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log\(_8\)x - 4log\(_8\)x = 2; log\(_8\)x - log\(_8\)x\(^4\) = 2 log\(_8\)(\(\frac{\text{x}}{x^4}\)) = 2 \(\frac{\text{x}}{x^4}\) = 8\(^2\) = \(\frac{1}{x^3}\) = 64 x\(^3\) = (\(\frac{1}{4}\))\(^3\); x = \(\frac{1}{4}\)

8x-2 = \(\frac{2}{25}\) = 200x-2 = 2 = 100x-2 = 1 x-2 = \(\frac{1}{100}\) x-2 = 10-2 x = 10

Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero \(\therefore (x - 1) = 0\) x = 1 Substitute in the polynomial the value x = 1 = \(p(1)^3 + q(1)^2 + 11(1) - 6 = 0\) p + q + 5 = 0 .....(i) Also since x - 3 is a factor, \(\therefore\) x - 3 = 0 x = 3 Substitute \(p(3)^3 + q(3)^2 + 11(3) - 6 = 0\) 27p + 9q = -27 ......(2) Combine eqns. (i) and (ii) Multiply equation (i) by 9 to eliminate q 9p + 9q = -45 Subtract (ii) from (i), \(18p = 18\) \(\therefore\) p = 1 Put p = 1 in (i), \(1 + q = -5 \implies q = -6\) \((p, q) = (1, -6)\)

\(\frac{a}{a - x}\) = \(\frac{b}{x - b}\) \(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\) ∴ 3(1 - x) = x - 3 3 - 3x = x - 3 Rearrange 6 = 4x; x = \(\frac {6}{4}\) = \(\frac{3}{2}\)

\(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\) Multiply through by r(r -1) which is the LCM = (r)(r + 1) + 2(r)(r - 1) = 3(r - 1)(r + 1) = r2 + r + 2r2 - 2r 3r2 - 3 = 3r2 r = 3r2 - 3 -r = -3 ∴ r = 3

In an AP, Tn = a + (n - 1)d T6 = a + 5d = 11 The first term = a = 1 ∴ T6 = 1 + 5d = 11 5d = 11 - 1 5d = 10 ∴ d = 2

\(\frac{(n + 1)(n + 2)}{2}\) If n = 1, the expression becomes 3 n = 2, the expression becomes 6 n = 4, the expression becomes 15 n = 5, the expression becomes 21

\(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) = x + 1 = 4 x = 4 - 1 = 3 3x + y =1 3(3) = y = 1 = 9 + y = 1 y = 1 - 9 = -8

Equation of curve; y = 2x2 - 2x + 3 gradient of curve; \(\frac{dy}{dx}\) = differential coefficient \(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2 ∴ \(\frac{dy}{dx}\) = 2 4x - 2 = 2 4x = 4 ∴ x = 1 When x = 1, y = 2(1)2 - 2(1) + 3 = 2 - 2 + 3 = 5 - 2 = 3 coordinate of the point where the curve; y = 2x2 - 2x + 3 has gradient equal to 2 is (1, 3)

y = x3 - 27, y = -27 \(\to\) (0, -27) when y = 0, x = 3 (3, 0)

2x - y - 2 \(\geq\) 0 = y \(\leq\) 2x - 2 when x = 0, y = -2, when y = 0, x = 1

Gradient of line = \(\frac{\text{Change in y}}{\text{Change in x}} = \frac{y_2 - y_1}{x_2 - x_1}\) y2 = 0, y1 = 4 x2 = 3 and x1 = 0 \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{3 - 0} = \frac{-4}{3}\) Equation of straight line = y = mx + c where m = gradient and c = y intercept = 4 y = \(\frac{-4x}{3}\) + 4, multiple through by 3 3y = - 4x + 12

(a) x -1 0 1 2 3 4 5 6 y 10 -1 -8 -11 -10 -5 4 17 (b) (c)(i) \(2x^{2} - 9x = 4\) \(2x^{2} - 9x - 1 = 4 - 1 = 3\) At y = 3, x = 0.2 and 4.3. (ii) Gradient of \(y = 2x^{2} - 9x - 1\) at x = 3 = \(\frac{BC}{AB} = \frac{9.5}{4.3}\) = \(2.2\)

(a) \(T_{n} = a + (n - 1)d\) (nth term of an A.P) \(T_{4} = a + 3d = 37 ... (1)\) \(T_{6} = a + 5d = 12 + 37 = 49 ... (2)\) (2) - (1) : \(2d = 12 \implies d = 6\) \(a + 3d = 37 \implies a + 3(6) = 37\) \(a + 18 = 37 \implies a = 37 - 18 = 19\) \(T_{7} = a + 6d = 19 + 6(6) = 19 + 36 = 55\) (b) \(P = {1, 2, 3, 4} ; Q = {3, 5, 6}\) (i) \(P \cap Q = {3}\) (ii) \(P \cup Q = {1, 2, 3, 4, 5, 6}\) (iii) \((P \cap Q) \cup Q = {3, 5, 6}\) (iv) \((P \cap Q) \cup P = {1, 2, 3, 4}\)

when n = 2 (-1)\(^{n-2}\) x 2\(^{n -1}\) = 2 When n = 3 (-1)\(^{n-2}\) x 2\(^{n - 1}\) = -4 Sum = 2 - 4 = - 2