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The roots are -3 and 1 x = -3 and x = 1 y = (x + 3)(x - 1) = x2 - x + 3x - 3 y = x2 + x - 6 = 0

(a) \(2x - y = \frac{9}{2} .... (1)\) \(x + 4y = 0 ....... (2)\) From (2), x = - 4y. Put into (1), we have \(2(- 4y) - y = \frac{9}{2}\) \(-8y - y = \frac{9}{2} \implies - 9y = \frac{9}{2}\) \(y = \frac{\frac{9}{2}}{-9} = -\frac{1}{2}\) \(x = - 4y = -4(-\frac{1}{2})\) \(x = 2\) \((x, y) = (2, -\frac{1}{2})\) (b) In \(\Delta ACD, \stackrel\frown{ACD} = 52^\circ \) \(\stackrel\frown{ACD} = \stackrel\frown{DAT}\) \(\therefore \stackrel\frown{BCD} = 40^\circ + 52^\circ = 92^\circ\) Note \(\stackrel\frown{BCD}\) and \(\stackrel\frown{BAD}\) are supplementary = 180^\circ \(\therefore \stackrel\frown{BAD} = 180^\circ - 92^\circ = 88^\circ\)

(a)(i) \(4x < 2 + 3x ..... (1)\) \(4x - 3x < 2\) \(x < 2\) \(x - 8 < 3x ..... (2)\) \(-8 < 2x\) \(x > -4\) \(x < 2 ; x > -4\). (ii) (b) Let the digits of the number be x and y. \(x + y = 9 .... (1)\) \((10y + x) - (10x + y) = 45 \) \(10y - y + x - 10x = 45\) \(9y - 9x = 45 .... (2)\) From (1), y = 9 - x. Putting it into (2), we have \(9(9 - x) - 9x = 45\) \(81 - 9x - 9x = 45\) \(36 = 18x \implies x = 2\) \(\therefore y = 9 - x\) = \(9 - 2 = 7\) \(\therefore\) The bill = £27.

(a) From the pyramid, construct \(\Delta BFG\) by drawing a line from B to F. \(|BF|^{2} = 4^{2} + 3^{2}\) \(|BF| = 25 \implies |BF| = 5m\) In calculating the height, \(h^{2} = 5^{2} - (\frac{5}{2})^{2}\) \(h^{2} = \frac{75}{4}\) \(h = \sqrt{75}{4}\) = \(4.33 m\) Volume of a pyramid = \(\frac{1}{3} \times 4 \times 3 \times 4.33\) = \(17.32 m^{3}\) Volume of a cuboid = \(l \times b \times h\) = \(3 \times 4 \times 2\) = \(24 m^{3}\) Total volume of the shape = \(17.32 + 24\) = \(41.32 m^{3}\) (b) \(5 - (x - 2) = d\) \((x + 2) - 5 = d\) \(\therefore (x + 2) - 5 = 5 - (x - 2)\) \(x - 3 = 7 - x\) \(2x = 10\) \(x = 5\)

(a) \(y = 2\sin x - \cos 2x\) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ y -1.0 0.5 2.2 3.0 2.2 0.5 -1.0 (b)

x = -2, x = -1 and x = 3 ∴x+2 = 0, x+1 = 0 and x-3 = 0 Product of the factors (x+2)(x+1)(x-3) = 0 (x2 + 3x + 2)(x-3) x(x2 + 3x + 2) -3(x2 + 3x + 2) = 0 x3 + 3x2 + 2x - 3x2 - 9x - 6 = 0 x3 - 7x - 6 = 0

The weight \( W \) kg of a metal bar varies jointly as its length \( L \) meters and the square of its diameter \( d \) meters. This relationship is expressed as: \(W = k \cdot L \cdot d^2\) where \( k \) is the constant of proportionality. Given \( W = 140 \), \( d = 4\frac{2}{3} = \frac{14}{3} \), and \( L = 54 \), substitute to find \( k \): \(140 = k \cdot 54 \cdot \left( \frac{14}{3} \right)^2\) Calculate \( \left( \frac{14}{3} \right)^2 = \frac{196}{9} \): \(140 = k \cdot 54 \cdot \frac{196}{9}\) Simplify: \( 54 \div 9 = 6 \), so: \(140 = k \cdot 6 \cdot 196 = k \cdot 1176\) Thus, \(k = \frac{140}{1176} = \frac{5}{42}\) Now rearrange the original equation to solve for \( d \): \(d^2 = \frac{W}{k \cdot L}\) \(d = \sqrt{ \frac{W}{k \cdot L} } = \sqrt{ \frac{W}{\frac{5}{42} \cdot L} } = \sqrt{ \frac{42W}{5L} }\) Therefore, \( d = \sqrt{ \dfrac{42W}{5L} } \).

8, 12, 16, .....96 a = 8, d = 4, l = 96, n = 20 \(S_{20} = \frac{n}{2}(2a + (n-1)d\) \(S_{20} = \frac{20}{2}((2\times 8) + (20-1)\times 4)\) \(S_{20} = 10(16 + (19\times 4)) = 10 \times 92\) =920

Gradient of the line y + 2x = 5 = -2 Gradient of the line perpendicular to y + 2x + 5 = 1/2 equation of a line perpendicular to y + 2x = 5 at the point (4, 3) y - y1 = m(x - x1) y - 3 = 1/2(x - 4) 2y - 6 = x -4 2y - x = 2

x2 (sum of roots)x + (product of roots) = 0 Sum of roots \(2+-3\frac{1}{2} = -1\frac{1}{2} = -\frac{1}{2}\) Product of roots \(= 2 \times - 3\frac{1}{2}\) = -7 x\(^2\) - \(\frac{-3}{2}\)x + (-7) = 0 ⇒ 2x\(^2\) + 3x - 14 = 0

2\(^7\) = 8\(^{5-x}\) 2\(^7\) = 2\(^{3(5 - x)}\) 7 = 15 - 3x 7 - 15 = - 3x -8 = -3x x = \(\frac{8}{3}\)

2x - 3y = 22 ---- eqn I 3x + 2y = 7 ---- eqn II multiply eqn I by 2 4x - 6y = 44 ---- eqn III multiply eqn II by 3 9x + 6y = 21 ---- eqn IV Adding eqn III and IV => 13x = 65 => x = 5 Substituting 5 for x in eqn II 3x5 + 2y = 7 => 2y = -8 y = -4

\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 8y - 4 - 9y + 3 = 12 \\ \Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\)

\(a = 2\frac{1}{2}, nth = a + (n-1)d \Rightarrow 25 = 2\frac{1}{2} + (n-1)2\frac{1}{2}\\ 25 = \frac{5}{2}+(n-1)\frac{5}{2} \Rightarrow 22\frac{1}{2} = \frac{5n-5}{2}\Rightarrow \frac{45}{2} = \frac{5n-5}{2}\\ 45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10\)

Substituting for x in the equation \(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4\)