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.
Simplify the given expression \(\sqrt{\frac{1 - cos x}{1 + cos x}}\)
\(\frac{1 - cos x}{sin x}\)
1 - cos x
sin x
1 + cos x
\(\frac{1 + cos x}{sin x}\)
Correct answer is A
\(\sqrt{\frac{1 - cos x}{1 + cos x}}\) = a
a2 = \(\frac{1 - cosx}{1 + cosx}\)
\(\frac{1 - cosx}{1 + cosx}\) = \(\frac{1 - cosx}{1 - cosx}\)
= \(\frac{(1 - cosx)^2}{cos^2 x}\)
a2 = \(\frac{(1 - cos x)^2}{sin^2 x}\)
a = \(\frac{1 - cos x}{sin x}\)
Solve the system of equation 2x + y = 32, 33y - x = 27
(3, 2)
(-3, 2)
(3, -2)
(-3, -2)
(2, 2)
Correct answer is A
2x + y = 32, 33y - x = 27
2x + y = 25
33y + x = 33
x + y = 5
\(\frac{3y - x = 3}{4y = 8}\)
y = 2
If it is given that 5x + 1 + 5x = 150, then the value of x is equal to
3
4
1
2
\(\frac{1}{2}\)
Correct answer is D
5x + 1 + 5x = 150
5(5x) + 5x = 150
6(5x) = 150
5x = \(\frac{150}{6}\)
= 25
= 52
= 2
Find the solution of the equation x + 2\(\sqrt{x} - 8\) = 0
(4, 16)
(2, 4)
(4, 1)
(1, 16)
(16, 16)
Correct answer is A
x + 2\(\sqrt{x} - 8 = 0, Let \sqrt{x} = y\)
x = \(y^2\)
\(y^2 + 2y\) - 8 = 0
(y + 4)(x - 2) = 0
y = -4 or 2
x = 16 or 4
What will be the value of k so that the quadratic equation kx2 - 4x + 1 = 0 has two equal roots?
2
3
4
8
\(\frac{1}{4}\)
Correct answer is C
kx2 - 4x + 1 = 0, comparing with ax2 + bx + c = 0
a = k, b = -4, c = 1 for equal root b2 = 4ac
(-4)2 = 4k
k = \(\frac{16}{4}\)
= 4