Find the value of $a$.

Show that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2y - {{\text{e}}^x}}}{{2(y - x)}}$.

Find the equation of the normal to $C$ at the point A.

Find the coordinates of the second point at which the normal found in part (c) intersects $C$.

Given that $v = {y^3},{\text{ }}y > 0$, find $\frac{{{\text{d}}v}}{{{\text{d}}x}}$ at $x = 0$.

${a^2} = 5 - 1$     (M1)

$a = 2$     A1

[2 marks]

$2y\frac{{{\text{d}}y}}{{{\text{d}}x}} - \left( {2x\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2y} \right) =  - {{\text{e}}^x}$     M1A1A1A1

Note:     Award M1 for an attempt at implicit differentiation, A1 for each part.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2y - {{\text{e}}^x}}}{{2(y - x)}}$     AG

[4 marks]

at $x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{3}{4}$     (A1)

finding the negative reciprocal of a number     (M1)

gradient of normal is $- \frac{4}{3}$

$y =  - \frac{4}{3}x + 2$     A1

[3 marks]

substituting linear expression     (M1)

${\left( { - \frac{4}{3}x + 2} \right)^2} - 2x\left( { - \frac{4}{3}x + 2} \right) + {{\text{e}}^x} - 5 = 0$ or equivalent

$x = 1.56$     (M1)A1

$y =  - 0.0779$     A1

$(1.56,{\text{ }} - 0.0779)$

[4 marks]

$\frac{{{\text{d}}v}}{{{\text{d}}x}} = 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}}$    M1A1

$\frac{{{\text{d}}v}}{{{\text{d}}x}} = 3 \times 4 \times \frac{3}{4} = 9$    A1

[3 marks]

Parts (a) to (c) were generally well done.

Parts (a) to (c) were generally well done.

Parts (a) to (c) were generally well done although a significant number of students found the equation of the tangent rather than the normal in part (c).

Whilst many were able to make a start on part (d), fewer students had the necessary calculator skills to work it though correctly.

There were many overly complicated solutions to part (e), some of which were successful.