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n = 3, U\(_3\) = 4x - 2y, U\(_n\) = a + (n-1)d, 4x - 2y = a + 2d... (1) U\(_9\) = 10x - 8y, 10x - 8y = a + 8d... (2) Solving (1) and (2), subtract eqn i from ii a-a + 8d - 2d = 10x -4x - 8y- (-2y) 6d = 6x - 8y+ 2y. 6d= 6x - 6y 6d = 6( x-y) d = x - y

If P\(^{-1}\) is the inverse of P and O is the identity, Then P\(^{-1}\) * P = P * P\(^{-1}\) = 0 i.e. P\(^{-1}\) + P - P\(^{-1}\).P = 0 P\(^{-1}\) - P\(^{-1}\).P = -P P\(^{-1}\)(1 - P) = -P P\(^{-1}\) =\(\frac{-P}{(P-1)}\) = \(\frac{P}{(P-1)}\)

S\(_{\infty}\) = \(\frac{\text{a}}{(1-r)}\), where a = \(\frac{1}{2}\), r = -\(\frac{1}{2}\). S\(_{\infty}\) = \(\frac{\frac{1}{2}}{1 - \frac{-1}{2}}\) = \(\frac{\frac{1}{2}}{\frac{3}{2}}\) = \(\frac{1}{2}\) x \(\frac{2}{3}\) \(\frac{1}{3}\) - 1 = -\(\frac{2}{3}\)

This is a polynomial of the 3rd order, thus x should have three answers. Use the factors given to get values of x as 1, -1 and -2. Form three equations, and carry out elimination and subsequent substitution to get a = -1/2, b = 1, and c = 1/2 Equate each factor to zero to get the values of x x-1 = 0, x = 1 x +1 = 0, x = -1 x - 2 = 0, x = 2 substitute these value of x in the given equation \(ax^3 + bx^2\) + cx - 1 when x = 1 \(ax^3 + bx^2\) + cx - 1 = a + b + c - 1= 0 ------------------------(1) when x = -1 \(ax^3 + bx^2\) + cx - 1 = -a + b - c - 1 = 0-------------------------(2) when x = 2 \(ax^3 + bx^2\) + cx - 1 = 8a + 4b + 2c - 1 = 0 ----------------------(3) add equations 1 and 2 a + b + c - 1= 0 + -a + b - c - 1 = 0 b = 1 put b = 1 into equation (1) a + b + c - 1= 0 = a + 1 + c -1 = 0 a = - c put a = - c into equation(3) 8a + 4b + 2c - 1 = 0, -8c + 4 + 2c - 1 = 0 ( b = 1 remember) -6c = -3 c = \(\frac{-3}{-6}\) = \(\frac{1}{2}\) put c = \(\frac{1}{2}\) and b = 1 into equation (1) a + b + c - 1= 0 a + 1 + \(\frac{1}{2}\) - 1 = 0 a + \(\frac{1}{2}\) = 0 then a = \(\frac{-1}{2}\)

y = 10x - x\(^2\) dy/dx = 10 - 2x As dy/dx = 0, 10 - 2x = 0 2x = 10 x = 5

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.

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