a) Let $f(x) = x - \frac{1}{2}$ for $x \leq 1$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2008 - Paper 1

From the calculations in part (i), we found the inverse function to be:
$f^{-1}(x) = x + \frac{1}{2}$

To evaluate $f^{-1}(\frac{3}{8})$, substitute $\frac{3}{8}$ into the expression we found in part (ii):
$f^{-1}(\frac{3}{8}) = \frac{3}{8} + \frac{1}{2} = \frac{3}{8} + \frac{4}{8} = \frac{7}{8}$

The maximum speed (v) in simple harmonic motion is given by:
$v = A \omega$
Given that $v = 2 \text{ m s}^{-1}$, we have

1. Let the amplitude be $A$ and $\omega$ be the angular frequency.

The maximum acceleration (a) is given by:
$a = A \omega^2$
Given that $a = 6 \text{ m s}^{-2}$, we set up the equations:

• From max speed:
  $2 = A \omega$
• From max acceleration:
  $6 = A \omega^2$

Dividing these two equations gives:
$\frac{6}{2} = \frac{A \omega^2}{A \omega} \Rightarrow 3 = \omega$

Substituting back to find amplitude:
$2 = A \times 3 \Rightarrow A = \frac{2}{3}$

The period (T) is given by:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{3}$

To show that $\triangle PKL$ is isosceles, we need to prove that $PK = PL$.

1. Since both $C_1$ and $C_2$ are circles, the radii $PK$ and $PL$ can be considered equal if they are drawn to the same point on the circumference.
2. Therefore, if $PK$ and $PL$ are radii from points $P$ to points $K$ and $L$ on circles $C_1$ and $C_2$, then they are equal.
3. Thus, $\triangle PKL$ is isosceles because $PK = PL$.