(a) The vectors $egin{pmatrix} a^2 \ 2 \ \\ a + 5 \ a - 4 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\}egin{pmatrix} a + 5 \ a - 4 \\ \\ \\ \\} are perpendicular - HSC - SSCE Mathematics Extension 1 - Question 12 - 2024 - Paper 1

To find the volume of the solid of revolution when the region $R$ is rotated about the x-axis, we use the formula for volume:

$V = \pi \int_1^2 (x^3)^2 \, dx$

Calculating this integral:

$V = \pi \int_1^2 x^6 \, dx$

Evaluating the integral:

$= \pi \left[ \frac{x^7}{7} \right]_1^2 = \pi \left( \frac{2^7}{7} - \frac{1^7}{7} \right) = \pi \left( \frac{128}{7} - \frac{1}{7} \right) = \frac{127\pi}{7}$

We denote the number of donations made as $X$. Given that there is a 0.31 chance of donations, the expected number of donations out of 100 people is:

$E(X) = n p = 100 \times 0.31 = 31$

The variance is calculated as:

$Var(X) = np(1 - p) = 100 \times 0.31 \times 0.69 = 21.39$

The standard deviation $\, \sigma = \sqrt{24.84} \approx 4.92$.

To find the probability that at least 35 people (i.e., 35%) made donations:

We apply the z-score formula:

$z = \frac{X - \mu}{\sigma} = \frac{35 - 31}{4.92} \approx 0.81$

Using standard normal distribution tables, we find:

$P(Z \geq 0.81) = 1 - P(Z < 0.81) \approx 1 - 0.7910 = 0.2090$

Thus, approximately 20.90% probability that at least 35% of people made donations.

Base case (n=1):

$2^1 + 13 = 15$, which is divisible by 7.

Inductive step: Assume true for $n = k$, that is, $2^k + 13$ is divisible by 7.

Now consider $n = k + 1$, we have:

$2^{k+1} + 13 = 2 \cdot 2^k + 13 = 2(2^k + 13) - 13$

Since $2^k + 13$ is divisible by 7, it follows that $2(2^k + 13)$ is also divisible by 7. Therefore, $2^{k+1} + 13$ is divisible by 7.

By induction, $2^n + 13$ is divisible by 7 for all integers $n \geq 1$.

To solve the inequality, we must consider two cases based on the absolute value.

Case 1: x > 5

In this scenario, the absolute value $|x - 5|$ simplifies to $(x - 5)$, yielding:

$\frac{x}{6} \geq \frac{1}{x - 5}$

Cross-multiplying gives:

$x(x - 5) \geq 6$

This simplifies to:

$x^2 - 5x - 6 \geq 0$

Factoring yields:

$(x - 6)(x + 1) \geq 0$

Therefore, the solution from this case is:

$x \in [6, \infty)$

Case 2: x < 5

Here, the absolute value eliminates as $|x - 5| = (5 - x)$, leading to:

$\frac{x}{6} \geq \frac{1}{5 - x}$

Cross-multiplying results in:

$x(5 - x) \geq 6$

This implies:

$-x^2 + 5x - 6 \geq 0$

Factoring gives:

$(x - 2)(x - 3) \leq 0$

Thus, from this case, we find:

$x \in [2, 3]$

Final Solution: Therefore, the complete solution set is:

$x \in [2, 3] \cup [6, \infty)$