A curve $C$ is given by the equation ${{\mathrm{e}}^{{{x}^{2}}}}y=\ln {{x}^{2}}$, where \(x

Question

A curve $C$ is given by the equation ${{\mathrm{e}}^{{{x}^{2}}}}y=\ln {{x}^{2}}$, where $x\ne 0$. Show that $2x\ln {{x}^{2}}+{{\mathrm{e}}^{{{x}^{2}}}}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{x}$.[2] Find the equation of the tangent to $C$ at the point $P$ where $x=1$. Given that the tangent at $P$ meets $C$ again at the point $Q$, find the coordinates of $Q$, leaving your answer correct to $5$ significant figures.[4] Determine the acute angle, in degrees, between the tangents to $C$ at the points $P$ and $Q$.[3]

Accepted Answer

(ii) $y=\dfrac2{\mathrm e}(x-1)$, $Q=(-0.49516,-1.1001)$ to $5$ s.f. (iii) $66.9^\circ$ to $1$ d.p.

2022 HCI P1 Q9

2022 HCI P1 Q9

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