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 the value of x is either at the extremities of the motion. The maximum occurs when the potential energy is minimum, corresponding to the maximum displacement. To find this, we observe that:

At these points, v^2 is maximal:

The maximum speed, therefore, is: $ext{max v} = n(a)$ Thus, the maximum speed of the particle is n times a.

In the equation, we can see that the constants must represent the amplitudes and scaling factors. Thus:

• a represents the amplitude of the motion,
• c represents the midpoint or equilibrium position,
• n represents the angular frequency.

By analyzing the physical situation of harmonic motion, we conclude:

• a = maximum displacement,
• c = equilibrium position,
• n = angular frequency.

Using the binomial expansion:

(2x + rac{1}{3x})^{18}

First, we calculate a_2, which corresponds to the term where the power of x is 14. This term arises from taking 2x to the power of 16 and the constant to the power of 2:

The general term formula for binomial expansion can be utilized:

a_k = \binom{n}{k} (2x)^{n-k} (1/3x)^k.

Thus, a_2 = \binom{18}{2} (2x)^{16}(1/3x)^2 ightarrow a_2 = \binom{18}{2} (2^{16}/3^2)x^{16 - 2}.

The term independent of x occurs for k = 18:

a_0 = \binom{18}{0} (2x)^{18 - 0}(1/3x)^{0} + \binom{18}{3} (2x)^{15}(1/3x)^3.

After calculating: Independently, we have:

• From k=0: 2^{18}
• From k=3 we sum up: \binom{18}{3} \cdot 2^{15} (1/3)^3. Thus, combine these to get the overall constant term.

To prove the statement by induction:

1. Base Case: For n = 1, $LHS = \frac{1}{2!} = 1 - \frac{1}{2!}$ which confirms the base case holds.
2. Inductive Step: Assume valid for n = k, then: rac{1}{2!} + \frac{2}{3!} + … + \frac{k}{(k + 1)!} = 1 - \frac{1}{(k + 1)!} For k + 1: $1 - \frac{1}{(k + 1)!} + \frac{k + 1}{(k + 2)!}$ Combine to simplify yielding : $= 1 - \frac{1}{(k + 2)!},$ which completes the induction.

To find the derivative of f(x): $f'(x) = \frac{d}{dx} \{cos^{-1}(x)\} + \frac{d}{dx} \{cos^{-1}(-x)\}$ Using the derivative of inverse cosine: And solving yields f'(x) = 0, proving that f(x) is constant.

From constancy, we know that: For x in the defined range, applying the properties of cosine functions: From f(x), we can deduce: $cos^{-1}(x)+cos^{-1}(-x)=π$ leading to: $cos^{-1}(-x)=π - cos^{-1}(x)$ confirming the required relationship.