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 value of $k$ for which the volume of the solid of revolution is $\pi^2$, we set up the integral for the volume:

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

Using the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ yields:

$V = \pi (k + 1) \left[ \frac{x}{2} - \frac{\sin(2kx)}{4k} \right]_0^{\frac{\pi}{2k}}.$

Evaluating it, we find:

$V = \frac{\pi^2 (k + 1)}{8}.$

Setting $V = \pi^2$ gives:

$\frac{\pi^2 (k + 1)}{8} = \pi^2 \Rightarrow k + 1 = 8 \Rightarrow k = 7.$

The function $f(x) = \sin(x)$ is a periodic function with values in the interval $[-1, 1]$. The function $g(x) = \arcsin(x)$ is defined as the inverse of the sine function in that same interval. However, the function $f^2(x) = \sin^2(x)$ is not one-to-one, as it will give the same output for multiple inputs.

Since $g(x) = \arcsin(x)$ does not satisfy the conditions for being an inverse (i.e., $g(f^2(x)) \neq x$ for all x), we conclude that $g$ is not the inverse of $f^2$.

Given the conditions: $\alpha^2 + \beta^2 + \gamma^2 = 85$ $P'(\alpha) + P'(\beta) + P'(\gamma) = 87,$

Using the relation of the sum of the roots and the derivatives, we find: $87 = 3\cdot(\alpha\beta + \beta\gamma + \gamma\alpha) + 2(\alpha + \beta + \gamma).$

Rearranging gives: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{87 - 2(\alpha + \beta + \gamma)}{3}.$
Applying the values to find the combination of these roots will yield the answer.

Using the normal approximation:

Let $p$ be the true proportion of bars that weigh less than 150 g. If the factory manager claims that 80% of bars weigh 150 g or more, then $p = 0.2$. The random variable $X$, representing the count of bars weighing less than 150 grams, follows a binomial distribution:

$X \sim B(n=16, p=0.2)$

To find $P(X \geq 8)$ using continuity correction: $P(X \geq 8) \approx P(Z \geq \frac{8 - 3.2}{\sqrt{16 \cdot 0.2 \cdot 0.8}}) = P(Z \geq 3.2) = 0.0007.$

The method used by the inspectors assumes that the sampling distribution is normal given $n=16$, but this may not hold if the probability of success is too high or low. The binomial distribution approximates well using the normal distribution when both $np$ and $n(1-p)$ are greater than 5. In this case, $np = 3.2$ and $n(1-p) = 12.8$, could lead to inaccuracies in the normal approximation, primarily due to the small sample size and lack of independence in results.