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.
Find the value of sin 45° - cos 30°
\(\frac{2+\sqrt{3}}{4}\)
\(\frac{\sqrt{2}+\sqrt{3}}{4}\)
\(\frac{\sqrt{2}+\sqrt{3}}{2}\)
\(\frac{\sqrt{2}-\sqrt{3}}{2}\)
Correct answer is D
\(Sin 45 - cos 30\\
\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\\
=\frac{\sqrt{2}-\sqrt{3}}{2}\)
If y = 3 cos 4x, dy/dx equals?
6 sin 8x
-24 sin 4x
12 sin 4x
-12 sin 4x
Correct answer is D
y = 3 cos 4x
Applying product rule:
\(\frac{dy}{dx}\) = u\(\frac{dv}{dx}\) + v\(\frac{du}{dx}\)
Where u = 3 and v = cos 4x
\(\frac{du}{dx}\) of 3 = 0 and
\(\frac{dv}{dx}\) of cos 4x = -4sin 4x
\(\frac{dy}{dx}\) = 3 \(\times\) -4 sin 4x + cos 4x \(\times\) 0
\(\frac{dy}{dx}\) = -12 sin 4x
What is the value of r if the distance between the point (4,2) and (1,r) is 3 units?
1
2
3
4
Correct answer is B
A(4,2) and B(1,r), AB = 3 units
3 = √ (x2- x1)2 + (y2-y1)2
3 = √ (4-1)2 + (2-r)2
3 = √ 32 + (2-r)2
3 = √ 9 + 4 – 4r + r2
3 = √ r2 - 4r + 13
9 = r2 - 4r + 13 By squaring both sides
r2 - 4r + 4 = 0
(r-2)(r-2) = 0
r = 2
What is the value of p if the gradient of the line joining (-1,p) and (p, 4) is
| 2 |
| 3 |
-2
-1
1
2
Correct answer is D
| Gradient = | y2 - y1 |
| x2 - x1 |
| = | 4-p | ||
| p-(-1) |
| = | 4-p | ||
| p+1 |
| p = | 10 |
| 5 |
A circle of radius 4 cm and the center 0
The perpendicular bisector of the chords
A straight line passing through 0
A circle of radius 6 cm and with center 0
Correct answer is A
No explanation has been provided for this answer.