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 derivative of the function y = 2x\(^2\)(2x - 1) at the point x = -1?
18
16
-4
-6
Correct answer is B
y = 2x\(^2\)(2x - 1)
y = 4x\(^3\) - 2x\(^2\)
dy/dx = 12x\(^2\) - 4x
at x = -1
dy/dx = 12(-1)\(^2\) - 4(-1)
= 12 + 4
= 16
234.00 cm3
526.50 cm3
166.00 cm3
687cm3
Correct answer is A
Let x represent total vol. 2 : 3 = 2 + 3 = 5
\(\frac{3}{5}\)x = 351
x = \(\frac{351 \times 5}{3}\)
= 585
Volume of smaller block = \(\frac{2}{5}\) x 585
= 234.00cm\(^3\)
\(\frac{-2}{7}\)
\(\frac{7}{6}\)
\(\frac{-6}{7}\)
2
Correct answer is D
Line: 2y+8x-17=0
recall y = mx + c
2y = -8x + 17
y = -4x + \(\frac{17}{2}\)
Slope m\(_1\) = 4
parallel lines: m\(_1\). m\(_2\) = -4
where Slope ( -4) = \(\frac{y_2 - y_1}{x_2 - x_1}\) at points (-1, -p) and (-2,2)
-4( \(x_2 - x_1\) ) = \(y_2 - y_1\)
-4 ( -2 - -1) = 2 - -p
p = 4 - 2 = 2
13 cm
4 cm
6 cm
7 cm
Correct answer is A
Area of Trapezium = 1/2(sum of parallel sides) * h
91 = \(\frac{1}{2}\) (5 + 9)h
cross multiply
91 = 7h
h = \(\frac{91}{7}\)
h = 13cm
Determine the maximum value of y=3x\(^2\) + 5x - 3
6
0
2
No correct option
Correct answer is D
y=3x\(^2\) + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = \(\frac{-5}{6}\)
Maximum value: 3 \( ^2{\frac{-5}{6}}\) + 5 \(\frac{-5}{6}\) - 3
3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3
Using the L.C.M. 36
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
No correct option