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a + 3d = 13 a + 9d = 31 6d = 18 = d = 3 a = 13 - 9 = 4 a + 20d = 4 + (20 x 3) = 64

\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a G.P) \(a = 1; r = \frac{9}{10}\) \(S_{\infty} = \frac{1}{1 - \frac{9}{10}}\) = \(\frac{1}{\frac{1}{10}} = 10\)

-2(-1 - 3k) - 1(-2 -k) + 1(6 - 1) = 23 7k = 14 k = 2

\(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\); y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\). = \(\begin{pmatrix} 2 + 8 & 1 + 6 \\ 0 + 12 & 0 + 9 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\)

Line PQ : x - 2y + 4 = 0 2y = x + 4 \(\implies y = \frac{x}{2} + 2 \) Slope = \(\frac{1}{2}\) Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\) Line QR: \(y = mx + b\) \(y = -2x + b\) Point of intersection: (1, -1) \(-1 = -2(1) + b \implies b = -1 + 2 = 1\) \(y = -2x + 1 \implies y + 2x - 1 = 0\) \(QR: 2x + y - 1 = 0\)

-1 \(\geq\) x \(\geq\) 1, 1 < x \(\geq\) 2 Combining solutions = x \(\leq\) 1; 1 \(\geq\) x \(\geq\) 2

(a) \(p = x + ym^{3}\) \(ym^{3} = p - x\) \(m^{3} = \frac{p - x}{y}\) \(m = \sqrt[3]{\frac{p - x}{y}}\) (b) \(x^{2} - 6x + 7 = 0\) \(\implies x^{2} - 6x = -7\) \(x^{2} - 6x + (-3)^{2} = -7 + (-3)^{2}\) \(x^{2} - 6x + 9 = 2\) \((x - 3)^{2} = 2 \implies x - 3 = \pm \sqrt{2}\) \(x = 3 \pm \sqrt{2} = 3 \pm 1.414\) \(x = 4.414 ; 1.586\) \(x \approxeq 4.4 ; 1.6\) (c) Let the first odd number be z. The next positive odd number = z + 2 \(\therefore z(z + 2) = 195\) \(z^{2} + 2z = 195\) \(z^{2} + 2z - 195 = 0 \implies z^{2} - 13z + 15z - 195 = 0\) \(z(z - 13) + 15(z - 13) = 0 \implies \text{z = 13 or -15}\) Since we are told the odd numbers are positive, then z = 13 is the suitable answer. Hence the second odd number = 13 + 2 = 15. The two odd numbers are 13 and 15.

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ \(y = 2\cos 2x - 1\) 1.0 0 -2 -3 -2 0 1.0 (b) (c) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ \(y = \frac{1}{180} (x - 360)\) -2.0 -1.83 -1.67 -1.5 -1.33 -1.17 -1.0 (d)(i) \(2 \cos 2x + \frac{1}{2} = 0\) \(2 \cos 2x + \frac{1}{2} - 1\frac{1}{2} = -1\frac{1}{2}\) \(2 \cos 2x - 1 = -1\frac{1}{2}\) The values of x at \(2 \cos 2x + \frac{1}{2} = 0\) are x = 51^\circ and 129^\circ. (ii) \(2 \cos 2x - \frac{x}{180} + 1 = 0\) \(2 \cos 2x - 1 = \frac{1}{180} (x - 360)\) The values of x is where the two graphs intersect; which are at x = 54^\circ and 132^\circ.

\(T_2\) = ar = 6 \(T_5\) = ar\(^{4}\) = 48 \(\frac{T_5}{T_2}\) = \(\frac{ar^{4}}{ar}\) = \(\frac{48}{6}\) = r\(^{3}\) = 8 => r = 2 \(S_n\) = \(\frac{a((r^n) - 1)}{r - 1}\) \(S_4\) = \(\frac{a((r^4) - 1)}{r - 1}\) \(S_4\) = \(\frac{3((2^4) - 1)}{2 - 1}\) = 3(16 -1) = 45

((1 x5x1) +( 3x6x2) + (3x4x0)) - ((3x5x2) + (1x6x0) + (3x4x1)) ( 5 + 36) - (30 +12) = 41 - 42 = -1

\(T_{n} = a + (n - 1) d\) (terms of an A.P) \(T_{1} = a\) \(T_{2} = a + d\) \(T_{10} = a + 9d\) \(a + a + d = 2a + d = 4 ... (i)\) \(a + 9d = 19 ... (ii)\) (ii) x 2: \(2a + 18d = 38 ... (iii)\) (iii) - (i) : \(17d = 34 \implies d = 2\) \(2a + 2 = 4 \implies 2a = 2\) \(a = 1\) \(T_{5} + T_{6}\) = \((a + 4d) + (a + 5d)\) = \(2a + 9d\) = \(2(1) + 9(2)\) = 20

The points P(-2, 3) and Q(2, 7) are on the line. Gradient of the line = \(\frac{7 - 3}{2 - (-2)}\) = \(\frac{4}{4}\) = 1 y = x + b To find the intercept, we have 7 = 2 + b or 3 = -2 + b. Solving for b, we get b = 5. \(\therefore\) The line is y = x + 5.

The points of intersection of the curve and the line are at x = -2 and x = 2. \(\therefore\) (x + 2) = 0; (x - 2) = 0. (x + 2)(x - 2) = 0 \(x^2 - 4 = 0\)

x = 3; x = \(\frac{2}{3}\). (x - 3)(x - \(\frac{2}{3}\)) = 0 \(x^2 - \frac{2x}{3} - 3x + 2 = 0\) \(x^2 - \frac{11x}{3} + 2 = 0\) \(3x^2 - 11x + 6 = 0\)

\(x^2 + 6x = 0\) \(x(x + 6) = 0\) \(x = 0\) or \(x = -6\)