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

Using the binomial expansion formula, we have:

[ (x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} ]

Applying this:

[ (2a - b)^{4} = \sum_{k=0}^{4} \binom{4}{k} (2a)^{4-k} (-b)^{k} ]

Calculating each term results in:

[ 16a^{4} - 32a^{3}b + 24a^{2}b^{2} - 8ab^{3} + b^{4} ]

Thus, the final expression is ( 16a^{4} - 32a^{3}b + 24a^{2}b^{2} - 8ab^{3} + b^{4} ).

Performing the substitution, we have ( u = x + 1 ) which gives us ( x = u - 1 ) and also ( dx = du ).

Thus, the integral becomes:

[ \int \sqrt{u} , du = \frac{2}{3}u^{3/2} + C ]

Substituting back gives:

[ \frac{2}{3}(x + 1)^{3/2} + C ]

The number of ways to select 5 men from 10 is given by:

[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = 252 ]

The number of ways to select 3 women from 8 is:

[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = 56 ]

Thus, the total number of ways to form the committee is:

[ 252 \times 56 = 14112 ]

The volume ( V ) of a sphere is given by:

[ V = \frac{4}{3}\pi r^{3} ]

Using the chain rule:

[ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi r^{2} \frac{dr}{dt} ]

Substituting ( r = 0.6 ) mm and ( \frac{dr}{dt} = 0.2 ) mm/s gives:

[ \frac{dV}{dt} = 4\pi (0.6)^{2}(0.2) = 0.9\pi \approx 2.8 , \text{mm}^3/s \text{ (rounded to one decimal place)} ]

This integral can be evaluated using trigonometric substitution. Let:

[ x = 2\sin(t) \quad \Rightarrow \quad dx = 2\cos(t)dt ]

The limits change accordingly (for ( x=0, t=0 ) and for ( x=\sqrt{3}, t=\frac{\pi}{3} )). Thus, we have:

[ \int_{0}^{\frac{\pi}{3}} \frac{2\cos(t)}{\sqrt{4 - (2\sin(t))^2}} , 2\cos(t)dt = 2 \int_{0}^{\frac{\pi}{3}} 1 , dt = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3} ]

Rearranging gives:

[ 2sin^{3}x + 2sin^{2}x - sin x - 1 = 0 ]

Factoring:

[ (sin(x) + 1)(2sin^{2}x - 1) = 0 ]

Thus, we have two cases:

1. ( sin(x) + 1 = 0 ) implies ( sin(x) = -1 ), giving ( x = \frac{3\pi}{2} ).

2. For ( 2sin^{2}x - 1 = 0 ), we have ( sin^{2}x = \frac{1}{2} ), yielding ( sin(x) = \pm \frac{\sqrt{2}}{2} ), which provides solutions: ( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} ).

Therefore, the complete set of solutions is ( x = \frac{3\pi}{2}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} ) within ( [0, 2\pi] ).

Using Vieta's formulas for the polynomial ( x^{4} - 3x^{3} + 6x = 0 ), we know:

[ \alpha + \beta + \gamma + \delta = 3 \text{ (the sum of roots)} ] [ \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = 0 \text{ (the sum of the product of roots taken two at a time)} ] [ \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = 6 \text{ (the negative coefficient of } x^{2}) ] [ \alpha \beta \gamma \delta = 0 \text{ (the product of roots)} ]

Thus,

[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} = \frac{\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta}{\alpha \beta \gamma \delta} = \frac{6}{0} \text{ (undefined, indicating that one root is zero)} ]