Further Mathematics Questions and Answers

\(Line: 2y + 5x - 6 = 0\)

\(2y = 6 - 5x  \implies y = 3 - \frac{5x}{2}\)

Gradient = \(\frac{-5}{2}\)

For the line perpendicular to the given line, Gradient = \(\frac{-1}{\frac{-5}{2}} = \frac{2}{5}\)

The midpoint of P(4, 3) and Q(-6, 1) = \((\frac{-6 + 4}{2}, \frac{3 + 1}{2})\)

= (-1, 2).

Therefore, the line = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)

\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)

Since f(3) = 0, then (x - 3) is a factor of f(x).

Dividing f(x) by (x - 3), we get \(2x^{2} + 3x - 2\).

\(2x^{2} + 3x - 2 = 2x^{2} - x + 4x - 2\)

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

Therefore, \(f(x) = (x - 3)(x + 2)(2x -1)\)

\(2x^{2} - 6x + 5 = 0 \implies a = 2, b = -6, c = 5\)

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{2} = 3\)

\(\alpha\beta = \frac{c}{a} = \frac{5}{2} \)

\(\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^{2} + \alpha^{2}}{\alpha\beta}\)

\(\frac{(\alpha + \beta)^{2} - 2\alpha\beta}{\alpha\beta} = \frac{3^{2} - 2(\frac{5}{2})}{\frac{5}{2}}\)

= \(\frac{4}{\frac{5}{2}} = \frac{8}{5}\)

\(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\)

Squaring both sides, we have

\((\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}\)

\(x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1\)

\(2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0\)

\((2\sqrt{x(x + 1)})^{2}= 0^{2}  \implies 4(x(x + 1)) = 0\)

\(\therefore x(x + 1) = 0\)

\(x = \text{0 or -1}\)

\((a - b)^{6} = ^{6}C_{0}(a)^{6}(-b)^{0} + ^{6}C_{1}(a)^{5}(-b)^{1} + ^{6}C_{2}(a)^{4}(-b)^{2} + ...\)

Third term = \(^{6}C_{2}(a)^{4}(-b)^{2} = \frac{6!}{(6-2)! 2!}(a^4)(b^2)\)

= \(15a^{4}b^{2}\)