(i) \(f'(x) = \frac{x\frac{1}{x} – \ln x}{x^2}\) M1A1

\( = \frac{1 – \ln x}{x^2}\)

so \(f'(x) = 0\) when \(\ln x = 1\), i.e. \(x = e\) A1

(ii) \(f'(x) > 0\) when \(x < e\) and \(f'(x) < 0\) when \(x > e\) R1

hence local maximum AG

Note: Accept argument using correct second derivative.

(iii) \(y \leq \frac{1}{e}\) A1

[5 marks]

\(f”(x) = \frac{x^2\frac{-1}{x} – (1 – \ln x)2x}{x^4}\) M1

\( = \frac{-x – 2x + 2x\ln x}{x^4}\)

\( = \frac{-3 + 2\ln x}{x^3}\) A1

Note: May be seen in part (a).

\(f”(x) = 0\) (M1)

\(-3 + 2\ln x = 0\)

\(x = e^{\frac{3}{2}}\)

since \(f”(x) < 0\) when \(x < e^{\frac{3}{2}}\) and \(f”(x) > 0\) when \(x > e^{\frac{3}{2}}\) R1

then point of inflexion \(\left(e^{\frac{3}{2}}, \frac{3}{2e^{\frac{3}{2}}}\right)\) A1

[5 marks]

A1A1A1

Note: Award A1 for the maximum and intercept, A1 for a vertical asymptote and A1 for shape (including turning concave up).

[3 marks]

(i)

A1A1

Note: Award A1 for each correct branch.

(ii) all real values A1

(iii)

(M1)(A1)

Note: Award (M1)(A1) for sketching the graph of h, ignoring any graph of g.

\(-e^2 < x < -1\) (accept \(x < -1\)) A1

[6 marks]