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3m + 3 > 9 3m > 9 - 3 3m > 6 m > 2

x\(^2\) - 2x - 3 = 0 x\(^2\) - 3x + x - 3 = 0 x(x - 3) + 1(x - 3) = 0 (x + 1)(x - 3) = 0 x = (-1, 3)

y = 3x ... (i); 4y - 5x =14 ... (ii) Put 3x for y in (ii). 4(3x) - 5x = 14 12x - 5x = 14 7x = 14 \(\implies\) x = 2. y = 3x = 3(2) = 6 (x, y) = (2, 6)

\(-2, -3, -4\frac{1}{2}, ...\) This is a G.P with r = 1\(\frac{1}{2}\). \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{5} = (-2)(\frac{3}{2})^{5 - 1}\) = \(-2 \times \frac{81}{16}\) = \(-\frac{81}{8}\) \(T_{7} = (-2)(\frac{3}{2})^{7 - 1}\) = \(-2 \times \frac{729}{64}\) = \(-\frac{729}{32}\)

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 \(\frac{2}{3}\)

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

2x + 3 \(\neq\) 2x - 3 for any value of x ∴ for the \(\bigtriangleup\) to be isosceles, either 2x - 3 = x + 3 or 2x + 3 = x + 3 solve the two equations we arrive at x = 6 or x = 0 When x = 6, the sides are 9, 15, 9 When x = 0, the sides are 3, 4, -3 since lengths of a \(\bigtriangleup\)can never be negative then the value of x = 6

The two equations are y = x(2 - x) and y = (x - 1)(x - 3) The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0 put y = 0; 0 = x(2 - x) 0 = (x - 1)(x - 3) x = 0 or x = 2; x = 1 or x = 3 The values of x at P, Q, R are increasing towards the positive direction of x-axis at P, x = 1 at Q, x = 2 at R, x = 3 P, Q, R are respectively (1, 2, 3)

(a) Draw RC produced to E parallel to DE. \(\therefore < BCR = 52^\circ\) (alternate angles BA//RC) also < RCD = 48^\circ (alternate angles RC//DE) \(x^\circ = 360^\circ - (52^\circ + 48^\circ) = 360^\circ - 100^\circ = 260^\circ\) (angles at a point = 360^\circ) (b) \(y - 2x < 5 ... (i)\) \(2y + 2x \geq 4 \implies y + x \geq 2 ... (ii)\) \(y + 2x \leq 10 ... (iii)\) From (i), if \(y - 2x = 5\), when x = 0, y = 5, when x = 2, y = 9. From (ii), if \(y + x = 2\). when x = 0, y = 2, when x = 3, y = -1 From (iii), if \(y + 2x = 10\), when x = 0, y = 10, when x = 2, y = 6.

(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4. \(T_{n} = a + (n - 1)d\) (terms of an A.P) \(T_{25} = 5 + (25 - 1) \times 4\) = \(5 + 24 \times 4\) = \( 5 + 96 = 101\) (b) \(\frac{8}{9}, \frac{-4}{3}, 2, ...\) is a G.P with first term = \(\frac{8}{9}\) \(r = \frac{T_{2}}{T_{1}} = -\frac{4}{3} \div \frac{8}{9} = \frac{-4}{3} \times \frac{9}{8} = -\frac{3}{2}\) \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{5} = (\frac{8}{9})(-\frac{3}{2})^{5 - 1}\) = \(\frac{8}{9} \times \frac{81}{16} = \frac{9}{2}\) (c) \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{3} = ar^{2} = 48 ... (1)\) \(T_{6} = ar^{5} = 14\frac{2}{9} ... (2)\) \((2) \div (1) : r^{3} = \frac{128}{9} \div 48 = \frac{128}{9} \times \frac{1}{48}\) \(r^{3} = \frac{8}{27}\) \(r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\) From (1), \(ar^{2} = 48 \implies a \times (\frac{2}{3})^{2} = 48\) \(4a = 48 \times 9 \implies a = \frac{48 \times 9}{4} = 108\) \(T_{2} = 108 \times \frac{2}{3} = 72\) \(T_{4} = 48 \times \frac{2}{3} = 32\) \(\therefore\) The first four terms of the sequence are \(108, 72, 48, 32\).

(a) \(\theta\) 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ y -1.0 1.732 2.098 0 -2.098 -1.732 1.0 (b) (c)(i) Solution of \(3 \sin 2 \theta - \cos \theta = 0\) is at y = 0 = \(9.30^\circ, 80^\circ, 174^\circ\) (ii) Maximum value of y = 2.098 \(\approxeq\) 2.1 (to 1 decimal place)

Given that 2, 6, 12, 20...? the nth term = n\(^2\) + n check: n = 1, u1 = 2 n = 2, u2 = 4 + 2 = 6 n = 3, u3 = 9 + 3 = 12 ∴ n = 4, u4 = 16 + 4 = 20

\(T_n = 3. 2^{n - 2} \\ T_{1} = 3. 2^{1 - 2} = 3. 2^{-1} \\ T_1 = \frac{3}{2} \) \(T_2 = 3. 2^{2 - 2} \\ T_2 = 3. 2^0 = 3\) \(T_3 = 3. 2^{3 - 2} = 3. 2^1 \\ T_3 = 6\) The first 3 terms of the sequence are \(\frac{3}{2}\), 3 and 6.

If x = \(-\frac{1}{2}\) and 2; then \(x + \frac{1}{2} = 0\) and \(x - 2 = 0\) \(\implies (x + \frac{1}{2})(x - 2) = 0\) \(x^2 - 2x + \frac{1}{2}x - 1 = 0\) \(x^2 - \frac{3}{2}x - 1 = 0\) \(2x^2 - 3x - 2 = 0\)