16N.1.AHL.TZ0.H_9

pestleMathematicsAAHLPaper 116N· ahl-5-14-implicit-functions-related-rates-optimisationsource ↗

A curve has equation $3x - 2{y^2}{{\text{e}}^{x - 1}} = 2$ .

Find an expression for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ in terms of $x$ and $y$ .

[5]

Find the equations of the tangents to this curve at the points where the curve intersects the line $x = 1$ .

[4]

Markscheme / solution

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

attempt to differentiate implicitly M1

$3 - \left( {4y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2{y^2}} \right){{\text{e}}^{x - 1}} = 0$ A1A1A1

Note: Award A1 for correctly differentiating each term.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 \bullet {{\text{e}}^{1 - x}} - 2{y^2}}}{{4y}}$ A1

Note: This final answer may be expressed in a number of different ways.

[5 marks]

$3 - 2{y^2} = 2 \Rightarrow {y^2} = \frac{1}{2} \Rightarrow y = \pm \sqrt {\frac{1}{2}}$ A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 - 2 \bullet \frac{1}{2}}}{{ \pm 4\sqrt {\frac{1}{2}} }} = \pm \frac{{\sqrt 2 }}{2}$ M1

at $\left( {1,{\text{ }}\sqrt {\frac{1}{2}} } \right)$ the tangent is $y - \sqrt {\frac{1}{2}} = \frac{{\sqrt 2 }}{2}(x - 1)$ and A1

at $\left( {1,{\text{ }} - \sqrt {\frac{1}{2}} } \right)$ the tangent is $y + \sqrt {\frac{1}{2}} = - \frac{{\sqrt 2 }}{2}(x - 1)$ A1

Note: These equations simplify to $y = \pm \frac{{\sqrt 2 }}{2}x$ .

Note: Award A0M1A1A0 if just the positive value of $y$ is considered and just one tangent is found.

[4 marks]

Examiners’ report

[N/A]