Three different points A, B and C are chosen on a circle centred at O - HSC - SSCE Mathematics Extension 1 - Question 13 - 2022 - Paper 1

To find the volume of revolution, we use:

$V = \pi \int_0^{\frac{\pi}{2k}} (k + 1)\sin(kx)^2 \, dx$

Setting this equal to $\pi^2$ allows us to solve for the value of $k$. After evaluating the integral and simplifying, we arrive at the required condition.

The function $f(x) = \sin(x)$ is defined for all real numbers and has a range of $[-1, 1]$. The function $g(x) = \arcsin(x)$ only applies over this range. However, because $f^2(x)$ maps $\mathbb{R}$ to $[0, 1]$, it doesn't have a well-defined inverse for all inputs. Therefore, while $g$ is potentially an inverse over restricted intervals, it is not valid universally.

Using the identity: $\alpha^2 + \beta^2 + \gamma^2 = 85$ and the derivative sums $P'(\alpha) + P'(\beta) + P'(\gamma) = 87$, we can derive a relevant cubic polynomial equation to find the relations of the roots, ultimately resulting in $\alpha\beta + \beta\gamma + \gamma\alpha = \text{computed value}$.

To calculate $P_p$, use the binomial probability formula for $n = 16$ and $k \geq 8$, given that $p = 0.2$:

$P_p = \sum_{k=8}^{16} \binom{16}{k} (0.2)^k (0.8)^{16-k}$

Evaluating this yields the required probability.

The normal approximation used might not be valid due to the small sample size and the assumption that the weights follow a normal distribution. If the actual distribution of weights is not normal, this could lead to incorrect probability calculations.