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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] = \(\frac{13}{3}\) + \(\frac{1}{3}\) = \(\frac{14}{3}\) = \(4 \frac{2}{3}\)

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

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

(a) \(3 \times 9^{1 + x} = 27^{-x}\) \(3 \times (3^{2})^{1 + x} = (3^{3})^{-x}\) \(3 \times 3^{2 + 2x} = 3^{-3x} \) \(3^{2x + 2 + 1} = 3^{-3x} \implies 3^{2x + 3} = 3^{-3x}\) \(\implies 2x + 3 = - 3x\) \(-5x = 3 \implies x = \frac{-3}{5}\) (b) \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\) = \(\log_{10} (\frac{\sqrt{35} \times \sqrt{2}}{\sqrt{7}}\) = \(\log_{10} (\frac{\sqrt{35 \times 2}{\sqrt{7}}\) = \(\log_{10} \sqrt{10}\) = \(\log_{10} 10^{\frac{1}{2}}\) = \(\frac{1}{2} \log_{10} 10 = \frac{1}{2}\)

(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two. I'll be using the latter method. \(11111111_{two} = 1 \times 2^{7} + 1 \times 2^{6} + 1 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0}\) = \(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\) \(101_{two} = 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0} \) = \(4 + 0 + 1 = 5_{ten}\) \(\frac{255}{5} = 51_{ten}\) 2 51 2 25 r 1 2 12 r 1 2 6 r 0 2 3 r 0 2 1 r 1 0 r 1 = \(110011_{two}\) (b) (i) Length of arc AXB = \(\frac{105}{360} \times 2 \times \frac{22}{7} \times 6cm\) = \(11cm\) \(\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\) (ii) Area of sector = \(\frac{105}{360} \times \frac{22}{7} \times 6cm \times 6cm\) = \(33 cm^{2}\)

(a) \(\log_{10} (3x - 1) - \log_{10} 2 = 3\) \(\log_{10} (\frac{3x - 1}{2}) = 3\) \(\implies \frac{3x - 1}{2} = 10^{3} = 1000\) \(\implies 3x - 1 = 2000\) \(3x = 2000 + 1 = 2001\) \(x = \frac{2001}{3} = 667\) (b) No Log 0.8972 \(\bar{1}.9528\) + 3.536 \(0.5485\) = 0.5013 - 0.00249 \(\bar{3}.3962\) Antilog - 35.70 \(3.1051 \div 2 = 1.5526\) \(\therefore \sqrt{\frac{0.8972 \times 3.536}{0.00249}} = 35.70\)

The shaded part exists on x \(\cap\) z but not in y

S = {1, 2, 3, 4, 5, 6}; T = {2, 4, 5, 7}; R = {1, 4, 5} (S∩T) ∪ R = {2, 4, 5} ∪ {1, 4, 5} = {1, 2, 4, 5}

\(\frac{3}{4} \div 1\frac{1}{4} \times (1\frac{1}{2} - \frac{2}{3})\) \(\frac{3}{4} \div \frac{5}{4} \times (\frac{9 - 4}{6})\) = \(\frac{3}{4} \times \frac{4}{5} \times \frac{5}{6}\) = \(\frac{1}{2}\)

\(89_{10}\) 2 89 2 44 r 1 2 22 r 0 2 11 r 0 2 5 r 1 2 2 r 1 2 1 r 0 0 r 1 \(89_{10} = 1011001_{2}\)

% error = \(\frac{error value}{true value}\) x 100% % error = \(\frac{1.80 - 1.75}{1.75}\) x 100% = \(\frac{0.05}{1.75}\) x 100% = \(\frac{20}{7}\)% = 2\(\frac{6}{7}\)%

\(\frac{4^{-\frac{1}{2}} \times 16^{\frac{3}{4}}}{4^{\frac{1}{2}}}\) = \(\frac{16^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\) = \(\frac{(2^4)^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\) = \(\frac{2^3}{4}\) = 2

\(\frac{\log \sqrt{27}}{\log 81}\) = \(\frac{\log \sqrt{3^3}}{\log 3^4}\) = \(\frac{\log 3^{\frac{3}{2}}}{\log 3^4}\) = \(\frac{\frac{3}{2} \log 3}{4 \log 3}\) = \(\frac{\frac{3}{2}}{4}\) = \(\frac{3}{8}\)

2s\(^2\) - 3st - 2t\(^2\) = 2s\(^2\) - 4st + st - 2t\(^2\) = 2s(s - 2t) + t(s - 2t) = (2s + t)(s - 2t)

\(\frac{5}{6r} - \frac{3}{4r}\) = \(\frac{10 - 9}{12r}\) = \(\frac{1}{12r}\)