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M = \(\begin{vmatrix} -2 & 0 & 4 \\ 0 & -1 & 6 \\ 5 & 6 & 3 \end{vmatrix}\) M\(^{T}\) = \(\begin{vmatrix} -2 & 0 & 5 \\ 0 & -1 & 6\\ 4 & 6 & 3 \end{vmatrix}\) 2M = \(\begin{vmatrix} -4 & 0 & 8\\ 0 & -2 & 12\\ 10 & 12 & 6\end{vmatrix}\) M\(^T\) + 2M = \(\begin{vmatrix} -6 & 0 & 13 \\ 0 & -3 & 18 \\ 14 & 18 & 9 \end{vmatrix}\)

Score (x) 0 1 2 3 4 5 6 Freq (f) 5 7 3 7 11 6 7 46 fx 0 7 6 21 44 30 42 150 Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{150}{46}\) = 3.26

x f fx (x - \(\bar{x}\)) (x - \(\bar{x}\))\(^2\) f(x - \(\bar{x}\))\(^2\) 0 5 0 -3.26 10.628 53.14 1 7 7 -2.26 5.108 35.756 2 3 6 -1.26 1.588 4.764 3 7 21 -0.26 0.068 0.476 4 11 44 0.74 0.548 6.028 5 6 30 1.74 3.028 18.168 6 7 42 2.74 7.508 52.556 46 150 170.888 Variance = \(\frac{\sum f(x - \bar{x})}{\sum f}\) = \(\frac{170.888}{46}\) = 3.72

The locus of a points equidistant from two intersecting straight lines is a pair of bisectors that bisect the angles formed by the two intersecting lines.

The sum of two opposite angles of a cyclic quadrilateral = 180^\circ \(\therefore\) (2x + 18)^\circ + 84^\circ = 180^\circ 2x + 102^\circ = 180^\circ \(\implies\) 2x = 78^\circ x = 39^\circ

\(4\sin^2 x - 3 = 0\) \(4 \sin^2 x = 3 \implies \sin^2 x = \frac{3}{4}\) \(\sin x = \frac{\sqrt{3}}{2}\) \(\therefore x = \sin^{-1} (\frac{\sqrt{3}}{2})\) x = 60^\circ

Area of rectangle ABCD = length x breadth = 7 x 4 = 28 cm\(^2\) Area of triangle CDE = \(\frac{1}{2}\) base x height = \(\frac{1}{2} \times 3 \times 4\) = 6 cm\(^2\) Area of the figure = 28 cm\(^2\) + 6 cm\(^2\) = 34 cm\(^2\)

Scores (Mark) 0 1 2 3 4 5 No of students 4 3 7 8 6 2 fx 0 3 14 24 24 10 75

The equation of a straight line is given as \(y = mx + b\) where m = the slope of the line b = intercept Given points A(3, 12) and B(5, 22), the slope = \(\frac{22 - 12}{5 - 3}\) = \(\frac{10}{2}\) = 5 Hence, the equation of the line is \(y = 5x + 2\).

Q(m, n) and R(n, -4) Midpoint : P(2, m) \(\implies (\frac{m + n}{2}, \frac{n - 4}{2}) = (2, m)\) \(m + n = 2 \times 2 \implies m + n = 4 ... (i)\) \(n - 4 = 2 \times m \implies n - 4 = 2m ... (ii)\) Solving (i) and (ii) simultaneously, m = 0 and n = 4.

\(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\) \(\implies (2 \times 9) - (-4 \times x) = 58\) \(18 + 4x = 58 \implies 4x = 58 - 18 = 40\) \(x = 10\)

\(y = 6x^3 + 2x^{-2} - x^{-3}\) \(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)

\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1) Let u = 4x\(^3\) - 2x. \(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\) \(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\) \(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\) \(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\) = \(\frac{12x^2 - 2}{4x^3 - 2x}\)

\(f(x) = 3x^3 + 4x^2 + x - 8\) \(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8\) = \(-24 + 16 - 2 - 8\) = -18

\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\) 2(4x - 6) \(\leq\) 3(3 + 2x) 8x - 12 \(\leq\) 9 + 6x 8x - 6x \(\leq\) 9 + 12 2x \(\leq\) 21 \(x \leq \frac{21}{2}\)