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.
Solve, correct to three significant figures, (0.3)\(^x\) = (0,5)\(^8\)
4.61
4.606
0.461
0.0130
Correct answer is A
(0,3)\(^x\) = (0.5)\(^8\)
xlog 0,3 = 9 log 0.5
x = \(\frac{8 \log 0.5}{\log 0.3}\)
= 4.606
70 kg ms\(^{-1}\)
55 kg ms\(^{-1}\)
50 kg ms\(^{-1}\)
35 kg ms\(^{-1}\)
Correct answer is A
Change in momentum
= F x t = 35N x 2
= 70kg ms\(^{-1}\)
Which of these inequalities is represented by the shaded portion of the graph?
2y + x - 3 < 0
2y - x - 3 < 0
2y - x + 3 < 0
2y + x +3 < 0
Correct answer is B
(0, 1.5), (-3, 0)
m = \(\frac{0 - 1.5}{-3, 0}\) = 0.5
0.5 = \(\frac{y - 1.5}{-3.0}\) = 0.5
y = 0.5x + 1.5
2y = x + 3
2y - x - 3 < 0
Find the constant term in the binomial expansion of (2x\(^2\) + \(\frac{1}{x^2}\))\(^4\)
10
12
24
42
Correct answer is B
6(2x\(^2\))\(^2\) (\(\frac{1}{x^2}\))\(^2\)
= 6 x 2
= 12
0 ms\(^{-2}\)
3 ms\(^{-2}\)
6 ms\(^{-2}\)
9 ms\(^{-2}\)
Correct answer is C
V = 3t\(^2\) - 6t
\(\frac{dx}{dt}\) = 6r - 6
at t = 4
\(\frac{dx}{dt}\) = 6 x 4 - 6 = 24 - 6
= 18
at t = 3
\(\frac{dy}{dt}\) = 6 x 4 - 6
= 24 - 6
= 18
at t = 3
\(\frac{dx}{dt}\) = 6 x 3 - 6 = 18 - 6
= 12
\(\frac{dy}{dt}\) = 18 - 12
= 6ms\(^{-2}\)