(a) (i) Use differentiation from first principles to show that d\left( x \right) = 1 - HSC - SSCE Mathematics Extension 1 - Question 7 - 2009 - Paper 1

To prove this by induction:

1. Base Case: For n = 1, we have: $\frac{d}{dx}(x^1) = 1 = 1 \cdot x^{1-1}.$ This is true.

2. Inductive Step: Assume it is true for n = k, i.e., ( \frac{d}{dx}(x^k) = kx^{k-1} ). For n = k + 1: $\frac{d}{dx}(x^{k+1}) = \frac{d}{dx}(x^k \cdot x) = x^k \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(x^k) = x^k + kx^k = (k + 1)x^k.$ Hence, it is true for n = k + 1. By induction, it holds for all positive integers n.

Given the triangle formed and applying the defined angles at points of the triangle, we can use the tangent subtraction formula:

If we let ( A = \tan^{-1}\left(\frac{a}{x}\right) ) and ( B = \tan^{-1}\left(\frac{h}{x}\right) ), we find:

To find the maximum of θ, we set the derivative ( \frac{d\theta}{dx} = 0. ) Using the previously found expression for θ:

1. Differentiate θ with respect to x: ( \frac{dθ}{dx} = \frac{d}{dx} \left( \tan^{-1}\left( \frac{ax}{x^2 + h(a + h)} \right) \right).

2. Set the derivative to zero and solve for x to find the critical points.

3. Evaluating the second derivative at these critical points will help determine if it is indeed a maximum.

By considering the geometric properties, when points P and T are distinct, φ captures a larger angle due to the exterior tangents from point T. Since θ is directly related to the height of the billboard and distances involved, as P approaches T, θ approaches φ, which implies that:

The points P, Q, and R create specific geometric properties due to the tangency of the circle. By applying properties of tangents and intercepts, we can derive:

1. Use the radius properties to relate the lengths TO and OQ involving T: $TO = OR \text{ or } OQ.$
2. This establishes the relationship of the tangential distance from T to the building based on triangle properties of points P and the center of the circle.