Let \(f(x) = k{x^4}\) . The point \({\text{P}}(1{\text{, }}k)\) lies on the curve of f . At P, the normal to the curve is parallel to \(y = - \frac{1}{8}x\) . Find the value of k.

gradient of tangent \(= 8\) (seen anywhere)     (A1)

\(f'(x) = 4k{x^3}\) (seen anywhere)     A1

recognizing the gradient of the tangent is the derivative     (M1)

setting the derivative equal to 8     (A1)

e.g. \(4k{x^3} = 8\) , \(k{x^3} = 2\)

substituting \(x = 1\) (seen anywhere)     (M1)

\(k = 2\)     A1    N4

[6 marks]

Candidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient of the normal line were not bothered by the lack of structure in the question, solving clearly with only a few steps, earning full marks. Those who were unclear often either gained a few marks for finding the derivative and substituting \(x = 1\) , or no marks for working that did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting the derivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among the candidates who demonstrated greater understanding, more used the gradient of the normal (the equation \( - \frac{1}{4}k = - \frac{1}{8}\) ) than the gradient of the tangent (\(4k = 8\) ) ; this led to more algebraic errors in obtaining the final answer of \(k = 2\) . A number of unsuccessful candidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.