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x\(^2\) > 0 (x+1)(x-1) > 0 x = 1 or -1 x < -1 or x > 1

3x - 7 \(\leq\) 0 and x + 5 > 0 3x \(\leq\) 7 and x > -5 x \(\leq\) \(\frac{7}{3}\) ∴ Range -5 < x \(\leq\) \(\frac{7}{3}\)

a = 5, d = 6, n = n Sn = n/2(2a + (n-1)d) = n/2(2(5) + (n-1)6) = n/2(10 + 6n-6) = n/2(6n+4) = 6n2/2 + 4n/2 = 32 + 2n = n(3n + 2)

a = 1, r =\(\frac{9}{10}\) S\(_n\) = \(\frac{a}{1 - r}\) S\(_n\) = \(\frac{1}{1-\frac{9}{10}}\) = 1\(\div \frac{1}{10}\) =1\(\times \frac{10}{1}\) = 10

m * n = n - (m+2) = -5 - (3+2) = -5-5 = -10

\(m \otimes n = m + n + mn\) Let the inverse of -5 be n. \(\therefore -5 \otimes n = 0\) \(-5 + n + (-5n) = 0\) \(n - 5n = 5 \implies -4n = 5\) \(n = -\frac{5}{4}\)

Q = \( \left[ \begin{array}{ccc} 9 & -2 \\ -7 & 4 \\ \end{array} \right]\) |Q| = 9x4 - (-2 x -7) = 36 - 14 = 22

\(P=\left[\begin{array}{cc}x+3 & x+2\\ x+1 & x-1\end{array}\right]\) evaluate x if |P| = -10 (x+3)(x-1) - {(x+1)(x+2)} = -10 x2 - x + 3x - 3 - {x2 + 2x + x + 2} = -10 x2 + 2x - 3 - {x2 + 3x + 2} = -10 -x - 5 = -10 -5 + 10 = x 5 = x ∴x = 5

y = x, Gradient (m1) = 1 y = √3x gradient (m2) = √3 \(tan\theta = \frac{m_2 - m_1}{1+m_2 m_1}\\ tan\theta = \frac{\sqrt{3}-1}{1+\sqrt{3}\times 1}\\ =\frac{1.73 -1 }{1+1.73}= \frac{0.73}{2.73}=0.268\) θ = Tan-1 (0.268) = 15o

Sum of interior angles of a polygon = each interior ∠ x n (n-2)180 = 150 x n 180n - 360 = 150n 180n-150n = 360 30n = 360 n = 360/30 n = 12 sides

Locus of a particle which moves in the first quadrant so that it is equidistant from the lines X = 0 and Y = 0 are the x and y axes as the lines bisecting the angle between x and y axes.

x\(^2\) + 3\(^3\) = 5\(^2\) x\(^2\) + 9 = 25 x\(^2\) = 25 – 9 x\(^2\) = 16 x = \sqrt 16 = 4 cm ∴ The locus is a circle of radius 4 cm with the center O

Gradient = \(\frac{y_2 - y_1}{x_2 - x_1}\) \(\frac{2}{3}\) = \(\frac{4 - p}{p - (-1)}\) \(\frac{2}{3}\) = \(\frac{4 - p}{p + 1}\) 3(4-p) = 2(p+1) 12 - 3p = 2p + 2 12 – 2 = 2p + 3p 10 = 5p p= \(\frac{10}{5}\) = 2

A(4,2) and B(1,r), AB = 3 units 3 = \sqrt (x2- x1)2 + (y2-y1)2 3 = \sqrt (4-1)2 + (2-r)2 3 = \sqrt 32 + (2-r)2 3 = \sqrt 9 + 4 – 4r + r2 3 = \sqrt r2 - 4r + 13 9 = r2 - 4r + 13 By squaring both sides r2 - 4r + 4 = 0 (r-2)(r-2) = 0 r = 2

y = 3 cos 4x Applying product rule: \(\frac{dy}{dx}\) = u\(\frac{dv}{dx}\) + v\(\frac{du}{dx}\) Where u = 3 and v = cos 4x \(\frac{du}{dx}\) of 3 = 0 and \(\frac{dv}{dx}\) of cos 4x = -4sin 4x \(\frac{dy}{dx}\) = 3 \(\times\) -4 sin 4x + cos 4x \(\times\) 0 \(\frac{dy}{dx}\) = -12 sin 4x