Use the Question 11 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2021 - Paper 1

Using the binomial expansion formula, we have:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.$$

For $(2a - b)^4$, we get:

$$(2a)^4 - 4(2a)^3b + 6(2a)^2b^2 - 4(2a)b^3 + b^4 = 16a^4 - 32a^3b + 24a^2b^2 - 8ab^3 + b^4.$$

Thus, the simplified form is:

$$16a^4 - 32a^3b + 24a^2b^2 - 8ab^3 + b^4.$$

Using the substitution $u = x + 1$, we have:

$$\int \sqrt{x + 1} \, dx = \int \sqrt{u} \, du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x + 1)^{3/2} + C.$$

To determine the number of ways to form this committee, we use combinations:

$\binom{10}{5} \times \binom{8}{3}.$

Calculating these: $\binom{10}{5} = 252\quad \text{and} \quad \binom{8}{3} = 56.$

Therefore, the total number of ways: $252 \times 56 = 14112.$

The volume $V$ of a spherical bubble is given by:

$V = \frac{4}{3}\pi r^3$

Differentiating with respect to time, we have:

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.$

Given that $\frac{dr}{dt} = 0.2 \, \text{mm/s}$ and when $r = 0.6 \, \text{mm}$:

$\frac{dV}{dt} = 4\pi(0.6^2)(0.2) = 4\pi(0.36)(0.2) = 0.144\pi.$

Rounding: $\frac{dV}{dt} \approx 0.5 \, \text{mm}^3/s.$

We use the identity:

$$$$

Here, $a = 2$. Hence, we have:

$\int_0^{\sqrt{3}} \frac{1}{\sqrt{4 - x^2}} \, dx = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) - \sin^{-1}(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3}.$

First, factor out:

$2\sin^2 x(\sin x + 1) - \sin x - 1 = 0.$

This can be arranged as: $\sin x(2\sin^2 x - 1) - 1 = 0.$

Solutions are: $\sin x = -1 \text{ or } 2\sin^2 x - 1 = 0.$

The solutions for $\sin x = -1$ are: $\frac{3\pi}{2}$. For the quadratic, we have: $\sin^2 x = \frac{1}{2} \Rightarrow \sin x = \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}.$ This gives solutions: $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{5\pi}{4}.$

Using Vieta's formulas for the roots of the polynomial $x^4 - 3x^3 + 6x^2 + \alpha x + \beta$, we know:

$\alpha + \beta + \gamma + \delta = 3.$

Thus, $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} = \frac{\alpha + \beta + \gamma + \delta}{\alpha \beta \gamma \delta} = \frac{3}{-6} = -\frac{1}{2}.$