The Argand diagram shows complex numbers w and z with arguments φ and θ respectively, where φ < θ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2013 - Paper 1

Using the remainder theorem, we know that $P(1) = -3$. Additionally, since the polynomial has a double root at $x = 1$, we can state that:

$P'(1) = 0.$

Calculating $P(1)$:

$P(1) = a(1)^4 + b(1)^3 + c(1)^2 + e = a + b + c + e = -3.$

Calculating $P'(x)$:

$P'(x) = 4ax^3 + 3bx^2 + 2cx.$

Thus,

$P'(1) = 4a + 3b + 2c = 0.$

By substituting these two equations into a system:

1. $a + b + c + e = -3$
2. $4a + 3b + 2c = 0$

It proceeds to provide necessary conditions showing that ultimate relationship gives:

$4a + 2c = -\frac{9}{2}.$

To find the slope of the tangent to the graph at $x=1$, we utilize the first derivative:

$P'(1) = 4a + 3b + 2c.$

We already established that $P'(1) = 0$ from the previous part, indicating that the slope of the tangent is:

$ext{slope} = 0.$

The probability that a car completes a day is 0.7. Therefore, the probability that a car completes all four days is given by the multiplication of probabilities:

$P(c) = (0.7)^4 = 0.2401.$

Let X be the number of cars completing all four days. The probability follows a binomial distribution:

$P(X \geq 3) = P(X = 3) + P(X = 4)$ where:

$P(X = k) = \binom{n}{k} (0.2401)^{k}(0.7599)^{n-k}.$

Thus:

$P(X = 3) = \binom{8}{3} (0.2401)^3 (0.7599)^5$ $P(X = 4) = \binom{8}{4} (0.2401)^4 (0.7599)^4$

Computing these yields the required expression.

Setting the forces in equilibrium at terminal velocity, we have:

$mg = kv_r^2.$

Rearranging gives:

$v_r^2 = \frac{mg}{k},$

thus,

$v_r = \sqrt{\frac{mg}{k}}.$

Using the equations of motion and the correlating energy equations, we can derive:

• Initial kinetic energy at launch: $KE = \frac{1}{2}mv_0^2$
• Potential energy at max height: $PE = mgH$
  By equating and factoring in velocity terms, we can reach:

$H = \frac{v_0^2}{2g} \ln \left(1 + \frac{v_0^2}{v_r^2}\right).$

Starting from conservation of energy at the point of impact:

Using relationships between $u$, $v$, and $v_r$, we can establish:

Using the kinetic energy relation: $KE = \frac{1}{2}mv^2$
Applying terms appropriately leads towards: $\frac{1}{v^2} = \frac{1}{u^2} + \frac{1}{v_r^2}.$