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\( = log_2 \frac{8 \times 16}{4}\) \( = log_2 32\) \( = log_2 2^5\) \( = 5log_2 2\) \( = 5 \times 1\) \( = 5 \)

Q = (P \(\cup\) Q) - P {6,7}

gt2 - k - w = 0 gt2 = k + w \(g = \frac{k + w}{t^2}\)

2y2 - 15xy + 18x2 2y2 - 12xy - 3xy + 18x2 2y(y - 6x) - 3x(y - 6x) (2y - 3x) (y - 6x)

if y - 1 is a factor of y3 + 4y2 + ky - 6, then f(1) = (1)3 + 4(1)2 + k(1) - 6 = 0 (factor theorem) 1 + 4 + k - 6 = 0 5 - 6 + k = 0 -1 + k = 0 k = 1

\(y \propto w^2\) y = kw2 8 = k(2)2 8 = k(4) k = 8/4 k = 2 Thus y = 2w2 When w = 3, y = 2(3)2 y = 2 x 9 = 18

Common ratio r of the G.P is \(r = \frac{T_n + 1}{T_n} = \frac{T_2}{T_1}\) \(r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}}\) \(r = \frac{\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5}}{\sqrt{10} - \sqrt{5}} \) \( = \frac{(\sqrt{10})(\sqrt{10}) + (\sqrt{10})(-\sqrt{5}) + (2\sqrt{5})(\sqrt{10}) + (2\sqrt{5})(-\sqrt{5})}{(\sqrt{10})^2 - (\sqrt{5})^2}\) \(\frac{10 - \sqrt{50} + 2\sqrt{50} - 10}{10 - 5}\) \(\frac{\sqrt{50}}{5}\) \(\frac{\sqrt{25 \times 2}}{5}\) \(\frac{5\sqrt{2}}{5}\) \(\sqrt{2}\)

x * y = xy x * 2 = 12 - x Thus by comparison, x = x, y = 2 But x * y = x * 2 xy = 12 - x x2 = 12 - x x2 + x - 12 = 0 x2 + 4x - 3x - 12 = 0 x(x + 4) - 3(x + 4) = 0 (x - 3)(x + 4) = 0 x - 3 = 0 or x + 4 = 0 So x = 3 or x = -4

Using \(V = \pi r^2 h\) 6160 = 22/7 x 28 x 28 x h \(h = \frac{6160}{22 \times 4 \times 28} \) \(h = 2.5m \)

Mid point of S(-5, 4) and T(-3, -2) is \([\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + (-2))]\) \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(-8), \frac{1}{2}(2)]\) \([-4, 1]\)

y - 4x + 3 = 0 When y = 0, 0 - 4x + 3 = 0 Then -4x = -3 x = 3/4 So the line cuts the x-axis at point (3/4, 0). When x = 0, y - 4(0) + 3 = 0 Then y + 3 = 0 y = -3 So the line cuts the y-axis at the point (0, -3) Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is; \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\) \([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\) \([\frac{3}{8}, \frac{-3}{2}]\)

If y = 4x3 - 2x2 + x, then; \(\frac{\delta y}{\delta x}\) = 3(4x2) - 2(2x) + 1 = 12x2 - 4x + 1

\(\int (2x + 3)^{\frac{1}{2}} \delta x\) let u = 2x + 3, \(\frac{\delta y}{\delta x} = 2\) \(\delta x = \frac{\delta u}{2}\) Now \(\int (2x + 3)^{\frac{1}{2}} \delta x = \int u^{\frac{1}{2}}.{\frac{\delta x}{2}}\) \( = \frac{1}{2} \int u^{\frac{1}{2}} \delta u\) \( = \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k\) \( = \frac{1}{3} u^{\frac{3}{2}} + k\) \( = \frac{1}{3} (2x + 3)^{\frac{3}{2}} + k\)

A team of 2 girls can be selected from 7 girls in \(^7C_3\) \( = \frac{7!}{(7 - 3)! 3!}\) \( = \frac{7!}{4! 3!} ways\)

No expanded working has been attached for this question yet. The verified answer is Option A.