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1 UM = 10\(^{-6}\)mm = 10\(^{-9}\)m 1.2 x 10\(^4\) x 10\(^{-9}\)m = 1.2 x 10\(^{-5}\)m = (0.000012)m

1 hr, = 60 mins, 60 mins = 360o 30 mins = \(\frac{360^o}{1}\) c \(\frac{30}{60}\) = 180o 90 mins = 360 + 180o = 540o

\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) \(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = \(\frac{\sqrt{2}}{\sqrt{3}}\) - \(\frac{\sqrt{3}}{\sqrt{2}}\) \(\frac{\sqrt{3} + \sqrt{2}}{3 + \sqrt{2}}\) = \(\sqrt{6}\) + 2 \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) = \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = 5 - 2\(\sqrt{6}\) \(\sqrt{6}\) + 2 - (5 - 2 \(\sqrt{6}\)) = \(\sqrt{6}\) + 2 - 5 + 2\(\sqrt{6}\) = 3\(\sqrt{6}\) - 3 = 3(\(\sqrt{6}\) - 1)

\(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\) \(\frac{x + 3 - x - 1}{(x + 1)(x + 3)}\) = \(\frac{1}{4}\) \(\frac{2}{x^2 + 4x + 3}\) = \(\frac{1}{4}\) = x2 + 4x + 3 = 8 x2 + 4x - 5 = 0 = (x - 1)(x + 5) = 0 x = 1 or -5

Let XW = a, a :7 \(\frac{a}{5 - a}\) = \(\frac{3}{7}\) 7a = 15 - 3a 10a = 15 a = \(\frac{15}{10}\) = \(\frac{3}{2}\) = 1.5cm

\(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\) When x = 4, \(\log_{2} y = 3 - \log_{2} 4^{\frac{3}{2}}\) \(\log_{2} y = 3 - \log_{2} 2^{3}\) \(\log_{2} y = 3 - 3 \log_{2} 2 = 3 - 3 = 0\) \(\log_{2} y = 0 \implies y = 2^{0} = 1\)

Given that log\(_{10}\)2 = 0.3010, If 10\(^x\) = 0.2 log\(_{10}\)10\(^x\) = log\(_{10}\) 0.2 = x = log\(_{10}\)0.2 But 0.2 = 2 x 10\(^{-1}\), so log\(_{10}\)0.2 = log\(_{10}(2 \times 10^{-1})\), x = log\(_{10}\)2 + log\(_{10}\)10\(^{-1}\) x = 0.3010 - 1 x = - 0.6990

Distance travelled by both x and y = 150 km. Speed of y = 60km per hour, Time = \(\frac{dist.}{Speed}\) Time spent by y = \(\frac{150}{60}\) = 2\(\frac{3}{6}\)hrs = \(\frac{1}{2}\) hrs if x arrived at p 25 minutes earlier than y, then time spent by x = 2 hour 5 mins = 125 mins x average speed = \(\frac{150}{125}\) x \(\frac{60}{1}\) = \(\frac{9000}{125}\) = 72 km\hr

\(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\) = \(\frac{(2 \times 3^{2n + 1} \times 3^{2n}) \times 2^{4n}}{(2 \times 9)^n \times 2^n \times (6 \times 2^{2n})}\) = \(\frac{(2^{2n + 1} \times 3^{2n + 1}) \times 3^{2n} \times 2^{4n}}{2^n \times 3^{2n} \times 2^n \times 2^{4n} \times 3^{2n}}\) = \(\frac{2^{2n} + 1 + 4^n \times 3^{2n} + 1 + 2^n}{2^{n + n + 4n} \times 3^{2n + 4n} \times 3^{2n + 2n}}\) = \(\frac{2^{6n + 1} \times 3^{4n + 1}}{2^{6n} x 3^{4n}}\) = 26n + 1 - 6n x 34n + 1 - 4n 2 x 3 = 6

\(\begin{array}{c|c} 2 & 25\\2 & 12 R 1\\2 & 6 R 0\\2 & 3 R 0\\2 & 1 R 1\\2&0 R 1\end{array}\) = (11001)

\(\frac{6.3 \times 10^5}{8.1 \times 10^3}\) \(\frac{7}{9} \times 10^{2}\) = \(0.77778 \times 10^{2}\) = \(77.778 \approxeq 77.80\)

\(4t^{2} + 7t - 1 = 0\) Using a = 4, b = 7, c = -1. \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) = \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\) = \(\frac{-7 \pm \sqrt{49 + 16}}{8}\) = \(\frac{-7 \pm \sqrt{65}}{8}\) = \(\frac{-7 \pm 8.0623}{8}\) The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\) \(\approxeq 0.1328\) (4 decimal place)

Area = L x B L - B = 6 L = 6 + B area = 135 B(6 + B) = B\(^2\) + 6B - 135 = 0 (B - 9)(B + 15) = 0 B = 9 or - 15 L = 6 + 9 = 15

Surface area of a sphere = 4\(\pi\)r2 \(\frac{1}{16}\) of 4\(\pi\)r2 = \(\frac{\pi r^2}{4}\) P = \(\frac{\pi r^2}{4}\) r = 2\(\frac{P}{\pi}\)

\(\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}\) \(\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}\) = \(\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}\) (Note: \(\sin^{2} x + \cos^{2} x = 1\)). = \(\frac{2 \sin^{2} x}{\sin^{2} x}\) = 2.