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)