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.
Evaluate \(\int_0^{\frac{\pi}{2}}sin2xdx\)
1
zero
-1/2
-1
Correct answer is A
\(\int_0^{\frac{\pi}{2}}\)sin 2x dx = [-1/2cos 2x + C]\(_0^{\frac{\pi}{2}}\)
=[-1/2 cos 2 * π/2 + C] - [-1/2 cos 2 * 0]
= [-1/2 cos π] - [-1/2 cos 0]
= [-1/2x - 1] - [-1/2 * 1]
= 1/2 -(-1/2) = 1/2 + 1/2 = 1
Find the area of the figure bounded by the given pair of curves y = x2 - x + 3 and y = 3
17/6 units (sq)
7/6 units (sq)
5/6 units (sq)
1/6 units (sq)
Correct answer is A
Area bounded by
y = x2 - x + 3 and y = 3
x2 - x + 3 = 3
x2 - x = 0
x(x -1) = 0
x= 0 and x = 1
\(\int^{1}_{0}\)(x2 - x + 3)dx = [1/3x3 - 1/2x2 + 3x]\(^1 _0\)
(1/3(1)3 - 1/2(1)2 + 3(1) - (1/3(0)3 - 1/2(0)2 + 3(0))
= (1/3 - 1/2 + 3)- 0
= \(\frac{2-3+18}{6}\)
= 17/6
The maximum value of the function
f(x) = 2 + x - x2 is
9/4
7/4
3/2
1/2
Correct answer is A
f(x) = 2 + x - x2
dy/dx = 1-2x
As dy/dx = 0
1-2x = 0
2x = 1
x = 1/2
At x = 1/2
f(x) = 2 + x - x2
= 2 + 1/2 -(1/2)2
= 2 + 1/2 - 1/4
= (8+2-1) / 4 = 9/4
36 π cm2/sec
18 π cm2/sec
6 π cm2/sec
3 π cm2/sec
Correct answer is C
Increase in radius dr/dt = 0.5
Area of the circular disc = πr2
Increase in area: dA/dr = 2πr
Rate of increase in area: dA/dt = dA/dr * dr/dt
when r = 6cm
= 2πr * 0.5
= 2 * π * 6 * 0.5
2 * π * 6 * 1/2 = 6 π cm2/sec
Find the derivative of y = sin(2x\(^3\) + 3x - 4)
cos (2x3 + 3x - 4)
-cos (2x3 + 3x - 4)
(6x2 + 3) cos (2x3 + 3x - 4)
-(6x2 + 3) cos (2x3 + 3x - 4)
Correct answer is C
y = sin (2x\(^3\) + 3x - 4)
let u = 2x\(^3\) + 3x - 4
∴du/dx = 6x\(^2\)
y = sin u
dy/du = cos u
dy/dx = du/dx * dy/du
∴dy/dx = (6x\(^2\) + 3) cos u
= (6x\(^2\) + 3)cos(2x\(^3\) + 3x - 4)