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

Given that the polynomial has a double root at $x=1$, we know:

1. $P(1) = 0$, which gives us: $a(1)^4 + b(1)^3 + c(1)^2 + e = 0$ This simplifies to: $a + b + c + e = 0. (1)$

2. The derivative must also equal zero at $x=1$: $P'(x) = 4ax^3 + 3bx^2 + 2cx,$ thus: $P'(1) = 4a + 3b + 2c = 0. (2)$

Substituting $e = -3$ into (1) yields: $a + b + c - 3 = 0 ext{ or } a + b + c = 3. (3)$

Now, we can solve the system of equations (2) and (3) to find: $4a + 2c = -\frac{9}{2}.$

To find the slope of the tangent, we use:

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

From our previous work, we can substitute for $b$ in terms of $a$ and $c$ to find that the slope at this point will be:

$\text{slope} = P'(1) = 0.$

Thus, the slope of the tangent is 0.

The probability of a car completing one day is 0.7. Hence, the probability that it completes all four days:

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

So, the probability is 0.2401.

Let X be the number of cars that complete all four days. We can model this using a binomial distribution:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k},$ where n is the number of cars (8) and p is the probability of one car completing all four days (0.2401). To find the probability of at least three cars completing all days, we find:

$P(X \geq 3) = 1 - P(X < 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2).$ Calculating these terms will yield the desired probability.

To show the terminal velocity, we equate the gravitational force and the resistive force:

$mg = kv_t^2,$

giving us: $v_t = \sqrt{\frac{mg}{k}}.$

To show the maximum height: When the ball goes up, we can write its energy conservation as:

$mgh = \frac{1}{2} mv_0^2.$ Setting this equal includes resistive forces as it rises. The derived relation shows:

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

Starting with the equation of motion:

We can equate kinetic energies at the fall's end:

Given that at height H just before hitting the ground, we see: $\frac{1}{v^2} = \frac{1}{u^2} + \frac{1}{v_t^2},$ which relates the velocities at impact and initial conditions.