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 equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)
7y + 4x + 5 = 0
7y - 4x - 5 = 0
3y - 5x + 2 = 0
3y + 5x - 2 = 0
Correct answer is A
Equation: 4y = 7x + 3
\(\implies y = \frac{7}{4} x + \frac{3}{4}\)
Slope = coefficient of x = \(\frac{7}{4}\)
Slope of perpendicular line = \(\frac{-1}{\frac{7}{4}}\)
= \(\frac{-4}{7}\)
The perpendicular line passes (-3, 1)
\(\therefore\) Using the equation of line \(y = mx + b\)
m = slope and b = intercept.
\(y = \frac{-4}{7} x + b\)
To find the intercept, substitute y = 1 and x = -3 in the equation.
\(1 = \frac{-4}{7} (-3) + b\)
\(1 = \frac{12}{7} + b\)
\(b = \frac{-5}{7}\)
\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)
\(7y + 4x + 5 = 0\)
Find the distance between the points C(2, 2) and D(5, 6).
13 units
7 units
12 units
5 units
Correct answer is D
= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\)
= \(\sqrt{3^2 + 4^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= 5 units
Differentiate \(\frac{2x}{\sin x}\) with respect to x.
\(2 \cot x \sec x (1 + \tan x)\)
\(2 \csc x - x \cot x\)
\(2x \csc x + \tan x\)
\(2\csc x(1 - x\cot x)\)
Correct answer is D
No explanation has been provided for this answer.
If y = 8x\(^3\) - 3x\(^2\) + 7x - 1, find \(\frac{\mathrm d^2 y}{\mathrm d x^2}\).
48x - 6
11x\(^2\) + 6x - 7
32x + 7
24x\(^2\) - 6x + 7
Correct answer is A
y = 8x\(^3\) - 3x\(^2\) + 7x - 1
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)
= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)
\(\begin{vmatrix} 6 & 2 \\ 1 & 6 \end{vmatrix}\)
\(\begin{vmatrix} \frac{2}{11} & \frac{1}{12}\\ \frac{3}{20} & \frac{1}{10} \end{vmatrix}\)
\(\begin{vmatrix} -3 & 2 \\ -1 & -6 \end{vmatrix}\)
\(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20}\end{vmatrix}\)
Correct answer is D
\(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\)
|A| = (3 x 6) - (-2 x 1)
= 18 + 2
= 20.
A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\)
= \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)
= \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)