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\(2 \ast (12 \ast 27)\) \(12 \ast 27 = + \sqrt{12 \times 27}\) = \(+ \sqrt{324} = 18\) \(2 \ast 18 = + \sqrt{2 \times 18}\) = \(+ \sqrt{36} = 6\)

\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a G.P) \(a = 1; r = \frac{9}{10}\) \(S_{\infty} = \frac{1}{1 - \frac{9}{10}}\) = \(\frac{1}{\frac{1}{10}} = 10\)

-2(-1 - 3k) - 1(-2 -k) + 1(6 - 1) = 23 7k = 14 k = 2

\(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\); y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\). = \(\begin{pmatrix} 2 + 8 & 1 + 6 \\ 0 + 12 & 0 + 9 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\)

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

Angular difference = 45^\circ + 15^\circ = 60^\circ. \(D = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7000\) = \(\frac{22000}{3} km\)

Line PQ : x - 2y + 4 = 0 2y = x + 4 \(\implies y = \frac{x}{2} + 2 \) Slope = \(\frac{1}{2}\) Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\) Line QR: \(y = mx + b\) \(y = -2x + b\) Point of intersection: (1, -1) \(-1 = -2(1) + b \implies b = -1 + 2 = 1\) \(y = -2x + 1 \implies y + 2x - 1 = 0\) \(QR: 2x + y - 1 = 0\)

< PYU = < YPQ = 40^\circ(Alternative) = PQ\(\parallel\)UV < XPQ = < YPQ = 40^\circ, Since QX = QY therefore < XPY = 80^\circ

h = 30 tan 60o = 30\(\sqrt{3}\)

\(\csc x = \frac{1}{\sin x}\) Using the quotient rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\) = \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\) = \(\frac{- \cos x}{\sin^{2} x}\) = \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\) = \(- \cot x \csc x\)

At the point where the tangent is parallel to the x- axis, the slope of the curve = 0. Given: \(y = x^{2} - 4x + 3\) \(\frac{\mathrm d y}{\mathrm d x} = 2x - 4\) \(2x - 4 = 0 \implies 2x = 4 \) \(x = 2\) When x = 2, the tangent of the curve is parallel to the x- axis.

∫dy = ∫(4x - 3)dx, y = 2x2 - 3x + C when y = 5, x = 2, C = 3 y = 2x2 - 3x + 3

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x + 1\) \(y = \int_{1} ^{3} (3x^{2} - 2x + 1) \mathrm d x\) \(y = [x^{3} - x^{2} + x]_{1} ^{3}\) = \([3^{3} - 3^{2} + 3] - [1^{3} - 1^{2} + 1]\) = \(21 - 1 = 20\)

Number of 15- year old pupils = 30. Total number of students: 3 + 10 + 30 + 42 + 15 = 100. Angle = \(\frac{30}{100} \times 360^\circ\) = \(108^\circ\)

\(x\) 1 2 3 4 5 sum =15 \(x - \bar{x}\) -2 -1 0 1 2 \((x - \bar{x})^{2}\) 4 1 0 1 4 10 Mean = \(\frac{15}{5} = 3\) \(Variance = \frac{\sum (x - \bar{x})^{2}}{n}\) = \(\frac{10}{5}\) = \(2.0\)