Use the principle of mathematical induction to show that for all integers $n \geq 1$, $$1 \cdot 2 + 2 \cdot 5 + 3 \cdot 8 + \dots + n(3n - 1) = n^2(n + 1)$$ - HSC - SSCE Mathematics Extension 1 - Question 12 - 2020 - Paper 1

The expected value for a binomial distribution is given by:

$E(X) = n \cdot p$

where ( n = 100 ) and ( p = \frac{3}{5} ), thus:

$E(X) = 100 \cdot \frac{3}{5} = 60$.

The variance for a binomial distribution is given by:

$Var(X) = n \cdot p \cdot (1 - p)$

Substituting:

$Var(X) = 100 \cdot \frac{3}{5} \cdot \frac{2}{5} = 24$

Therefore, the standard deviation ( \sigma ) is:

$\sigma = \sqrt{Var(X)} = \sqrt{24} \approx 4.9 \approx 5.$

Using the normal approximation:

1. Convert to standard normal variable ( Z ):

   $Z = \frac{X - E(X)}{\sigma}$

2. For ( X = 55 ):

   $Z_{55} = \frac{55 - 60}{5} = -1$

3. For ( X = 65 ):

   $Z_{65} = \frac{65 - 60}{5} = 1$

Using the standard normal distribution table:

$P(-1 < Z < 1) \approx 0.6826 \approx 68\%$

There are ( \binom{8}{3} = 56 ) possible combinations of topics.

With 400 students and only 56 combinations available, by the pigeonhole principle, at least:

$\left\lceil \frac{400}{56} \right\rceil \approx 8$

students must have chosen the same combination of topics.

Using the product-to-sum identities:

$\cos A \sin B = \frac{1}{2}(\sin(A + B) - \sin(A - B))$

We can express the integral as:

$\int_0^\frac{\pi}{2} \cos 5x \sin 3x \, dx = \frac{1}{2} \left( \int_0^\frac{\pi}{2} \sin(8x) \, dx - \int_0^\frac{\pi}{2} \sin(2x) \, dx \right)$

Calculating these integrals yields:

$= \frac{1}{2} \left( -\frac{1}{8}(\cos(8 \times \frac{\pi}{2}) - \cos(0)) + \frac{1}{2} \right) = \frac{1}{2} \left( -\frac{1}{8}(0 - 1) + \frac{1}{2} \right) = 0.5$.

Separating variables:

$\frac{dy}{-x - y} = dx$

Integrating both sides gives:

$\int \frac{1}{-x - y} \, dy = \int dx$

This results in an equation in terms of ( y ):

( y^2 + x^2 = d ), where ( d = 1 ) since it passes through (1, 0).

Thus, the equation of the curve is:

$x^2 + y^2 = 1$.