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Date	May 2018	Marks available	3	Reference code	18M.2.AHL.TZ2.H_11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Show that	Question number	H_11	Adapted from	N/A

Question

A curve C is given by the implicit equation $x + y - {\text{cos}}\left( {xy} \right) = 0$ .

The curve $xy = - \frac{\pi }{2}$ intersects C at P and Q.

Show that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \left( {\frac{{1 + y\,{\text{sin}}\left( {xy} \right)}}{{1 + x\,{\text{sin}}\left( {xy} \right)}}} \right)$ .

[5]

a.

Find the coordinates of P and Q.

[4]

b.i.

Given that the gradients of the tangents to C at P and Q are m₁ and m₂ respectively, show that m₁ × m₂ = 1.

[3]

b.ii.

Find the coordinates of the three points on C, nearest the origin, where the tangent is parallel to the line $y = - x$ .

[7]

c.

Markscheme

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

attempt at implicit differentiation M1

$1 + \frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {y + x\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right){\text{sin}}\left( {xy} \right) = 0$ A1M1A1

Note: Award A1 for first two terms. Award M1 for an attempt at chain rule A1 for last term.

$\left( {1 + x\,{\text{sin}}\left( {xy} \right)} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 1 - y\,{\text{sin}}\left( {xy} \right)$ A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \left( {\frac{{1 + y\,{\text{sin}}\left( {xy} \right)}}{{1 + x\,{\text{sin}}\left( {xy} \right)}}} \right)$ AG

[5 marks]

a.

EITHER

when $xy = - \frac{\pi }{2},\,\,{\text{cos}}\,xy = 0$ M1

$\Rightarrow x + y = 0$ (A1)

OR

$x - \frac{\pi }{{2x}} - {\text{cos}}\left( {\frac{{ - \pi }}{2}} \right) = 0$ or equivalent M1

$x - \frac{\pi }{{2x}} = 0$ (A1)

THEN

therefore ${x^2} = \frac{\pi }{2}\left( {x = \pm \sqrt {\frac{\pi }{2}} } \right)\left( {x = \pm 1.25} \right)$ A1

${\text{P}}\left( {\sqrt {\frac{\pi }{2}} ,\, - \sqrt {\frac{\pi }{2}} } \right),\,\,{\text{Q}}\left( { - \sqrt {\frac{\pi }{2}} ,\,\sqrt {\frac{\pi }{2}} } \right)$ or $P\left( {1.25,\, - 1.25} \right),\,Q\left( { - 1.25,\,1.25} \right)$ A1

[4 marks]

b.i.

m₁= $- \left( {\frac{{1 - \sqrt {\frac{\pi }{2}} \times - 1}}{{1 + \sqrt {\frac{\pi }{2}} \times - 1}}} \right)$ M1A1

m₂= $- \left( {\frac{{1 + \sqrt {\frac{\pi }{2}} \times - 1}}{{1 - \sqrt {\frac{\pi }{2}} \times - 1}}} \right)$ A1

m₁m₂= 1 AG

Note: Award M1A0A0 if decimal approximations are used.
Note: No FT applies.

[3 marks]

b.ii.

equate derivative to −1 M1

$\left( {y - x} \right){\text{sin}}\left( {xy} \right) = 0$ (A1)

$y = x,\,{\text{sin}}\left( {xy} \right) = 0$ R1

in the first case, attempt to solve $2x = {\text{cos}}\left( {{x^2}} \right)$ M1

(0.486,0.486) A1

in the second case, ${\text{sin}}\left( {xy} \right) = 0 \Rightarrow xy = 0$ and $x + y = 1$ (M1)

(0,1), (1,0) A1

[7 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.

Syllabus sections

Topic 5—Calculus » SL 5.1—Introduction of differential calculus

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