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.