Find the equation of the normal to the curve \(y = \frac{{{{\text{e}}^x}\cos x\ln (x + {\text{e}})}}{{{{({x^{17}} + 1)}^5}}}\) at the point where \(x = 0\).

In your answer give the value of the gradient, of the normal, to three decimal places.

\(x = 0 \Rightarrow y = 1\)     (A1)

\(y'(0) = 1.367879 \ldots \)     (M1)(A1)

Note:     The exact answer is \(y'(0) = \frac{{{\text{e}} + 1}}{{\text{e}}} = 1 + \frac{1}{{\text{e}}}\).

so gradient of normal is \(\frac{{ - 1}}{{1.367879 \ldots }}\;\;\;( =  - 0.731058 \ldots )\)     (M1)(A1)

equation of normal is \(y =  - 0.731058 \ldots x + c\)     (M1)

gives \(y =  - 0.731x + 1\)     A1

Note:     The exact answer is \(y =  - \frac{{\text{e}}}{{{\text{e}} + 1}}x + 1\).

Accept \(y - 1 =  - 0.731058 \ldots (x - 0)\)

[7 marks]

Surprisingly many candidates ignored that fact that paper 2 is a calculator paper, attempted an algebraic approach and wasted lots of time. Candidates that used the GDC were in general successful and achieved 7/7. A number of candidates either found the equation of the tangent or used the positive reciprocal for the normal and many did not find the value of \(y\) corresponding to \(f(0)\).