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.
If \(3x - \frac{1}{4})^{\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is
x > \(\frac{1}{3}\)
x < \(\frac{1}{3}\)
x < \(\frac{1}{4}\)
x < \(\frac{9}{16}\)
x > \(\frac{9}{16}\)
Correct answer is E
\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)
= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\)
= \(3x - 2 > \frac{1}{4} - x\)
= \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\)
\(x > \frac{9}{16}\)
The minimum point on the curve y = x2 - 6x + 5 is at
(1, 5)
(2, 3)
(-3, -4)
(3, -4)
Correct answer is D
Given the curve \(y = x^{2} - 6x + 5\)
At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\)
\(2x - 6 = 0 \implies x = 3\)
Since 3 > 0, it is a minimum point.
When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\)
Hence, the turning point has coordinates (3, -4).
\(\sqrt{3}\) units
3 units
\(\sqrt{41 + 20 \sqrt{2}}\) units
\(\sqrt{41 - 20 \sqrt{2}}\) units
Correct answer is C
Force to the east = 5 units
force to the North - east = 4 units.
Resultant of the two forces is the square root
52 + 42 = 41 and plus the sum of its resistance
5 x 4\(\sqrt{2}\) = 20\(\sqrt{2}\)
= \(\sqrt{41 + 20 \sqrt{2}}\) units
Simplify \(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\)
\(\frac{4ab}{a - b}\)
\(\frac{-4ab}{a^2 - b^2}\)
\(\frac{-4ab}{a^{-2} - b}\)
\(\frac{4ab}{a^{-2} - b^{-2}}\)
Correct answer is B
\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\)
applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b)
= (a = b + a = b)(a = b - a = b)
2a(-2b) = -4ab
= \(\frac{-4ab}{(a + b)(a - b)}\)
= \(\frac{-4ab}{a^2 - b^2}\)
125o
82o
135o
105o
60o
Correct answer is B
Sum of interior angles of a polygon = (2n - 4) x 90
where n is the no. of sides
sum of interior angles of the quad. = ({2 x 4} -4) x 90
(8 - 4) x 90 = 4 x 90o
= 360o
(p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o
p + 10o + p - 30o + 2p - 45o + p + 15 = 360o
5p = 360o + 50o
= 82o