(a) Find the particular solution to the differential equation $(x - 2) \frac{dy}{dx} = xy$ that passes through the point (0, 1) - HSC - SSCE Mathematics Extension 1 - Question 14 - 2022 - Paper 1

Let ( \mathbf{p} = \lambda_0 \mathbf{v} ) and consider the expression:

$|\mathbf{i} - \lambda \mathbf{v}|^2 = |\mathbf{i}|^2 - 2\lambda \mathbf{i} \cdot \mathbf{v} + \lambda^2 |\mathbf{v}|^2$

To minimize this, we take the derivative with respect to ( \lambda ) and set it to zero:

$\frac{d}{d\lambda} (|\mathbf{i} - \lambda \mathbf{v}|^2) = -2 \mathbf{i} \cdot \mathbf{v} + 2\lambda |\mathbf{v}|^2 = 0$

Thus:

$\lambda = \frac{\mathbf{i} \cdot \mathbf{v}}{|\mathbf{v}|^2}$

This shows that ( |\mathbf{i} - \lambda \mathbf{v}| ) is minimized at ( \lambda = \lambda_0 ).

The range ( R ) of the projectile is given by:

$R = \frac{u^2 \sin(2\theta)}{g}$

For the target to be hit, the distance ( d ) must satisfy:

$d < R/3$

Solving for ( d ) gives:

$d < \frac{u^2 \sin(2\theta)}{3g}$

Using the maximum value of ( \sin(2\theta) = 1 ), we obtain:

$d < \frac{u^2}{3g}$

To find the upper limit, we relate the initial speed and the parameters, leading to a conclusion that confirms the requirement that ( d ) must indeed be less than 37% of the maximum range.

Let ( n ) be the number of tickets sold. The probability of a passenger showing up is ( p = 0.95 ). Thus, the expected number of passengers is ( 0.95n ).

To ensure not more than 10% of flights are oversold, we set:

$0.95n \leq 350 \Rightarrow n \leq \frac{350}{0.95} \approx 368.42$

Thus, the maximum number of tickets that can be sold is approximately 368.