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 particle is at the maximum displacement, where x = c. Plugging x = c into the velocity equation:

$v^{2} = n^{2}(a^{2} - (c - c)^{2}) = n^{2}a^{2}$

Thus, the maximum speed v_max is given by:

$v_{max} = n a$

From the structure of the given equation v² = n²(a² - (x - c)²), we can compare it to the standard form of harmonic motion. Thus, we conclude that:

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

Hence, a = amplitude, c = equilibrium position, and n = angular frequency (in appropriate units).

To find a_{2}, we will use the binomial expansion formula to identify coefficients. The binomial expansion is given by:

$(x + y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k}$

In our case, we have:

$(2x + \frac{1}{3x})^{18}$,

To find a_{2}, set k = 2:

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

Calculating this gives:

$a_{2} = \binom{18}{2} \frac{(2^{16})}{9} x^{16 - 2} = \frac{153}{9} 2^{16} x^{14}$

The term independent of x can be found by considering the case where k = 9 (as we want the total exponent of x to be 0):

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

This simplifies to:

$a_{9} = \binom{18}{9} \frac{2^{9}}{3^{9}}$

This gives us the term independent of x as:

$\binom{18}{9} \frac{2^{9}}{3^{9}}$

To prove by induction, we start with the base case n=1:

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

Now assume it holds for n = k:

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

We must show it holds for k + 1:

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

Adding the k + 1 inductive hypothesis gives us:

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

Thus, by the principle of mathematical induction, the statement is proven.

The function is given as:

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

Taking the derivative f'(x):

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

Since f'(x) = 0 for -1 ≤ x ≤ 1, it follows that f(x) is constant over this interval.

From the relationship established by the derivative, we gather that:

$f(x) = f(-x)$

This implies:

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

Since f(x) is a constant, we have:

If \cos^{-1}(-x) = c - \cos^{-1}(x), we can deduce:

$\cos^{-1}(-x) = \pi - \cos^{-1}(x)$

This completes the deduction.