Further Mathematics Questions and Answers

\((\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...\)

\(\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}\)

= \(\frac{8!}{6!2!}(\frac{x^{6}}{64})(1) \)

= \(28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}\)

\(\log_{3} x = \log_{9} 3  \implies \log_{3} x = \log_{9} 9^{\frac{1}{2}} = \frac{1}{2}\log_{9} 9\)

\(\log_{3} x = \frac{1}{2} \)

\(\therefore x = 3^{\frac{1}{2}}\)

\(4(2^{x^2}) = 8^{x}  \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}\)

\(\implies 2^{2 + x^{2}} = 2^{3x}\)

Comparing bases, we have

\(2 + x^{2} = 3x \implies x^{2} - 3x + 2 = 0\)

\(x^{2} - 2x - x + 2 = 0 \)

\(x(x - 2) - 1(x - 2) = 0\)

\((x - 1) = 0\) or \((x - 2) = 0\)

\(x = \text{1 or 2}\)

\(2\cos x - 1 = 0 \implies 2\cos x = 1\)

\(\cos x = \frac{1}{2}\)

\(x = \cos^{-1} (\frac{1}{2})\)

= \(\frac{\pi}{3}\) = \(\frac{5\pi}{3}\)

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

= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{4 - 6\sqrt{2} + 6\sqrt{2} - 18}\)

= \(\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{-14}\)

= \(\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 2 - 6\sqrt{10})\) (dividing through with the minus sign)