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CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 12 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Applications Of Integrals
Quiz 12
If the line
x
=
α
divides the area of region
R
=
{
(
x
,
y
)
∈
R
2
:
x
3
≤
y
≤
x
,
0
≤
x
≤
1
}
into two equal parts, then
Report Question
0%
2
α
4
−
4
α
2
+
1
=
0
0%
α
4
+
4
α
2
−
1
=
0
0%
0
<
α
≤
1
2
0%
1
2
<
α
<
1
Explanation
Area
(
x
=
0
→
x
=
α
)
=
Area of
△
O
A
B
−
Area of
△
O
C
B
=
1
2
α
⋅
α
−
∫
α
0
x
2
d
x
=
α
2
2
−
α
2
4
Area
(
0
=
α
→
x
=
1
)
=
Area of
◻
B
A
E
F
−
Area of
◻
B
C
E
F
1
2
(
α
+
1
)
(
1
−
α
)
−
∫
1
α
x
3
d
x
1
−
α
2
2
−
(
1
4
−
α
2
4
)
According to the question:
α
2
2
−
α
4
4
=
1
2
−
α
2
2
−
1
4
+
α
4
4
⇒
2
α
4
−
4
α
2
+
1
=
0
... Option A
f
(
a
)
=
2
α
2
−
4
α
2
+
1
=
0
At
α
=
0
,
f
(
a
)
=
1
At
α
=
1
,
f
(
α
)
=
−
1
At
α
=
1
2
,
f
(
a
)
=
1
8
>
0
Therefore, Root lies in
α
∈
(
1
2
,
1
)
.. Option D.
The area bounded by the parabolas
y
2
=
4
a
(
x
+
a
)
and
y
2
=
−
4
a
(
x
−
a
)
is
Report Question
0%
16
3
a
2
sq units
0%
8
3
sq units
0%
4
3
a
2
sq units
0%
None of these
Explanation
y
2
=
4
a
(
x
+
a
)
y
=
±
√
4
a
x
+
4
a
2
As the area bounded by this curve is in -ve
x
quadrant . Therfore
y
=
−
√
−
4
a
x
+
4
a
2
Area bounded by this curve will be
∫
0
−
a
√
4
a
x
+
4
a
2
d
x
Area bounded by this curve will be
8
3
a
2
Similarly , The area bounded by
y
2
=
−
4
a
(
x
−
a
)
curve will be
∫
a
0
√
−
4
a
x
+
4
a
2
d
x
Area =
8
3
a
2
thus , total bounded area will be
16
3
a
2
The area of the region bounded by the curves
y
=
2
x
,
y
=
2
x
−
x
2
and
x
=
2
is
Report Question
0%
3
log
2
−
4
3
0%
3
log
2
−
4
9
0%
3
2
−
log
2
9
0%
None of these
Explanation
Given curves are
y
=
2
x
,
y
=
2
x
−
x
2
and
x
=
2
∴
Required area
=
∫
2
0
[
2
x
−
(
2
x
−
x
2
)
]
d
x
=
∫
2
0
(
2
x
−
2
x
+
x
2
)
d
x
=
[
2
x
log
2
−
x
2
+
x
3
3
]
2
0
=
4
log
2
−
4
+
8
3
−
1
log
2
=
3
log
2
−
4
3
.
The area of the portion of the circle
x
2
+
y
2
=
64
which is exterior to the parabola
y
2
=
12
x
, is
Report Question
0%
(
8
π
−
√
3
)
sq units
0%
16
3
(
8
−
√
3
)
sq units
0%
16
3
(
8
π
−
√
3
)
sq units
0%
None of the above
Explanation
Required shaded area
=
=
A
r
e
a
o
f
c
i
r
c
l
e
−
2
[
∫
4
0
2
√
3
√
x
d
x
+
∫
8
4
√
64
−
x
2
d
x
]
=
64
π
−
2
[
(
2
√
3
x
3
/
2
×
2
3
)
4
0
+
(
x
2
√
64
−
x
2
+
64
2
sin
−
1
x
8
)
8
4
]
=
64
π
−
64
√
3
−
32
π
+
16
√
3
+
32
π
3
=
128
π
3
−
16
√
3
3
=
16
3
(
8
π
−
√
3
)
sq units.
The area bounded byu the curve y = cos x, the line joining
(
−
π
/
4
,
cos
(
−
π
/
4
)
)
and (0, 2) and the line joining
(
π
/
4
,
cos
(
π
/
4
)
)
and (0, 2) is :
Report Question
0%
4
+
√
2
8
π
−
√
2
0%
4
+
√
2
8
π
+
√
2
0%
4
+
√
2
4
π
−
√
2
0%
4
+
√
2
4
π
+
√
2
Explanation
Area between the curves
f
(
x
)
and
g
(
x
)
is given by
∫
(
f
(
x
)
−
g
(
x
)
)
d
x
Thus, the area between the given curves is
A
=
∫
0
−
π
4
(
(
2
+
2
−
1
√
2
π
4
x
)
−
cos
x
)
d
x
+
∫
π
4
0
(
(
2
−
2
−
1
√
2
π
4
x
)
−
cos
x
)
d
x
⇒
A
=
2
+
1
√
2
2
π
2
−
√
2
=
\drac
4
+
√
2
8
π
−
√
2
Area common to the curves
y
2
=
a
x
and
x
2
+
y
2
=
4
a
x
is equal to
Report Question
0%
(
9
√
3
+
4
π
)
a
2
3
0%
(
9
√
3
+
4
π
)
a
2
0%
(
9
√
3
−
4
π
)
a
2
3
0%
None of these
Explanation
Area common to the curves
y
2
=
a
x
x
2
+
y
2
=
4
a
x
is equal to :
x
2
+
y
2
=
4
a
x
y
2
=
a
x
x
=
0
,
3
a
y
=
0
,
±
√
3
a
Shaded region
∫
3
0
√
a
x
d
x
+
∫
4
a
3
a
√
4
a
x
−
x
2
d
x
=
[
√
a
x
3
2
3
2
]
3
a
0
+
∫
4
a
3
a
√
(
2
a
)
2
−
(
x
−
2
a
)
2
d
x
=
2.
√
a
.3
√
3
.
a
√
a
3
+
[
x
−
2
a
2
√
4
a
x
−
x
2
+
4
a
2
2
sin
−
1
x
−
2
a
2
a
]
4
a
3
a
=
2
√
3
a
3
+
[
0
+
2
a
2
(
π
2
)
−
√
3
a
2
2
−
2
a
2
(
π
b
)
]
=
2
√
3
a
3
−
√
3
a
2
2
+
π
a
2
−
π
3
a
2
=
3
√
3
a
2
2
+
2
π
a
2
3
=
[
9
√
3
+
4
π
]
a
2
6
Required area is :
=
2
(
s
h
a
d
e
d
a
r
e
a
)
=
2
(
[
9
√
3
+
4
π
]
a
2
6
)
=
[
9
√
3
+
4
π
]
a
2
6
s
q
.
u
n
i
t
s
Let function
f
n
be the number of way in which a positive integer n can be written as an ordered sum of several positive integers. For example, for
n
=
3
,
f
3
=
3
,
s
i
n
c
e
,
3
=
3
,
3
=
2
+
1
and
3
=
1
+
1
+
1
. Then
f
5
=
Report Question
0%
4
0%
5
0%
6
0%
7
Area bounded by the curves y=
[
x
2
64
+
2
]
,
y
=
x
−
1
and x=0 above x-axis is where
[
.
]
denotes greatest
x
.
Report Question
0%
2 sq. unit
0%
3 sq. unit
0%
4 sq. unit
0%
none of these
The area bounded by circles
x
2
+
y
2
=
r
2
,
r
=
1
,
2
and rays given by
2
x
2
−
3
x
y
−
2
y
2
=
0
(
y
>
0
) is?
Report Question
0%
π
0%
3
π
4
0%
π
2
0%
π
4
On the real line R, we define two functions f and g as follows:
f
(
x
)
=
m
i
n
[
x
−
[
x
]
,
1
−
x
+
[
x
]
]
,
g
(
x
)
=
m
a
x
[
x
−
[
x
]
,
1
−
x
+
[
x
]
]
,
where [x] denotes the largest integer not exceeding x.
The positive integer n for which
∫
n
0
(
g
(
x
)
−
f
(
x
)
)
d
x
=
100
is?
Report Question
0%
100
0%
193
0%
200
0%
202
Explanation
Given,
f
(
x
)
=
m
i
n
[
x
−
[
x
]
,
1
−
x
+
[
x
]
]
=
m
i
n
[
f
,
1
−
f
]
and
g
(
x
)
=
m
a
x
[
x
−
[
x
]
,
1
−
x
+
[
x
]
]
=
m
a
x
[
f
,
1
−
f
]
where
f
is the fractional part of the function.
and
f
is always
0
≤
f
<
1
.
∴
if
0
<
f
<
0.5
, then
f
(
x
)
=
f
g
(
x
)
=
1
−
f
and If
0.5
<
f
<
1
, then
f
(
x
)
=
1
−
f
g
(
x
)
=
f
∫
n
0
(
g
(
x
)
−
f
(
x
)
)
d
x
=
n
∫
1
0
(
g
(
x
)
−
f
(
x
)
)
d
x
=
n
∫
0.5
0
(
1
−
2
x
)
d
x
+
n
∫
1
0.5
(
2
x
−
1
)
d
x
=
n
(
0.5
−
0.25
)
+
n
(
0.75
−
0.5
)
0.5
n
=
100
⟹
n
=
200
The parabola
y
2
=
4
x
+
1
divides the disc
x
2
+
y
2
≤
1
into two regions with areas
A
1
and
A
2
. Then
|
A
1
−
A
2
|
equals.
Report Question
0%
1
3
0%
2
3
0%
π
4
0%
π
3
Explanation
⇒
Area of semi-circle
=
π
/
2
suppose the area of the parabola between
−
1
/
4
to
0
=
P
.
Thus,
A
1
=
π
/
2
−
P
and
A
2
=
π
/
2
+
P
⇒
A
2
−
A
1
=
2
P
=
2
×
2
∫
0
−
1
/
4
(
√
4
x
+
1
)
d
x
⇒
A
2
−
A
1
=
4
×
2
(
4
x
+
1
)
3
/
2
3
×
4
=
2
/
3
(
1
−
0
)
=
2
/
3
The area bounded by the curves
y
=
sin
x
,
y
=
cos
x
and x-axis from
x
=
0
to
x
=
π
/
2
is
Report Question
0%
2
+
√
2
0%
√
2
0%
2
0%
2
−
√
2
The area bounded by min (|x|, |y|) = 2 and max (|x|, |y|) = 4 is
Report Question
0%
8 sq unit
0%
16 sq unit
0%
24 sq unit
0%
32 sq unit
Explanation
→
represents
m
a
x
(
|
x
|
,
|
y
|
)
=
4
→
represents
m
i
n
(
|
x
|
,
|
y
|
)
=
2
To solve this question we first need to understand the meaning of
m
i
n
(
|
x
|
,
|
y
|
)
=
2
and
m
a
x
(
|
x
|
,
|
y
|
)
=
4
m
i
n
(
|
x
|
,
|
y
|
)
=
2
means that either
|
x
|
=
2
and
|
y
|
≥
2
or
|
y
|
=
2
and
|
x
|
≥
2
Similarly,
m
a
x
(
|
x
|
,
|
y
|
)
=
4
means that either
|
x
|
=
4
and
|
y
|
≤
4
or
|
y
|
=
4
and
|
x
|
≤
4
As can be seen from the image of the graph plotted for this, we can see 4 squares each of size
2
×
2
So, total area=
4
×
(
2
×
2
)
=
16
Hence, correct answer is option
B
Area bounded by the curves
y
=
[
x
2
64
+
2
]
([.] denotes the greatest integer function)
y
=
x
−
1
and
x
=
0
above the x-axis is
Report Question
0%
2
0%
3
0%
4
0%
none of these
The area bounded by the curves
y
=
1
4
|
4
−
x
2
|
and
y
=
7
−
|
x
|
is
Report Question
0%
18
0%
32
0%
36
0%
64
Explanation
The graphs can be drawn as shown in the figure.
We need to find the area of the shaded portion. Since the area is symmetric about the y-axis, we will find the area on the right-hand side of the y-axis and later, multiply it by 2.
First, we find the area bounded by the straight line and the x-axis. Then, we subtract the area bounded by the parabola and the x-axis from our previous result.
Now, point A is the intersection point of both the graphs for
x
>
0
. So, we have
x
2
4
−
1
=
7
−
x
⟹
x
2
−
4
=
28
−
4
x
⟹
x
2
+
4
x
−
32
=
0
⟹
(
x
+
8
)
(
x
−
4
)
=
0
⟹
x
=
4
(since
x
>
0
)
⟹
y
=
7
−
x
=
7
−
4
=
3
So,
A
≡
(
4
,
3
)
Area bounded by the straight line and the x-axis
=
∫
4
0
(
7
−
x
)
d
x
=
(
7
x
−
x
2
2
)
4
0
=
28
−
4
2
2
=
20
Area bounded by the parabola
=
∫
2
0
(
1
−
x
2
4
)
d
x
+
∫
4
2
(
x
2
4
−
1
)
d
x
=
(
x
−
x
3
12
)
2
0
+
(
x
3
12
−
x
)
4
2
=
4
So, the shaded area on the right hand side becomes
(
20
−
4
)
=
16
∴
The total area
=
2
×
16
=
32
Consider two curves
C
1
:
(
y
−
√
3
)
2
=
4
(
x
−
√
2
)
and
C
2
:
x
2
+
y
2
=
(
6
+
2
√
2
)
x
+
2
√
3
y
−
6
(
1
+
√
2
)
then
Report Question
0%
C
1
a
n
d
C
−
2
touch each other only at one point.
0%
C
1
a
n
d
C
−
2
touch each other exactly at two points.
0%
C
1
a
n
d
C
−
2
intersect (but do not touch) at exactly two points.
0%
C
1
a
n
d
C
−
2
neither intersect nor touch each other.
The area bounded by
x
2
+
y
2
−
2
x
=
0
&
y
=
sin
π
x
2
in the upper half of the circle is?
Report Question
0%
π
2
−
4
π
0%
π
4
−
2
π
0%
π
−
8
π
0%
None
Explanation
The area bonded by
x
2
+
y
2
−
2
x
=
0
y
=
sin
π
x
2
In the upper half of the circle is
Required Area area = shaded area
=
1
2
π
r
2
−
∫
2
0
y
d
x
=
1
2
π
(
1
)
2
−
∫
2
0
sin
π
x
2
d
x
=
π
2
+
[
cos
π
x
2
π
2
]
2
0
=
π
2
+
2
π
[
cos
π
x
2
]
2
0
=
π
2
+
2
π
[
cos
π
−
cos
0
]
=
π
2
+
2
π
−
2
=
π
2
+
4
π
s
q
.
u
n
i
t
s
Hence the correct answer is
π
2
+
4
π
s
q
.
u
n
i
t
s
Consider two curves
C
1
:
y
=
1
x
a
n
d
C
2
:
y
=
ℓ
n
x
on the
x
y
plane. Let
D
1
denotes the region surrounded by
C
1
,
C
2
and the lines
x
=
1
and
D
2
denotes the region the region surrounded by
C
1
,
D
2
and the line
x
=
a
. If a
D
1
=
D
2
then the value of
′
a
′
-
Report Question
0%
e
2
0%
e
0%
1
0%
2
(
e
−
1
)
What is the area of the region bounded by the parabola
y
2
=
6
(
x
−
1
)
and
y
2
=
3
x
Report Question
0%
√
6
3
0%
2
√
6
3
0%
4
√
6
3
0%
5
√
6
3
Explanation
Solving
y
2
=
6
(
x
−
1
)
and
y
2
=
3
x
we get
6
(
x
−
1
)
=
3
x
⇒
x
=
2
Hence
y
=
±
√
6
y
2
=
6
(
x
−
1
)
⇒
x
=
1
+
y
2
6
and
y
2
=
3
x
⇒
x
=
y
2
3
Area
=
∫
√
6
−
√
6
(
1
+
y
2
6
−
y
2
3
)
d
y
=
2
∫
√
6
0
(
1
−
y
2
6
)
d
y
=
2
[
y
−
y
3
18
]
√
6
0
=
2
×
2
√
6
3
=
4
√
6
3
Area
=
4
√
6
3
Area common to the circle
x
2
+
y
2
=
64
and the parabola
y
2
=
4
x
is
Report Question
0%
16
3
(
4
π
+
√
3
)
0%
16
3
(
8
π
+
√
3
)
0%
16
3
(
4
π
−
√
3
)
0%
n
o
n
e
o
f
t
h
e
s
e
The area bounded by the curves
x
=
a
cos
3
t
,
y
=
a
sin
3
t
is
Report Question
0%
3
π
a
2
8
0%
3
π
a
2
16
0%
3
π
a
2
32
0%
None of the above
Explanation
x
=
a
cos
3
t
,
y
=
a
sin
2
t
x
2
3
+
y
2
3
=
a
2
3
A
=
∫
2
π
0
x
d
y
A
=
a
2
∫
2
π
0
cos
3
t
×
3
sin
2
t
×
cos
t
d
t
=
3
a
2
∫
2
π
0
cos
2
t
×
sin
2
t
d
t
→
(
1
)
similarily
A
=
∫
2
π
0
y
d
x
=
3
a
2
∫
2
π
0
sin
2
t
×
cos
2
t
d
t
→
(
2
)
Adding (1) and (2)
$$2A=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt
A
=
3
a
2
8
∫
2
π
0
sin
2
2
d
t
→
(
3
)
Similarily
A
=
3
a
2
8
∫
2
π
0
cos
2
t
d
t
→
(
4
)
By putting
t
=
t
+
π
4
in (3)
Adding (3) and (4)
2
A
=
3
a
2
8
∫
2
π
0
d
t
A
=
3
a
2
8
π
L
e
t
f
(
x
)
=
2
−
|
x
−
1
|
a
n
d
g
(
x
)
=
(
x
−
1
)
2
,
t
h
e
n
Report Question
0%
area bounded by
f
(
x
)
and
g
(
x
)
is
7
6
0%
area bounded by
f
(
x
)
and
g
(
x
)
is
7
3
0%
area bounded by
f
(
x
)
g
(
x
)
and
x
−
axis is
5
3
0%
area bounded by
f
(
x
)
g
(
x
)
and
x
−
axis is
5
6
Explanation
Area of parabola w.r.t x axis
=
g
(
X
)
∫
2
0
(
x
−
1
)
2
d
x
⇒
[
(
x
−
1
)
3
3
]
2
0
=
1
3
+
1
2
=
2
3
Area of
f
(
x
)
=
2
×
1
+
1
2
×
2
×
1
⇒
2
+
1
=
3
Area bounded by
f
(
x
)
and
g
(
x
)
=
3
−
2
3
=
7
3
In the square
ABCD
, the "shaded" region is the intersection of two circular regions centered at
B
and
D
respectively. If
AB= 10
, then what is the area of the shaded region?
Report Question
0%
25
(
π
−
2
)
0%
50
(
π
−
2
)
0%
25
π
0%
50
π
0%
40
π
(
5
−
√
2
)
The area bounded by the curves
y
=
(
x
−
1
)
2
,
y
=
(
x
+
1
)
2
and
y
=
1
4
is
Report Question
0%
1
3
s
q
u
n
i
t
0%
2
3
s
q
u
n
i
t
0%
1
4
s
q
u
n
i
t
0%
1
5
s
q
u
n
i
t
Find area curved by three circles
Report Question
0%
(
5
π
−
3
√
3
)
u
n
i
t
s
2
0%
(
5
π
+
4
√
3
)
u
n
i
t
s
2
0%
(
5
π
+
3
√
3
)
u
n
i
t
s
2
0%
(
5
π
+
3
√
2
)
u
n
i
t
s
2
Explanation
Simple area concept
If
f
(
x
)
=
max
{
sin
x
,
cos
x
,
1
2
}
, then the area of the region bounded by the curves
y
=
f
(
x
)
,
x
−
axis
y
−
axis and
x
=
2
π
is
Report Question
0%
(
5
π
12
+
3
)
.sq.unit
0%
(
5
π
12
+
√
2
)
.sq.unit
0%
(
5
π
12
+
√
3
)
.sq.unit
0%
(
5
π
12
+
√
2
+
√
3
)
.sq.unit
Explanation
∵
f
(
x
)
=
max
{
sin
x
,
cos
x
,
1
2
}
Interval value of
f
(
x
)
For
0
≤
x
<
π
4
,
cos
x
For
π
4
≤
x
<
5
π
6
,
sin
x
For
5
π
6
≤
x
<
5
π
3
,
1
2
For
5
π
3
≤
x
<
2
π
,
cos
x
Hence, required area
=
∫
π
4
0
cos
x
d
x
+
∫
5
π
6
π
4
sin
x
d
x
+
∫
5
π
3
5
π
6
1
2
d
x
+
∫
2
π
5
π
3
cos
x
d
x
=
[
sin
x
]
π
4
0
−
[
cos
x
]
5
π
6
π
4
+
1
2
[
x
]
5
π
3
5
π
6
+
[
sin
x
]
2
π
5
π
3
=
(
1
√
2
−
0
)
−
(
−
√
3
2
−
1
√
2
)
+
1
2
(
5
π
3
−
5
π
6
)
+
(
0
+
√
3
2
)
On simplification, we get
=
(
5
π
12
+
√
2
+
√
3
)
.sq.unit.
The parabola
y
=
x
2
2
divides the circle
x
2
+
y
2
=
8
into two parts. Find the area of both parts.
Report Question
0%
6
π
+
4
3
,
2
π
−
4
3
0%
6
π
−
4
3
,
2
π
+
4
3
0%
6
π
+
2
3
,
2
π
−
2
3
0%
6
π
−
2
3
,
2
π
+
2
3
Explanation
Let us first find the x-values of the points of intersection using the given equations of parabola
y
=
x
2
2
and circle
x
2
+
y
2
=
8
as follows:
x
2
+
y
2
=
8
⇒
x
2
+
(
x
2
2
)
2
=
8
(
∵
y
=
x
2
2
)
⇒
x
2
+
x
4
4
=
8
⇒
4
x
2
+
x
4
=
32
⇒
x
4
+
4
x
2
−
32
=
0
⇒
x
4
+
8
x
2
−
4
x
2
−
32
=
0
⇒
x
2
(
x
2
+
8
)
−
4
(
x
2
+
8
)
=
0
⇒
(
x
2
−
4
)
(
x
2
+
8
)
=
0
⇒
(
x
2
−
4
)
=
0
,
(
x
2
+
8
)
=
0
⇒
x
2
=
4
,
x
2
=
−
8
⇒
x
=
±
√
4
⇒
x
=
±
2
Let
A
1
be the area of the region inside the circle and above the parabola and
A
2
be the area of the region inside the circle and below the parabola. Then we have,
A
1
=
∫
2
−
2
(
√
8
−
x
2
−
1
2
x
2
)
d
x
=
2
∫
2
0
(
√
8
−
x
2
−
1
2
x
2
)
d
x
=
2
[
1
2
×
8
sin
−
1
(
2
√
8
)
+
1
2
×
2
√
8
−
2
2
−
1
2
[
1
3
x
3
]
2
0
]
=
8
sin
−
1
(
1
√
2
)
+
2
√
4
−
8
3
=
8
×
π
4
+
4
−
8
3
=
2
π
+
4
3
We know that the area of the circle is
π
r
2
, therefore the area of the circle
x
2
+
y
2
=
8
with radius
r
=
√
8
is:
π
(
√
8
)
2
=
8
π
Thus, we have
A
2
=
8
π
−
(
2
π
+
4
3
)
=
6
π
−
4
3
Hence, the area of parabola and circle is
2
π
+
4
3
and
6
π
−
4
3
respectively.
The area enclosed by the curve
y
=
√
(
4
−
x
2
)
,
y
≥
√
2
sin
(
x
π
2
√
2
)
and x-axis is divided by y-axis in the ratio.
Report Question
0%
π
2
−
8
π
2
+
8
0%
π
2
−
4
π
2
+
4
0%
π
−
4
π
+
4
0%
2
π
2
π
2
+
2
π
−
8
Explanation
y
=
√
4
−
x
2
…
(given)
⋯
(
i
)
⇒
y
2
−
4
x
2
⇒
x
2
+
y
2
=
4
y
≥
√
2
sin
(
π
x
2
√
2
)
(given) ... (ii)
To find the period,
=
2
π̸
π̸
×
2
√
2
=
4
√
2
⇒
period.
=
4
√
2
2
=
2
√
2
=
2
×
1.732
[
3.4
]
To find the intersection point using (i) and (ii)
x
=
√
2
Area of circle
=
π
r
2
=
π
×
y
=
4
π
Area of circle
=
π
=
A
1
.. (iii)
from (-2,0) to (0,0)
A
2
=
∫
√
2
0
√
4
−
x
2
−
√
2
sin
(
π
2
√
2
x
)
d
x
⇒
∫
√
2
0
√
4
−
x
2
d
x
−
√
2
∫
√
2
0
sin
[
π
x
2
√
2
]
d
x
[
∴
√
a
2
−
x
2
d
x
=
x
2
√
a
2
−
x
2
+
a
2
2
sin
−
1
x
a
]
=
[
x
2
√
4
−
x
2
+
4
2
sin
−
1
x
2
]
√
2
0
−
√
2
(
π
2
√
2
)
[
−
cos
(
π
x
2
√
2
)
]
√
2
0
=
[
2
2
π
√
2
+
4
2
sin
−
1
[
1
√
2
]
]
+
2
×
2
π
|
cos
(
π
x
2
√
2
)
|
0
=
[
1
+
4
2
sin
−
1
[
1
√
2
]
]
+
4
π
[
cos
(
π
2
√
2
×
√
2
)
−
cos
0
]
⇒
[
1
+
π
/
2
−
4
/
π
]
⇒
2
π
+
π
2
−
8
2
π
.. (iv)
Using equation (iii) and (iv).
A
1
A
2
=
π
×
2
π
2
π
+
π
2
−
8
=
2
π
2
2
π
+
π
2
−
8
Answer (D)
If
k
=
2
then
f
(
x
)
attains point of inflection at
Report Question
0%
0
0%
√
2
0%
−
√
2
0%
None of these
Area bounded by
|
x
−
1
|
≤
2
and
x
2
−
y
2
=
1
, is
Report Question
0%
6
√
2
+
1
2
In
|
3
+
2
√
2
|
0%
6
√
2
+
1
2
In
|
3
−
2
√
2
|
0%
6
√
2
−
In
|
3
+
2
√
2
|
0%
n
o
n
e
o
f
t
h
e
s
e
Explanation
Given :
|
x
−
1
|
≤
2
and
x
2
−
y
2
=
1
We have to find area bounded by both curve.
As we know
|
x
|
≤
a
means
−
a
≤
x
≤
a
So
|
x
−
1
|
≤
2
−
2
≤
x
−
1
≤
2
−
2
≤
x
−
1
≤
2
−
2
+
1
≤
x
≤
2
+
1
−
1
≤
x
≤
3
Area of curve
|
x
−
1
|
≤
2
Area
=
2
∫
1
3
√
x
2
−
1
d
x
∵
∫
√
x
2
−
a
2
d
x
=
x
2
=
a
2
2
l
n
(
x
+
√
x
2
−
a
2
)
=
2
[
x
2
√
x
2
−
1
−
1
2
i
n
(
x
+
√
x
2
−
1
)
]
3
1
2
[
3
2
√
8
−
1
2
i
n
(
3
+
√
8
)
]
−
2
[
1
2
√
0
−
1
2
i
n
(
1
+
0
)
]
6
√
2
i
n
(
3
+
2
√
2
)
−
0
A
=
6
√
2
−
i
n
(
3
+
2
√
2
)
Consider the two curves
C
1
:
y
2
=
4
x
C
2
:
x
2
+
y
2
−
6
x
+
1
=
0
Then, the area of region between these curves?
Report Question
0%
20
3
−
2
π
0%
10
3
−
2
π
0%
20
3
−
π
0%
10
3
−
π
The area enclosed between the curve
y
=
x
3
and
y
=
√
x
is
Report Question
0%
5
3
0%
5
4
0%
5
12
0%
None of these
Explanation
Given two curves are
y
=
x
3
and
y
=
√
x
Both curves intersect at point
(
1
,
1
)
Now equating the curves we get,
x
3
=
√
x
;
when
x
=
1
∴
The curve represented by
y
=
√
x
is upward
to the curve represented by
y
=
x
3
∴
Area enclosed by the curves= Area of the shaded region
⇒
Area enclosed by the curves
=
∫
0
1
(
√
x
−
x
3
)
d
x
=
i
n
t
0
1
x
1
2
d
x
−
∫
0
2
x
3
d
x
=
[
x
3
/
2
3
2
]
1
0
−
[
x
4
4
]
1
0
=
[
(
1
)
3
/
2
3
2
−
0
]
−
[
(
1
)
4
4
−
0
$
]
=
1
3
2
−
1
4
$$=\dfrac{2}{3}-\dfrac{1}{4}$
$
=
8
−
3
12
∴
Area enclosed by the curves
=
5
12
sq. units
The area between the curves y=tan x, cot x and axis in the interval
[
0
,
π
/
2
]is ?
Report Question
0%
log
2
0%
log
3
0%
log
5
0%
none of these
Explanation
y
=
t
a
n
x
,
y
=
c
o
t
x
are symmetrical about
x
=
π
4
,where they intersect
π
4
R
e
q
A
r
e
a
=
2
|
∫
t
a
n
x
d
x
|
π
4
|
2
∫
c
o
t
x
d
x
|
=
2
×
|
(
l
n
s
e
c
π
4
)
−
l
n
s
e
c
0
|
π
22
×
|
l
n
√
2
−
0
|
=
2
×
1
2
×
l
n
2
=
l
n
2
The area bounded by the curves
y
=
sin
(
x
−
[
x
]
)
,
y
=
sin
1
and the x-axis is
Report Question
0%
sin
1
0%
1
−
sin
1
0%
1
+
sin
1
0%
None of these
The area (in square units) bounded by the curves
y
=
cos
−
1
|
cos
x
|
and
y
=
(
cos
−
1
|
cos
x
|
)
2
,
x
∈
[
0
,
π
]
is
Report Question
0%
4
3
+
π
2
4
(
π
3
−
1
)
0%
4
3
+
π
2
4
(
π
3
+
1
)
0%
2
3
+
π
2
4
(
π
3
−
1
)
0%
2
3
+
π
2
4
(
π
3
+
1
)
The area bounded by the curves
y
=
s
i
n
−
1
|
s
i
n
x
|
and
y
=
(
s
i
n
−
1
|
s
i
n
x
|
)
2
,
0
≤
x
≤
2
π
is
Report Question
0%
(
π
3
3
+
4
3
)
sq. unit
0%
(
π
3
6
−
π
2
2
+
4
3
)
sq. unit
0%
(
π
2
2
−
4
3
)
sq. unit
0%
(
π
2
6
−
π
4
+
4
3
)
sq. unit
Explanation
y
=
sin
−
1
|
sin
n
|
and
y
=
(
sin
−
1
|
sin
n
|
2
)
2
Required area
=
4
∫
0
1
[
x
−
(
sin
−
1
(
sin
n
)
)
2
]
d
n
+
4
∫
1
π
1
[
(
sin
−
1
(
sin
n
)
)
2
−
n
]
d
n
=
4
(
1
2
)
−
4
∫
0
1
(
sin
−
1
(
sin
x
)
)
2
d
n
+
4
∫
1
π
2
(
sin
−
1
(
sin
n
)
)
2
d
n
−
4
2
(
π
2
4
−
1
)
1
=
4
−
π
2
−
4
∫
0
1
(
sin
−
1
(
sin
n
)
)
2
d
n
+
4
∫
0
π
2
(
sin
−
1
(
sin
n
)
)
2
d
n
=
4
−
π
2
2
−
4
∫
0
1
n
2
d
n
+
4
∫
1
π
2
n
2
d
n
=
4
−
π
2
2
−
4
3
+
4
3
[
π
3
8
−
1
1
]
=
4
−
π
2
2
−
8
3
+
π
3
6
=
(
4
3
−
π
2
2
+
π
3
6
)
s
q
u
n
i
t
The area bounded by the curves
y
2
=
4
x
and
x
2
=
4
y
is :
Report Question
0%
32
3
0%
16
3
0%
8
3
0%
0
Explanation
x
2
=
4
y
y
2
=
4
x
x
=
2
√
y
=
4
×
2
√
y
y
2
=
8
√
y
⇒
y
4
−
8
y
=
0
then
y
=
0
,
4
x
2
=
4
y
we get
x
=
4
,
0
So, the points of intersection are
(
0
,
0
)
&
(
4
,
4
)
Area
=
∫
4
0
∫
x
2
/
4
2
√
x
d
y
d
x
=
∫
4
0
[
y
]
x
2
/
4
2
√
x
d
x
=
∫
4
0
[
x
2
4
−
2
√
x
]
d
x
=
[
x
3
12
−
4
3
⋅
x
3
/
2
]
4
0
=
[
4
3
12
−
4
3
(
4
)
3
/
2
]
=
16
3
sq. units.
The area bounded by
y
=
2
−
|
2
−
x
|
and
y
=
3
|
x
|
is :
Report Question
0%
4
+
3
ℓ
n
3
2
0%
4
−
3
ℓ
n
3
2
0%
3
2
+
ℓ
n
3
0%
1
2
+
ℓ
n
3
Explanation
y
=
2
−
|
2
−
x
|
y
=
3
|
x
|
1
can be rewriiten as
y
=
x
i
f
x
<
+
2
On solving these two equation, we get
x
=
√
3
and
x
=
3
A
r
e
a
=
∫
3
√
3
(
y
1
−
y
2
)
d
x
=
∫
3
√
3
(
2
−
|
2
−
x
|
)
−
3
|
x
|
d
x
=
∫
2
√
3
x
−
3
|
x
|
+
∫
3
2
4
−
x
−
3
|
x
|
d
x
[
x
2
2
−
3
log
e
x
]
2
√
3
+
[
4
x
−
x
2
2
−
3
log
e
x
]
3
2
=
1
2
+
3
log
e
√
3
2
+
12
−
9
2
−
3
log
e
3
−
8
+
2
+
3
log
e
2
=
1
2
+
3
ln
√
3
2
+
3
2
3
ln
2
3
2
+
3
ln
√
3
2
=
2
−
3
ln
√
3
=
4
−
3
ln
3
2
Area of the region defined by
1
≤
|
x
|
+
|
y
|
and
x
2
−
2
x
+
1
≤
1
−
y
2
is
k
π
then
k
=
.
.
.
.
.
s
q
units
Report Question
0%
3
4
0%
7
6
0%
128
5
0%
10
3
Area bounded by the curves
y
=
log
e
x
and
y
=
(
log
e
x
)
2
is ?
Report Question
0%
e
−
2
0%
3
−
e
0%
e
0%
e
−
1
The maximum area bounded by the curves
y
2
=
4
a
x
,
y
=
a
x
a
and
y
=
x
a
,
1
≤
a
≤
2
is
Report Question
0%
44
s
q
.
u
n
i
t
s
0%
74
s
q
.
u
n
i
t
s
0%
84
s
q
.
u
n
i
t
s
0%
114
s
q
.
u
n
i
t
s
The area bounded by the curves
y
=
x
e
x
,
y
=
x
e
−
x
and the line
x
=
1
, is
Report Question
0%
2
e
0%
1
−
2
e
0%
1
e
0%
1
−
1
e
The area (in sq.units) of the region
{
(
x
,
y
)
:
y
2
≥
2
x
and
x
2
+
y
2
≤
4
x
,
x
≥
0
,
y
≥
0
is
Report Question
0%
π
−
8
3
0%
π
−
4
√
2
3
0%
π
2
−
2
√
2
3
0%
π
−
4
3
Explanation
Consider
y
2
=
2
x
and
x
2
+
y
2
=
4
x
, on solving,
we get
x
2
+
2
x
=
4
x
x
2
=
2
x
⇒
x
=
0
,
2
The integral lies between
0
and
2
y
2
=
2
x
y
2
+
x
2
=
4
x
⇒
y
=
√
2
x
⇒
y
=
√
4
x
−
x
2
Area
=
∫
2
0
√
4
x
−
x
2
−
√
2
x
d
x
=
∫
2
0
√
2
2
−
(
2
−
x
)
2
−
∫
2
0
√
2
x
d
x
=
[
x
−
2
2
√
4
x
−
x
2
−
4
2
sin
−
1
(
x
−
2
2
)
]
2
0
−
[
√
2
3
/
2
x
3
/
2
]
2
0
=
|
2
sin
−
1
(
−
1
)
|
−
2
√
2
3
⋅
2
√
2
=
π
−
8
3
.
The area bounded by the curves
√
x
+
√
y
=
1
and
x
+
y
=
1
is ?
Report Question
0%
1
3
0%
1
6
0%
1
2
0%
5
6
0%
1
4
Explanation
Step -1: Finding the point of intersection.
Find the points of intersection of the curves
⟹
√
x
=
1
−
√
y
⟹
x
=
(
1
−
√
y
)
2
=
1
+
y
−
2
√
y
⟹
1
+
y
−
2
√
y
+
y
=
1
⟹
2
y
=
2
√
y
⟹
y
=
0
,
1
⟹
Point of intersections are
(
1
,
0
)
and
(
0
,
1
)
Step -2: Calculating the area .
√
y
=
1
−
√
x
⟹
y
=
1
+
x
−
2
√
x
⟹
f
1
(
x
)
=
x
−
2
√
x
+
1
⟹
f
2
(
x
)
=
1
−
x
⟹
Area
=
∫
1
0
(
f
1
(
x
)
−
f
2
(
x
)
)
d
x
⟹
∫
1
0
(
2
x
−
2
√
x
)
d
x
⟹
x
2
−
2
x
3
2
3
2
|
1
0
⟹
|
1
−
4
3
|
=
|
−
1
3
|
=
1
3
Hence, The required area is
1
3
and correct answer is option A .
In a system of three curves
C
1
,
C
2
and
C
3
,
C
1
is a circle whose equation is
x
2
+
y
2
=
4
.
C
2
is the locus of orthogonal tangents drawn on
C
1
.
C
3
is the intersection of perpendicular tangents drawn on
C
2
. Area enclosed between the curve
C
2
and
C
3
is-
Report Question
0%
8
π
s
q
.
u
n
i
t
s
0%
16
π
s
q
.
u
n
i
t
s
0%
32
π
s
q
.
u
n
i
t
s
0%
N
o
n
e
o
f
t
h
e
s
e
Area bounded between the curves
y
=
√
4
−
x
2
and
y
2
=
3
|
x
|
is/are?
Report Question
0%
π
−
1
√
3
0%
2
π
−
1
3
√
3
0%
2
π
−
√
3
3
0%
2
π
−
√
3
3
√
3
Area bounded by the curves
y
=
cos
−
1
(
sin
x
)
and
y
=
sin
−
1
(
sin
x
)
in the interval
[
0
,
π
]
is
Report Question
0%
π
2
16
0%
π
2
32
0%
π
2
4
0%
π
2
8
Area bounded between asymptomes of curves
f
(
x
)
and
f
−
1
(
x
)
is
Report Question
0%
4
0%
9
0%
16
0%
25
The area of the region bounded by the X-axis and the curves defined by
y
=
t
a
n
x
(
−
π
3
≤
x
≤
π
3
)
a
n
d
y
=
c
o
t
x
(
π
6
≤
x
≤
3
π
2
)
Report Question
0%
l
o
g
3
2
0%
l
o
g
√
3
2
0%
2
l
o
g
3
2
0%
l
o
g
(
3
√
2
)
The area bounded by the curves is
√
|
x
|
+
√
|
y
|
=
√
a
and
x
2
+
y
2
=
a
2
(where
a
>
0
) is
Report Question
0%
(
π
−
2
3
)
a
2
s
q
u
n
i
t
s
0%
(
π
+
2
3
)
a
2
s
q
u
n
i
t
s
0%
(
π
+
2
3
)
a
3
s
q
u
n
i
t
s
0%
(
π
−
2
3
)
a
3
s
q
u
n
i
t
s
0:0:1
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Answered
1
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Incorrect : 0
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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