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\(\frac{a^{3x} - 26a^{2x} + 156a^{x} - 216}{a^{2x} - 24a^{x} + 108}\) Let \(a^{x} = z\) \(\therefore = \frac{z^{3} - 26z^{2} + 156z - 216}{z^{2} - 24y + 108} ... (i)\) Dividing (i) above, we get \(z - 2\) = \(a^{x} - 2\)

\(y = px^{2} + q ... (i)\) \(y = 2x^{2} - 1 ... (ii)\) At x = 2, (i): \(y = p(2^{2}) + q = 4p + q\) (ii): \(y = 2(2^{2}) - 1 = 7\) \(\therefore \text{The coordinates of the point of intersection = (2, 7)}\) (i): \(7 = 4p + q \implies p = \frac{7 - q}{4}\)

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)

By definition a*b = a + b + 1. Let the inverse of the element 2 be x, Therefore 2*x = -1 i.e. 2 + x + 1 = -1 3 + x = -1 x = -1 - 3 x = -4

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

Given: 4x2 - 9y2 + 20x + 25 Collect like terms: 4x2 + 20x + 25 - 9y2 (2x + 5)(2x + 5) - 9y2 (2x + 5)2 - (3y)2 (2x - 3y +5)(2x + 3y + 5)

(r/L) = (\theta/360^\circ) Given \theta = 300, and L = 7.2cm, => r = (300 x 7.2)/360 r = 6cm

Hint: Make a sketch of an isosceles triangle with two of its sides and angles angles. : PQ (r) = PR (q) = 8cm : R^\circ = Q^\circ = 30^\circ Sum of angles in a triangle = 180^\circ P^\circ + Q^\circ + R^\circ = 180^\circ P^\circ + 30^\circ + 30^\circ = 180^\circ P^\circ = 180^\circ - 60^\circ p^\circ = 120^\circ. : PQ = r, PR = q, QR = p Using sine rule: \(\frac{p}{sinP}\) = \(\frac{q}{sinq}\) \(\frac{p}{sin120^\circ}\) = \(\frac{8}{sin30^\circ}\) cross multiply p = \(\frac{8 X sin120^\circ}{sin30^\circ}\) p = \(\frac{8 X \sqrt3/2 }{1/2}\) p = 8\sqrt3

Cos 30 = \(\frac{5}{\text{x}}\) x cos 30 = 5, => x = 5\(\sqrt{3}\) Coordinates of P = - 5\(\sqrt{3}\), 0 Coordinates of Q = 0, 5 Gradient of PQ = (y\(_2\) - y\(_1\)) (X\(_2\) - X\(_1\)) = \(\frac{(5 - 0)}{(0 - 5\sqrt{3}}\)) = \(\frac{5}{5\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\) Equation of PQ = y - y\(_1\) = m (x -x\(_1\)) y - 0 = \(\frac{1}{\sqrt{3}}\) (x -( -5\(\sqrt{3}\))) Thus: \(\sqrt{3}\)y = x + 5\(\sqrt{3}\)

The line joining (P, 4) and (6, -2). Gradient: \(\frac{-2 - 4}{6 - P} = \frac{-6}{6 - P}\) The line joining (2, P) and (-1, 3) Gradient: \(\frac{3 - P}{-1 - 2} = \frac{3 - P}{-3}\) For perpendicular lines, the product of their gradient = -1. \((\frac{-6}{6 - P})(\frac{3 - P}{-3}) = -1\) \(\frac{6 - 2P}{6 - P} = -1 \implies 6 - 2P = P - 6\) \(6 + 6 = P + 2P \implies P = \frac{12}{3} = 4\)

Let the ext. angle = x Thus int. angle = 2x But sum of int + ext = 180 (angle of a straight line). 2x + x = 180 3x = 180 x = 180/3 = 60 Each ext angle = 360/n => 60 = 360/n n = 360/60 = 6

PQ\(^2\) = (x\(_2\) - x\(_1\))\(^2\) + (y\(_2\) - y\(_1\))\(^2\) = 12\(^2\) + 5\(^2\) = 144 + 25 = 169 PQ = \(\sqrt{169}\) = 13 But PQ = diameter = 2r, r = \(\frac{\text{PQ}}{2}\) = 6.5 units

From the diagram above,

Source: Myschool Past Questions Archive (Record #2122)