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.
Evaluate without using tables sin(-1290º)
\(\frac{3}{2}\)
-\(\frac{3}{2}\)
\(\frac{2}{2}\)
1
\(\frac{1}{2}\)
Correct answer is E
sin(-1290º) = -sin(1290º)
sin([3*360] + 210)
where sin 360 = 0 and sin 210 = 180 + 30 ⇔ -30º
-sin ([3* 0] + [-30])
-sin(-30)
sin30 = \(\frac{1}{2}\)
k = -7, 1 = -15
k = -15, 1 = -7
k = \(\frac{15}{3}\) , 1 = -7
k = \(\frac{7}{3}\) , 1 = -17
Correct answer is A
If k = -7 is put as -15, the equation x4 - kx3 + 10x2 + 1x - 3 becomes x4 - (7x3) + 10x2 + (15)-3 = x4 + 7x3 + 10x2 - 15x - 3
This equation is divisible by (x - 1) and (x + 2) with the remainder as 27
k = -7, 1 = -15
Simplify \(\frac{(a^2 - \frac{1}{a}) (a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)
a\(\frac{2}{3}\)
a-\(\frac{1}{3}\)
\(a^{2}\) + 1
a
a\(\frac{1}{3}\)
Correct answer is E
\(\frac{(a^2 - \frac{1}{a})(a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)
= \(\frac{(\frac{a^2 - 1}{a})(\frac{a^2 + 1}{a^{\frac{2}{3}}})}{a^{4} - \frac{1}{a^2}}\)
= \(\frac{a^4 - 1}{a^{\frac{5}{3}}}\) x \(\frac{a^2}{a^4 - 1}\)
= a\(\frac{1}{3}\)
Without using tables, simplify \(\frac{1n \sqrt{216} - 1n \sqrt{125} - 1n\sqrt{8}}{2(1n3 - 1n5)}\)
-3
3
\(\frac{3}{5}\)
\(\frac{-3}{2}\)
1n 6 - 2 1n 5
Correct answer is C
No explanation has been provided for this answer.
The locus of all points having a distance of 1 unit from each of the two fixed points a and b is
A line parallel to the line ab
A line perpendicular to the line ab through the mid-point of ab
A circle through a and b with centre at the mid-point of ab
A circle with centre at a and passes through b
A circle in a plane perpendicular to ab and centre at the mid-point of the line ab
Correct answer is B
The locus of all points having a distance of 1 unit from each of the two fixed points. a and b is: a perpendicular to the (ab) through the mid-points ab