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Tn = a + (n - 1)d; AP; a = 2d; Tn = 2d + (n - 1)d T5 = 2d + (5 - 1)d = 2d + 4d = 6d

\(3x + y = 8 ... (i)\) \(x^{2} + xy = 6 ... (ii)\) From (i), \(y = 8 - 3x\) From (ii), \(xy = 6 - x^{2} \implies y = \frac{6 - x^{2}}{x}\) Equating the two values of y, we have \(8 - 3x = \frac{6 - x^{2}}{x} \implies x(8 - 3x) = 6 - x^{2}\) \(8x - 3x^{2} = 6 - x^{2} \implies 6 - x^{2} - 8x + 3x^{2} = 0\) \(2x^{2} - 8x + 6 = 0\) \(x^{2} - 4x + 3 = 0\) \(x^{2} - 3x - x + 3 = 0 \implies x(x - 3) - 1(x - 3) = 0\) \((x - 1)(x - 3) = 0 \therefore \text{x = 1 or 3}\) \(y = 8 - 3x \) When x = 1, \(y = 8 - 3(1) = 5\) When x = 3, \(y = 8 - 3(3) = -1\) \(\therefore \text{y = -1 or 5}\)

\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1 = \(\frac{6x + 4x + 3x < 12}{12}\) i.e. 13 x < 12 = x < \(\frac{12}{13}\)

Given 27, 9, 3,......this is a G.P r = \(\frac{9}{27}\) = \(\frac{1}{3}\) T = arn - 1 = 27\(\frac{1}{3}\) n - 1

(x - 2) (x - 3) = 12 x\(^2\) - 3x - 2x + 6 = 12 x\(^2\) - 5x - 6 = 0 (x +1)(x - 6) = 0 x = -1 or 6

Given the first and last term of an A.P, the sum of the terms is given by \(S_{n} = \frac{n}{2} [a + l]\) where a = first term; l = last term and n = number of terms. \(\therefore S_{20} = \frac{20}{2} [7 + 117]\) = \(10 (124)\) = 1240

(a) \(2x^{2} + 7x - 11 = 0\) Using the quadratic formula, \(a = 2, b = 7, c = -11\) \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-7 \pm \sqrt{7^{2} - 4(2)(-11)}}{2(2)}\) \(x = \frac{-7 \pm \sqrt{49 + 88}}{4}\) \(x = \frac{-7 \pm 11.705}{4}\) \(x = \frac{-7 + 11.705}{4} ; \frac{-7 - 11.705}{4}\) \(x = \frac{4.705}{4} = 1.17625\) or \(x = \frac{-18.705}{4} = -4.67625\) \(x \approxeq 1.18\) or \(x \approxeq -4.68\) (b) \(\frac{2}{x} + \frac{1}{y} = 3\) \(\frac{1}{x} - \frac{5}{y} = 7\) Substituting using \(P = \frac{1}{x}; Q = \frac{1}{y}\), we have \(2P + Q = 3 ... (i)\) \(P - 5Q = 7 ... (ii)\) Multiply (ii) by 2, we have \(2P - 10Q = 14 ... (iii)\) (iii) - (i) : \(-10Q - Q = 14 - 3 \implies -11Q = 11\) \(Q = -1\) Substitute Q = -1 in (i), \(2P - 1 = 3 \implies 2P = 3 + 1 = 4\) \(\implies P = 2\) P = 2; Q = -1 But \(P = \frac{1}{x} \implies 2 = \frac{1}{x}\) \(\therefore x = \frac{1}{2}\) Also, \(Q = \frac{1}{y} \implies -1 = \frac{1}{y}\) \(y = -1\) \(\therefore x = \frac{1}{2} ; y = -1\)

\(y = 3 + 2x - x^{2}\) and \(y = 2x - 3\) Table of values for the equation for \(-3 \leq x \leq 4\) x -3 -2 -1 0 1 2 3 4 \(y = 3 + 2x - x^{2}\) -12 -5 0 3 4 3 0 5 \(y = 2x - 3\) -9 -7 -5 -3 -1 1 3 5 (i) \(6 - x^{2} = 0\) \(6 - 3 + 2x - x^{2} = 2x - 3\) \(3 + 2x - x^{2} = 2x - 3\) \(\therefore y = 2x - 3\) Read the point where the two equations intersect on the graph. x = -2.6 and x = 2.5. (ii) Maximum value of \(3 + 2x - x^{2}\) is at y = 4. (iii) Range for which \(3 + 2x - x^{2} \leq 1\) is represented by the shaded portion in the graph.

\(U_{n} = a + (n - 1)d\) \(U_{4} = 2 + (4 - 1) \times 0.5\) = \(2 + 1.5\) = 3.5

Where the curve cut x-axis is the root of the equation i.e. x = -2.0 and x = 1.5

\(S_{n} = \frac{a(r^n - 1)}{r - 1}\) a = 2; r = 6/2 = 3. \(S_{5} = \frac{2(3^5 - 1)}{3 - 1}\) = \(243 - 1 = 242\)

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

Vertical Line x = 3: This line indicates a constraint on x, specifically that x \le 3.(behind is shaded) The line y + 2x = -3 or y = - 2x - 3 This line has a negative slope of -2 and intersects the y-axis at -3. The region below this line corresponds to y \le − 2x − 3. Line y = x: .This line has a slope of 1 and passes through the origin. The region below this line corresponds to y \le x The lines are all thick = strict inequalities.

7y\(^2\) = 3y 7y\(^2\) - 3y = 0 y(7y - 3) = 0 y = 0 or y = \(\frac{3}{7}\)

2a\(^2\) - 3a - 27 = 0 2a\(^2\) - 9a + 6a - 27 = 0 a(2a - 9) + 3(2a - 9) = 0 (a + 3)(2a - 9) = 0 a = -3 or a = \(\frac{9}{2}\)