(a) (M1A1A1A1, N4).

Working:
Either
\(\alpha = \arctan\frac{7}{10} – \arctan\frac{5}{10}\) ( = \(34.992 \ldots^\circ – 26.5651 \ldots^\circ\)) (M1)(A1)(A1).
Note: Award (M1) for \(\alpha = \hat{A PT} – \hat{B PT}\), (A1) for a correct \(\hat{A PT}\) and (A1) for a correct \(\hat{B PT}\).
Or
\(\alpha = \arctan 2 – \arctan\frac{10}{7}\) ( = \(63.434 \ldots^\circ – 55.008 \ldots^\circ\)) (M1)(A1)(A1).
Note: Award (M1) for \(\alpha = \hat{P BT} – \hat{P AT}\), (A1) for a correct \(\hat{P BT}\) and (A1) for a correct \(\hat{P AT}\).
Or
\(\alpha = \arccos\left(\frac{125 + 149 – 4}{2 \times \sqrt{125} \times \sqrt{149}}\right)\) (M1)(A1)(A1).
Note: Award (M1) for use of cosine rule, (A1) for a correct numerator and (A1) for a correct denominator.
Then
\(\alpha = 8.43^\circ\) (A1).

[4 marks]

(b) (M1A1A1M1AG, N4).

Working:
Either
\(\tan \alpha = \frac{\frac{7}{x} – \frac{5}{x}}{1 + \left(\frac{7}{x}\right)\left(\frac{5}{x}\right)}\) (M1A1A1).
Note: Award M1 for use of \(\tan(A – B)\), A1 for a correct numerator and A1 for a correct denominator.
\( = \frac{\frac{2}{x}}{1 + \frac{35}{x^2}}\) (M1).
Or
\(\tan \alpha = \frac{\frac{x}{5} – \frac{x}{7}}{1 + \left(\frac{x}{5}\right)\left(\frac{x}{7}\right)}\) (M1A1A1).
Note: Award M1 for use of \(\tan(A – B)\), A1 for a correct numerator and A1 for a correct denominator.
\( = \frac{\frac{2x}{35}}{1 + \frac{x^2}{35}}\) (M1).
Or
\(\cos \alpha = \frac{x^2 + 35}{\sqrt{(x^2 + 25)(x^2 + 49)}}\) (M1A1).
Note: Award M1 for either use of the cosine rule or use of \(\cos(A – B)\).
\(\sin \alpha = \frac{2x}{\sqrt{(x^2 + 25)(x^2 + 49)}}\) (A1).
\(\tan \alpha = \frac{\frac{2x}{\sqrt{(x^2 + 25)(x^2 + 49)}}}{\frac{x^2 + 35}{\sqrt{(x^2 + 25)(x^2 + 49)}}}\) (M1).
Then
\(\tan \alpha = \frac{2x}{x^2 + 35}\) (AG).

[4 marks]

(c) (M1A1A1M1A1M1A1R1AG, N11).

Working:
(i) \(\frac{\text{d}}{\text{d}x}(\tan \alpha) = \frac{2(x^2 + 35) – (2x)(2x)}{(x^2 + 35)^2}\) ( = \(\frac{70 – 2x^2}{(x^2 + 35)^2}\)) (M1A1A1).
Note: Award M1 for attempting product or quotient rule differentiation, A1 for a correct numerator and A1 for a correct denominator.
(ii) Method 1
Either
\(\frac{\text{d}}{\text{d}x}(\tan \alpha) = 0 \Rightarrow 70 – 2x^2 = 0\) (M1).
\(x = \sqrt{35} \, \text{m}\) ( = \(5.9161 \ldots \, \text{m}\)) (A1).
\(\tan \alpha = \frac{1}{\sqrt{35}}\) ( = \(0.16903 \ldots\)) (A1).
Or
Attempting to locate the stationary point on the graph of \(\tan \alpha = \frac{2x}{x^2 + 35}\) (M1).
\(x = 5.9161 \ldots \, \text{m}\) ( = \(\sqrt{35} \, \text{m}\)) (A1).
\(\tan \alpha = 0.16903 \ldots\) ( = \(\frac{1}{\sqrt{35}}\)) (A1).
Then
\(\alpha = 9.59^\circ\) (A1).
Method 2
Either
\(\alpha = \arctan\left(\frac{2x}{x^2 + 35}\right) \Rightarrow \frac{\text{d}\alpha}{\text{d}x} = \frac{70 – 2x^2}{(x^2 + 35)^2 + 4x^2}\) (M1).
\(\frac{\text{d}\alpha}{\text{d}x} = 0 \Rightarrow x = \sqrt{35} \, \text{m}\) ( = \(5.9161 \, \text{m}\)) (A1).
Or
Attempting to locate the stationary point on the graph of \(\alpha = \arctan\left(\frac{2x}{x^2 + 35}\right)\) (M1).
\(x = 5.9161 \ldots \, \text{m}\) ( = \(\sqrt{35} \, \text{m}\)) (A1).
Then
\(\alpha = 0.1674 \ldots\) ( = \(\arctan\frac{1}{\sqrt{35}}\)) (A1).
\(\alpha = 9.59^\circ\) (A1).
(iii) \(\frac{\text{d}^2}{\text{d}x^2}(\tan \alpha) = \frac{(x^2 + 25)^2(-4x) – (2)(2x)(x^2 + 35)(70 – 2x^2)}{(x^2 + 35)^4}\) ( = \(\frac{4x(x^2 – 105)}{(x^2 + 35)^3}\)) (M1A1).
Substituting \(x = \sqrt{35}\) ( = \(5.9161 \ldots\)) into \(\frac{\text{d}^2}{\text{d}x^2}(\tan \alpha)\) (M1).
\(\frac{\text{d}^2}{\text{d}x^2}(\tan \alpha) < 0\) ( = \(-0.004829 \ldots\)) and so \(\alpha = 9.59^\circ\) is the maximum value of \(\alpha\) (R1).
\(\alpha\) never exceeds 10° (AG).

[11 marks]

(d) (M1A1A1, N3).

Working:
Attempting to solve \(\frac{2x}{x^2 + 35} \geqslant \tan 7^\circ\) (M1).
Note: Award (M1) for attempting to solve \(\frac{2x}{x^2 + 35} = \tan 7^\circ\).
\(x = 2.55\) and \(x = 13.7\) (A1).
\(2.55 \leqslant x \leqslant 13.7 \, \text{m}\) (A1).

[3 marks]

Total [22 marks]