Question Library

\(\frac{\sqrt{2}}{x + 2}\) = x - \(\frac{1}{\sqrt{2}}\) x\(\sqrt{2}\) (x - \(\sqrt{2}\)) = x + \(\sqrt{2}\) (cross multiply) x\(\sqrt{2}\) - 2 = x + \(\sqrt{2}\) = x\(\sqrt{2}\) - x = 2 + \(\sqrt{2}\) x (\(\sqrt{2}\) - 1) = 2 + \(\sqrt{2}\) = \(\frac{2 + \sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}\) x = \(\frac{2 \sqrt{2} + 2 + 2 + \sqrt{2}}{2 - 1}\) = 3\(\sqrt{2}\) + 4

\(\begin{bmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{bmatrix}\) = 0 To find the value of x, we will have to calculate for the determined of the given matrix and equate it to 0 x[x\(^2\) - 1] - 1[x - 0] + 0[0 - 1] = 0 x\(^3\) - x] - x + 0 = 0 x\(^3\) - x - x = 0 x\(^3\) - 2x = 0 x\(^3\) = 2x x\(^2\) = 2 take the square root of both sides x = \(\sqrt{2}\)

[(\(\frac{16}{9}\))\(^{\frac{-3}{2}}\) x 16\(^{\frac{-3}{4}}\)]\(^{\frac{1}{3}}\) = [(\(\frac{9}{16}\))]\(^{\frac{3}{2}}\) x [(\(\frac{1}{16}\))\(^{\frac{3}{4}}\)]\(^{\frac{1}{3}}\) = [(\(\sqrt{\frac{9}{16}}\))\(^3\) x (4\(\sqrt{\frac{1}{16}})^3\)]\(^{\frac{1}{3}}\) = [(\(\frac{3}{4})^3 \times (\frac{1}{2})^3\)]\(^\frac{1}{3}\) (\(\frac{27}{64} \times \frac{1}{8}\))\(^\frac{1}{3}\) = \({3}\sqrt{\frac{27}{64} \times \frac{1}{8}}\) = \(\frac{3}{4} \times \frac{1}{2}\) = \(\frac{3}{8}\)

1 + 2log\(_{10}^3\) = log\(_{10}^{10} + log_{10}^{3^2}\) = log\(_{10}^{10} + log_{10}^{9}\) = log\(_{10}^{10 \times 90}\) = log\(_{10}^{90}\)

\(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}\) = x + y\(\sqrt{15}\) cross multiply to have: \(\sqrt{3}\) + \(\sqrt{5}\) = x\(\sqrt{5}\) + 5y\(\sqrt{3}\) Collect like roots : x\(\sqrt{5}\) = \(\sqrt{5}\) → x = 1 5y\(\sqrt{3}\) = \(\sqrt{3}\) → y = \(\frac{1}{5}\) ∴ ( x + y ) = 1 + \(\frac{1}{5}\) = 1\(\frac{1}{5}\)

ZT = \(\sqrt{6^2 - 3^2}\) ZT = \(\sqrt{27}\) = \(3\sqrt{3}\) tan (< XZT) = \(\frac{3}{3\sqrt{3}}\) = - \(\frac{1}{\sqrt{3}}\)

(a) C\(B\)D = C\(D\)B = \(\frac{180^o - 108^o}{2}\) = \(\frac{72^o}{2}\) = 36\(^o\) < CDE = < BDE + 36\(^o\) < CDE = 90\(^o\) + 36 = 126\(^o\) (b) tan x = \(\sqrt{3}\) y\(^2\) (\(\sqrt{3}\))\(^2\) - (l)\(^2\) y\(^2\) 3 + 1, Cos x = \(\frac{1}{2}\) y = \(\sqrt{4}\) = 2 sin x = \(\frac{\sqrt{3}}{2}\) \(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\) = \(\frac{(\frac{1}{2})^2 - \frac{\sqrt{3}}{2}} {(\frac{\sqrt{3}}{2})^2 + \frac{1}{2}}\) = \(\frac{\frac{1}{4} - \frac{\sqrt{3}}{2}}{\frac{3}{4} + \frac{1}{2}}\) = \(\frac{1 - 2\sqrt{3}}{4}\) \(\div\) \(\frac{5}{4}\) = \(\frac{1 - 2 \sqrt{3}}{4}\) x \(\frac{4}{5}\) = \(\frac{1 - 2\sqrt{3}}{5}\)

< AOB = 108\(^o\) - (90\(^o\) + 40\(^o\)) = 180\(^o\) - 130\(^o\) = 50\(^o\) < DCO = 40\(^o\) (Alternate angle) < ODE = \(\frac{180^o - 80^o}{2}\) = \(\frac{100^o}{2}\) = 50\(^o\) (b)(i) (b)(ii) Area of the parallelogram = ab sin \(\theta\) = 5cm x 7cm sin 125\(^3\) = 35 sin 125 = 28.7cm\(^2\) (c) x = \(\frac{1}{2}(1 - \sqrt{2})\) evaluate (2x\(^2\) - 2x) = 2[\(\frac{1}{2}(1 - \sqrt{2})^2\)] - 2[\(\frac{1}{2} (1 - \sqrt{2}\))] = 2[\(\frac{1}{4}(1 - 2\sqrt{2} + 2)\)] - 1 + \(\sqrt{2}\) = 2[\(\frac{1}{4}(3 - 2\sqrt{2})\) - 1 + \(\sqrt{2}\)] 2[\(\frac{3}{4} - \frac{1}{2} \sqrt{2} - 1 + \sqrt{2}\)] = \(\frac{3}{2} - \sqrt{2} - 2 + 2\sqrt{2}\) = \(\frac{1}{2} + \sqrt{2}\)

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

\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1) Let u = 4x\(^3\) - 2x. \(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\) \(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\) \(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\) \(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\) = \(\frac{12x^2 - 2}{4x^3 - 2x}\)

81\(^{\frac{-3}{4}}\) x 25\(^{\frac{1}{2}}\) x 243\(^{\frac{2}{5}}\) = \((\sqrt[4]{81})^{-3} \times \sqrt{25} \times (\sqrt[5]{243})^2\) = \(\frac{5 \times 3^2}{3^{3}}\) = \(\frac{5}{3}\)

2\(^{x + y}\) = 16 ; 4\(^{x - y}\) = \(\frac{1}{32}\). \(\implies 2^{x + y} = 2^4\) \(x + y = 4 ... (1)\) \(2^{2(x - y)} = 2^{-5} \) \(2^{2x - 2y} = 2^{-5}\) \(\implies 2x - 2y = -5 ... (2)\) Solving the equations (1) and (2) simultaneously, we have x = \(\frac{3}{4}\) and y = \(\frac{13}{4}\)

\(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\) \((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\) \(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\) \(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\) Equating bases, we have \(2 - 2x + x + 2 + 3x = -4\) \(4 + 2x = -4 \implies 2x = -4 - 4\) \(2x = -8\) \(x = -4\)

\(\sqrt{28} + \sqrt{112} - \sqrt{k} = \sqrt{175}\) \(\sqrt{4 \times 7} + \sqrt{16 \times 7} - \sqrt{k} = \sqrt{25 \times 7}\) \(2\sqrt{7} + 4\sqrt{7} - \sqrt{k} = 5\sqrt{7}\) \(6\sqrt{7} - 5\sqrt{7} = \sqrt{k}\) \(\sqrt{k} = \sqrt{7}\) \(\implies k = 7\)

\(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\) = \(\frac{2\log_3 3^2 \times \log_3 (3^4)^{-2}}{\log_5 (5^4)}\) = \(\frac{4\log_3 3 \times -8\log_3 3}{4\log_5 5}\) = -8.