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2 - x > x\(^2\) 2 - x - x\(^2\) > 0 (x+2)(x-1) < 0 Either (x + 2) > 0 and (x - 1) < 0. x + 2 > 0 \(\implies\) x > -2. -2 < x < 1

1/\alpha + 1/\beta = \beta+\alpha/\alpha\beta 3x\(^2\) + 5x - 2 = 0 x\(^2\) + 5x/3 - 2/3 = 0 \alpha\beta = -2/3 \beta+\alpha = -5/3 Thus; \beta+\alpha/\alpha\beta = -5/3 \div -2/3 = 5/2

The lateral view of the frustrum \(\tan \theta = \frac{6}{1.5} = 4\) \(\theta = \tan^{-1} 4 = 75.96^\circ\) Extending to the full height of the pyramid, we have The height of the pyramid which formed the frustrum = (6 + x)m, \(\tan 75.96 = \frac{6 + x}{2.5}\) \(6 + x = 2.5 \times 4 = 10.0\) \(\therefore \text{The height of the pyramid} = 10.0m\)

See Figure above. Since ST is parallel to UV => angle< U = 50^\circ(alternate to Angle UST = 180^\circ - 50^\circ = 130^\circ (angle on a straight line)

Since the inscribed triangle is equilateral, therefore the angles at all the points = 60^\circ Using the formula for inscribed circle, 2R = \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) where R = radius of the circle; a, b and c are the sides of the triangle. ⇒ 2R = \(\frac{\sqrt{3}}{\sin 60}\) 2R = \(\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}\) 2R = 2 R = 1cm

Grad of 3y = 4x - 1 y = \(\frac{4x}{3}\) - \(\frac{1}{3}\) Grad = \(\frac{4}{3}\) Grad of Ky = x + 3 y = \(\frac{\text{x}}{\text{k}}\) + \(\frac{3}{4}\) Grad = \(\frac{1}{\text{k}}\) Since the two lines are perpendicular, \(\frac{1}{\text{k}}\) = -\(\frac{3}{4}\) -3k = 4 k = -\(\frac{4}{3}\)

A perpendicular bisector always divides a line segment through its midpoint. The term bisect itself means dividing equally or uniformly.

2x + x = 180^\circ, => 3x = 180^\circ, and thus x = 60^\circ Each exterior angle = 60^\circ but size of ext. angle = 360^\circ/n Therefore 60^\circ = 360^\circ/n n = 360^\circ/60^\circ = 6 sides

x2 + y2 = (\sqrt2)2 x2 + y2 = 2 but y = x Thus; x2 + x2 = 2 2x2 = 2 x2 = + or - 1 But x2 + y2 = 2 12 + y2 = 2 1 + y2 = 2 y2 = 2 - 1 y2 = 1 y = + or - 1 Thus point (x,y) = (1,1) only.

\(\int^{\pi}_{0}\frac{cos^{2}\theta-1}{sin^{2}\theta}d\theta\) = \(\int^{\pi}_{0}\frac{-sin^{2}\theta}{sin^{2}\theta}\) = \(\int^{\pi}_{0}d\theta = -\pi\)

y = 2x cos2x - sin2x \(\frac{dy}{dx}\) = 2 cos2x +(-2x. 2sin2x) - 2cos2x = 2 cos2x - 2x sin2x - cos2x = -2x 2sin2x = -4 x (\(\frac{\pi}{4}\)) sin2 x (\(\frac{\pi}{4}\)) = -(\pi) x 1 = - \pi

∫30 \pi(y+1) dy = \pi[y\(^2 + y]^3_0\) = \pi(9/2 + 3) = 15\pi/2

\(V = \frac{2}{3} \pi r^{3}\) Given: \(\frac{\mathrm d V}{\mathrm d t} = 18\pi m^{3} s^{-1}\) \(\frac{\mathrm d V}{\mathrm d t} = \frac{\mathrm d V}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(18\pi = 2\pi r^{2} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d r}{\mathrm d t} = \frac{18\pi}{2\pi r^{2}} = \frac{9}{r^{2}}\) The rate of change of the radius when r = 6m, \(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\) = \(0.25 ms^{-1}\)

P (X or Y) = P(X) + P(Y), when they are independent as given. 0.7 = 0.4 + P(Y) P(Y) = 0.7 - 0.4 = 0.30

Mean = \(\frac{{0 + (x+2) + (3x+6) + (4x+8)}}{4}\) = 4 => 0 + (x+2) + (3x+6) + (4x+8) = 16 8x + 16 = 16 x = 0 Now prepare a table showing the deviation of each of 0, (x+2), (3x+6), and (4x+8). Adding the deviations will give 12. Thus M.D = \(\frac{12}{4}\) = 3