WAEC Further Mathematics Past Questions & Answers - Page 78

$\int \frac{4}{x^{3}} \mathrm {d} x = \int 4x^{-3} \mathrm {d} x$

$\frac{4x^{-3 + 1}}{-2} = -2x^{-2}|_{1}^{2} = \frac{-2}{x^{2}}|_{1}^{2}$

= $\frac{-2}{2^{2}}  - \frac{-2}{1^{2}} = -\frac{1}{2}  + 2 = 1\frac{1}{2}$

$\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}$

$m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}$

$m_{1} = \frac{3}{4}$

$m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x$

$m_{2} = -\frac{2}{3}$

$\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}$

$\tan \theta = \frac{17}{6}$

$\theta \approxeq 70.6°$

$y = 2(2x + \sqrt{x})^{2}$

Let $u = 2x + \sqrt{x}$

$y = 2u^{2}$

$\frac{\mathrm d y}{\mathrm d u} = 4u$

$\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}$

$\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})$

= $4u(2 + \frac{1}{2\sqrt{x}})$

= $4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})$

$XY = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$ is the distance between a point $X(x_{1}, y_{1})$ and $Y(x_{2}, y_{2})$.

$XY = \sqrt{(3 - 5)^{2} + (5 - 1)^{2}} = \sqrt{20}$

= $2\sqrt{5} = 4. 467 \approxeq 4.5$

An fraction is undefined when the denominator has value = 0.

$\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}$ is undefined when $x^{2} + 2x - 35 = 0$

$x^2 + 7x - 5x - 35 = 0 \implies (x + 7)(x - 5) = 0$

$x = \text{-7 or 5}$