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JEE Questions for Maths Applications Of Integrals Quiz 3 - MCQExams.com
JEE
Maths
Applications Of Integrals
Quiz 3
Area bounded by the curve
x
= 0 and
x
+ 2|y| = 1 is
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0%
1/4
0%
1/2
0%
1
0%
2
The area (in sq units) of the region bounded by
x
= - 1,
x
= 2, y =
x
2
+ 1 and y = 2
x
- 2 is
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0%
10
0%
7
0%
8
0%
9
The area enclosed between the curves y =
x
3
and y = √
x
is
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0%
5/3
0%
5/4
0%
5/12
0%
12/5
The area enclosed by the curves y = sin
x
+ cos
x
and y = |cos
x
- sin
x
| over the interval [0, π/2] is
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0%
4(√2 - 1)
0%
2√2(√2 - 1)
0%
2(√2 + 1)
0%
2√2(√2 + 1)
The area (in sq units) bounded by the curves y = √
x
, 2y -
x
+ 3 = 0, X - axis and lying in the first quadrant is
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0%
9
0%
6
0%
18
0%
27/4
The area (in sq units) of the region bounded by the curves y
2
= 4a
x
and
x
2
= 4ay, a > 0 is
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0%
0%
2)
0%
0%
0%
None of these
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0%
0%
2)
0%
0%
None of these
The area bounded by the curves y = e
x
, y = e
-
x
, the ordinates
x
= 0 and
x
= 1 is given by
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0%
e + e-1 - 2
0%
e - e-1
0%
e + e-1
0%
e + e-1 + 2
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0%
0%
2)
0%
0%
The area bounded between the parabolas
x
2
= y/4 and
x
2
= 9y and the straight line y = 2 is
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0%
0%
2)
0%
0%
The area of the region bounded by the curves y =
x
3
, y = 1/
x
,
x
= 2 is
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0%
4 - loge 2
0%
1/4 + loge 2
0%
3 - loge 2
0%
15/3 - loge 2
If the straight line
x
= b divide the area enclosed by y = (1 -
x
)
2
, y = 0 and
x
= 0 into two parts R
1
(0 ≤
x
≤ b) and R
2
(b ≤
x
≤such that R
1
- R
2
= 1/4. Then, b equals
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0%
3/4
0%
1/2
0%
1/3
0%
1/4
The area in the positive quadrant enclosed by the circle
x
2
+ y
2
= 4, the line
x
= y √3 and X - axis is
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0%
π/2
0%
π/4
0%
π/3
0%
π
Find the area of a curve
x
y = 4, bounded by the lines
x
= 1 and
x
= 3 and X - axis
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0%
log 12
0%
log 64
0%
log 81
0%
log 27
The area (in sq units) bounded by the curve y = 1 + log
e
x
, the X - axis and the straight line
x
= e is equal to
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0%
3e - 2
0%
e
0%
e - 1/e
0%
e + 1/e
The area enclosed by y = 3
x
- 5, y = 0,
x
= 3 and
x
= 5 is
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0%
12 sq units
0%
13 sq units
0%
13 1/2 sq units
0%
14 sq units
The area of the region bounded by the curves y
2
=
x
and y = |
x
| is
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0%
1/3 sq units
0%
1/6 sq units
0%
2/3 sq units
0%
1 sq units
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0%
0%
2)
0%
0%
The given figure shows a ∆AOB and the parabola y =
x
2
. The ration of the area of the ∆AOB to the area of the region AOB of the parabola y =
x
2
is equal to
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0%
3/5
0%
3/4
0%
7/8
0%
5/6
0%
2/3
The area bounded by = |sin
x
|, X - axis and the lines |
x
| = π is
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0%
2 sq units
0%
3 sq units
0%
4 sq units
0%
None of these
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