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1 + sec2 30o = sec 30o = \(\frac{2}{\sqrt{3}}\) \(\frac{(2)^2}{3}\) = \(\frac{4}{3}\) 1 + sec2 30o = sec 30o = 1 + \(\frac{4}{3}\) = 2\(\frac{1}{3}\)

Area of a regular hexagon = 8 x 8 x sin 60o = 32 x \(\frac{\sqrt{3}}{2}\) Area = 16\(\sqrt{3}\) x 6 = 96 \(\sqrt{3}\)cm2

\(\frac{\cot (90 - \theta)}{sin^2\theta}\) \(\cot (90 - \theta) = \tan \theta\) \(\therefore \frac{\cot (90 - \theta)}{\sin^{2} \theta} = \frac{\tan \theta}{\sin^{2} \theta}\) \(\tan \theta = \frac{\sqrt{3}}{3}\) \(\sin \theta = \frac{1}{2} \implies \sin^{2} \theta = \frac{1}{4}\) \(\frac{\cot(90 - \theta)}{\sin^{2} \theta} = \frac{\sqrt{3}}{3}\div\frac{1}{4}\) = \(\frac{4}{\sqrt{3}}\)

By Pythagoras, KZ2 = 252 - 52 KZ2 = (25 + 15)(25 - 15) = 400 KZ = \(\sqrt{400}\) = 20 area of XYZ = \(\frac{1}{2} \times 28 \times 15\) = 210 sq. cm

HJ2 = a2 + a2 = 2a2 HJ = \(\sqrt{2a^2} = a \sqrt{2}\) HN2 = a2 + (a\(\sqrt{2}\))2 = a2 + 2a2 = 3a2 HN = \(\sqrt{3a^2}\) = a\(\sqrt{3}\)cm

Area of sector = \(\frac{120}{360} \times \pi \times (3)^2 = 3 \pi\) Area of triangle = \(\frac{1}{2} \times 3 \times 3 \times \sin 120^o\) = \(\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9\sqrt {3}}{4}\) Area of shaded portion = 3\(\pi - \frac{9\sqrt {3}}{4}\) = 3 (\(\pi - 3 \frac{\sqrt{3}}{4}\))

Let x represent the price of an orange and y represent the number of oranges that can be bought xy = 240k, y = \(\frac{240}{x}\).....(i) If the price of an oranges is raised by \(\frac{1}{2}\)k per orange, number that can be bought for N240 is reduced by 16 Hence, y - 16 = \(\frac{240}{x + \frac{1}{2}\) = \(\frac{480}{2x + 1}\) = \(\frac{480}{2x + 1}\).....(ii) subt. for y in eqn (ii) \(\frac{240}{x}\) - 16 = \(\frac{480}{2x + 1}\) = \(\frac{240 - 16x}{x}\) = \(\frac{480}{2x + 1}\) = (240 - 16x)(2x + 1) = 480x = 480x + 240 - 32x2 - 16 480x = 224 - 32x2 x2 = 7 x = \(\sqrt{7}\) = 2.5 = 2\(\frac{1}{2}\)k

\(\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}\) \(\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3\) \(\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3\) \(\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4\) \(\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}\) = \(\frac{6}{-4} = \frac{-3}{2}\)

\(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) = \(\frac{5\sqrt{7} - 7\sqrt{5}}{\sqrt{7} - \sqrt{5}}\) x \(\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\) = \(\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (-7 \times 5)}{(\sqrt{7})^2}\) = \(\frac{5 \sqrt{35} - 7\sqrt{35}}{2}\) = \(\frac{-2\sqrt{35}}{2}\) = - \(\sqrt{35}\)

\(\frac{3^n - 3^{n - 1}}{3^3 \times 3^n - 27 \times 3^{n - 1}}\) = \(\frac{3^n - 3^{n - 1}}{3^3(3^n - 3^{n - 1})}\) = \(\frac{3^n - 3^{n - 1}}{27(3^n - 3^{n - 1})}\) = \(\frac{1}{27}\)

f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 = (2 + 3) (x - 3) + 4 = 5(x - 3) + 4 = 5x - 15 + 4 = 5x - 11 f(3) = 5 x 3 - 11 = 4 g(f(3)) = g(4) = \(\sqrt{5 + 4}\) = \(\sqrt{9}\) = 3 g(4) = 3 f(g(4)) = f(3) = 4 g[f(3)] and f[g(4)] = 3 and 4 respectively.

Asector = \(\frac{60}{360} \times \pi r^2\) = \(\frac{1}{6} \pi r^2\) A\(\bigtriangleup\) = \(\frac{1}{2}r^2 \sin 60^o\) \(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\) A\(\text{shaded portion}\) = Asector - A\(\bigtriangleup\) = (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\) = \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\) = \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)

by Pythagoras's theorem, XW = 3cm Also by Pythagoras theorem, XY2 = 62 + 32 XY2 = 36 + 9 = 45 XY = \(\sqrt{45} = 3 \sqrt{5}\)

From the diagram, sin 60o = \(\frac{PR}{8}\) PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\) = 4\(\sqrt{3}\) Cos 45o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\) PS Cos45o = 4\(\sqrt{3}\) PS = 4\(\sqrt{3}\) x 2 = 4\(\sqrt{6}\)

0.0000152 x 0.00042 = A x 10\(^{B}\) 1 \(\leq\) A < 10, it means values of A include 1 - 9 0.0000152 = 1.52 x 10\(^{-5}\) 0.00042 = 4.2 x 10\(^{-4}\) 1.52 x 4.2 = 6.384 10\(^{-5}\times 10^{-4}\) = 10\(^{-5-4}\) = 10\(^{-9}\) = 6.38 x 10\(^{-9}\) A = 6.38, B = -9