Let \(f(x) = \frac{{\ln (4x)}}{x}\) for \(0 < x \le 5\).

Points \({\text{P}}(0.25,{\text{ }}0)\) and \(Q\) are on the curve of \(f\). The tangent to the curve of \(f\) at \(P\) is perpendicular to the tangent at \(Q\). Find the coordinates of \(Q\).

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

eg\(\;\;\;f'\)

finding the gradient of \(f\) at \(P\)     (A1)

eg\(\;\;\;f'(0.25) = 16\)

evidence of taking negative reciprocal of their gradient at \(P\)     (M1)

eg\(\;\;\;\frac{{ - 1}}{m},{\text{ }} - \frac{1}{{f'(0.25)}}\)

equating derivatives     M1

eg\(\;\;\;f'(x) = \frac{{ - 1}}{{16}},{\text{ }}f' =  - \frac{1}{m},{\text{ }}\frac{{x\left( {\frac{1}{x}} \right) - \ln (4x)}}{{{x^2}}} = 16\)

finding the \(x\)-coordinate of \(Q\), \(x = 0.700750\)

\(x = 0.701\)     A1     N3

attempt to substitute their \(x\) into \(f\) to find the \(y\)-coordinate of \(Q\)     (M1)

eg\(\;\;\;f(0.7)\)

\(y = 1.47083\)

\(y = 1.47\)     A1     N2

[7 marks]

Few candidates were completely successful with this question. Students using an analytical approach were aware of the relationship between the gradients of the tangent and normal, but were often unable to find a correct derivative initially. The solution was made significantly easier when the GDC was used effectively and the few candidates who used this approach, were generally successful. Attempting to find the equation of the tangent and/or the normal were common, ineffective approaches.