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\(\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x\) = \(\frac{4x^{-3 + 1}}{-3 + 1} - \frac{7x^{2 + 1}}{2 + 1} + \frac{5x^{1 + 1}}{1 + 1} - 6x\) = \(-2x^{-2} - \frac{7}{3} x^3 + \frac{5}{2} x^2 - 6x\)

S = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34} n(S) = 14 multiples of 3 = {21, 24, 27, 30, 33} n(multiples of 3) = 5 Prob( picking a multiple of 3) = 5/14

\(T_n = ar^{n - 1}\) (terms of a G.P) \(T_3 = ar^2 = 9\) ... (i) \(T_7 = ar^6 = \frac{1}{9}\) ... (ii) Divide (i) by (ii); \(\frac{ar^6}{ar^2} = \frac{\frac{1}{9}}{9}\) \(r^4 = \frac{1}{81}\) \(r^4 = (\frac{1}{3})^4\) \(r = \frac{1}{3}\)

\mu = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {3, 6, 9, 12, 15, 18} B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} A ∩ B = {3, 9, 15}

For the committee to include 2 females, we must have 3 males, so that there should be 5 members. That is, \(^4C_3 \times ^3C_2\) = \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\) = 4 \times 3 = 12 ways

\(\sqrt{28} + \sqrt{112} - \sqrt{k} = \sqrt{175}\) \(\sqrt{4 \times 7} + \sqrt{16 \times 7} - \sqrt{k} = \sqrt{25 \times 7}\) \(2\sqrt{7} + 4\sqrt{7} - \sqrt{k} = 5\sqrt{7}\) \(6\sqrt{7} - 5\sqrt{7} = \sqrt{k}\) \(\sqrt{k} = \sqrt{7}\) \(\implies k = 7\)

\(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\) = \(\frac{2\log_3 3^2 \times \log_3 (3^4)^{-2}}{\log_5 (5^4)}\) = \(\frac{4\log_3 3 \times -8\log_3 3}{4\log_5 5}\) = -8.

\(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\) \(\frac{2 + 2x - 6}{3} \geq \frac{4x - 6}{5}\) \(\frac{2x - 4}{3} \geq \frac{4x - 6}{5}\) \(5(2x - 4) \geq 3(4x - 6)\) \(10x - 20 \geq 12x - 18\) \(10x - 12x \geq -18 + 20\) \(-2x \geq 2\) \(x \leq -1\)

\(x^2 - 7x + 10 \leq 0\) Solve for \(x^2 - 7x + 10 = 0\) We have, (x - 5)(x - 2) \(\leq\) 0. Conditions: Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0. \(\implies\) x \(\leq\) 5; x \(\geq\) 2. 2 \(\leq\) x \(\leq\) 5. Choosing x = 3, 3\(^2\) - 7(3) + 10 = 9 - 21 + 10 = -2 \(\leq\) 0. \(\therefore\) 2 \(\leq\) x \(\leq\) 5.

Given q(x) [quotient], d(x) [divisor] and r(x) [remainder], the polynomial is gotten by multiplying the quotient and the divisor and adding the remainder. i.e In this case, the polynomial = (x\(^2\) - x - 5)(3x - 1) + 7. = (3x\(^3\) - x\(^2\) - 3x\(^2\) + x - 15x + 5) + 7 = (3x\(^3\) - 4x\(^2\) - 14x + 5) + 7 = 3x\(^3\) - 4x\(^2\) - 14x + 12

When you divide a polynomial p(x) by (x - a), the remainder = p(a) i.e. In the case of 2x\(^2\) + x - 3 \(\div\) (x - 2), the remainder = p(2). = 2(2)\(^2\) + 2 - 3 = 8 + 2 - 3 = 7.

\(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\) \(\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132\) \(2(2 - 12) + 5(2x) + 3(3x) = 132\) \(-20 + 10x + 9x = 132\) \(19x = 152\) \(x = 8\)

\(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\) |A| = (3 x 6) - (-2 x 1) = 18 + 2 = 20. A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\) = \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\) = \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)

y = 8x\(^3\) - 3x\(^2\) + 7x - 1 \(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\) = \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\) \(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)

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