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 when v is at its peak. From the parabola's nature, the maximum value of v² is found at the vertex, corresponding to the value of x when the displacement is at its midpoint. Thus:

At maximum speed: $v_{max} = n imes a$

Therefore, the maximum speed of the particle is given by: $v_{max} = n imes a$

From the equation provided: $v^2 = n^2(a^2 - (x - c)^2)$ where a, c, and n are constants.

By comparing it with the standard form of motion, we can derive the values. We understand that:

• 'a' represents the amplitude of motion in simple harmonic motion.
• 'c' is the mean position around which the particle oscillates.
• 'n' is related to the angular frequency.

Thus:

• Value of a: Amplitude of the oscillation,
• Value of c: Midpoint of oscillation,
• Value of n: Frequency related parameter.

The exact numerical values can be determined based on additional context or specific constants set in a defined problem.

Using the binomial theorem:

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

For the expression given, we identify:

• $a = 2x$
• $b = \frac{1}{3}$
• $n = 18$

To find $a_2$, we need the coefficient of $x^{16}$:

$a_2 = {18 \choose 2} (2x)^{16} \left(\frac{1}{3}\right)^2$

Thus: a_2 = 153 \times (2^{16} x^{16}) \times rac{1}{9}

Simplifying gives the final expression for $a_2$.

To find the term independent of x, we look for the term where the powers of x cancel out. This occurs when: $(2x)^{18-k} \left( \frac{1}{3} \right)^k$

Set $n - k = 0$ implies: $18 - k = 0 \Rightarrow k = 18$

Inserting this into our binomial expansion gives: $a_{0} = {18 \choose 18} (2x)^0 \left( \frac{1}{3} \right)^{18} = \frac{1}{387420489}$

To prove via induction:

1. Base Case: For n=1,

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

Thus holds.

2. Induction Hypothesis: Assume true for n = k: $\frac{1}{2!} + \frac{3}{3!} + \cdots + \frac{k}{(k+1)!} = \frac{1}{(k+1)!}$

3. Inductive Step: Show true for n = k+1: $\frac{1}{2!} + \frac{3}{3!} + \cdots + \frac{k}{(k+1)!} + \frac{k+1}{(k+2)!} = \frac{1}{(k+2)!}$

Adjust accordingly to show equal.

By induction principle, the statement is true for all n ≥ 1.

To show f(x) is constant, we must demonstrate that its derivative is zero:

$f(x) = cos^{-1}(x) + cos^{-1}(-x)$

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

Since the derivative is zero, f(x) is constant throughout its domain.

Since f(x) = cos⁻¹(x) + cos⁻¹(-x) is constant and taking the range into account yields:

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

Thus, $cos^{-1}(-x) = π - cos^{-1}(x)$ as required.