How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
48 cm
40 cm
43 cm
45 cm
Correct answer is C
Perimeter of the sector = \(2r + \frac{\theta}{360°} \times 2\pi r\)
= \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\)
= \(21 + 22\)
= 43 cm
1
2
3
4
Correct answer is C
No explanation has been provided for this answer.
Evaluate \(\int_{0} ^{\frac{\pi}{2}} \sin x \mathrm d x\)
-2
2
1
-1
Correct answer is C
\(\int_{0} ^{\frac{\pi}{2}} \sin x \mathrm d x\)
= \(- \cos x |_{0} ^{\frac{\pi}{2}\)
= \(- \cos (\frac{\pi}{2}) - (- \cos 0)\)
= \(0 + 1\)
= 1
Integrate \(\frac{1 + x}{x^{3}} \mathrm d x\)
\(2x^{2} - \frac{1}{x} + k\)
\(-\frac{1}{2x^{2}} - \frac{1}{x} + k\)
\(-\frac{x^{2}}{2} - \frac{1}{x} + k\)
\(x^{2} - \frac{1}{x} + k\)
Correct answer is B
\(\int \frac{1 + x}{x^{3}} \mathrm d x\)
= \(\int (\frac{1}{x^{3}} + \frac{x}{x^{3}}) \mathrm d x\)
= \(\int (x^{-3} + x^{-2}) \mathrm d x\)
= \(\frac{-1}{2x^{2}} - \frac{1}{x} + k\)
If \(\tan \theta = \frac{3}{4}\), find the value of \(\sin \theta + \cos \theta\).
\(1\frac{1}{3}\)
\(1\frac{2}{3}\)
\(1\frac{3}{5}\)
\(1\frac{2}{5}\)
Correct answer is D
\(\tan \theta = \frac{opp}{adj} = \frac{3}{4}\)
\(hyp^{2} = opp^{2} + adj^{2}\)
\(hyp = \sqrt{3^{2} + 4^{2}}\)
= 5
\(\sin \theta = \frac{3}{5}; \cos \theta = \frac{4}{5}\)
\(\sin \theta + \cos \theta = \frac{3}{5} + \frac{4}{5}\)
= \(\frac{7}{5} = 1\frac{2}{5}\)