a) A particle is moving along the x-axis in simple harmonic motion - HSC - SSCE Mathematics Extension 1 - Question 13 - 2015 - Paper 1

The maximum speed occurs at the vertex of the parabola representing v², which is the maximum point. In the equation:

$v² = n²(a² - (x - c)²)$

The maximum value of v² is reached when (x - c) = 0, hence:

$v²_{max} = n²a²$

Thus, the maximum speed v is:

$v_{max} = ext{max}(v) = n a$.

From the equation:

$v² = n²(a² - (x - c)²)$

we can compare it with the standard form of a parabola. Here, we can deduce that:

• n is the coefficient representing the amplitude of the motion,
• a is the amplitude that defines the displacement reaching maximum peak,
• c is the axis of symmetry within the oscillation.

Therefore, the parameters can be determined based on the context or initial conditions provided in a problem (not included here), with a, c, n all being positive constants.

Using the binomial theorem:

$(a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}$

In our case, let (a = 2x) and (b = \frac{1}{3x}), thus:

$a_{2} = \binom{18}{2} (2x)^{16} \left( \frac{1}{3x} \right)^{2}$

resulting in:

$a_{2} = \frac{18!}{2!(18-2)!} \cdot 2^{16} \cdot \frac{1}{9x^2}$

On simplification:

$a_{2} = \frac{18 \times 17}{2} \cdot \frac{2^{16}}{9}.$

The term independent of x can be derived from considering the coefficients where the combined degree of x sums to zero:

Thus, we need the k such that:

$2k + \left(18 - 2 - k\right) = 0$,

which implies:

$k = 9$

Using the binomial coefficient,

the expression will yield:

$\binom{18}{9}(2x)^{9} \left(\frac{1}{3x}\right)^{9}$

This leads to the result which can be evaluated directly. Thus, substituting into the polynomial and simplifying yields the term independent of x.

We use the principle of mathematical induction:

1. Base case (n=1):

   $\frac{1}{2!} = 1 - \frac{1}{2!}$, which holds true.

2. Induction step: Assume true for n=k. Show for n=k+1:

$1/2! + 3/3! + ... + k/(k+1)! + (k+1)/(k+2)! = 1 - \frac{1}{k+2}!$

Reorganizing gives us:

$1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1 - \frac{1}{(k+2)!}$,

matching with the inductive hypothesis; thus, proven by induction.

To show that f(x) is constant, we compute the derivative:

$f'(x) = \frac{d}{dx}\left[\cos^{-1}(x) + \cos^{-1}(-x)\right]$

differentiating gives:

$f'(x) = -\frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-(-x)^2}}$

Both terms are equal, hence:

$f'(x) = -\frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x^2}} = 0$,

showing that f(x) is constant.

From the property of the cosine function and the results of f(x):

Since f(x) is constant, we can calculate:

$f(1) = \cos^{-1}(1) + \cos^{-1}(-1) = 0 + \pi = \pi$

thus applying:

$f(x) = 0 + \pi - \cos^{-1}(x) ⇒ \cos^{-1}(-x) = π - \cos^{-1}(x).$