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(a) \(y = k + \frac{c}{x^{2}}\) where c and k are constants. (b) When x = 1, y = 11 \(11 = k + \frac{c}{1^{2}} \implies 11 = k + c ... (1)\) When x = 2, y = 5 \(5 = k + \frac{c}{2^{2}} \implies 5 = k + \frac{c}{4}\) \(\equiv 20 = 4k + c ... (2)\) (2) - (1) : \(4k - k = 20 - 11 \implies 3k = 9\) \(k = 3\) Put k = 3 in (1), we have \(11 = 3 + c \implies c = 11 - 3 = 8\) \(\therefore y = 3 + \frac{8}{x^{2}}\) When x = 4, \(y = 3 + \frac{8}{4^{2}} = 3 + \frac{1}{2}\) = \(3\frac{1}{2}\)

(a)(i) \(\sin 63.35^\circ = 0.8938\) \(2 \sin 63.35^\circ = 2 \times 0.8938 = 1.7876\) (ii) \(\log \cos 44.74^\circ = \bar{1}.8514\) (b) \(\log K - \log (K - 2) = \log 5\) \(\log (\frac{K}{K - 2}) = \log 5\) \(\therefore \frac{K}{K - 2} = 5 \implies K = 5(K - 2)\) \(K = 5K - 10 \implies 10 = 5K - K = 4K\) \(K = 2\frac{1}{2}\) (c) \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\) No Log \((3.68)^{2}\) \(0.5658 \times 2 = 1.316\) \(6.705\) \(0.8264 = 0.8264\) = \(1.9580\) \(\sqrt{0.3581}\) \(\bar{1}.5540 \div 2 = \bar{1}.7770\) \(151.7\) \(\gets 2.1810\) \(\therefore \frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}} = 151.7\)

\(\frac{72}{125} = \frac{8 \times 9}{5 \times 5 \times 5}\) = \(\frac{2^3 \times 3^2}{5^3}\)

\(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \) = \(\frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}}\) = \(\frac{3}{2^{-1}}\) = 6

16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\) = \((2^4)^{\frac{5}{4}} \times 2^{-3} \times 1\) = \(2^5 \times 2^{-3} \times 1\) = \(2^2\) = 4.

2lo\(g_3\) 6 + lo\(g_3\) 12 - lo\(g_3\) 16 = \(\log_3 6^2 + \log_3 12 - \log_3 16\) = \(\log_{3} (\frac{36 \times 12}{16})\) = \(\log_{3} (27)\) = \(\log_{3} 3^3\) = \(3 \log_{3} 3\) = 3

log\(_8\)x - 4log\(_8\)x = 2; log\(_8\)x - log\(_8\)x\(^4\) = 2 log\(_8\)(\(\frac{\text{x}}{x^4}\)) = 2 \(\frac{\text{x}}{x^4}\) = 8\(^2\) = \(\frac{1}{x^3}\) = 64 x\(^3\) = (\(\frac{1}{4}\))\(^3\); x = \(\frac{1}{4}\)

\(\frac{r}{6} = \sin 60 \) \(r = 6 \sin 60\) = \(6 \times \frac{\sqrt{3}}{2}\) = \(3\sqrt{3}\) Chord = 2r = \( 2 \times 3\sqrt{3}\) = \(6\sqrt{3}\)

\(25^{x} + 3(5^{x}) = 4\) Let \(5^{x}\) = y. \((5^{2})^{x} + 3(5^{x}) - 4 = 0\) \(y^{2} + 3y - 4 = 0\) \(y^{2} - y + 4y - 4 = 0\) \(y(y - 1) + 4(y - 1) = 0\) \((y + 4)(y - 1) = 0\) \(y = -4 ; y = 1\) y = -4 is not possible. y = 1 \(\implies\) x = 0.

\(\frac{log_5 0.04}{log_3 18 - log_3 2}\) = \(\frac{log_5 0.04}{log_3(\frac{18}{2})}\) = \(\frac{log_5 0.04}{log_3 9}\) = \(\frac{-2}{2}\) = -1 Let log5 0.04 = x 5x = 0.04 x = \(\frac{4}{100}\) = 5\(^{-2}\) Let log\(_3\) 9 = z 3\(^2\) = 3\(^z\) z = 3

\(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\) Rearrange = \(\sqrt{48}\) + \(\sqrt{75}\) - \(\frac{9}{\sqrt{3}}\) = (\(\sqrt{16}\) x \(\sqrt{3}\)) + (\(\sqrt{25}\) x \(\sqrt{3}\)) - \(\frac{9}{\sqrt{3}}\) = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) - \(\frac{9}{\sqrt{3}}\) Rationalize \(\frac{9}{\sqrt{3}}\) \(\to\) = \(\frac{9}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = 3\(\sqrt{3}\) = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) - 3\(\sqrt{3}\) = 6\(\sqrt{3}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{\sqrt{2}}{2}\) = \(\frac{1.414}{2}\) = 0.707

A\(\bigtriangleup\) = \(\sqrt{S(S - a) (S - b)(S - c)}\) (Hero's Formula) S = \(\frac{a + b + c}{2}\) = \(\frac{5 + 6 + 7}{2}\) \(\frac{18}{2} = 9\) A\(\bigtriangleup\) = \(\sqrt{9 \times 4 \times 3 \times 2}\) \(\sqrt{216} = 6 \sqrt{6}cm^3\) A\(\bigtriangleup\) = \(\frac{1}{2} \times 6 \times h\) 6\(\sqrt{6} = \frac{1}{2} \times 7 \times h\) h = \(\frac{12}{7} \sqrt{6}\)cm

(a) \(3 \times 9^{1 + x} = 27^{-x}\) \(3 \times (3^{2})^{1 + x} = (3^{3})^{-x}\) \(3 \times 3^{2 + 2x} = 3^{-3x} \) \(3^{2x + 2 + 1} = 3^{-3x} \implies 3^{2x + 3} = 3^{-3x}\) \(\implies 2x + 3 = - 3x\) \(-5x = 3 \implies x = \frac{-3}{5}\) (b) \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\) = \(\log_{10} (\frac{\sqrt{35} \times \sqrt{2}}{\sqrt{7}}\) = \(\log_{10} (\frac{\sqrt{35 \times 2}{\sqrt{7}}\) = \(\log_{10} \sqrt{10}\) = \(\log_{10} 10^{\frac{1}{2}}\) = \(\frac{1}{2} \log_{10} 10 = \frac{1}{2}\)

(a) \(17.57^{2} - 12.43^{2}\) Using the difference of two squares method, = \((17.57 + 12.43)(17.57 - 12.43)\) = \((30.00)(5.14)\) = \(154.2\) (b) Given: D and C are points on the major arc of circle ADCB. To prove that < ADB = < ACB. Construction : Join A and B to O, the centre of the circle . Proof: With the lettering \(< AOB = 2x_{1}\) (angle at the centre is twice that subtended at the circumference) But \(< AOB = 2x_{2}\) (the same theorem applies here) \(\therefore 2x_{1} = 2x_{2} \implies x_{1} = x_{2}\) \(\therefore < ADB = < ACB\)