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

To make \( p \) the subject of the formula \( y = \frac{a + p}{a - p} \): Start with the equation: \(y = \frac{a + p}{a - p}\) Cross-multiply: \(y(a - p) = a + p\) Distribute \( y \): \(ya - yp = a + p\) Rearrange to isolate terms involving \( p \): \(ya - a = p + yp\) Factor out \( p \) on the right side: \(ya - a = p(1 + y)\) Finally, solve for \( p \): \(p = \frac{ya - a}{1 + y}\) Thus, the formula with \( p \) as the subject is: \(p = \frac{a(y - 1)}{y + 1}\)

The roots of a quadratic equation are -\(\frac{1}{4}\) and 3. The quadratic equation in x x = -\(\frac{1}{4}\) or 3 (x + \(\frac{1}{4}\)) = 0 or (x - 3) = 0 (x + \(\frac{1}{4}\))(x - 3) = 0 x\(^2\) - 3x + \(\frac{x}{4}\) - 3(\(\frac{1}{4}\)) = 0 x\(^2\) - \(\frac{11x}{4}\) - \(\frac{3}{4}\) = 0 multiply thru by 4 4x\(^2\) - 11x - 3 = 0

(a) \(6y + 5x = -12 ... (1)\) \(4y - 3x = 11 ... (2)\) Multiply (1) by 2 and (2) by 3, \(12y + 10x = -24 ... (1a)\) \(12y - 9x = 33 ... (2a)\) \((1a) - (2a) : 19x = -57 \implies x = -3\) \(4y - 3(-3) = 11 \implies 4y + 9 = 11\) \(4y = 2 \implies y = \frac{1}{2}\) (b) \(/BC/^{2} = 36^{2} + 48^{2}\) \(/BC/^{2} = 1296 + 2304 = 3600\) \(/BC/ = \sqrt{3600} = 60 cm\) \(/AC/^{2} = /BC/^{2} + /AB/^{2}\) \((y + 48)^{2} = 60^{2} + 36^{2} + y^{2}\) \(y^{2} + 96y + 2304 = 3600 + 1296 + y^{2}\) \(96y = 3600 + 1296 - 2304 = 2592\) \(96y = 2592 \implies y = \frac{2592}{96} = 27 cm\)

(a) x -3 -2 -1 0 1 2 3 4 \(y = 3x^{2} - 5x - 7\) 35 15 1 -7 -9 -5 5 21 (b) (c) \(y + 3x + 2 = 0 \implies y = -3x - 2\) when x = -2, y = 6 - 2 = 4. when x = 3, y = -9 - 2 = -11. (d) \(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\) The roots of the equation \(3x^{2} - 2x - 5 = 0\) is when the line \(y = -3x - 2\) cuts the equation i.e. x = 1.0 or 1.7.

A.P 11, 4, -3 1st term = 11 A.P = a, a + d, a + 2d ...... a + (n - 1)d If a = 11 a + d = 4 d = 4 - 11 = -7 nth term = a + (n-1)d          = 11 + (n-1)(-7)          = 11 - 7n + 7          = 18 - 7n

16/9, x, 1, y => a, ar, ar2, ar3 ar2 = 1 => 16r2/9 = 1 => 16r2 9 => r2 = 9/16 => r = 3/4 ar2 = y = ar2 x r = 1 x 3/4 = 3/4 x xy = 4/3 x 3/4 = 1

Gradient of the line = \(\frac{3 - 0}{0 - 1}\) = -3 y = -3x + b. Using (1,0), we have 0 = -3(1) + b 0 = -3 + b b = 3 y = -3x + 3 \(\therefore\) The graph illustrates y \(\leq\) -3x + 3.

\(\frac{1}{3}(2x - 1) < 5\) \(2x - 1 < 15\) \(2x < 16\) \(x < 8\)

3x\(^2\) + 25x - 18 = 0 3x\(^2\) + 27x - 2x - 18 = 0 3x(x + 9) - 2(x + 9) = 0 (3x - 2)(x + 9) = 0 x = \(\frac{2}{3}\) or x = -9.

(x + 2)(x - 7) = 0 x + 2 = 0 \(\implies\) x = -2 x - 7 = 0 \(\implies\) x = 7 x = -2 or 7

x + y = \(\frac{3}{2}\) ... (i) x - y = \(\frac{5}{2}\) ... (ii) (i) - (ii): 2y = \(\frac{-2}{2}\) = -1 y = \(-\frac{1}{2}\) x + y = \(\frac{3}{2}\) x - \(\frac{1}{2}\) = \(\frac{3}{2}\) x = \(\frac{3}{2} + \frac{1}{2}\) = \(\frac{4}{2}\) x = 2 \(\therefore\) 2y + x = 2(\(-\frac{1}{2}\)) + 2 = -1 + 2 = 1.

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

\(x^{3} + px^{2} + qx + 6 = 0\) f(x - 1) = 0; f(1) = 0. \(1^{3} + p(1^{2}) + q(1) + 6 = 0 \implies p + q = -7 ... (1)\) f(x + 1) = 0; f(-1) = 0. \((-1)^{3} + p(-1^{2}) + q(-1) + 6 = 0 \implies p - q = -5 ... (2)\) Subtract (2) from (1). \(2q = -2 \implies q = -1\) \(p - (-1) = -5 \implies p = -5 - 1 = -6\) \((p, q) = (-6, -1)\)

(x - 3)(x - 4) \(\leq\) 0 Case 1 (+, -) = x - 3 \(\geq\) 0, X - 4 \(\geq\) 0 = X \(\leq\) 3, x \(\geq\) 4 = 3 < x \(\geq\) 4 (solution) Case 2 = (-, +) = x - 3 \(\leq\) 0, x - 4 \(\geq\) 0 = x \(\leq\) 3, x \(\geq\) 4 therefore = 3 \(\leq\) x \(\leq\) 4