Let a and b be real numbers with a ≠ b - HSC - SSCE Mathematics Extension 2 - Question 4 - 2011 - Paper 1

From part (i), we determined that the locus of complex numbers z satisfying this equation is a vertical line in the complex plane. Specifically, it can be represented by the line with the equation:

$x = \frac{a + b}{2} + \frac{1}{2(b - a)}.$

As the value of y varies, z traces out this vertical line.

To show that FADG is a cyclic quadrilateral, we need to demonstrate that the opposite angles, namely $\angle FAD$ and $\angle FGD$, sum to 180 degrees. By the properties of cyclic quadrilaterals and the inscribed angle theorem, we can use the fact that quadrilateral ABCD is cyclic, and the angles formed by intersecting tangents and secants will provide the necessary relationships between the angles, confirming that FADG must be cyclic.

From the properties of cyclic quadrilaterals and the equal angles formed by alternate segments, we have that angle GFD is equal to angle ZED due to their inscribing arc being the same. This occurs because E lies on the circumcircle of quadrilateral ABCD, which supports the equality of these angles.

To prove that GA is a tangent, we can show that the angle between the tangent GA and the chord AE is equal to the angle subtended by the chord at the opposite side, i.e., angle GAE is equal to angle GCE. This can be shown using the properties of tangents and the fact that points A, B, C, D lie on the circumference of the circle, which further validates that GA indeed acts as a tangent.

Given the linearity of the differential equation, we can substitute $y = Af(t) + Bg(t)$ into the equation and use the linearity property:

$\frac{d^2(Af(t) + Bg(t))}{dt^2} + \frac{d(Af(t) + Bg(t))}{dt} + 2(Af(t) + Bg(t)) = 0.$

Since both $f(t)$ and $g(t)$ are solutions, the equation holds true, confirming that this form is also a solution.

Starting with the assumed solution $y = Ae^{kt},$ we find:

$\frac{d^2y}{dt^2} = A k^2 e^{kt}, \quad \frac{dy}{dt} = Ak e^{kt}.$

Replacing these back into the differential equation:

$A k^2 e^{kt} + A k e^{kt} + 2A e^{kt} = 0.$

Factoring out $Ae^{kt}$ gives:

$A e^{kt}(k^2 + k + 2) = 0.$

Given that A and e^{kt} are non-zero, we require:

$k^2 + k + 2 = 0,$

The solution to this quadratic gives the only possible values as $k = -1$ and $k = -2.$

Given the conditions $y(0) = 0$ and $\frac{dy}{dt}(0) = 1$, we first evaluate:

1. When t = 0: $y(0) = A + B = 0.$

2. Now for the derivative: $\frac{dy}{dt} = -2Ae^{-2t} - Be^{-t}.$
   Substituting t = 0 gives: $\frac{dy}{dt}(0) = -2A - B = 1.$

We now solve the system of equations:

1. $A + B = 0$
2. $-2A - B = 1$

Solving gives: $B = -A,$ thus substituting into the second equation: