Further Mathematics Questions and Answers

Given the endpoints of the diameter |EF|, the midpoint is the centre of the circle

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

The radius is the distance from the centre to any point on the circle. Using \((\frac{1}{2}, \frac{1}{2})\) and \((3, 2)\);

\(r^{2} = (3 - \frac{1}{2})^{2} + (2 - \frac{1}{2})^{2} = \frac{25}{4} + \frac{9}{4}\)

\(r^{2} = \frac{34}{4}\)

The equation of a circle is given as:

\((x - a)^{2} + (y - b)^{2} = r^{2}\), (a, b) as the centre of the circle.

\(= (x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{34}{4}\)

\(x^{2} - x + \frac{1}{4} + y^{2} - y + \frac{1}{4} = \frac{17}{2}\)

= \(x^{2} - y^{2} - x - y - 8 = 0\) 

The gradient of the line y = 2x + 5 ;

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

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

\(y - 3 = 2(x + 4) \implies y = 2x + 11\)

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

\(\tan x = \frac{5}{12} ; \tan y = \frac{3}{4}\)

\(\tan (x + y) = \frac{\frac{5}{12} + \frac{3}{4}}{1 - (\frac{5}{12} \times \frac{3}{4}})\)

= \(\frac{\frac{14}{12}}{\frac{33}{48}}\)

= \(\frac{56}{33}\)

\(\cos(a + b) = \cos a\cos b - \sin a\sin b\)

\(\cos75° = \cos(30 + 45) = (\cos30)(\cos45) - (\sin30)(\sin45)\)

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

= \(\frac{\sqrt{6} - \sqrt{2}}{4}\)

= \(\frac{\sqrt{2}(\sqrt{3} - 1)}{4}\)

\(\begin{pmatrix} 1 & 2 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix}\)

= \(\begin{pmatrix} 1\times 0 + 2\times 1 & 1\times 1 + 2\times3 \\ 5\times0 + 1\times1 &  5\times1 + 1\times 3 \end{pmatrix}\)

= \(\begin{pmatrix} 2 & 7 \\ 1 & 8 \end{pmatrix}\)