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(a)(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {1, 2, 5, 7} A' = U - A = {3, 4, 6, 8, 9, 10} (ii) \(A \cap B = {1, 2, 5, 7} \cap {1, 3, 6, 7} = {1, 7}\) \((A \cap B)' = {2, 3, 4, 5, 6, 8, 9, 10}\) (iii) \(A \cup B = {1, 2, 5, 7} \cup {1, 3, 6, 7}\) = \({1, 2, 3, 5, 6, 7}\) \((A \cup B)' = {4, 8, 9,10}\) (iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7} and {3, 6, 7}. (b) The nth term of the sequence is given as \(U_{n} = \frac{n + 1}{n(n + 2)}, n = 1, 2,3\) For the 15th term, \(U_{15} = \frac{16}{15 \times 17}\) (c)(i) \(T_{n} = a + (n - 1)d\) \(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\) = \(4n - 1\) (ii) \(4n - 1 > 100\) \(4n > 100 + 1 \implies 4n > 101\) \(n > 25.25\) The least term greater than 100 = 26th term.

(a) Class Boundaries Tally Frequency 59.5 - 64.5 || 2 64.5 - 69.5 ||| 3 69.5 - 74.5 |||| | 6 74.5 - 79.5 |||| |||| | 11 79.5 - 84.5 |||| ||| 8 84.5 - 89.5 |||| || 7 89.5 - 94.5 || 2 94.5 - 99.5 | 1 (b) An histogram to represent the distribution of 40 students in an examination. (c) Modal mark : \(\frac{74.5 + 79.5}{2} = \frac{154}{2} = 77\) (d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18. Total number of students = 40 \(\therefore\) P(mark greater than 79) = \(\frac{18}{40} = \frac{9}{20}\)

(a) < QRT = < RQS (alternate angles) < RPS = < RQS (angles in the same segment) \(\therefore\) < RPS = < QRT (ii) < QRT = < PQS (corresponding angles) < PQS = < PRS (angles in the same segment) \(\therefore\) < QTR = < PRS. (b) \(< ABC = \hat{B} = 180^\circ - 132^\circ = 48^\circ\) \(< ACB = \hat{C} = 180^\circ - 96^\circ = 84^\circ\) \(< BAC = \hat{A} = 180^\circ - (48^\circ + 84^\circ) = 48^\circ\) \(< BAC = < ABC = 48^\circ\) \(\therefore \Delta ABC = isosceles\).

(a) \(\frac{r}{R} = \cos 50\) \(r = 6400 \cos 50 = 4,113.84 km\) \(\approxeq 4,114km\) The distance PQ along the parallel of latitude = speed (km/h) \(\times\) time (h) = 600km/h x 3h = 1800km \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times r = 1800\) \(\theta = \frac{1800 \times 7 \times 360}{44 \times 4113.84}\) \(\theta = 25.059^\circ \approxeq 25.1^\circ\) \(\therefore\) The longitude of Q = \(60 + \theta\) = \(60 + 25.1 = 85.1^\circW\) (b) The distance (Q) along the meridian \(\frac{\theta}{360} \times 2\pi R\) \(500 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\theta = \frac{500 \times 360 \times 7}{44 \times 6400}\) \(\theta = 4.47^\circ\) \(\therefore \text{The latitude of Y} = 50^\circ - 4.47^\circ = 45.53^\circ\) \(\approxeq 45.5^\circN\)

(d) /PM/ = 9.7 cm.

(a) x -4 -3 -2 -1 0 1 2 3 \(x^{2}\) 16 9 4 1 0 1 4 9 \(11\) 11 11 11 11 11 11 11 11 \(-2x\) 8 6 4 2 0 -2 -4 -6 \(-2x^{2}\) -32 -18 -8 -2 0 -2 -8 -18 y -13 -1 7 11 11 7 -1 -13 (b) (c)(i) x = -2.9 or 1.9 (ii) \(y + (3 - 2x - 2x^{2}) = 11 - 2x - 2x^{2}\) \(y = 11 - 2x - 2x^{2} - 3 + 2x + 2x^{2}\) \(y = 8\) \(\therefore x = \text{-1.8 or 0.8}\) (iii) \(y = 11 - 2x - 2x^{2}\) \(\frac{\mathrm d y}{\mathrm d x} = -2 - 4x\) Gradient at x = 1 : \(-2 - 4(1) = -2 - 4 = -6\)

\(\left| \begin{array}{cc} -x & 2 \\ 4x &1 \end{array} \right|\) = \(\left| \begin{array}{cc} 3 & 3x \\ 4 & -5 \end{array} \right|\), Means that the determinants of both Matrices are equal - x - 8x = -15 - 12x -9x = -15 - 12x 12x - 9x = - 15 3x = -15 x = \(\frac{-15}{3}\) = -5

Let \( A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix} \) (given as \( P^{-1} \)). Determinant of \( A \) = \( (1)(1) - (-3)(0) = 1 \). Inverse of \( A \) is \( \frac{1}{1} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \), which is exactly \( P \).

POM is a diameter, so arc MP (lower semicircle via Q) = 180^\circ. ∠MNQ = 42^\circ is an inscribed angle subtending arc MQ. Therefore, arc MQ = 2 \times 42^\circ = 84^\circ Arc MQ + arc QP = lower semicircle: arc QP = 180^\circ − 84^\circ = 96^\circ - ∠QMP is an inscribed angle subtending arc QP. Therefore, ∠QMP = \(\frac{1}{2}\) \times 96^\circ = 48^\circ. (Alternatively: in right triangle PQM, ∠PQM = 90^\circ (angle in a semicircle), so the angles at M and P sum to 90^\circ; the arc method directly gives the result.)

The given line is 2y + 8x - 17 = 0. Rewrite in slope form: 2y = -8x + 17 y = -4x + \(\frac{17}{2}\) Slope = −4. The line through points \((-1, -p)\) and \((-2p, 2)\) must have the same slope (parallel lines). Slope = \(\frac{2 - (-p)}{-2p - (-1)} = \frac{2 + p}{1 - 2p}\) \(\frac{2 + p}{1 - 2p} = -4\) Cross-multiply: 2 + p = -4(1 - 2p) 2 + p = -4 + 8p 2 + 4 = 8p - p 6 = 7p p = \(\frac{6}{7}\)

To find the maximum of \(f(\theta) = \frac{4}{6 + 2\cos \theta}\) for \(\frac{\pi}{2} \leq \theta \leq 2\pi\), simplify to \(f(\theta) = \frac{2}{3 + \cos \theta}\). The maximum occurs when \(3 + \cos \theta\) is minimized. In the interval, \(\cos \theta\) ranges from \(-1\) to \(1\), with minimum \(-1\) at \(\theta = \pi\). Thus, minimum denominator is \(3 - 1 = 2\), so maximum \(f(\theta) = \frac{2}{2} = 1\).

\((343)^{\frac{1}{3}} \times (0.14)^{-1} \times (25)^{-\frac{1}{2}}\) = \(\sqrt[3]{343} \times (\frac{14}{100})^{-1} \times (\sqrt{25})^{-1}\) = \(7 \times \frac{100}{14} \times \frac{1}{5}\) = 10.

Hint: Represent the question in a venn diagram, such that n(B ∩ M) = x n(B) only = 125-x n(M) only = 110-x => 125-x + 110-x + x = 220 => x = 15. n(B) only = 125 - 15 = 110

52.4 - 5.7 - 3.45 - 1.75 = 52.4 - (5.7 + 3.45 + 1.75) = 52.4 - 10.90 = 41.5

\((\sqrt{0.7} + \sqrt{70})^{2}\) = \((\sqrt{0.7} + \sqrt{70})(\sqrt{0.7} + \sqrt{70})\) = \(0.7 + 2\sqrt{0.7 \times 70} + 70\) = \(0.7 + 14 + 70 \) = 84.7