C3 June 2014 (R) Q6
6. The function f is defined by\[\mathrm{f}:x\to\mathrm{e}^{2x}+k^2,\qquad x\in\mathbb{R},\qquad k\text{ is a positive constant.}\]
The function g is defined by\[\mathrm{g}:x\to\ln(2x),\qquad x>0\]
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x)>k^2\) | B1 |
| (1) |
Notes
B1 States the correct range for f. Accept \(\mathrm{f}(x)>k^2,\ \mathrm{f}>k^2,\ \text{Range}>k^2,\ \left(k^2,\infty\right),\ y>k^2\) Range is greater than \(k^2\)
Do not accept \(\mathrm{f}(x)\geqslant k^2,\ x>k^2,\ [k^2,\infty)\)
| Scheme | Marks |
|---|---|
| \(y=\mathrm{e}^{2x}+k^2\Rightarrow\mathrm{e}^{2x}=y-k^2\) | M1 |
| \(\Rightarrow x=\dfrac{1}{2}\ln(y-k^2)\) | dM1 |
| \(\Rightarrow\mathrm{f}^{-1}(x)=\dfrac{1}{2}\ln(x-k^2),\quad x>k^2\) | A1 |
| (3) |
Notes
M1 Attempts to make \(x\) or a swapped \(y\) the subject of the formula. Score for \(y=\mathrm{e}^{2x}+k^2\Rightarrow\mathrm{e}^{2x}=y\pm k^2\) and proceeding to \(x=\ln\ldots\) The minimum expectation is that \(\mathrm{e}^{2x}\) is made the subject before taking ln's
dM1 Dependent upon the previous M having been scored. It is for proceeding by firstly taking ln's of the whole rhs, not the individual elements, and then dividing by 2. Score M1, dM1 for writing down \(x=\dfrac{1}{2}\ln(y\pm k^2)\) or alternatively \(y=\dfrac{1}{2}\ln(x\pm k^2)\). Condone missing brackets for this mark.
A1 The correct answer in terms of \(x\) including the bracket and the domain \(\mathrm{f}^{-1}(x)=\dfrac{1}{2}\ln(x-k^2),\quad x>k^2\).
Accept equivalent answers like \(y=0.5\ln\left|x-k^2\right|\), Domain greater than \(k^2\), \(\left(k^2,\infty\right)\)
| Scheme | Marks |
|---|---|
| \(\ln 2x+\ln 2x^2+\ln 2x^3=6\) | M1 |
| \(\Rightarrow\ln 8x^6=6\) | M1 |
| \(\Rightarrow 8x^6=\mathrm{e}^6\Rightarrow x=..\) | M1 |
| \(\Rightarrow x=\left(\dfrac{\mathrm{e}}{\sqrt[6]{8}}\right)=\dfrac{\mathrm{e}}{\sqrt{2}}\) (Ignore any reference to \(-\dfrac{\mathrm{e}}{\sqrt{2}}\)) | A1 |
| (4) |
(alt c)
| Scheme | Marks |
|---|---|
| \(\ln 2x+\ln 2x^2+\ln 2x^3=6\) \(\Rightarrow\ln 2+\ln x+\ln 2+2\ln x+\ln 2+3\ln x=6\) \(\Rightarrow 3\ln 2+6\ln x=6\) | M1 |
| \(\Rightarrow\ln x=1-\dfrac{1}{2}\ln 2\) | M1 |
| \(\Rightarrow x=\mathrm{e}^{1-\frac{1}{2}\ln 2},=\dfrac{\mathrm{e}}{\sqrt{2}}\) (Ignore any reference to \(-\dfrac{\mathrm{e}}{\sqrt{2}}\)) | M1, A1 |
| (4) |
Notes
M1 Attempts to solve equation by writing down \(\ln 2x+\ln 2x^2+\ln 2x^3=6\)
M1 Uses addition laws of logs to write in the form \(\ln Ax^n=6\)
M1 Takes exp's (correctly) and proceeds to a solution for \(x=..\)
A1 Correct solution and correct answer. \(x=\dfrac{\mathrm{e}}{\sqrt{2}}\). You may ignore any reference to \(x=-\dfrac{\mathrm{e}}{\sqrt{2}}\)
Special cases. Candidate who solve (and treat it as though it was bracketed)S. Case 1 \(\ln 2x+\ln 2x^2+\ln 2x^3=6\Rightarrow\ln 2x+2\ln 2x+3\ln 2x=6\Rightarrow 6\ln 2x=6\Rightarrow x=\dfrac{\mathrm{e}}{2}\)
S. Case 2 \(\ln 2x+\ln(2x)^2+\ln(2x)^3=6\Rightarrow 6\ln 2x=6\Rightarrow\ln 2x=1\Rightarrow x=\dfrac{\mathrm{e}}{2}\)
S. Case 3 \(\ln 2x+\ln(2x)^2+\ln(2x)^3=6\Rightarrow\ln 2x+\ln 4x^2+\ln 8x^3=6\Rightarrow\ln 64x^6=6\Rightarrow 64x^6=\mathrm{e}^6\Rightarrow x=\dfrac{\mathrm{e}}{2}\)
scores M0 (Incorrect statement/ may be implied by subsequent work), M1 (Correct ln laws), M1 (Correct method of arriving at \(x=\)), A0
| Scheme | Marks |
|---|---|
| \(\mathrm{fg}(x)=\mathrm{e}^{2\times\ln(2x)}+k^2\) | M1 |
| \(\Rightarrow\mathrm{fg}(x)=(2x)^2+k^2=4x^2+k^2\) | A1 |
| (2) |
Notes
(d) For the purposes of marking you can score (d) and (e) together
M1 Correct order of applying g before f to give a correct unsimplified answer. Accept \(y=\)
Accept versions of \(\mathrm{fg}(x)=\mathrm{e}^{2\times\ln(2x)}+k^2,\ y=\mathrm{e}^{\ln(2x)^2}+k^2\)
A1 A correct simplified answer \(\mathrm{fg}(x)=(2x)^2+k^2\), or \(\mathrm{fg}(x)=4x^2+k^2\). Accept \(y=\)
| Scheme | Marks |
|---|---|
| \(\mathrm{fg}(x)=2k^2\Rightarrow 4x^2+k^2=2k^2\) | |
| \(\Rightarrow 4x^2=k^2\Rightarrow x=..\) | M1 |
| \(\Rightarrow x=\dfrac{k}{2}\) only | A1 |
| (2) | |
| (12 marks) |
(alt e)
| Scheme | Marks |
|---|---|
| \(\Rightarrow 2\ln(2x)=\ln(2k^2-k^2)\) | |
| \(\Rightarrow\ln(2x)^2=\ln(k^2),\Rightarrow 4x^2=k^2\Rightarrow x=\dfrac{k}{2}\) | M1, A1 |
Notes
M1 Sets the answer to (d) in the form \(Ax^2+k^2=2k^2\), where \(A=2\) or \(4\) and proceeds in the correct order to reach an equation of the form \(Ax^2=k^2\).
In the alternative method it would be for reaching \(\ln(Ax^2)=\ln(k^2)\), \(A=2\) or \(4\) or any equivalent form ln…= ln…
A1 \(x=\dfrac{k}{2}\) only. The answer \(x=\pm\dfrac{k}{2}\) is A0.