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.
20
28
54
58
Correct answer is A
Since 2 students must be included, we have to arrange the remaining 3 students from the 6 students left
= \(^{6}C_{3} = \frac{6!}{(6-3)!3!}\)
= 20 ways
2
4
6
8
Correct answer is D
The gradient of the equation = \(\frac{y_{1} - y_{0}}{x_{1} - x_{0}} = \frac{\frac{5}{2} - 0}{\frac{5}{4} - 0}\)
= \(\frac{\frac{5}{2}}{\frac{5}{4}} = 2\)
The line passes through the origin, therefore the equation of the line is y = mx
\(y = 2x\)
When x = 4, y = 2 x 4 = 8.
1
2
3
4
Correct answer is B
\(Gradient = \frac{y_{1} - y_{0}}{x_{1} - x_{0}}\)
Line passing through \((0, 0)\) & \((1\frac{1}{4}, 2\frac{1}{2})\),
\(Gradient = \frac{2\frac{1}{2} - 0}{1\frac{1}{4} - 0} = \frac{\frac{5}{2}}{\frac{5}{4}} \)
Find the minimum value of \(y = x^{2} + 6x - 12\).
-21
-12
-6
-3
Correct answer is A
\(y = x^{2} + 6x - 12\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x + 6 = 0\)
\(2x = -6 \implies x = -3\)
\(y(-3) = (-3)^{2} + 6(-3) - 12 = 9 - 18 - 12 = -21\)
\(\frac{9}{16}\)
\(\frac{81}{16}\)
\(9\)
\(9\frac{9}{16}\)
Correct answer is D
\(T_{n} = ar^{n-1}\) (for an exponential sequence)
\(T_{1} = 36 = a\)
\(T_{2} = ar = 36r = p\)
\(T_{3} = ar^{2} = 36r^{2} = \frac{9}{4}\)
\(T_{4} = ar^{3} = 36r^{3} = q\)
\(36r^{2} = \frac{9}{4} \implies r^{2} = \frac{\frac{9}{4}}{36} = \frac{1}{16}\)
\(r = \sqrt{\frac{1}{16}} = \frac{1}{4}\)
\( p = 36 \times \frac{1}{4} = 9 ; q = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}\)
\(p + q = 9 + \frac{9}{16} = 9\frac{9}{16}\)