EXN.2.AHL.TZ0.6

pestleMathematicsAAHLPaper 2EXN· sl-5-4-tangents-and-normalsource ↗

The curve $C$ has equation $e^{2 y} = x^{3} + y$ .

Show that $\frac{d y}{d x} = \frac{3 x^{2}}{2 e^{2 y} - 1}$ .

[3]

The tangent to $C$ at the point Ρ is parallel to the $y$ -axis.

Find the $x$ -coordinate of Ρ.

[4]

Markscheme / solution

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

attempts implicit differentiation on both sides of the equation M1

$2 e^{2 y} \frac{d y}{d x} = 3 x^{2} + \frac{d y}{d x}$ A1

$(2 e^{2 y} - 1) \frac{d y}{d x} = 3 x^{2}$ A1

so $\frac{d y}{d x} = \frac{3 x^{2}}{2 e^{2 y} - 1}$ AG

[3 marks]

attempts to solve $2 e^{2 y} - 1 = 0$ for $y$ (M1)

$y = - 0.346 \dots (= \frac{1}{2} \ln \frac{1}{2})$ A1

attempts to solve $e^{2 y} = x^{3} + y$ for $x$ given their value of $y$ (M1)

$x = 0.946 (= {(\frac{1}{2} (1 - \ln \frac{1}{2}))}^{\frac{1}{3}})$ A1

[4 marks]

Examiners’ report

[N/A]