How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
26.7m
14.2m
\(1.7+(25\frac{\sqrt{3}}{3}m\)
\(1.7+(25\frac{\sqrt{2}}{2}m\)
Correct answer is B
Hint: Make a sketch forming a right angled triangle. Let x = height of the tree above the man. Such that x/25 = sin 30.
x = 12.5m
The height of the tree = 12.5 + 1.7
= 14.2 m
1
2
3
4
Correct answer is C
\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\)
r = 4 units; centre (\(\alpha\), -2).
\((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\)
\((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\)
\(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\)
\(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\)
\((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)
\(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\)
\(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\)
\(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\)
Divide 4x\(^3\) - 3x + 1 by 2x - 1
2x2-x+1
2x2-x-1
2x2+x+1
2x2+x-1
Correct answer is D
No explanation has been provided for this answer.
8/5
8/3
72/25
56/9
Correct answer is C
Let the common ratio be r so that the first term is 2r.
Sum, s = \(\frac{a}{1-r}\)
ie. 8 = \(\frac{2r}{1-r}\)
8(1-r) = 2r,
8 - 8r = 2r
8 = 2r + 8r
8 = 10r
r = \(\frac{4}{5}\).
where common ratio (r) = \(\frac{second term(n_2)}{first term(a)}\),
r = \(\frac{n_2}{2r}\)
r = \(\frac{4}{5}\) and a = 2r or \(\frac{8}{5}\)
\(\frac{4}{5}\) * \(\frac{8}{5}\) = n\(_2\)
\(\frac{32}{25}\) = n\(_2\)
The sum of the first two terms = a + n\(_2\)
= \(\frac{8}{5}\) + \(\frac{32}{25}\)
= \(\frac{40 + 32}{25}\)
= \(\frac{72}{25}\)
Express \(\frac{1}{x^{3}-1}\) in partial fractions
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{x - 2}{x^{2} + x + 1})\)
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 2)}{x^{2} + x + 1})\)
\(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 1)}{x^{2} - x - 1})\)
Correct answer is A
\(\frac{1}{x^{3} - 1}\)
\(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\)
\(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\)
\(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\)
Comparing the two sides of the equation,
\(A + B = 0 ... (1)\)
\(A - B + C = 0 ... (2)\)
\(A - C = 1 ... (3)\)
From (3), \(C = A - 1\), putting that in (2),
\(A - B = -C \implies A - B = 1 - A\)
\(2A - B = 1 ... (4)\)
(1) + (4): \(3A = 1 \implies A = \frac{1}{3}\)
\(A = -B \implies B = -\frac{1}{3}\)\(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\)
= \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)