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Date November 2015 Marks available 4 Reference code 15N.1.hl.TZ0.7
Level HL only Paper 1 Time zone TZ0
Command term Show that Question number 7 Adapted from N/A

Question

A curve is defined by \(xy = {y^2} + 4\).

Show that there is no point where the tangent to the curve is horizontal.

[4]
a.

Find the coordinates of the points where the tangent to the curve is vertical.

[4]
b.

Markscheme

\(x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     M1A1

a horizontal tangent occurs if \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) so \(y = 0\)     M1

we can see from the equation of the curve that this solution is not possible \((0 = 4)\) and so there is not a horizontal tangent     R1

[4 marks]

a.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{2y - x}}\) or equivalent with \(\frac{{{\text{d}}x}}{{{\text{d}}y}}\)

the tangent is vertical when \(2y = x\)     M1

substitute into the equation to give \(2{y^2} = {y^2} + 4\)     M1

\(y =  \pm 2\)     A1

coordinates are \((4,{\text{ }}2),{\text{ }}( - 4,{\text{ }} - 2)\)     A1

[4 marks]

Total [8 marks]

b.

Examiners report

[N/A]
a.
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b.

Syllabus sections

Topic 6 - Core: Calculus » 6.1 » The derivative interpreted as a gradient function and as a rate of change.

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