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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 \(y\).

[4]
a.

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

[3]
b.

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]

a.

\({m_T} = \frac{{{{\text{e}}^2}}}{{1 - {\text{e}} \times \frac{2}{{\text{e}}}}}\)     (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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 6 - Core: Calculus » 6.2 » Implicit differentiation.

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