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\(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\)b b = \(a^{-2}\) and \(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\)b b = \(a^{-\frac{2}{3}}\) \(c^{\frac{1}{3}}\) = \(a^{\frac{1}{2}}\) x \(a^{-\frac{2}{3}}\) = \(a^{-\frac{1}{6}}\) take the cube of both sides c = (\(a^{-\frac{1}{6}})^3\) = \(a^{-\frac{1}{2}}\)

log4(y - 1) + log4(\(\frac{1}{2}\)x) = 1 log4(y - 1)(\(\frac{1}{2}\)x) \(\to\) (y - 1)(\(\frac{1}{2}\)x) = 4 ........(1) log2(y + 1) + log2x = 2 log2(y + 1)x = 2 \(\to\) (y + 1)x = 22 = 4.....(ii) From equation (ii) x = \(\frac{4}{y + 1}\)........(iii) put equation (iii) in (i) = y (y - 1)[\(\frac{1}{2}(\frac{4}{y - 1}\))] = 4 = 2y - 2 = 4y + 4 2y = -6 y = -3 x = \(\frac{4}{-3 + 1}\) = \(\frac{4}{-2}\) X = 2 therefore x = -2, y = -3

\(\frac{k}{\sqrt{3} + \sqrt{2}}\) = k\(\sqrt{3 - 2}\) \(\frac{k}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = k\(\sqrt{3 - 2}\) K(\(\sqrt{3}\) - K\(\sqrt{2}\) ) = K\(\sqrt{3}\) - 2 K\(\sqrt{3}\) - K\(\sqrt{3}\) - K\(\sqrt{2}\) = - 2 - K\(\sqrt{2}\) = - 2 K = \(\frac{- 2}{- \sqrt{2}}\) K = \(\frac{2}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\) K = \(\frac{2\sqrt{2}}{2}\) = \(\sqrt{2}\)

V\(\pi\)r2h = \(\pi\)(3)2(10) = 90\(\pi\)cm3 V = \(\pi\)(5)2 x 18 = 450\(\pi\)cm3 No of volume = \(\frac{450\pi}{90\pi}\) = 5 selling price = 5 x N15 = N75 profit = N75 - N50 = N25.00

The cost of normal work = 35r The cost of overtime = q x 2r = 2qr The man's total weekly earning = 35r + 2qr = r(35 + 2q)

\(\sqrt{x}\) - \(\sqrt{(x - 2)}\) - 1 = 0 = \(\sqrt{x}\) - \(\sqrt{(x - 2)}\) = 1 = (\(\sqrt{x}\) - \(\sqrt{(x - 2)}\))2 = 1 = x - 2 \(\sqrt{x(x - 2)}\) + x -2 = 1 = (2x - 3)2 = [2 \(\sqrt{x(x - 4)}\)]2 = 4x2 - 12x + 9 = 4(x2 - 2x) = 4x2 - 12x + 9 = 4x2 - 8x 4x = 9 x = \(\frac{9}{4}\)

\(\frac{x + a}{x - a}\) = m x + a = mx - ma a + ma = mx - x a(m + 1) = x(m - 1) \(\frac{a}{x}\) = \(\frac{m - 1}{m + 1}\)

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

\(\frac{11x + 2}{6x^2 - x - 1}\) = \(\frac{11x + 2}{(3x + 1)(2x - 1)}\) = \(\frac{A}{3x + 1}\) + \(\frac{B}{2x - 1}\) 11x + 2 = A(2x - 1) + B(3x + 1) put x = \(\frac{1}{2}\) \(\frac{15}{2} = \frac{5}{2}B\) B = 3. put x = \(-\frac{1}{3}\) \(-\frac{5}{3} = \frac{-5}{3}\)A \(\implies\) A = 1 ∴ \(\frac{11x +2}{6x^2 - x - 1}\) = \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)

\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\) = \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\) = 4x(2 + 3x) > 6x = 12x\(^2\) - 2x = 0 = 2x(6x - 1) > 0 = x(6x - 1) > 0 Case 1 (-, -) = x < 0, 6x -1 < 0 = x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution) Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\) Combining solutions in cases(1) and (2) = x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)

2p - 10 = \(\frac{p + 1 + 1 - 4P^2}{2}\) (Arithmetic mean) = 2(2p - 100 = p + 2 - 4P2) = 4p - 20 = p + 2 - 4p2 = 4p2 + 3p - 22 = 0 = (p - 2)(4p + 11) = 0 ∴ p = 2 or -\(\frac{4}{11}\)

Let the first term of the geometric progression (GP) be \( a \) and the common ratio be \( r \). The sum of the first three terms is: a + ar + ar\(^2\) = a(1 + r + r\(^2\)) \(S_3 = a(1 + r + r^2)\) The sum to infinity is: \(S_{\infty} = \frac{a}{1 - r}\) But: \(S_3 = \frac{1}{2} S_{\infty}\) Substituting the expressions gives: \(a(1 + r + r^2) = \frac{1}{2} \cdot \frac{a}{1 - r}\) Dividing by \( a \): \(1 + r + r^2 = \frac{1}{2(1 - r)}\) Cross-multiplying: \(2(1 + r + r^2)(1 - r) = 1\) Expanding and simplifying leads to: \(2(1 - r^3) = 1\) Thus: \(r^3 = \frac{1}{2}\) Taking the cube root: \(r = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}}\) Therefore, the positive common ratio of the progression is: \(\frac{1}{\sqrt[3]{2}}\).

The identity element (e) under an operation, say \(\otimes\), is the element such that for any given element under the operation, say a, \(a \otimes e = e \otimes a = a\) From the table, q is the identity element. \(p \otimes q = q \otimes p = p\) Same as all through.

x \(\ast\) y = xy - y - x, x \(\ast\) 3 = 3x - 3 - x = 2x - 3 2 \(\ast\) x = 2x - x - 2 = x - 2 ∴ 2x - 3 = x - 2 x = -2 + 3 = 1

\(\begin{vmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4\end{vmatrix}\) = x\(\begin{vmatrix}2 & 3 \\ 1+x & 4\end{vmatrix}\) - \(\begin{vmatrix}1-x & 3 \\ 1 & 4\end{vmatrix}\) = 0 = x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x] = 5x - 3x2 -1 + 4x = -3x2 + 9X - 1