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 determine if (g) is the inverse of (f^2), we check if (g(f(x)) = x) holds true for all (x) in the domain of (f).

Since (f(x) = \sin(x)) and (g(x) = \arcsin(x)), we find that:

For (y = f(x) = \sin(x)):
[ g(y) = g(f(x)) = \arcsin(\sin(x)) = x \text{ iff } |x| \le 1 ]

Thus, (g) is not the inverse of (f^2) over the reals, since (f^2) is restricted to (x\ in \mathbb{R}). However, it is for a restricted domain that excludes non-invertible intervals.

Using the identity: [ \alpha^2 + \beta^2 + \gamma^2 = 85 ],
we also know: [ P' (\alpha) + P' (\beta) + P' (\gamma) = 87 ].

From the polynomial roots, we derive: [ \alpha \beta + \beta \gamma + \gamma \alpha = p = \frac{P'}{2} = 87-85/2 \text{ leading to the value} ]

Using the normal approximation: [ P_{p} \text{ can be calculated based on } \text{mean } = 0.8 \times 16 = 12.8 \text{ and standard deviation } = \sqrt{16 \times 0.8 \times 0.2} ] We find: [ P = P(X \geq 8) \text{ using the cumulative distribution function.} ]

The inspectors' use of a normal approximation in a discrete distribution context may not hold due to the small sample size, which can lead to inaccuracies. Assuming a binomial distribution while approximating might ignore the variability among individual weights causing misjudgment in estimation.