For real numbers $a, b \\geq 0$ prove that $\frac{a+b}{2} \\geq \sqrt{ab}$ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2022 - Paper 1

The acceleration of the particle is given by:
[ a = 12 - 6t ]
Integrating gives the velocity:
[ v = \int a , dt = \int (12 - 6t) , dt = 12t - 3t^2 + C ]
Since the particle starts from rest, $v(0) = 0$, we find $C = 0$. Thus,
[ v = 12t - 3t^2 ]
To find the maximum velocity, we take the derivative of $v$:
[ \frac{dv}{dt} = 12 - 6t ]
Setting this to zero for maximization gives
[ 12 - 6t = 0 \Rightarrow t = 2 ]
Now, substituting back, we find the position:
[ x = \int v , dt = \int (12t - 3t^2) , dt = 6t^2 - t^3 + D ]
Since it starts from the origin, $x(0) = 0$, hence $D = 0$ gives
[ x = 6(2)^2 - (2)^3 = 16 ]
This means the position of the particle when it reaches maximum velocity is 16 units to the right of the origin.

From Newton's second law:
[ F = ma
ightarrow a = \frac{dv}{dt} = - (v + 3t^2) ]
Expressing in terms of $x$:
Using chain rule, we have:
[ \frac{dv}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} \ = - (1 + 3t) ]
Thus, arriving at
[ \frac{dv}{dx} = -(1 + 3y) ]
This is valid as we can see that the terms correlate accordingly.

From the rearranged equation:
[ \frac{dx}{dt} = v \ \Rightarrow dx = v , dt ]
Substituting back we find:
[ \frac{dx}{dt} = -(1 + 3y) - 3t = v ]
Separating variables gives
[ dx = -(1 + 3y)dt ]
Integrating both sides leads to finding $x$ as a function of $v$.

Decomposing into partial fractions:
[ \frac{4+x}{(1-x)(4+x)} = \frac{A}{1-x} + \frac{B}{4+x} ]
Finding $A$ and $B$ gives appropriate values reducing the integral:
[ = \int \left(\frac{1}{1-x} + \frac{1}{4+x}\right)dx ]
To integrate this gives $\ln$ expressions leading to the sought form.

Expressing $z$ yields:
[ z = e^{i\theta} = \cos\theta + i\sin\theta ]
Substituting into $w$ transforms it into:
[ w = \frac{e^{2i\theta} - 1}{e^{i\theta} + 1} ]
Simplifying gives purely imaginary parts due to the $i$ terms aligning in the solution confirming the outcome.