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 displacement x is at the mean position, where (x - c) = 0. Substituting this in the velocity equation gives:

$v² = n² a²$

Thus the maximum speed is: $v_{max} = n a$

The values of a, c, and n can usually be determined from the specific constants given in the problem context or graph. Without specific numerical values or additional context, we can only state that:

• a is the amplitude of motion,
• c is the equilibrium position, and
• n is related to the angular frequency of the motion.

In the binomial expansion, we seek the coefficient a₂. For the term involving x¹⁴, we use the general formula for the binomial coefficients:

a_k = inom{n}{k} (a + b)^{n-k}

Here, we need to find: a₂ = inom{18}{2} (2x)^{16} (1/3x)^{2}

Calculating this yields:

a₂ = inom{18}{2} * 2^{16} * (1/3)^{2}

To find the term independent of x, we consider when the powers of x = 0 in:

(2x + rac{1}{3x})^{18}. We have the term given by:

inom{n}{k} (2x)^k (1/3x)^{n-k}

Setting the powers of x to sum to zero, we find: $k + (n - k) = 0$ implying n = k$and requires:$k = 9, n - k = 9.$$

Thus the constant term is: inom{18}{9} (2)^{9} (1/3)^{9}

To prove the given identity, we use induction:

1. Base Case (n = 1): rac{1}{2!} = 1 - rac{1}{2!}\ ext{True}.

2. Induction Step: Assume true for n = k: rac{1}{2!} + rac{3}{3!} + ... + rac{k}{(k + 1)!} = 1 - rac{1}{(k + 1)!}.

For n = k + 1: rac{1}{2!} + rac{3}{3!} + ... + rac{k}{(k + 1)!} + rac{k + 1}{(k + 2)!} = 1 - rac{1}{(k + 2)!}.

Using the assumption to show: 1 - rac{1}{(k + 2)!} = (1 - rac{1}{(k + 1)!}) + rac{k + 1}{(k + 2)!}.

Upon simplifying, this proves the statement for n = k + 1.

To show that f(x) is constant, we differentiate:

f'(x) = rac{d}{dx}(cos^{-1}(x) + cos^{-1}(-x)).

Using the chain rule: f'(x) = -rac{1}{ ext{ extsqrt{1-x²}}} + rac{1}{ ext{ extsqrt{1-x²}}} rac{d}{dx}(-x).

The two terms cancel each other, yielding: $f'(x) = 0$ which means f(x) is constant.

Using the property of inverse cosine, we can establish that: $f(x) = cos^{-1}(x) + cos^{-1}(-x) = constant$.

Therefore: $cos^{-1}(-x) = constant - cos^{-1}(x).$

Given the range of inverse cosine, this must equal: rac{ ext{π}}{2} - cos^{-1}(x), leading to the conclusion that: $cos^{-1}(-x) = π - cos^{-1}(x).$