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 y = 4x3 - 2x2 + x, find \(\frac{\delta y}{\delta x}\)
8x2 - 2x + 1
8x2 - 4x + 1
12x2 - 2x + 1
12x2 - 4x + 1
Correct answer is D
If y = 4x3 - 2x2 + x, then;
\(\frac{\delta y}{\delta x}\) = 3(4x2) - 2(2x) + 1
= 12x2 - 4x + 1
If \(\sin\theta = \frac{12}{13}\), find the value of \(1 + \cos\theta\)
\(\frac{25}{13}\)
\(\frac{18}{13}\)
\(\frac{8}{13}\)
\(\frac{5}{13}\)
Correct answer is B
No explanation has been provided for this answer.
\(\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}\)
\(\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}\)
\(\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}\)
\(\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}\)
Correct answer is A
y - 4x + 3 = 0
When y = 0, 0 - 4x + 3 = 0
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y - 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, -3)
Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;
\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)
\([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\)
\([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\)
\([\frac{3}{8}, \frac{-3}{2}]\)
The gradient of a line joining (x,4) and (1,2) is \(\frac{1}{2}\). Find the value of x
5
3
-3
-5
Correct answer is A
\(\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}\)
\(\frac{1}{2} = \frac{2 - 4}{1 - x}\)
1 - x = 2(2 - 4)
1 - x = 4 - 8
1 - x = -4
-x = -4 - 1
x = 5
Find the mid point of S(-5, 4) and T(-3, -2)
-4, 2
4, -2
-4, 1
4, -1
Correct answer is C
Mid point of S(-5, 4) and T(-3, -2) is
\([\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + 2)]\)
\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)
\([\frac{1}{2}(-8), \frac{1}{2}(2)]\)
\([-4, 1]\)