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)