Date | May 2017 | Marks available | 4 | Reference code | 17M.2.hl.TZ1.2 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The curve \(C\) is defined by equation \(xy - \ln y = 1,{\text{ }}y > 0\).
Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of \(x\) and \(y\).
Determine the equation of the tangent to \(C\) at the point \(\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)\)
Markscheme
\(y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1A1A1
Note: Award A1 for the first two terms, A1 for the third term and the 0.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2}}}{{1 - xy}}\) A1
Note: Accept \(\frac{{ - {y^2}}}{{\ln y}}\).
Note: Accept \(\frac{{ - y}}{{x - \frac{1}{y}}}\).
[4 marks]
\({m_T} = \frac{{{{\text{e}}^2}}}{{1 - {\text{e}} \times \frac{2}{{\text{e}}}}}\) (M1)
\({m_T} = - {{\text{e}}^2}\) (A1)
\(y - {\text{e}} = - {{\text{e}}^2}x + 2{\text{e}}\)
\( - {{\text{e}}^2}x - y + 3{\text{e}} = 0\) or equivalent A1
Note: Accept \(y = - 7.39x + 8.15\).
[3 marks]