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log5(62.5) - log5(\(\frac{1}{2}\)) = log5\(\frac{(62.5)}{\frac{1}{2}}\) - log5(2 x 62.5) = log5(125) = log553 - 3log55 = 3

Principal = N255.00, Interest = 27.00 year = x Rate: 4% ∴ 1 = \(\frac{PRT}{100}\) 27 = \(\frac{225 \times 4 \times T}{100}\) 2700 = 900T T = 3 years

\(\sqrt{x^2 + 9}\) = x + 1 x\(^2\) + 9 = (x + 1)\(^2\) 0 = x\(^2\) + 2x + 1 - x\(^2\) - 9 = 2x - 8 = 0 2(x - 4) = 0 x = 4

\(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication, q(1 + ax) = p(1 - ax) q + qax = p - pax qax + pax = p - q ∴ x = \(\frac{p - q}{a(p + q)}\)

Factorize 15 + 7x - 2x2 (5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0 ∴ (5 - x)(3 + 2x) = 0 x = 5 or x = -\(\frac{3}{2}\) Since 5 is a root, then (x - 5) is a factor

\(\left(x + \frac{1}{x} + 1\right)^2 - \left(x - \frac{1}{x} - 1\right)^2,\) We can use the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b),\) where \(a = x + \frac{1}{x} + 1 \quad \text{and} \quad b = x - \frac{1}{x} - 1.\) Calculating \( a - b \): \(a - b = \left(x + \frac{1}{x} + 1\right) - \left(x - \frac{1}{x} - 1\right) = x + \frac{1}{x} + 1 - x + \frac{1}{x} + 1 = 2 \cdot \frac{1}{x} + 2 = 2\left(\frac{1}{x} + 1\right).\) Calculating \( a + b \): \(a + b = \left(x + \frac{1}{x} + 1\right) + \left(x - \frac{1}{x} - 1\right) = x + \frac{1}{x} + 1 + x - \frac{1}{x} - 1 = 2x.\) Substituting back into the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b) = \left(2\left(\frac{1}{x} + 1\right)\right)(2x) = 4x\left(\frac{1}{x} + 1\right).\) Distributing \(4x\): \(= 4x \cdot \frac{1}{x} + 4x \cdot 1 = 4 + 4x.\) Thus, the evaluated expression is: 4 + 4x = 4(1 + x)

x2 + y - 5 = 0.....(i) y - 7x + 3 = 0.........(ii) y = 7x - 3, substituting the value of y in equation (i) x2 + (7x - 3) - 5 = 0 x2 + 7x + 3 = 0 (x + 8)(x - 1) = 0 x = -8 or 1

(3x - 2)(5x - 4) = (3x - 2)2 = 5x2 - 22x + 6 = 9x2 = 12x + 4 6x2 - 10x + 4 = 0 6x2 - 6x - 4x + 4 = 0 6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0 x = 1 or \(\frac{2}{3}\)

f(x + 2) = f(-1) f(-1) is gotten when x = -3. \(2x^{2} + 7x - 5\) at x = -3. f(-1) = \(2(-3)^{2} + 7(-3) - 5\) = \(18 - 21 - 5 = -8\)

No expanded working has been attached for this question yet. The verified answer is Option D.

\(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) = \(\frac{q - p}{pq}\) \div \(\frac{p^2 - q^2}{pq}\) \(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 - q^2}\) \(\frac{-(p - q)}{(p + q)(p - q)}\) = \(\frac{-1}{p + q}\)

y\(^2\) - 3y > 18 = y\(^2\) - 3y - 18 > 0 y\(^2\) - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0 = (y + 3) (y - 6) > 0 Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0 = y > -3, y > 6 By simply substituting the value of y into the inequality, y > -3 is invalid, and y < 6 is also invalid. Therefore, y < - 3 or y > 6

\(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\) = (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3) x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0 = x(x + 3)9x + 4) > 0 Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0 = x > -4 (solution only) Case 2 (+, -, -) = x > 0, x + 4 0, x -3, x 0 = x -4 = x < -0, -4 < x < -3(solution) combining the solutions -4 < x < -3

x \(\ast\) y = xy x \(\ast\) 2 = x2 x \(\ast\) 2 = x ∴ x2 - x = 0 x(x - 1) = 0 x = 0 or 1

\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\) \((k + 1)(3k + 1) = (2k - 1)(2k - 1)\) \(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\) \(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\) \(k^{2} - 8k = 0\) \(k(k - 8) = 0\) \(\therefore \text{k = 0 or 8}\) The terms of the sequence given k = 0: (1, -1, 1) \(\implies \text{The common ratio r = -1}\) The terms of the sequence given k = 8: (9, 15, 25) \(\implies \text{The common ratio r = } \frac{5}{3}\) The possible values of the common ratio are -1 and \(\frac{5}{3}\).