JAMB Mathematics Past Questions & Answers - Page 58

$\int \frac{1 + x}{x^{3}} \mathrm d x$

= $\int (\frac{1}{x^{3}} + \frac{x}{x^{3}}) \mathrm d x$

= $\int (x^{-3} + x^{-2}) \mathrm d x$

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

$\tan \theta = \frac{opp}{adj} = \frac{3}{4}$

$hyp^{2} = opp^{2} + adj^{2}$

$hyp = \sqrt{3^{2} + 4^{2}}$

= 5

$\sin \theta = \frac{3}{5}; \cos \theta = \frac{4}{5}$

$\sin \theta + \cos \theta = \frac{3}{5} + \frac{4}{5}$

= $\frac{7}{5} = 1\frac{2}{5}$

Using the cosine rule, we have

$p^{2} = q^{2} + r^{2} - 2qr \cos P$

$p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})$

= $64 + 36 - 8$

$p^{2} = 92 \therefore p = \sqrt{92} cm$

Given P(2, -3) and Q(-5, 1)

Midpoint = $(\frac{2 + (-5)}{2}, \frac{-3 + 1}{2})$

= $(\frac{-3}{2}, -1)$

Slope of the line PQ = $\frac{1 - (-3)}{-5 - 2}$

= $-\frac{4}{7}$

The slope of the perpendicular line to PQ = $\frac{-1}{-\frac{4}{7}}$

= $\frac{7}{4}$

The equation of the perpendicular line: $y = \frac{7}{4}x + b$

Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).

$-1 = (\frac{7}{4})(\frac{-3}{2}) + b$

$-1 + \frac{21}{8} = \frac{13}{8} = b$

$\therefore$ The equation of the perpendicular bisector of the line PQ is $y = \frac{7}{4}x + \frac{13}{8}$

$\equiv 8y = 14x + 13 \implies 8y - 14x - 13 = 0$

Midpoint of a line PQ where P has coordinates (x$_{1}$, y$_{1}$) and Q has coordinates (x$_{2}$, y$_{2}$) is given as 

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$.

$\therefore$ If Q has coordinates (r, s), then

$\frac{-2 + r}{2} = 2$ and $\frac{1 + s}{2} = 3$

$-2 + r = 4 \implies r = 6$

$1 + s = 6 \implies s = 5$

Q = (6, 5)