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\(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) = \(\frac{1}{(x + 2)(x + 3)}\) + \(\frac{1}{(x + 1)(x + 2)}\) \(\frac{(x + 1)+ (x + 3)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{x + 1 + x + 3}{(x + 1)(x + 2)(x + 3)}\) \(\frac{2x + 4}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2(x + 2)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2}{(x + 1)(x + 3)}\)

\(\cot \theta = \frac{x}{y}\) \(\implies \tan \theta = \frac{y}{x}\) \(opp = y; adj = x\) Using Pythagoras theorem, \(Hyp^{2} = Opp^{2} + Adj^{2}\) \(Hyp^{2} = y^{2} + x^{2}\) \(Hyp = \sqrt{y^{2} + x^{2}}\) \(\sin \theta = \frac{y}{\sqrt{y^{2} + x^{2}}}\) \(\therefore \csc \theta = \frac{\sqrt{y^{2} + x^{2}}}{y}\) = \(\frac{1}{y}(\sqrt{y^{2} + x^{2}})\)

PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o PR2 = 12 + 22 - 2(1)(2) x - cos 60o = 5 - 2(1)(2) x -\(\frac{1}{2}\) = 5 + 2 = 7 PR = \(\sqrt{7}\)cm

The volume of the pipe is equal to the area of the cross section and length. let outer and inner radii be R and r respectively. Area of the cross section = (R2 - r2) where R = 6 and r = 6 - 1 = 5cm Area of the cross section = (62 - 52)\(\pi\) = (36 - 25)\(\pi\)cm sq vol. of the pipe = \(\pi\) (R2 - r2)L where length (L) = 10 volume = 11\(\pi\) x 10 = 110\(\pi\)cm3

\(\bigtriangleup\)PUS is right angled \(\frac{US}{5}\) = sin60o US = 5 x \(\frac{\sqrt{3}}{2}\) = 2.5\(\sqrt{3}\) = 4.33cm

Mean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6 Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\) \(\frac{31}{2}\) = 15.5 \(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065 1.0(1 d.p)

The total sample space when two dice are thrown together is 6 x 6 = 36 \(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\) At least 10 means 10 and above P(at least 10) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

SRT is a straight line, where QRT = 120 SRQ = 180o - 120o = 60o - (angle on a straight line) also angle QRS = 180o - 100o (angle on a straight line) . In angles where QR = SR and angle SRQ = 60o x = 100 - 60 = 40o

From the figure, 3x + x - 40o = 180o 4x = 180o + 40o 4x = 220o x = \(\frac{220}{4}\) = 55o

From the figure, PS = QR = YT = 7cm Area of parallelogram PQRS = 56cm 56 = base x height, where base = 7 7 x h = 56cm, h = \(\frac{56}{7}\) = 8cm Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8 \(\frac{1}{2}\) x 30 x 8 = 120cmsq

From similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\) QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\) = \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\) = \(\frac{47}{5}\)

In the figure PS = RS = QS, they will have equal base QR = RP In angle SQR, angle S = 50º In angle QRP, 65 + 65 = 130º Since RQP = angle RPQ = \(\frac{180 - 130}{2}\) = \(\frac{50}{2} = 25^o\) QPR = 25^\circ

Given: ∠ZXR = 45º (tangent XR), ∠YZX = 55º (inscribed angle). ∠YZX subtends arc YX, so arc YX = 2 \times 55^\circ = 110º. Major arc YZX = 360º - 110º = 250º. By the alternate segment theorem, ∠ZXR subtends arc XZ, so arc XZ = 2 \times 45º = 90º. Arc YZ = 250º - 90º = 160º. Major arc YZ = 360º - 160º = 200º. ∠ZYQ (tangent YQ) = \(\frac{1}{2}\) \times major arc YZ = \(\frac{1}{2}\) \times 200º = 100º (by alternate segment theorem, larger arc for obtuse angle).

The volume of the solid = vol. of cone + vol. of hemisphere volume of cone = \(\frac{1}{3} \pi r^2 h\) = \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\) vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\) = \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\) vol. of solid = 18\(\pi\) + 18\(\pi\) = 36\(\pi\)cm3

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