0606 P12 - Mar 2019 - Q10 - 10 marks
8239
A curve is such that \(\dfrac{d^2y}{dx^2}=4e^{2x}+3\). When \(x=0\), \(y=-5\) and \(\dfrac{dy}{dx}=10\).
(i) Find the equation of the curve.
(ii) Find the equation of the normal to the curve at the point where \(x=\dfrac14\).
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