17M.2.AHL.TZ1.H_2

pestleMathematicsAIHLPaper 217M· sl-5-1-introduction-of-differential-calculussource ↗

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 $y$ .

[4]

Determine the equation of the tangent to $C$ at the point $\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ 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]

Examiners’ report

[N/A]