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3x2 + 6x - 2 = 0 Using almighty formula i.e. x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2a}\) a = 3, b = 6, c = -2 x = \(\frac{-6 \pm \sqrt{6^2 - 4(3)(-2)}}{2(3)}\) x = \(\frac{-6 \pm \sqrt{36 + 24}}{6}\) x = \(\frac{-6 \pm \sqrt{60}}{6}\) x = \(\frac{-6 \pm \sqrt{4 \times 15}}{6}\) x = \(-1 \pm \frac{\sqrt{15}}{3}\)

\(\frac{1}{5x + 5}\) + \(\frac{1}{7x+ 7}\) = \(\frac{1}{5(x + 1)}\) + \(\frac{1}{7(x + 1)}\) = \(\frac{7 + 5}{35(x + 1)}\) = \(\frac{12}{35(x + 1)}\)

(4a + 3)2 - (3a - 2)2 = a2 - b2 = (a + b)(a - b) = [(4a + 3) + (3a - 2)][(4a + 3) + (3a - 2)] = [(4a + 3 + 3a - 2)][(4a + 3 - 3a + 2)] = (7a + 1)(a + 5) ∴ (a + 5)(7a + 1)

\(5^{(x + 2y)} = 5\) ∴ x + 2y = 1.....(i) \(4^{(x + 3y)} = 16 = 4^2\) x + 3y = 2 .....(ii) x + 2y = 1.....(i) x + 3y = 2......(ii) y = 1 Substitute y = 1 into equation (i) \(x + 2y = 1 \implies x + 2(1) = 1\) \(x + 2 = 1 \implies x = -1\) \(\therefore 3^{(x + y)} = 3^{(-1 + 1)}\) \(3^{0} = 1\)

\(\frac{1}{x - 2}\) + \(\frac{1}{x + 2}\) + \(\frac{2x}{x^2 - 4}\) = \(\frac{(x + 2) + (x - 2) + 2x}{(x + 2)(x - 2)}\) = \(\frac{4x}{x^2 - 4}\)

S = \(\sqrt{\frac{6}{v} - \frac{w}{2}}\) S = \(\frac{12 - vw}{2v}\) 2vs\(^2\) = 12 - vw v(2s\(^2\) + w) = 12 v = \(\frac{12}{2s^2 + w}\)

16x - 5 x 4x + 4 = 0 (4x)2 - 5(4x) + 4 = 0 let 4x = y y2 - 5y + 4 = 0 (y - 4)(y - 1) = 0 y = 4 or 1 4x = 4 x = 1 4x = 1 i.e. 4x = 4o, x = 0 ∴ x = 1 or 0

Given that \(\frac{a}{b} = \frac{c}{d} = k\) we can express \(a\) and \(c\) as: \(a = kb, \quad c = kd\) We want to find the value of \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\) Substituting \(a\) and \(c\) into the expression gives: \(3a^2 - ac + c^2 = 3k^2b^2 - k^2bd + k^2d^2 = k^2(3b^2 - bd + d^2)\) Thus, the expression simplifies to: \(\frac{k^2(3b^2 - bd + d^2)}{3b^2 - bd + d^2} = k^2\) Therefore, the value is: \(k^2\)

line = y = 2x + 1, curve = y = \(2x^2\) + 5x - 1 2x + 1 = \(2x^2\) + 5x - 1 \(2x^2\) + 3x - 2 = 0 factorizing (2x - 1)( x + 2) = 0 x = 1/2 or -2 when x = 1/2 y = 2 * 1/2 + 1 = 2 ( using the equation of a straight line ) when x = -2, y = -3 therefore, the points are ( 1/2, 2) and ( -2, -3)

cos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\) Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\) 1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\) = \(\frac{a^2 + b^2 - a^2}{a^2}\) = \(\frac{b^2}{a^2}\)

No expanded working has been attached for this question yet. The verified answer is Option A.

QY = 452 + 242 = 2025 + 576 = 2601 QY = \(\sqrt{2601}\) = 51

Radius of the circle r = 6cm, Length of the arc = 8cm Area of sector = \(\frac{\theta}{360}\) x \(\pi\)r2........(i) Length of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r........(ii) from eqn. (ii) \(\theta\) = \(\frac{240}{\pi}\), subt. for \(\theta\) in eqn (i) Area x \(\frac{240}{1}\) x \(\frac{1}{360}\) x \(\frac{\pi 6}{1}\) = 24cm\(^2\)

Total surface area of a solid cone r = 2\(\sqrt{3}\) = \(\pi r^2\) + \(\pi\)rH H = 4\(\sqrt{3}\), \(\pi\)r(r + H) ∴ Area = \(\pi\)2\(\sqrt{3}\) [2\(\sqrt{3}\) + 4\(\sqrt{3}\)] = \(\pi\)2\(\sqrt{3}\)(6\(\sqrt{3}\)) = 12\(\pi\) x 3 = 36\(\pi \)cm2

No expanded working has been attached for this question yet. The verified answer is Option D.