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.
The solution of the quadratic inequality (x3 + x - 12) ≥ 0 is
x ≥ -3 or x ≤ 4
x ≥ 3 or x ≥ -4
x ≤ 3 or x v -4
x ≥ 3 or x ≤ -4
Correct answer is B
(x3 + x - 12) ≥ 0
(x + 4)(x - 3) ≥ 0
Either x + 4 ≥ 0 implies x ≥ -4
Or x - 3 ≥ 0 implies x ≥ 3
∴ x ≥ 3 or x ≥ -4
Make L the subjects of the formula if \(\sqrt{\frac{42w}{5l}}\)
\(\sqrt{\frac{42w}{5d}}\)
\(\frac{42W}{5d^2}\)
\(\frac{42}{5dW}\)
\(\frac{1}{d}\sqrt{\frac{42w}{5}}\)
Correct answer is B
\(\sqrt{\frac{42w}{5l}}\)
square both side of the equation
\(d^2 = \left(\sqrt{\frac{42W}{5l}}\right)^2\\
d^2 = \frac{42W}{5l}\\
5ld^2=42W\\
l = \frac{42W}{5d^2}\)
3/4
-9/2
45/4
-3/4
Correct answer is A
x⊕y = xy + x + y
= -3/4 (6) + (-3/4) + 6
= -9/2 - 3/4 + 6
= (-18-3+3+24) / 4
= 3/4
Determine the value of \(\int_0 ^{\frac{\pi}{2}
}(-2cos x)dx\)
-2
-1/2
-3
-3/2
Correct answer is A
\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\\
=(-2sin\frac{\pi}{2}+c+2sin0-c)\\
=-2sin90+c+2sin0-c\\
=-2(1)+2(0)\\
=-2\)
Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value
2/3
1
- 2/3
-1
Correct answer is B
f(x) = 2x3 - x2 - 4x – 4
f’(x) = 6x2 - 2x – 4
As f’(x) = 0
Implies 6x2 - 2x – 4 = 0
3x – x – 2 = 0 (By dividing by 2)
(3x – 2)(x + 1) = 0
3x – 2 = 0 implies x = -2/3
Or x + 1 = 0 implies x = -1
f’(x) = 6x2 - 2x – 4
f’’(x) = 12x – 2
At max point f’’(x) < 0
∴f’’(x) = 12x – 2 at x = -1
= 12(-1) – 2
= -12 – 2 = -14
∴Max at x = 1