarrow\)
A1* cso. \(x = \ln(6 - x) + 1\) Note that this is a given answer (and a proof).
‘Invisible’ brackets are allowed for the M but not the A
Do not accept recovery from earlier errors for the A mark. The solution below scores 0 marks.
\(0 = e^{x - 1} + x - 6 \Rightarrow \quad 0 = x - 1 + \ln(x - 6) \Rightarrow x = \ln(6 - x) + 1\)
Alternative solution to (a) working backwards
M1 Proceeds from \(x = \ln(6 - x) + 1\) using correct exp work to …….=0
A1 Arrives correctly at \(e^{x - 1} + x - 6 = 0\) and makes a statement to the effect that this is g(x)=0
| Scheme | Marks |
|---|---|
| Sub \(x_0 = 2\) into \(x_{n+1} = \ln(6 - x_n) + 1 \Rightarrow x_1 = 2.3863\) | M1, A1 |
| AWRT 4 dp. \(x_2 = 2.2847\) \(x_3 = 2.3125\) | A1 |
| (3) |
Notes
M1 Sub \(x_0 = 2\) into \(x_{n+1} = \ln(6 - x_n) + 1\) to produce a numerical value for \(x_1\).
Evidence for the award could be any of \(\ln(6 - 2) + 1\), \(\ln 4 + 1\), 2.3….. or awrt 2.4
A1 Answer correct to 4 dp \(x_1 = 2.3863\).
The subscript is not important. Mark as the first value given/found.
A1 Awrt 4 dp. \(x_2 = 2.2847\) and \(x_3 = 2.3125\)
The subscripts are not important. Mark as the second and third values given/found
| Scheme | Marks |
|---|---|
| Chooses interval [2.3065,2.3075] | M1 |
| \(g(2.3065) = -0.0002(7)\), \(g(2.3075) = 0.004(4)\) | dM1 |
| Sign change , hence root (correct to 3dp) | A1 |
| (3) | |
| (8 marks) |
Notes
M1 Chooses the interval [2.3065,2.3075] or smaller containing the root 2.306558641
dM1 Calculates \(g(2.3065)\) and \(g(2.3075)\) with at least one of these correct to 1sf.
The answers can be rounded or truncated
\(g(2.3065) = -0.0003\) rounded, \(g(2.3065) = -0.0002\) truncated
\(g(2.3075) = (+)\ 0.004\) rounded and truncated
A1 Both values correct (rounded or truncated),
A reason which could include change of sign, >0 <0, \(g(2.3065) \times g(2.3075) \lt 0\)
AND a minimal conclusion such as hence root, \(\alpha\)=2.307 or □
Do not accept continued iteration as question demands an interval to be chosen.
Alternative solution to (c) using \(\mathrm{f}(x) = \ln(6 - x) + 1 - x\) {Similarly \(\mathrm{h}(x) = x - 1 - \ln(6 - x)\)}
M1 Chooses the interval [2.3065,2.3075] or smaller containing the root 2.306558641
dM1 Calculates f(2.3065) and f(2.3075) with at least 1 correct rounded or truncated
f(2.3065) = 0.000074. Accept 0.00007 rounded or truncated. Also accept 0.0001
f(2.3075) = - 0.0011.. Accept -0.001 rounded or truncated