SSCE HSC Mathematics Extension 2 Polynomial equations: The Argand diagram shows complex

Question

The Argand diagram shows complex numbers w and z with arguments φ and θ respectively, where φ < θ. The area of the triangle formed by 0, w and z is A.

Show that $zw = wz = 4iA.$

The polynomial $P(x) = ax^4 + bx^3 + cx^2 + e$ has remainder -3 when divided by $x-1$. The polynomial has a double root at $x=1$.

(i) Show that $4a + 2c = -\frac{9}{2}.$

(ii) Hence, or otherwise, find the slope of the tangent to the graph $y = P(x)$ when $x=1.$

Eight cars participate in a competition that lasts for four days. The probability that a car completes a day is 0.7. Cars that do not complete a day are eliminated.

(i) Find the probability that a car completes all four days of the competition.

(ii) Find an expression for the probability that at least three cars complete all four days of the competition.

A ball of mass m is projected vertically into the air from the ground with initial velocity v. After reaching the maximum height H it falls back to the ground. While in the air, the ball experiences a resistive force $k v^2$, where v is the velocity of the ball and k is a constant.

The equation of motion when the ball falls can be written as $m\frac{d^2y}{dt^2} = mg - kv^2$. (Do NOT prove this.)

(i) Show that the terminal velocity $v_t$ of the ball when it falls is $\sqrt{\frac{mg}{k}}.$

(ii) Show that when the ball goes up, the maximum height H is
$$H = \frac{v^2}{2g} \ln\left( 1 + \frac{v^2}{v_t^2} \right).$$

(iii) When the ball falls from height H it hits the ground with velocity w.

Show that $\frac{1}{v^2} = \frac{1}{u^2} + \frac{1}{v_t^2}.$

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

Worked Solution & Example Answer: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

Show that $\frac{1}{v^2} = \frac{1}{u^2} + \frac{1}{v_t^2}.$

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