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

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