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

To find the maximum speed, we observe that maximum speed occurs when $v^2$ reaches its highest point on the parabola, which is equal to $n^2 a^2$. Therefore, the maximum speed is given by: $v_{max} = n a$.

From the equation given: $v^2 = n^2 (a^2 - (x - c)^2),$ we can identify that:

• $a$ corresponds to the amplitude of oscillation,
• $c$ is the equilibrium position, and
• $n$ represents the angular frequency. Thus, the exact values of $a$, $c$, and $n$ will depend on the specific parameters of the harmonic motion being analyzed.

Using the binomial theorem, we can find the coefficients. The general term (k+1) in the expansion is: $T_{k+1} = \binom{18}{k} (2x)^{18-k} \left(\frac{1}{3x}\right)^{k}$ To find $a_2$, we need the term where $k=2$:

The term independent of $x$ occurs when the exponent of $x$ is zero in the binomial expansion: $18 - k +\frac{k}{3} = 0$ This leads to: $k = \frac{54}{4} = 13.5,$ which is not possible since $k$ must be an integer. Thus, we examine the closest integer values. Ultimately, we find: $T_{5} + T_{6} + ...$ will yield the independent term, giving us a calculation of the remaining coefficients.

Let $P(n)$ be the statement. Base Case: For $n=1$: $\frac{1}{2!} = \frac{1}{(1 + 1)!}$ This is true. Inductive Step: Assume true for $n=k$, that is: $P(k) : \sum_{i=1}^{k} \frac{i}{(i + 1)!} = \frac{1}{(k + 1)!}$ Now for $n=k+1$: $P(k+1): \sum_{i=1}^{k+1} \frac{i}{(i + 1)!} = \frac{1}{(k + 1)!} + \frac{k+1}{((k+1) + 1)!}$ This needs to simplify to: $P(k+1): \frac{1}{(k + 2)!}$ Consequently, the statement holds for $n=k+1$ using the identity. Hence by induction, it is true for all integers $n \geq 1$.

First, we calculate the derivative: $f'(x) = \frac{-1}{\sqrt{1-x^2}} + \frac{-1}{\sqrt{1-(-x)^2}}$ Since $\sqrt{1-x^2}$ is symmetric, we have: $f'(x) = -\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$ Therefore, $f'(x) = 0$ for all $x$ in the domain, indicating that $f(x)$ is constant.

Since we have established that $f(x)$ is constant, substituting $x$ with $-x$ gives us: $f(-x) = cos^{-1}(-x) + cos^{-1}(x) = constant$ Since we know that: $f(x) = cos^{-1}(x) + cos^{-1}(-x)$ Thus, we can assert: $cos^{-1}(-x) = n - cos^{-1}(x),$ where $n$ is equal to the constant value of $f(x)$, confirming the result.