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\(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\) \(\frac{m}{3} - \frac{m}{4} = \frac{3}{4} - \frac{1}{2}\) \(\frac{m}{12} = \frac{1}{4}\) \(4m = 12 \implies m = 3\)

\(\frac{4^{2x}}{4^{3x}}\) = 2 42x - 3x = 2 4-x = 2 (22)-x = 21 Equating coefficients: -2x = 1 x = -\(\frac{1}{2}\)

\(y - 11\sqrt{y} + 24 = 0 \implies y + 24 = 11\sqrt{y}\) Squaring both sides, \(y^{2} + 48y + 576 = 121y\) \(y^{2} + 48y - 121y + 576 = 0 \implies y^{2} - 73y + 576 = 0\) \(y^{2} - 64y - 9y + 576 = 0\) \(y(y - 64) - 9(y - 64) = 0\) \((y - 9)(y - 64) = 0\) \(\therefore \text{y = 64 or y = 9}\)

To solve -11 \(\leq\) 4 - 3x \(\leq\) 28 -11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28 - 11 - 4\(\leq\) - 3x and -3x \(\leq\) 28 - 4 -15\(\leq\) -3x and -3x \(\leq\) 24 i.e. x \(\leq\) 5 and x \(\geq\) -8 ∴ -8 \(\leq\) x \(\leq\) 5

3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + ..... a = 3 r = \(\frac{2}{3}\) s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\) = \(\frac{3}{\frac{1}{3}}\) = 3 x 3 = 9

a = 2, d = 3 and n = 11 To find Sn/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)] = \(\frac{11}{2}\) [2(2) + (11 - 1) 3] = \(\frac{11}{2}\)n [4 + 10(3)] = \(\frac{11}{2}\)(34) = 11 x 17 = 187

p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\) PT = \(\begin{vmatrix}0 & 2 & 4 \\ 2 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\) /pT/ = \(\begin{vmatrix}0 & 2 & 4 \\ 3 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\) = 0[2 - 6] - 2[6 - 0] + 4[9 - 0] = 0 - 12 + 36 = 24

p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\) Q = \(\begin{vmatrix} 0 & 2 & 4 \\ 3 & 1 & 2\\ 0 & 3 & 2 \end{vmatrix}\) = pT pq = ppT

Equation (5, 7) parallel to the line 7x + 5y = 12 5Y = -7x + 12 y = \(\frac{-7x}{5}\) + \(\frac{12}{5}\) Gradient = \(\frac{-7}{5}\) ∴ Required equation = \(\frac{y - 7}{x - 5}\) = \(\frac{-7}{5}\) i.e. 5y - 35 = -7x + 35 5y + 7x = 70

Given the graph and the curve y = ax2 + bx + c the roots are x - 2 and 2 while its equation (x + 2)(x - 2) = y y = x2 - 4 i.e. y = x2 + 0x - 4 a = 1, b = 0 and c = -4

(a) \(2x + 5y = \frac{13}{2} \implies 4x + 10y = 13 ... (1)\) \(5x - 2y = 9 ... (2)\) \((2) \times 5 : 25x - 10y = 45 ... (2a)\) \((1) + (2a) : 29x = 58 \implies x = 2\) \(4(2) + 10y = 13 \implies 10y = 13 - 8 = 5\) \(y = \frac{5}{10} = 0.5\) (b) \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\) \(\log_{10} (\frac{2x + 1}{3x - 2}) = 1\) \(\frac{2x + 1}{3x - 2} = 10^{1} = 10\) \(2x + 1 = 10(3x - 2) \implies 2x + 1 = 30x - 20\) \(21 = 28x \implies x = \frac{3}{4}\)

(a) \(17x = 375^{2} - 356^{2}\) \(17x = (375 + 356)(375 - 356)\) \(17x = (731)(19)\) \(x = \frac{731 \times 19}{17} = 817\) (b) \(4^{x} = 2^{\frac{1}{2}} \times 8\) \(2^{2x} = 2^{\frac{1}{2}} \times 2^{3}\) \(2^{2x} = 2^{3\frac{1}{2}}\) \(\implies 2x = 3\frac{1}{2} \implies x = \frac{7}{4}\) (c) \(S_{n} = \frac{n}{2} (2a + (n - 1)d)\) (sum of terms of an A.P) \(S_{9} = \frac{9}{2} [2a + (9 - 1) d] = \frac{9}{2} [2a + 8d]\) \(72 = 9(a + 4d) \implies 8 = a + 4d ... (1)\) \(S_{9 + 4} = S_{13} = \frac{13}{2} [2a + (13 - 1)d] = \frac{13}{2} [2a + 12d]\) \(72 + 71 = 143 = 13(a + 6d) \implies 11 = a + 6d ... (2)\) \((2) - (1) : 2d = 3 \implies d = \frac{3}{2}\) \(a + 4(1\frac{1}{2}) = 8 \implies a + 6 = 8\) \(\implies a = 8 - 6 = 2\) \(\therefore \text{The A.P is } 2, 3\frac{1}{2}, 5, 6\frac{1}{2}, 8,...\)

(a) x -2.0 -1.5 -1.0 -0.5 0 0.5 1 1.5 2.0 y -15.5 -8.0 -2.5 1 2.5 2.0 -0.5 -5.0 -11.5 (b) see the diagram above (c) The maximum value of y = 2.5. (d) Obtaining the root of \(8x^{2} - 2x - 5 = 0\) Divide both sides by -2, i.e \(-4x^{2} + x + \frac{5}{2} = 0\) \(\therefore \text{Roots are } x = -0.7 ; x = 0.8.

3a + 10 = a\(^2\) a\(^2\) - 3a - 10 = 0 a\(^2\) - 5a + 2a - 10 = 0 a(a - 5) + 2(a - 5) = 0 (a - 5)(a + 2) = 0 a = 5 or a = -2.

x = \(-\frac{2}{3}\) and \(-\frac{1}{4}\) \(\implies (x + \frac{2}{3}) = 0; (x + \frac{1}{4}) = 0\) \((x + \frac{2}{3})(x + \frac{1}{4}) = 0\) \(x^{2} + \frac{1}{4}x + \frac{2}{3}x + \frac{1}{6} = 0\) \(x^2 + \frac{11}{12}x + \frac{1}{6} = 0\) \(12x^2 + 11x + 2 = 0\)