Further Mathematics Questions and Answers

\(P = \begin{pmatrix} 2 & 1 \\ 5 & -3 \end{pmatrix} ; Q = \begin{pmatrix} 4 & -8 \\ 1 & -2 \end{pmatrix}\)

\(2P = \begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix}\)

\(2P - Q = \begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix} - \begin{pmatrix} 4 & -8 \\ 1 & -2 \end{pmatrix}\)

= \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\)

At turning point, \(\frac{\mathrm d y}{\mathrm d x} = 0\).

Given \(x^{3} - x^{2} - x + 6 \)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x - 1 = 0 \)

\(3x^{2} - 3x + x - 1 = 0 \implies (3x + 1)(x - 1) = 0\)

\(x = \frac{-1}{3}, 1\)

\(\int (3x^{2} - 2x - 12) \mathrm {d} x = \frac{3x^{2 + 1}}{2 + 1} - \frac{2x^{1 + 1}}{2} - 12x\)

= \(x^{3} - x^{2} - 12x\)

\((x^{3} - x^{2} - 12x)|_{-2}^{3} = ((3^{3}) - (3^{2}) - 12(3)) - ((-2^{3}) - (-2^{2}) - 12(-2))\)

= \((27 - 9 - 36) - (-8 - 4 + 24) = -18 - 12 = -30\)

Midpoint between \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) = \((\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})\).

\((-k, k) = (\frac{2 + (1 + k)}{2}, \frac{-4 + (k + 1)}{2})\)

\(-k = \frac{k + 3}{2} \implies -2k = k + 3\)

\(-3k = 3 \implies k = -1\)

The equation of a circle is given as: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have: \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

\(\implies x^{2} + y^{2} - 2ax - 2by = r^{2} - a^{2} - b^{2}\)

Comparing with the given equation: \(3x^{2} + 3y^{2} + 24x - 12y = 15\)

Making the coefficients of \(x^{2}\) and \(y^{2}\) = 1, we have

\(x^{2} + y^{2} + 8x - 4y = 5\)

\(2a = -8 \implies a = -4\)

\(2b = 4 \implies b = 2\)

\(r^{2} - a^{2} - b^{2} = 5 \implies r^{2} = 5 + (-4)^{2} + (2)^{2} = 5 + 16 + 4 = 25\)

\(\therefore r = 5\)