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\(y = 1 - 2x - 3x^2\) At maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\) \(\frac{\mathrm d y}{\mathrm d x} = -2 - 6x\) \(-2 - 6x = 0 \implies -2 = 6x\) \(x = -\frac{1}{3}\) At x = \(-\frac{1}{3}\), y = \(1 - 2(-\frac{1}{3}) - 3(-\frac{1}{3})^{2}\) = \(1 + \frac{2}{3} - \frac{1}{3}\) = \(\frac{4}{3}\)

\(T_{n} = a + (n - 1) d\) (nth term of an AP). \(T_{9} = 5T_{5}\) \(a + 8d = 5(a + 4d) \implies a + 8d = 5a + 20d\) \(5a - a + 20d - 8d = 0 \implies 4a + 12d = 0\) \(a + 3d = 0\)

-2 is the solution implies x = -2. 2x + 1 - 3c = 2c + 3x - 7 2(-2) + 1 - 3c = 2c + 3(-2) - 7 -4 + 1 - 3c = 2c - 6 - 7 -3 + 13 = 2c + 3c \(\implies\) c = 2.

The line equidistant from x = 1 and x = 7 is \(x = \frac{1 + 7}{2} \implies x = 4\)

\(T_{1} = 4; T_2 = 10; T_3 = 16\) \(T_{2} - T_1 = T_3 - T_1 = 6\) \(T_n = a + (n - 1) d\) = \(4 + (n - 1) \times 6\) = \(4 + 6n - 6\) = \(6n - 2\) = 2(3n - 1)

From the graph, the zeros of the equation exist at x = -2 and x = 3 \(\therefore\) (x + 2) = 0 and (x - 3) = 0 \(\implies (x + 2)(x - 3) = 0\) \(x^2 - 3x + 2x - 6 = 0\) \(x^2 - x - 6 = 0\) is the equation represented on the graph.

The equation \(\frac{3x - 2}{4x^2 + 9x - 9}\) is undefined when the denominator = 0. \(4x^2 + 9x - 9 = 0\) \(4x^2 + 12x - 3x - 9 = 0\) \(4x(x + 3) - 3(x + 3) = 0\) \((4x - 3)(x + 3) = 0\) x = \(\frac{3}{4}\) or x = -3.

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

\(\frac{1}{3}x\) + \(\frac{2}{5}\) = \(\frac{3}{4}x\) - \(\frac{1}{6}\) collect like terms \(\frac{2}{5}\) + \(\frac{1}{6}\) = \(\frac{3}{4}x\) - \(\frac{1}{3}x\) \(\frac{17}{30}\) = \(\frac{5x}{12}\) cross multiply 150x = 204 x = \(\frac{204}{150}\) = \(\frac{34}{25}\)

(a) (i) First discount = 20% of N25,000 = \(\frac{20}{100} \times N25,000 = N5,000\) Further discount for prompt payment = \(2\frac{1}{2}%\) of N25,000. = \(\frac{5}{200} \times N25,000 = N625\) \(\therefore \text{Total amount saved by distributor} = N(5,000 + 625)\) = \(N5625\) (ii) Total %age discount for prompt payment = \((20 + 2\frac{1}{2})% = 22\frac{1}{2}%\) Therefore, the distributor paid \((100% - 22\frac{1}{2}%) \times Nx\) = \(77\frac{1}{2}% \times Nx\) \(\implies 77\frac{1}{2}% \times x = 11,700\) \(x = \frac{11,700 \times 200}{155}\) = \(N15,096.77\) (b) \(6y^{2} - 149y - 102 \) = \(6y^{2} - 153y + 4y - 102\) = \(3y(2y - 51) + 2(2y - 51)\) = \((3y + 2)(2y - 51)\) \(\therefore 6y^{2} - 149y - 102 = 0 \implies (3y + 2)(2y - 51) = 0\) \(3y + 2 = 0 \implies 3y = -2 \) \(y = -\frac{2}{3}\) \(2y - 51 = 0 \implies 2y = 51\) \(y = 25\frac{1}{2}\) \(\therefore y = -\frac{2}{3} ; y = 25\frac{1}{2}\)

(a) \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\) = \(\log_{10} (25^{\frac{1}{2}}) - \log_{10} (3^{2}) + \log_{10} 18\) = \(\log_{10} (\frac{5 \times 18}{9})\) = \(\log_{10} 10\) = 1 (b) \(123_{y} = (1 \times y^{2}) + (2 \times y^{1}) + (3 \times y^{0})\) \(y^{2} + 2y + 3 = 83\) \(y^{2} + 2y + 3 - 83 = 0 \implies y^{2} + 2y - 80 = 0\) Equation : \(y^{2} + 2y - 80 = 0\) \(y^{2} - 8y + 10y - 80 = 0\) \(y(y - 8) + 10(y - 8) = 0\) \((y + 10)(y - 8) = 0 \implies y = -10 ; 8\) y = 8 since y cannot be negative. (c) \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\) \(\implies 9^{2x - 3} = 3^{x + 3}\) \((3^{2})^{2x - 3} = 3^{x + 3}\) \(3^{4x - 6} = 3^{x + 3}\) \(\implies 4x - 6 = x + 3\) \(4x - x = 3 + 6 \implies 3x = 9\) \(x = 3\)

(a) \(a = 3; d = 4 ; n = 28\) \(S_{n} = \frac{n}{2}(2a + (n - 1)d)\) \(S_{28} = \frac{28}{2}(2(3) + (28 - 1) 4)\) = \(14(6 + 108)\) = \(14(114)\) = \(1596\) (b) \(x = \frac{2m}{1 - m^{2}} ; y = \frac{2m}{1 + m}\) \(2x - y = 2(\frac{2m}{1 - m^{2}}) - (\frac{2m}{1 + m})\) = \(\frac{4m}{1 - m^{2}} - \frac{2m}{1 + m}\) = \(\frac{4m - 2m(1 - m)}{1 - m^{2}}\) = \(\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\) = \(\frac{2m^{2} + 2m}{1 - m^{2}}\) = \(\frac{2m(m + 1)}{(1 + m)(1 - m)}\) = \(\frac{2m}{1 - m}\) (c) Sum of interior angles = (2n - 4)90^\circ When n = 5 (pentagon), = \((2(5) - 4) \times 90^\circ\) = \(540^\circ\) \(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\) = \(11x - 10 = 540^\circ\) = \(11x = 550^\circ\) = \(x = 50^\circ\)

(a) The roots of the equation are -1.5 or 2. (b) C = 6 (intercept on the positive vertical axis) \(ax^{2} + bx + c = 0\) When x = 2, c = 6 \(4a + 2b + 6 = 0 \implies 4a + 2b = -6 ....(1)\) When x = -1.5, c = 6 \(a(-1.5)^{2} + b(-1.5) = -6\) \(\frac{9a}{4} - \frac{3b}{2} = -6 \implies 9a - 6b = -24 ... (2)\) Multiply (1) by 3 : \(12a + 6b = -18 ... (1a)\) (1a) + (2) : \(12a + 9a = -18 - 24 \implies 21a = -42\) \(a = -2\) \(4a + 2b = -6 \implies 4(-2) + 2b = -6\) \(2b = -6 + 8 = 2 \implies b = 1\) (a, b, c) = (-2, 1, 6) Hence the relation illustrated in the graph is \(y = -2x^{2} + x + 6\). (c) When \(x = 0.2, y_{max} = 6.2\) (d) When \(y = 2, x = \text{ -1.2 or 1.7}\)

m2 + n2 = 29 .......(1) m + n = 7 ............(2) From (2), m = 7 - n but m2 + n2 = 29, substituting; (7-n)2 + n2 = 29 49 - 14n + n2 + n2 = 29 => 2n2 -14n + 20 = 0 Thus n2 -7n + 10 = 0 Factorizing; (n-5)(n-2) = 0 n - 5 = 0, => n = 5 n - 2 = 0, => n = 2. When n = 5, m + n = 7, => m = 2, When n = 2, m + n = 7, => m = 5. Thus (m,n) = (5,2) and (2,5)

1st statement: U6 = 1/2(U12) a + (n -1)d = 1/2[a + (n-1)d] a + 5d = a + 11d 2(a + 5d) = a + 11d 2a + 10d = a + 11d Solving, => a = d Hence the first term is equal to the common difference