Further Mathematics Questions and Answers

Let the number of mixed schools be S

The number of boys only = 42 - S

The number of girls only = 35 - S

(42 - S) + S + (35 - S) = 77 - S = 70

\(\implies\) S = 7

\(Probability = \frac{7}{70} = \frac{1}{10}\)

 \(\int_{-1}^{1} (x + 1)^{2}\mathrm {d} x \equiv \int_{-1}^{1} (x^{2} + 2x + 1)\mathrm {d} x\)

= \(\left. \frac{x^{3}}{3} + x^{2} + x \right |_{-1}^{1}\)

= \((\frac{1^{3}}{3} + 1^{2} + 1) - (\frac{(-1)^{3}}{3} + (-1)^{2} + (-1)) = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}\)

\(\mu = \frac{24+25+28+29+30+32}{6} = \frac{168}{8} = 28\)

\(S.D = \sqrt{\frac{\sum{(x - \mu)^{2}}}{n}} = \sqrt{\frac{46}{6}}\)

= \(\sqrt{7.67} \approxeq 2.8\)

\(\int_{\frac{1}{2}}^{1} \frac{x^{3} - 4}{x^{3}} \mathrm {d} x \equiv \int_{\frac{1}{2}}^{1} 1 - \frac{4}{x^{3}} \mathrm {d} x\).

= \(\int_{\frac{1}{2}}^{1} 1 - 4x^{-3} \mathrm {d} x\).

= \(\left. x - \frac{4x^{-2}}{-2}\right |_{\frac{1}{2}}^{1} = \left. x + \frac{2}{x^{2}}\right |_{\frac{1}{2}}^{1}\)

= \((1 + 2) - (\frac{1}{2} + 8) = 3 - 8.5 = -5.5\)

\(y = 3x^{2} - 2x^{3}\)

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 6x^{2} = 0 \implies 6x(1 - x) = 0\)

\(x = 0, 1\), when x = 1, y = 0.

when x = 1, y = 0.

The stationary point is (1, 0)