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.
Given n = 3, evaluate \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!}\)
\(12\)
\(2\frac{1}{2}\)
\(2\)
\(\frac{11}{24}\)
Correct answer is D
n = 3, \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!} = \frac{1}{(3-1)!} - \frac{1}{(3+1)!}\)
= \(\frac{1}{2} - \frac{1}{24} = \frac{12 -1}{24}\)
= \(\frac{11}{24}\)
Simplify \(\frac{^{n}P_{5}}{^{n}C_{5}}\)
80
90
110
120
Correct answer is D
\(\frac{^{n}P_{5}}{^{n}C_{5}} = \frac{n!}{(n-5)!} ÷ \frac{n!}{(n-5)!5!}\)
= \(\frac{n!}{(n-5)!} \times \frac{(n-5)!5!}{n!} = 5! = 120\)
If \(\log_{10}y + 3\log_{10}x \geq \log_{10}x\), express y in terms of x.
\(y \geq \frac{1}{x}\)
\(y \leq \frac{1}{x}\)
\(y \leq \frac{1}{x^{2}}\)
\(y \geq \frac{1}{x^{2}}\)
Correct answer is D
\(\log_{10}y + 3\log_{10}x \geq \log_{10}x\)
\(\implies \log_{10}y \geq \log_{10}x - 3 \log_{10}x \)
\(\log_{10}y \geq -2\log_{10}x = \log_{10}y \geq \log_{10}x^{-2}\)
\(\log_{10}y \geq \log_{10}(\frac{1}{x^{2}}) \implies y \geq \frac{1}{x^{2}}\)
Solve for x in the equation \(5^{x} \times 5^{x + 1} = 25\)
\(-2\)
\(\frac{-1}{2}\)
\(\frac{1}{2}\)
\(2\)
Correct answer is C
\(5^{x} \times 5^{x+1} = 25\)
\(5^{x} \times 5^{x+1} = 5^{2}\)
\(5^{x+x+1} = 5^{2}\), equating powers,
\(2x + 1 = 2 \implies 2x = 1\)
\(\therefore x = \frac{1}{2}\)
If (2t - 3s)(t - s) = 0, find \(\frac{t}{s}\)
\(\frac{3}{2}\) or \(1\)
\(\frac{3}{2}\) or \(-1\)
\(\frac{-3}{2}\) or \(-1\)
\(\frac{-3}{2}\) or \(1\)
Correct answer is A
\((2t - 3s)(t - s) = 0 \implies (2t - 3s) = \text{0 or} (t - s) = 0\)
\(2t - 3s = 0 \implies 2t = 3s \therefore \frac{t}{s} = \frac{3}{2}\)
\(t - s = 0 \implies t = s \therefore \frac{t}{s} = 1\)
\(\frac{t}{s} = \frac{3}{2} or 1\)