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.
\(p \vee q\)
\(p \vee \sim q\)
\(p \wedge \sim q\)
\(p \wedge q\)
Correct answer is C
No explanation has been provided for this answer.
\(\frac{-1}{2}\)
\(0\)
\(\frac{2}{3}\)
\(2\)
Correct answer is A
Given the formula for p * q as: \(p + q + 2pq\) and its identity element is 0, such that if, say, t is the inverse of p, then
\(p * t = 0\), then \(p + t + 2pt = 0 \therefore p + (1 + 2p)t = 0\)
\(t = \frac{-1}{1 + 2p}\) is the formula for the inverse of p and is undefined on R when
\(1 + 2p) = 0\) i.e when \(2p = -1; p = \frac{-1}{2}\).
Express \(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\) in the form \(p\sqrt{3} + q\sqrt{2}\)
\(7\sqrt{3} - \frac{17\sqrt{2}}{3}\)
\(7\sqrt{2} - \frac{17\sqrt{3}}{3}\)
\(-7\sqrt{2} + \frac{17\sqrt{3}}{3}\)
\(-7\sqrt{3} - \frac{17\sqrt{2}}{3}\)
Correct answer is B
Given \(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\),
first, we rationalise by multiplying through with \(2\sqrt{3} - 3\sqrt{2}\) (the inverse of the denominator).
\((\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}})(\frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}})\)
= \(\frac{16\sqrt{3} - 24\sqrt{2} - 18\sqrt{2} + 18\sqrt{3}}{4(3) - 6\sqrt{6} + 6\sqrt{6} - 9(2)}\)
= \(\frac{34\sqrt{3} - 42\sqrt{2}}{-6} = 7\sqrt{2} - \frac{17\sqrt{3}}{3}\)
\({x : -5 < x < 5}\)
\({x : -5 \leq x \leq 5}\)
\({x : -5 \leq x < 5}\)
\({x : -5 < x \leq 5}\)
Correct answer is A
\(P = {x : -2 < x < 5}\) and \(Q = {x: -5 < x < 2}\)
\((P \cup Q) = {x : -5 < x < 5}\)
\(x^{2} + 4x - 5\)
\(x^{2} - 4x + 5\)
\(x^{2} - 1\)
\(x - 1\)
Correct answer is B
\(f(x) = x^{2} + 1\) and \(g(x) = x - 2\)
\(f o g = f(g(x)) = f(x - 2) = (x - 2)^{2} + 1 \)
= \(x^{2} - 4x + 4 + 1 = x^{2} - 4x + 5\)