Further Mathematics Questions and Answers
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
Find the remainder when \(5x^{3} + 2x^{2} - 7x - 5\) is divided by (x - 2).
-51
-23
29
49
Correct answer is C
Using remainder theorem, the remainder when \(5x^{3} + 2x^{2} - 7x -5\) is divided by (x - 2) = f(2)
\(f(2) = 5(2^{3}) + 2(2^{2}) - 7(2) -5 = 40 + 8 - 14 - 5\)
= 29
Evaluate \(\cos (\frac{\pi}{2} + \frac{\pi}{3})\)
\(\frac{-2}{\sqrt{3}}\)
\(\frac{-\sqrt{3}}{2}\)
\(\frac{\sqrt{3}}{4}\)
\(\frac{4}{\sqrt{3}}\)
Correct answer is B
\(\cos (x + y) = \cos x \cos y - \sin x \sin y\)
\(\cos (\frac{\pi}{2} + \frac{\pi}{3}) = \cos \frac{\pi}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{2} \sin \frac{\pi}{3}\)
= \((0 \times \frac{1}{2}) - (1 \times \frac{\sqrt{3}}{2})\)
= \(0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}\)
If \(\log_{9} 3 + 2x = 1\), find x.
\(\frac{-1}{2}\)
\(\frac{-1}{4}\)
\(\frac{1}{4}\)
\(\frac{1}{2}\)
Correct answer is C
\(\log_{9} 3 = \log_{9} (9^{\frac{1}{2}}) = \frac{1}{2}\log_{9} 9 = \frac{1}{2}\)
\(\frac{1}{2} + 2x = 1 \implies 2x = \frac{1}{2}\)
\(x = \frac{1}{4}\)
Express 75° in radians, leaving your answer in terms of \(\pi\).
\(\frac{5\pi}{12}\)
\(\frac{3\pi}{4}\)
\(\frac{5\pi}{6}\)
\(\frac{7\pi}{6}\)
Correct answer is A
\(180° = \pi rads\)
\(1° = \frac{\pi}{180}\)
\(\therefore 75° = \frac{\pi}{180} \times 75 \)
= \(\frac{5\pi}{12}\)
-1
0
1
2
Correct answer is A
\(a * a^{-1} = e = -1\)
\(\therefore a^{-1} = -1\)