WAEC Further Mathematics Past Questions & Answers - Page 94

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

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

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

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

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

$x^{2} + y^{2} + 8x - 4y = 5$

$2a = -8 \implies a = -4$

$2b = 4 \implies b = 2$

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

$\therefore r = 5$