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(a) By summing 2y + x + 6y - 2x + 1 + 4y = 28 and simplified to arrive at 12y - x = 27 ...............(1). Similarly, IPRI = IRQI, which implied that 4y = 6y - 2x + 1. This simplified to 2y - 2x - 1 ............... (2) Solving equations (1) and (2) simultaneously gave 2y - 2(12y - 27) = -1. so that y = 2(\(\frac{1}{2}\)). Then, x = 12(\(\frac{5}{2}\)) - 27 = 30 - 27 = 3 (b) The lengths of the sides of the triangle were also obtained as |PO| = 2y + x = 2(\(\frac{5}{2}\)) + 3 = 8 cm. |QR| = 4y = 4(\(\frac{5}{2}\)) = 10 cm and IPRI = 6y - 2x + 1 = 6(\(\frac{5}{2}\)) - 2(3) + 1 = 10 cm

(a) x < 4 or -1 \(\geq\) x. The range of values of x is -1 \(\geq\) x < 4. (b) First, < PQR + 56^\circ= 180^\circ so that < PQR = 180^\circ- 56^\circ= 124^\circ From triangle PQR, < PQR + < QRP+ < QPR = 180^\circ. But IRQI = IQPI so that < QRP = < QPR. 124^\circ + 2 < QRP = 180^\circ and < QRP =28^\circ. But < PRS = 90^\circ (angle on a semi-circle), therefore, < QRS = 28^\circ+ 90^\circ = 118^\circ.

(a) They were expected to obtain 20,000 = \(\frac{x^2}{8} + 5x\) and thereafter multiply through by 8 and simplify to get \(x ^2\) + 40x - 16,000. Using the quadratic formula x = \(\frac {-b \pm \sqrt{b^2 - 4ac} }{2a}\) where a = 1, b = 40 and c = - 160,000 x = \(\frac{-40 \pm \sqrt{(40)^2 - 4(1) (-160000)}} {2(1)}\) = \(\frac{-40 \pm \sqrt{(40)^2 - 4(1) (-160000)}} {2}\) = \(\frac{-40 \pm \sqrt{1600 + 640000}}{2}\) = 38.4997 or -420.4997 The number of books required to make the profit approximately 380,500, correct to the nearest whole number. (b) The required diagram is From triangle PQR, tan x = \(\frac{y}{2}\)W y = 2w tan x..........(1) Also, from \(\bigtriangleup\) PTQ, tan 28\(^o\) = \(\frac{y}{w}\) y = w tan 28\(^o\).........(2) Equating (1) and (2) yielded 2w tan X = w tan 28\(^o\) so that tanX = \(\frac{\text{w} tan 28^o}{2w}\) = \(\frac{0.5317}{2}\) which implies that tan X = 0.2659 X = tan\(^{-1}\)(0.2659) = 15.42\(^o\). The angle of elevation of R from P = 15\(^o\), correct to the nearest degree.

(a). They integrated the given function correctly to obtain (\(\frac{x^4}{2} - 2x^2 + 6x\)). Substituting the limits 2 and 1 for x and simplifying yielded 12 - \(\frac{9}{2}\) = 7\(\frac{1}{2}\) (b). Find the determinant of which was P(\(^{-1}\))\(^{-1}\) = |P=\(^{-1}\)| (-3 x -1) - (4 x 1) = -1. Using the concept of P(\(^{-1}\))\(^{-1}\) = P, then, P = -1 \(\begin{pmatrix} -3 & -1 \\ 4 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\)

2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\)..... There are 4 terms in the series Therefore the next number will be the 5th term Tn = ar\(^{n − 1}\) (formular for geometric series) a = first term = 2 r = common rate = \(\frac{\text{next term}}{\text{previous term}}\) = \(\frac{1}{2}\) n = number of terms T5 = 5th term = ? T5 = ar\(^{5 - 1}\) = ar\(^4\) = 2 × (ar\(^{n − 1}\))4 = 2 × \(\frac{1}{16}\) = \(\frac{1}{8}\)

Sum to infinity ∑ = arn − 1 = \(\frac{a}{1}\) − r a = \(\frac{1}{4}\) r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\) r = \(\frac{1}{s}\) × \(\frac{4}{1}\) = \(\frac{1}{2}\) S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\) = \(\frac{1}{4}\) ÷ \(\frac{1}{2}\) = \(\frac{1}{4}\) × \(\frac{2}{1}\) = \(\frac{1}{2}\)

Y ∝ \(\frac{1}{2}\) Y = 6, X = 7 Y = \(\frac{k}{x}\) where k is constant 6 = \(\frac{k}{7}\) k = 42

The locus of a point p(x, y) such that pv = pw where v = (1, 1) and w = (3, 5). This means that the point p moves so that its distance from v and w are equidistance \(\sqrt{(x − x_1)^2 + (y − y_1)^2}\) = \(\sqrt{(x − x_2)^2 + (y − y_2)^2}\) \(\sqrt{(x -1)^2 + (y - 1)^2}\) = \(\sqrt{(x - 3)^2 + (y - 5)^2}\) square both sides (x - 1)2 + (y - 1)2 = (x - 3)2 + (y - 5)2 x2 - 2x + 1 + y2 - 2y + 1 = x2 - 6x + 9 + y2 - 10y + 25 x2 + y2 -2x -2y + 2 = x2 + y2 - 6x - 10y + 34 Collecting like terms x2 - x2 + y2 - y2 - 2x + 6x -2y + 10y = 34 - 2 4x + 8y = 32 Divide through by 4 x + 2y = 8

1/\alpha + 1/\beta = \beta+\alpha/\alpha\beta 3x2 + 5x - 2 = 0 x2 + 5x/3 - 2/3 = 0 \alpha\beta = -2/3 \beta+\alpha = -5/3 Thus; \beta+\alpha/\alpha\beta = -2/3 -2/3 = -5/2

\(\frac{1}{5x} + \frac{1}{x}\)= 3 \(\frac{1 + 5}{5x}\) = 3 6 = 15x x = \(\frac{6}{15}\) = \(\frac{2}{5}\)

a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\) = 7 - 10 = \(\frac{-3}{15}\) d = - \(\frac{-1}{5}\) T6 = a + 5d = \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\) = \(\frac{2}{3}\) - 1 = \(\frac{2 - 3}{3}\) = \(\frac{-1}{3}\)

(x + \(\frac{1}{2}\)) (n - \(\frac{2}{3}\)) \(x^2 - \frac{2}{3^x} + \frac{x}{2} - \frac{1}{3}\) \(6x^2 - 4n + 3n - 2 = 0\) \(6x^2 - x - 2 = 0\)

(a) \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\) \(\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\) \(30x - 50 - 18x + 27 = 3 \times 15 = 45\) \(12x = 45 + 23 = 68\) \(x = \frac{68}{12} = 5\frac{2}{3}\) (b) \(n + r = 80^\circ .... (1)\) \(180^\circ - m + r = 88^\circ .... (2)\) From (1), r = 80^\circ - n. Putting into (2), we have \(180 - m + 80 - n = 88 \implies - m - n = 88 - 260 = - 172^\circ\) \(-(m + n) = - 172 \implies m + n = 172^\circ\).

(a) \(T_{n} = a + (n - 1)d\) (For an AP series) \(T_{6} = 37 = a + (6 - 1) d = a + 5d\) \(a + 5d = 37 .... (1)\) \(S_{n} = \frac{n}{2}(2a + (n - 1) d)\) or \(S_{n} = \frac{n}{2}(a + l)\) where a = first term; and l = last term. Since, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case. \(S_{6} = 147 = \frac{6}{2} (a + 37)\) \(147 = 3(a + 37) \implies 147 = 3a + 111 \) \(3a = 147 - 111 = 36 \implies a = 12\) (b) From equation (1) above, \(a + 5d = 37\) \(12 + 5d = 37\) \(5d = 37 - 12 = 25\) \(d = 5\) \(S_{15} = \frac{15}{2} (2(12) + (15 - 1)(5))\) \(\frac{15}{2} (24 + 70) = \frac{15}{2} (94)\) = \(15 \times 47\) = \(705\).

(a) x -3 -2 -1 0 1 2 3 4 5 6 y 30 13 0 -9 -14 -15 -12 -5 6 21 (b) (c) (i) From the graph, the roots of the equation 2x\(^2\) - 7x = 26 can be obtained by drawing a line through y = 17 to meet the curve 2x\(^2\) - 7x - 9. The roots of the equation 2x\(^2\) - 7x = 26 : -2.2 and 5.8. (ii) The coordinates of the minimum point of y : 1.8, -15. (iii) The range of values for which 2x\(^{2}\) - 7x < 9 : - 0.9 < x < 4.5.