\(P = {x : -2 < x < 5}\) and \(Q = {x: -5 < x < 2}\)

\((P \cup Q) = {x : -5 < x < 5}\)

Given \(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\),

first, we rationalise by multiplying through with \(2\sqrt{3} - 3\sqrt{2}\) (the inverse of the denominator).

\((\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}})(\frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}})\)

= \(\frac{16\sqrt{3} - 24\sqrt{2} - 18\sqrt{2} + 18\sqrt{3}}{4(3) - 6\sqrt{6} + 6\sqrt{6} - 9(2)}\)

= \(\frac{34\sqrt{3} - 42\sqrt{2}}{-6} = 7\sqrt{2} - \frac{17\sqrt{3}}{3}\)

Given the formula for p * q as: \(p + q + 2pq\) and its identity element is 0, such that if, say, t is the inverse of p, then

\(p * t = 0\), then \(p + t + 2pt = 0 \therefore p + (1 + 2p)t = 0\)

\(t = \frac{-1}{1 + 2p}\) is the formula for the inverse of p and is undefined on R when

\(1 + 2p) = 0\) i.e when \(2p = -1; p = \frac{-1}{2}\).

Explanation not provided, please join discuss, your contributions are welcomed

\(f(x) = \frac{4}{x} - 1\). Let y = f(x)

\(y = \frac{4 - x}{x} \implies xy + x = 4\)

\(x(y + 1) = 4 \therefore x = \frac{4}{y + 1}\)

\(f^{-1}(7) = \frac{4}{7 + 1} = \frac{1}{2}\)

The elements of x are {-2, -1, 0, 1, 2}

\(y = 4x - 1\) = 4(-2) - 1 = -9; 4(-1) - 1 = -5; 4(0) - 1 = -1; 4(1) - 1 = 3; 4(2) - 1 = 7.

The range of x is {-9, -5, -1, 3 7}.

\(x^{2} + x^{2}y + 3x - 10y + 3xy -10\)

= \(x^{2} + x^{2}y + 3x + 3xy - 10y - 10 = x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)\)

= \((x^{2} + 3x - 10)(y + 1)\)

= \((x^{2} + 3x - 10) = x^{2} - 2x + 5x - 10\)

= \(x(x - 2) + 5(x - 2) = (x - 2)(x +5)\)

\(\therefore x^{2} + x^{2}y + 3x - 10y + 3xy -10 = (x - 2)(x + 5)(y + 1)\).

Given x = (-1, 5) for the equation \(x^{2} + kx - 5 = 0\)

\(x = -1 \implies x + 1 = 0\); \(x = 5 \implies x - 5 = 0\)

\((x + 1)(x - 5) = 0\), expanding,

\(x^{2} - 5x + x - 5 = 0 \therefore x^{2} - 4x - 5 = 0\)

\(\therefore\) k = -4.

The remainder theorem states that if f(x) is divided by (x - a), the remainder is f(a).

\(f(x) = x^{3} - 2x + m\) divided by (x - 1), so that a = 1.

Remainder = \(f(1) = 1^3 - 2(1) + m = -1 + m\)

\(f(x) = 2x^{3} + x - m\) divided by (2x + 1), so that a = \(\frac{-1}{2}\)

\(f(\frac{-1}{2}) = 2(\frac{-1}{2}^{3}) + (\frac{-1}{2}) - m = \frac{-3}{4} - m\)

\(\implies m - 1 = \frac{-3}{4} - m\), collecting like terms,

\(2m = \frac{1}{4} \therefore m = \frac{1}{8}\)

\((2t - 3s)(t - s) = 0 \implies (2t - 3s) = \text{0 or} (t - s) = 0\)

\(2t - 3s = 0 \implies 2t = 3s \therefore \frac{t}{s} = \frac{3}{2}\)

\(t - s = 0 \implies t = s \therefore \frac{t}{s} = 1\)

\(\frac{t}{s} = \frac{3}{2} or 1\)