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\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\) = \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\) = \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\) = [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1] = \(\frac{13}{3}\) + \(\frac{1}{3}\) = \(\frac{14}{3}\) = \(4 \frac{2}{3}\)

x 1 2 3 4 5 Total f y + 2 y - 1 2y - 3 y + 4 3y - 4 8y - 2 fx y + 2 2y - 2 6y - 9 4y + 16 15y - 20 28y - 13 Mean = \(\frac{\sum fx}{\sum f}\) \(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\) \(\implies 14(28y - 13) = 43(8y - 2)\) \(392y - 182 = 344y - 86\) \(392y - 344y = -86 + 182 \implies 48y = 96\) \(y = 2\)

x = \(\frac{\sum x}{N}\) = \(\frac{18}{3}\) = 6 \(\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}\) M.D = \(\frac{|x - x|}{N}\) = \(\frac{6}{3}\) = 2

Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h N = Sum of frequencies L = lower class boundary of median class f = sum of all frequencies below L fm = frequency of modal class and h = class width of median class Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5 = 11 + (\(\frac{25 - 21}{20}\))5 = 11 + (\(\frac{(4)}{20}\)) 11 + 1 = 12

\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\) x = \(\frac{\sum fx}{\sum f}\) = \(\frac{24}{8}\) = 3 Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\) = \(\frac{18}{8}\) = \(\frac{9}{4}\)

40 = 20 - x + x + 35 - x 40 = 55 - x x = 55 - 40 = 15 ∴ P(both) \(\frac{15}{40}\) = \(\frac{3}{8}\)

y = x3 - 27, y = -27 \(\to\) (0, -27) when y = 0, x = 3 (3, 0)

2x - y - 2 \(\geq\) 0 = y \(\leq\) 2x - 2 when x = 0, y = -2, when y = 0, x = 1

SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR < STP = < RQV = 30o But SQR + 30o + 50o = 180o SQR = 180 - 80 = 100o

o (angle in the same segment of a circle are equal) But o (angle in a semicircle) also o (angles in a triangle = 180o) o + 90o = 190o o = 180o PSQ = 180o - 128o PSQ = 52o

RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180 RQT = 180o - (50 + 60) = 180o - 110o = 70o Since RQT = RTS = 70o

A\(\bigtriangleup\) = \(\sqrt{S(S - a) (S - b)(S - c)}\) (Hero's Formula) S = \(\frac{a + b + c}{2}\) = \(\frac{5 + 6 + 7}{2}\) \(\frac{18}{2} = 9\) A\(\bigtriangleup\) = \(\sqrt{9 \times 4 \times 3 \times 2}\) \(\sqrt{216} = 6 \sqrt{6}cm^3\) A\(\bigtriangleup\) = \(\frac{1}{2} \times 6 \times h\) 6\(\sqrt{6} = \frac{1}{2} \times 7 \times h\) h = \(\frac{12}{7} \sqrt{6}\)cm

\(\frac{x}{r}\) = \(\frac{x + h}{2r}\) 2 x r = r (x + h) Total height of cone = x + h but x = h total height = 2h

Gradient of line = \(\frac{\text{Change in y}}{\text{Change in x}} = \frac{y_2 - y_1}{x_2 - x_1}\) y2 = 0, y1 = 4 x2 = 3 and x1 = 0 \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{3 - 0} = \frac{-4}{3}\) Equation of straight line = y = mx + c where m = gradient and c = y intercept = 4 y = \(\frac{-4x}{3}\) + 4, multiple through by 3 3y = - 4x + 12

Percentage of students who obtained C4 = \(\frac{43.2}{360}\) x 100% = = \(\frac{4320}{360}\) = 12%