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

Find the coordinates of P and Q.

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

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

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]

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]

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

m₁ m₂ = 1     AG

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

[3 marks]

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]