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

We need to calculate the volume of the solid formed by revolving the region bounded by the x-axis and the curve $y = (k + 1) ext{sin}(kx)$ around the x-axis.

The formula for the volume $V$ of revolution is:

$V = \pi \int_{a}^{b} [f(x)]^2 \, dx$

For our case:

$V = \pi \int_{0}^{\frac{\pi}{2k}} ((k + 1) \text{sin}(kx))^2 \, dx$

Solving this integral will lead to an equation in terms of $k$. Set this equal to $\pi^2$ and solve for the value of $k$. This involves simplification and potentially numerical methods to arrive at the correct $k$.

To determine if $g(x) = \text{arcsin}(x)$ is the inverse of $f(x) = \text{sin}(x)$ under the restriction of the domain, we proceed as follows:

1. Calculate $f(g(x))$ and $g(f(x))$ to verify the identity relationship:

   • $f(g(x)) = \text{sin}(\text{arcsin}(x)) = x$
   • $g(f(x)) = \text{arcsin}(\text{sin}(x))$, valid only for $x$ in appropriate domains.

2. Since $g(x)$ is only defined for values in $[-1, 1]$, $g$ cannot be the inverse for all of $f$, as the range of $f$ extends beyond this interval.

Thus, while they are inverses on their restricted domains, $g$ is not the inverse of $f^2$ due to the failure in the full set of reals.

From the given information that:

1. $\alpha^2 + \beta^2 + \gamma^2 = 85$
2. The polynomial $P(x) = (x - \alpha)(x - \beta)(x - \gamma)$ and from Vieta's relationships, we have:
   • The sum of the roots $\alpha + \beta + \gamma$ is the negative coefficient of the $x^2$ term divided by the leading coefficient.
   • The sum of the products of the roots taken two at a time is given as $\alpha\beta + \beta\gamma + \gamma\alpha$.

By applying these relationships and solving the equations formed, you can isolate and calculate $\alpha\beta + \beta\gamma + \gamma\alpha$ using the equation derived from $P'(\alpha)$ sums to $87$.

Using the normal approximation to the binomial distribution:

1. Set up the binomial distribution with parameters $n = 16$ and $p = 0.2$ (the proportion of bars weighing less than 150 grams).

$P(X \geq 8) = 1 - P(X < 8) = 1 - \sum_{x=0}^{7} \binom{16}{x} (0.2)^x (0.8)^{16-x}$

2. Calculate using the normal approximation of parameters:

   • Mean $\mu = np = 16 \times 0.2 = 3.2$.
   • Standard deviation $\sigma = \sqrt{np(1-p)} = \sqrt{16 \times 0.2 \times 0.8} = 1.6$.

3. Apply continuity correction for the discrete binomial distribution, leading to:

$P(X \geq 8) = P(Z \geq (8 - 3.2)/1.6)$

4. Find the corresponding z-value and look it up in the standard normal distribution table to retrieve $P(X \geq 8)$.

The method employed by the inspectors to apply the normal approximation assumes that the sample size is large enough and that $np$ and $n(1-p)$ are both greater than 5. Given the proportion claimed by the factory manager (80% bars weighing >= 150g), the probability of selecting a sample that meets this might not align well with the actual weights distributed.

Also, the underlying distribution of the weights might not be normal, especially if the sample shows significant skewness due to the lower weight bars outnumbering the heavier bars in a smaller sample size. Thus, the normal approximation may provide inaccurate results leading to potential misinterpretation of the factory's actual proportion of heavier bars.