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

To find the inverse, we set $y = f(x)$, resulting in:

$y = x - \frac{1}{2}$
Swapping $x$ and $y$, we get:

$x = y - \frac{1}{2}$
Thus,
$y = x + \frac{1}{2}$
Therefore, the expression for the inverse function is:
$f^{-1}(x) = x + \frac{1}{2}.$

To evaluate $f^{-1}(\frac{3}{8})$, we substitute $\frac{3}{8}$ into the expression for the inverse function:

$f^{-1}(\frac{3}{8}) = \frac{3}{8} + \frac{1}{2} = \frac{3}{8} + \frac{4}{8} = \frac{7}{8}.$

In simple harmonic motion, the maximum speed ($v_{max}$) and the maximum acceleration ($a_{max}$) are given as 2 m/s and 6 m/s² respectively. The relationships between these quantities are:

• Maximum speed: $v_{max} = A \omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.
• Maximum acceleration: $a_{max} = A \omega^2$.

Using these two equations, we can express $\omega$ in terms of $A$:

$\omega = \frac{v_{max}}{A} = \frac{2}{A}$
Substituting this into the equation for acceleration:

$a_{max} = A \left( \frac{2}{A} \right)^2 = \frac{4}{A}$
Setting this equal to 6:

$\frac{4}{A} = 6 \Rightarrow A = \frac{4}{6} = \frac{2}{3}.$
The period ($T$) is given by:

$T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{2}{A}} = \frac{2\pi A}{2} = \pi A = \pi \left(\frac{2}{3}\right) = \frac{2\pi}{3}.$

To prove that $\triangle PKL$ is isosceles, we note that $PK$ and $PL$ are tangents to the circles from point $P$. By the property of tangents to a circle from a single external point, we have:

• $PK = PL$, which directly shows that the triangle is isosceles. Additionally, the angles at points $K$ and $L$ will also be equal due to the tangent properties.