Further Mathematics Questions and Answers

$\frac{\mathrm d y}{\mathrm d x} = \sqrt{x} = x^{\frac{1}{2}}$

$y = \int x^{\frac{1}{2}} \mathrm {d} x$

= $\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c$

= $\frac{2}{3}x^{\frac{3}{2}} + c$

No explanation has been provided for this answer.

$T_{n} = ar^{n - 1}$

$T_{4} = ar^{4 - 1} = ar^{3} = 192$

$T_{9} = ar^{9 - 1} = ar^{8} = 6$

Dividing $T_{9}$ by $T_{4}$, 

$r^{8 - 3} = \frac{6}{192}$

$r^{5} = \frac{1}{32} = (\frac{1}{2})^{5}$

$r = \frac{1}{2}$

$\frac{\mathrm d}{\mathrm d x}(x^2 + xy - 5) = \frac{\mathrm d (x^{2})}{\mathrm d x} + \frac{\mathrm d (xy)}{\mathrm d x} - \frac{\mathrm d (5)}{\mathrm d x} = 0$

= $2x + x\frac{\mathrm d y}{\mathrm d x} + y = 0$

$\implies x\frac{\mathrm d y}{\mathrm d x} = -(2x + y)$

$\frac{\mathrm d y}{\mathrm d x} = \frac{-(2x + y)}{x}$

$Line: 2y + 5x - 6 = 0$

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

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 = $\frac{y - 2}{x + 1} = \frac{2}{5}$

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