Application of Integrals Test

Result

Application of Integrals Test - 8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The area bounded by the angle bisectors of the lines x²− y²+ 2y = 1 and x + y = 3 is

A
2 sq. units

B
6 sq. units

C
3 sq.units

D
4 sq. units

Solution

The angle bisectors of the line given by x²− y²+ 2y = 1 are x = 0 , y = 1. Required area : = 1/2 .2.2 = 2 sq. units.
Question 2
1 / -0

The area bounded by x = sin t and y = cos t +3 , where -2008 < t < 2008 , is equal to

A
3π

B
π

C
2π

D
none of these

Solution

We have : sin t = x , cos t = y – 3 ., therefore , sin2t + cos2t = 1, ⇒ x²+ (y − 3)²= 1 It is a circle with centre ( 0, 3 ) and radius is 1. Therefore , area = πr²= π × 1 = π.