a. [4 marks]

METHOD 1:
\( \alpha = \angle APT – \angle BPT \). For \( x = 10 \), in \( \triangle APT \): \( \tan \angle APT = \frac{7}{10} \), so \( \angle APT = \arctan \frac{7}{10} \approx 34.992^\circ \) (A1). In \( \triangle BPT \): \( \tan \angle BPT = \frac{5}{10} \), so \( \angle BPT = \arctan \frac{5}{10} \approx 26.565^\circ \) (A1). Thus, \( \alpha = 34.992^\circ – 26.565^\circ \approx 8.43^\circ \) (M1)(A1).
Answer: \( \alpha = 8.43^\circ \).

METHOD 2:
\( \alpha = \angle PBT – \angle PAT \). In \( \triangle PBT \): \( \tan \angle PBT = \frac{10}{5} = 2 \), so \( \angle PBT \approx 63.434^\circ \) (A1). In \( \triangle PAT \): \( \tan \angle PAT = \frac{10}{7} \), so \( \angle PAT \approx 55.008^\circ \) (A1). Thus, \( \alpha = 63.434^\circ – 55.008^\circ \approx 8.43^\circ \) (M1)(A1).
Answer: \( \alpha = 8.43^\circ \).

METHOD 3:
Use cosine rule in \( \triangle APB \): \( \cos \alpha = \frac{PA^2 + PB^2 – AB^2}{2 \cdot PA \cdot PB} \). Compute: \( PA = \sqrt{10^2 + 7^2} = \sqrt{149} \), \( PB = \sqrt{10^2 + 5^2} = \sqrt{125} \), \( AB = 2 \). Numerator: \( 149 + 125 – 4 = 270 \). Denominator: \( 2 \cdot \sqrt{149} \cdot \sqrt{125} \) (A1)(A1). Thus, \( \cos \alpha = \frac{270}{2 \cdot \sqrt{149} \cdot \sqrt{125}} \), so \( \alpha \approx 8.43^\circ \) (M1)(A1).
Answer: \( \alpha = 8.43^\circ \).

b. [4 marks]

METHOD 1:
Use \( \tan \alpha = \tan (\angle APT – \angle BPT) = \frac{\tan \angle APT – \tan \angle BPT}{1 + \tan \angle APT \cdot \tan \angle BPT} \). In \( \triangle APT \): \( \tan \angle APT = \frac{7}{x} \) (A1). In \( \triangle BPT \): \( \tan \angle BPT = \frac{5}{x} \) (A1). Thus, \( \tan \alpha = \frac{\frac{7}{x} – \frac{5}{x}}{1 + \frac{7}{x} \cdot \frac{5}{x}} = \frac{\frac{2}{x}}{1 + \frac{35}{x^2}} = \frac{2x}{x^2 + 35} \) (M1)(A1).
Answer: \( \tan \alpha = \frac{2x}{x^2 + 35} \).

METHOD 2:
Use \( \tan \alpha = \tan (\angle PBT – \angle PAT) = \frac{\tan \angle PBT – \tan \angle PAT}{1 + \tan \angle PBT \cdot \tan \angle PAT} \). In \( \triangle PBT \): \( \tan \angle PBT = \frac{x}{5} \) (A1). In \( \triangle PAT \): \( \tan \angle PAT = \frac{x}{7} \) (A1). Thus, \( \tan \alpha = \frac{\frac{x}{5} – \frac{x}{7}}{1 + \frac{x}{5} \cdot \frac{x}{7}} = \frac{\frac{2x}{35}}{1 + \frac{x^2}{35}} = \frac{2x}{x^2 + 35} \) (M1)(A1).
Answer: \( \tan \alpha = \frac{2x}{x^2 + 35} \).

METHOD 3:
In \( \triangle APB \), use cosine rule: \( \cos \alpha = \frac{PA^2 + PB^2 – AB^2}{2 \cdot PA \cdot PB} = \frac{x^2 + 49 + x^2 + 25 – 4}{2 \cdot \sqrt{x^2 + 49} \cdot \sqrt{x^2 + 25}} = \frac{x^2 + 35}{\sqrt{(x^2 + 25)(x^2 + 49)}} \) (M1)(A1). Then, \( \sin \alpha = \frac{2x}{\sqrt{(x^2 + 25)(x^2 + 49)}} \) (A1). Thus, \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{2x}{x^2 + 35} \) (M1).
Answer: \( \tan \alpha = \frac{2x}{x^2 + 35} \).

c. [11 marks]

(i) Differentiate \( \tan \alpha = \frac{2x}{x^2 + 35} \): Use quotient rule: \( \frac{\text{d}}{\text{d}x}(\tan \alpha) = \frac{(2)(x^2 + 35) – (2x)(2x)}{(x^2 + 35)^2} = \frac{2x^2 + 70 – 4x^2}{(x^2 + 35)^2} = \frac{70 – 2x^2}{(x^2 + 35)^2} \) (M1)(A1)(A1).
Answer: \( \frac{\text{d}}{\text{d}x}(\tan \alpha) = \frac{70 – 2x^2}{(x^2 + 35)^2} \).

(ii) METHOD 1:
Set \( \frac{\text{d}}{\text{d}x}(\tan \alpha) = 0 \): \( 70 – 2x^2 = 0 \implies x^2 = 35 \implies x = \sqrt{35} \approx 5.916 \) m (M1)(A1). Then, \( \tan \alpha = \frac{2 \cdot \sqrt{35}}{35 + 35} = \frac{1}{\sqrt{35}} \approx 0.16903 \) (A1). Thus, \( \alpha = \arctan \frac{1}{\sqrt{35}} \approx 9.59^\circ \) (A1).
Answer: \( \alpha = 9.59^\circ \).

METHOD 2:
For \( \alpha = \arctan \left( \frac{2x}{x^2 + 35} \right) \), compute derivative: \( \frac{\text{d}\alpha}{\text{d}x} = \frac{1}{1 + \left( \frac{2x}{x^2 + 35} \right)^2} \cdot \frac{70 – 2x^2}{(x^2 + 35)^2} \). Set \( \frac{\text{d}\alpha}{\text{d}x} = 0 \): \( 70 – 2x^2 = 0 \implies x = \sqrt{35} \) (M1)(A1). Then, \( \alpha = \arctan \frac{2 \cdot \sqrt{35}}{35 + 35} = \arctan \frac{1}{\sqrt{35}} \approx 9.59^\circ \) (A1)(A1).
Answer: \( \alpha = 9.59^\circ \).

(iii) Differentiate \( \frac{\text{d}}{\text{d}x}(\tan \alpha) = \frac{70 – 2x^2}{(x^2 + 35)^2} \): Use quotient rule: \( \frac{\text{d}^2}{\text{d}x^2}(\tan \alpha) = \frac{(-4x)(x^2 + 35)^2 – (70 – 2x^2) \cdot 2(x^2 + 35) \cdot 2x}{(x^2 + 35)^4} = \frac{-4x(x^2 + 35)^2 – 4x(70 – 2x^2)(x^2 + 35)}{(x^2 + 35)^4} = \frac{4x(x^2 – 105)}{(x^2 + 35)^3} \) (M1)(A1). At \( x = \sqrt{35} \), \( \frac{\text{d}^2}{\text{d}x^2}(\tan \alpha) = \frac{4 \sqrt{35} (35 – 105)}{(35 + 35)^3} < 0 \), confirming a maximum (M1)(R1). Since \( \alpha \approx 9.59^\circ < 10^\circ \), \( \alpha \) never exceeds 10° (A1).
Answer: \( \alpha \leq 9.59^\circ \).

d. [3 marks]

Solve \( \tan \alpha \geq \tan 7^\circ \): \( \frac{2x}{x^2 + 35} \geq \tan 7^\circ \approx 0.1228 \) (M1). Solve \( \frac{2x}{x^2 + 35} = 0.1228 \): \( 2x = 0.1228 (x^2 + 35) \implies 0 = 0.1228 x^2 – 2x + 4.298 \). Quadratic formula gives \( x \approx 2.55, 13.7 \) (A1). Since \( \tan \alpha \) is maximized at \( x = \sqrt{35} \), test interval: \( 2.55 \leq x \leq 13.7 \) (A1).
Answer: \( 2.55 \leq x \leq 13.7 \) m.