Use the Question 11 Writing Booklet (a) Find \( (i + 6j) + (2i - 7j) \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2021 - Paper 1

[ (2a - b)^4 = (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 ]

Let ( u = x + 1 ) so that ( x = u - 1 ) and ( dx = du ). Then, the integral becomes: [ \int \sqrt{u} , du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x + 1)^{3/2} + C ]

The number of ways to choose 5 men from 10 is given by ( \binom{10}{5} ) and the number of ways to choose 3 women from 8 is ( \binom{8}{3} ). Therefore, the total number of ways to form the committee is: [ \binom{10}{5} \times \binom{8}{3} ]

The volume ( V ) of a sphere is given by ( V = \frac{4}{3} \pi r^3 ). Taking the derivative with respect to time ( t ) gives: [ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} ] Substituting ( r = 0.6 ) mm and ( \frac{dr}{dt} = 0.2 ) mm/s: [ \frac{dV}{dt} = 4\pi (0.6)^2 (0.2) \approx 0.5 , mm^3/s ]

The integral corresponds to the arcsin function: [ \int \frac{1}{\sqrt{4 - x^2}} , dx = \sin^{-1} \left( \frac{x}{2} \right) + C ] Evaluating from 0 to ( \sqrt{3} ): [ \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) - \sin^{-1}(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3} ]

Rearranging, we have: [ 2\sin^2 x - \sin x - 1 = 0 ] Factoring gives: [ (2\sin x + 1)(\sin x - 1) = 0 ] Thus, ( \sin x = -\frac{1}{2} ) or ( \sin x = 1 ). From ( 0 \leq x \leq 2\pi ), the solutions are: [ x = \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{\pi}{2} ]

The equation can be rearranged as follows: [ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} = \frac{\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta}{\alpha\beta\gamma\delta} ] Using Vieta's formulas, we know that the sum of the roots is given by the coefficient of ( x^{n-1} ). Thus, in this case, it simplifies to: [ 1 = 6 ]