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Diagonal |AC| = (2x + 1)cm In the diagram,area of \(\Delta\)ABC is \(\frac{110}{2}\) = \(\frac{1}{2}\) x |AC| x |HB| 55 = \(\frac{1}{2}\) x (2x + 1) x 10 55 = (2x + 1)5 55 = 10x + 5 55 - 5 = 10x 50 = 10x x = \(\frac{50}{10}\) = 5.0

Two word mathematics is first rewritten as mathematics, since every letter of mathematics appears once in the english alphabet. Hence the probability that letter selected is from the word mathematics is prob. (matheics) = \(\frac{8}{26}\) = \(\frac{4}{13}\)

In the diagram |AM| = |MB| - \(\frac{|AB|}{2}\) = \(\frac{16\sqrt{3}}{2}\)cm = 8\(\sqrt{3}\)cm in \(\Delta\) AMO, r2 = |AM|Z + |MO|2 r2 = (8\(\sqrt{3}\))2 + 102 = 64 x 3 + 100 = 192 + 100 = 292 r = \(\sqrt{292}\) 17.088cm 17cm

(a) Area of trapezium PQRS = \(\frac{1}{2}(PQ + RS) \times 40\) = \(\frac{1}{2}(20 + 60) \times 40\) = \(40 \times 40\) = \(1600 cm^{2}\) (b) In \(\Delta\) SPT, \(|SP|^{2} = |ST|^{2} + |TP|^{2}\) (Pythagoras theorem) \(|ST|^{2} = |SP|^{2} - |TP|^{2}\) = \(50^{2} - 40^{2}\) = \(2500 - 1600\) \(|ST|^{2} = 900\) \(\therefore |ST| = 30 cm\) \(|UR| = 60 cm - (30 + 20) cm\) = 10 cm Let < QRS = \(\alpha\), \(\tan \alpha = \frac{40}{10} = 4\) \(\alpha = \tan^{-1} (4)\) = \(75.96^\circ \approxeq 76^\circ\)

(a) (i) , (ii) (b) (i) |QX| = 2.7 cm ; (ii) < XWZ = 75^\circ

< URT = < TRQ (angle alternate a tangent and a chord equal to angle in the alternate segment) < RUT = xo In \(\bigtriangleup\) URT < RUT + < RUT + < UTR = 180o (sum of int. < s of \(\bigtriangleup\)) < URT + x + x = 180o < URT = 180o - 2x

0.00043 x 2000 = 43 x 10-5 x 2 x 103 = 43 x 2 x 10-5+3 = 86 x 10-2 = 8.6 x 101 x 10-2 = 8.6 x 10-1

Consider the range -3 < x < -1 = { -2, -1, 0}, for instance When x = -2, \(\frac{-2 - 5}{-2 + 3} < -1\) \(\frac{-7}{1} < -1\) When x = -1, \(\frac{-1 - 5}{-1 + 3} < -1\) \(\frac{-6}{2} < -1\) = -3 < -1 When x = 0, \(\frac{0 - 5}{0 + 3} < -1\) \(\frac{- 5}{3} < -1\) Hence -3 < x < 1

If each interior angle of the polygon is 135^\circ, then each exterior angle is 180^\circ - 135^\circ = 45^\circ. Hence, number of sides = \(\frac{360^\circ}{\text{one exterior angle}}\) \(\frac{360^\circ}{45^\circ}\) = 8

\(\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}\) \(\frac{1}{2} = \frac{2 - 4}{1 - x}\) 1 - x = 2(2 - 4) 1 - x = 4 - 8 1 - x = -4 -x = -4 - 1 x = 5

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

y = cos 3x Let u = 3x so that y = cos u Now, \(\frac{\delta y}{\delta x} = 3\), \(\frac{\delta y}{\delta x} = -sin u\) By the chain rule, \(\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}\) \(\frac{\delta y}{\delta x} = (-\sin u) (3)\) \(\frac{\delta y}{\delta x} = -3 \sin u\) \(\frac{\delta y}{\delta x} = -3 \sin 3x\)

\(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k\) \(- \frac{1}{2} \cos 2x + k\)

Mean x = \(\frac{\sum x}{n}\) \( = \frac{5 + 4 + 3 + 2 + 1}{5}\) \( = \frac{15}{5}\) = 3 \(\begin{array}{c|c} x & d = x - 3 & d^2 \\ \hline 5 & 2 & 4 \\ 4 & 1 & 1 \\ 3 & 0 & 0 \\ 2 & -1 & 1 \\ 1 & -2 & 4 \\ \hline & & \sum d^2 + 10 \end{array}\) Hence, standard deviation; \( = \sqrt{\frac{\sum d^2}{n}} = \sqrt{\frac{10}{5}}\) \( = \sqrt{2}\)

Sample space S = {10, 11, 12, ... 30} Let E denote the event of choosing a number divisible by 3 Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7 Prob (E) = \(\frac{n(E)}{n(S)}\) Prob (E) = \(\frac{7}{21}\) Prob (E) = \(\frac{1}{3}\)