Question Library

f(x) = Lx3 + 2kx2 + 24 f(-2) = -8L + 8k = -24 4L - 4k = 12 f(1):L + 2k = -24 L - 4k = 3 3k = -27 k = -9 L = -6

O\(\hat{P}\)T = 90º (The radius is perpendicular to the tangent at the point of contact.) Sum of angles in \(\triangle\) PQT = 30 + ( 90 + x ) + (x + 2x) = 180 = 120 + 4x = 180 4x = 180 - 120 = 60º x = \(\frac{60}{4}\) = 15^\circ Therefore, P\(\hat{T}\)O = 2x = 2 x 15 = 30º

Let the bearing of H from P be represented by x^\circ. \(\tan \theta = \frac{30}{60} = 0.5\) \(\theta = \tan^{-1} (0.5) = 26.565^\circ\) \(x = 180^\circ + (90 - 26.565)\) \(x = 180 + 63.435 = 243.435^\circ\) = \(243^\circ 26'\)

First re-arrange the given data in the form 3, 5, 5, 5, 7, 8, 9. 3, 5, 5,|5| 7, 8, 9 median(M) = 5 mode(D) = 5 (M, D) = (5, 5)

Total sharing ratio is 9 X has 4, Y has 4 + 1 If 1 is N5000 Total profit = 5000 x 9 = N45,000

Sum of interior angles of polygon = (2n - 4)rt o = (12 - 4) x 90o = 8 x 90o = 720o each interior angle will have \(\frac{720^o}{6}\) = 120o

\(x \propto y^{3} z^{4}\) \(x = ky^{3} z^{4}\) If y is halved and z is doubled, we have \(x = k (\frac{1}{2} y)^{3} (2z)^{4}\) \(x = k (\frac{y^{3}}{8}) (16z^{4})\)\) \(x = 2k y^{3} z^{4}\) \(\therefore \text{x is increased in the ratio 2 : 1}\)

4x - 3 = 3x + y = x - y = 3.......(i) 3x + y = 2y + 5x - 12.........(ii) eqn(ii) + eqn(i) 3x = 15 x = 5 substitute for x in equation (i) 5 - y = 3 y = 2

From the diagram, sin 60o = \(\frac{PR}{8}\) PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\) = 4\(\sqrt{3}\) Cos 45o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\) PS Cos45o = 4\(\sqrt{3}\) PS = 4\(\sqrt{3}\) x 2 = 4\(\sqrt{6}\)

Given Cos z = L, z is an acute angle \(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z = \(\frac{\text{cos z}}{\text{sin z}}\) cosec z = \(\frac{1}{\text{sin z}}\) cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\) cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\) sec z = \(\frac{1}{\text{cos z}}\) tan z = \(\frac{\text{sin z}}{\text{cos z}}\) sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\) = \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\) the original eqn. becomes \(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = \(\frac{L - \frac{1}{\text{sin z}}}{1 + sin \frac{z}{L}}\) = \(\frac{L(L - 1)}{\text{sin z}(1 + \text{sin z})}\) = \(\frac{L(L - 1)}{\text{sin z} + 1 - cos^2 z}\) = sin z + 1 = 1 + \(\sqrt{1 - L^2}\) = \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)

0.0000152 x 0.00042 = A x 10\(^{B}\) 1 \(\leq\) A < 10, it means values of A include 1 - 9 0.0000152 = 1.52 x 10\(^{-5}\) 0.00042 = 4.2 x 10\(^{-4}\) 1.52 x 4.2 = 6.384 10\(^{-5}\times 10^{-4}\) = 10\(^{-5-4}\) = 10\(^{-9}\) = 6.38 x 10\(^{-9}\) A = 6.38, B = -9

\(\frac{av}{1 - v}\) = \(\sqrt[3]{\frac{2v + T}{a + 2T}}\) \(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) \(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) = \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)

Using the method of difference of two cubes, \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\) \((0.03)^{3} - (0.02)^{3} = (0.03 - 0.02)((0.03)^{2} + (0.03)(0.02) + (0.02)^{2}\) = \((0.01)(0.0009 + 0.0006 + 0.0004)\) = \(0.01 \times 0.0019\) = \(0.000019\)

y = kx2 + \(\frac{c}{\sqrt{x}}\) y = 2when x = 1 2 = k + \(\frac{c}{1}\) k + c = 2 y = 6 when x = 4 6 = 16k + \(\frac{c}{2}\) 12 = 32k + c k + c = 2 32k + c = 12 = 31k + 10 k = \(\frac{10}{31}\) c = 2 - \(\frac{10}{31}\) = \(\frac{62 - 10}{31}\) = \(\frac{52}{31}\) y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)

\(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\) = \(\frac{x - 7}{(x - 3)(x + 3)}\) x \(\frac{x(x - 3)}{(x - 7)(x + 7)}\) = \(\frac{x}{(x + 3)(x + 7)}\)