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The two equations are y = x(2 - x) and y = (x - 1)(x - 3) The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0 put y = 0; 0 = x(2 - x) 0 = (x - 1)(x - 3) x = 0 or x = 2; x = 1 or x = 3 The values of x at P, Q, R are increasing towards the positive direction of x-axis at P, x = 1 at Q, x = 2 at R, x = 3 P, Q, R are respectively (1, 2, 3)

The total angle in a regular pentagon = \((2(5) - 4) \times 90\) = \(6 \times 90 = 540^\circ\) Each interior angle = \(\frac{540}{5} = 108^\circ\) While the interior of a quadrilateral = \(\frac{360}{4} = 90^\circ\) \(PTU + TPU + TUP = 180^\circ\) \(108^\circ + (180^\circ - 90^\circ) + TUP = 180^\circ\) \(TUP = 180^\circ - (108^\circ + 18^\circ) = 54^\circ\)

< PSO = \(\frac{1}{2}\) < SOQ = \(\frac{1}{2}\)(180) = 90^\circ < RTS = < PQS = 32^\circ (Alternative angle) < PSQ = 90 - < PSQ = 90^\circ - 32^\circ = 58^\circ

OQ = OR = radii < ROQ = 180 - 96 = 84º \(\bigtriangleup\)OQR = Isosceles R = Q R + Q + 84 = 180(angle in a \(\bigtriangleup\)) 2R = 96 since R = Q R = 48º ORQ = 48º

Construction joins R to P such that SRP = straight line R = 180o - 107o < p = 180o - (107o - 34o) 108 - 141o = 39o Angle < S = 39o (corr. Ang.) But in \(\bigtriangleup\)SRT < S = < T = 39o SRT = 180 - (39o + 39o) = 180o - 78o = 102o

Since QRU= Xo RSU = Xo, But RSU = 180o - (50o + 35o) = 180o - 85o = 95o x = 95o

Using angle bisector theorem: line PS bisects angle QPR QS/QP = SR/PR 3/4 = SR/9 4SR = 27 SR = \(\frac{27}{4}\) = 6\(\frac{3}{4}\)cm

SQ2 + OS2 = OQ2 + 52 = 142 SQ2 = 142 - 52 196 - 25 = 171 ST2 + TQ2 = SQ2 ST2 + 22 = 171 ST2 = 171 - 4 = 167 ST = \(\sqrt{167}\) = 12.92 = 12.9cm

Since area of square PQRS = 100cm2 each lenght = 10cm Also TUYS : XQVU = 1 : 16 lengths are in ratio 1 : 4, hence TU : UV = 1: 4 Let TU = x UV = 1: 4 hence TV = x + 4x = 5x = 10cm x = 2cm TU = 2cm UV = 8cm But TU = SY and UV = YR YR = 8cm

Number f students that took the test = \(\sum f\). Where f is the frequencies = 2 + 5 + 0 + 3 + 4 + 0 + 0 + 2 + 3 + 1 + 2 = 20

(a) \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16}) - \log_{10} (\frac{5}{9})^{2} + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \div \frac{25}{81})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25})\) = \(\log_{10} (10) = 1\) (b) \(\sqrt{\frac{P}{Q}} = \sqrt{\frac{3.6 \times 10^{-3}}{2.25 \times 10^{6}}\) = \(\sqrt{\frac{36 \times 10^{-4}}{2.25 \times 10^{6}}\) = \(\frac{6 \times 10^{-2}}{1.5 \times 10^{3}}\) = \(4 \times 10^{-2 - 3} = 4 \times 10^{-5}\)

\(P = {..., -5, -4, -3, -2, -1}\) \(Q = {..., -5, -3, -1, 1, 3, 5, ...}\) \(R = {-2, -1, 0, 1, 2, 3, 4, 5, 6}\) (a) \(Q \cap R = {-1, 1, 3, 5}\) (b) \(R' = {..., -5, -4, -3, 7, 8, ...}\) (c) \(P' = {0, 1, 2, 3, ...}\) \(P' \cup R' = {-5, -4, -3, 0, 1, 2, 3, ...}\) (d) \(Q = {..., -5, -3, -1, 1, 3, 5,...}\) \(P \cap Q = {..., -7, -5, -3, -1}\) \((P \cap Q)' = {..., -8, -6, -4, -2, 0, 1, 2, 3, ...}\)

In \(\Delta\) BPY, \(\tan 35^\circ = \frac{h}{x}\) \(h = x \tan 35 = 0.7 x ... (1)\) In \(\Delta\) BPX, \(\frac{h}{x + 4000} = \tan 15\) \(\frac{h}{x + 4000} = 0.3\) \(h = 0.3 (x + 4000) = 0.3x + 1200 ... (2)\) Equating (1) and (2) as the values of h, we have \(0.7x = 0.3x + 1200 \implies 0.7x - 0.3x = 1200\) \(0.4x = 1200 \times x = \frac{1200}{0.4} = 3000m\) \(\therefore h = 0.7x = 0.7 (3000) = 2100m\) \(\text{The total height = } 5200m + 2100m = 7300m\)

(a) \(\frac{3}{m + 2n} - \frac{2}{m - 3n}\) = \(\frac{3(m - 3n) - 2(m + 2n)}{(m + 2n)(m - 3n)}\) = \(\frac{3m - 9n - 2m - 4n}{(m + 2n)(m - 3n)}\) = \(\frac{m - 13n}{(m + 2n)(m - 3n)}\) (b) Let the numbers in the digit be x and z. Hence the number is \(10x + z\). When it is interchanged, we have \(10z + x\). \(10z + x = 10x + z + 9\) \(10z - z + x - 10x = 9\) \(9z - 9x = 9 \implies z - x = 1 ... (1)\) \(x + z = 11 ... (2)\) From (1), \(z = 1 + x\) \(\therefore x + 1 + x = 11\) \(2x + 1 = 11 \implies 2x = 11 - 1 = 10\) \(\therefore x = 5\) \(z = 1 + x = 1 + 5 = 6\) \(\therefore\) The original number = 56.

n(R) = number of red balls = 12; n(W) = number of white balls = 16; n(B) = number of blue balls = 8 Total number of balls = 36 (a) P(three are red) = \(\frac{12}{36} \times \frac{11}{35} \times \frac{10}{34}\) = \(\frac{11}{357}\) (b) P(first is blue and other two red) = \(\frac{8}{36} \times \frac{12}{35} \times \frac{11}{34}\) = \(\frac{44}{1785}\) (c) P(two are white and one is blue) = \(\frac{16}{36} \times \frac{15}{35} \times \frac{8}{34}\) = \(\frac{16}{357}\)