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CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Application Of Integrals
Quiz 8
Find the area of bounded by
y
=
sin
x
from
x
=
π
4
to
x
=
π
2
Report Question
0%
√
2
−
1
√
2
0%
1
2
0%
1
4
0%
None of these
Explanation
The area bounded is given as
∫
π
/
2
π
/
4
sin
x
d
x
−
cos
x
|
π
/
2
π
/
4
−
cos
π
2
+
cos
π
4
√
2
−
1
√
2
The area of the region by curves
y
=
x
log
x
and
y
=
2
x
−
2
x
2
=
Report Question
0%
1
/
12
0%
3
/
12
0%
7
/
12
0%
N
o
n
e
o
f
t
h
e
s
e
The area bounded by x-axis the curve
y
=
f
(
x
)
and the lines
x
=
1
,
x
=
b
equal to
(
√
(
b
2
+
1
)
−
√
2
)
f
o
r
a
l
l
b
>
1
,
t
h
e
n
f
(
x
)
Report Question
0%
√
(
x
−
1
)
0%
√
(
x
+
1
)
0%
√
(
x
2
+
1
)
0%
x
√
1
+
x
2
The area bounded by the curve
y
=
e
x
and the lines y = |x - 1|, x = 2 is given by :
Report Question
0%
e
2
+
1
0%
4
e
2
−
1
0%
e
2
−
2
0%
e
−
2
Explanation
Given,
y
=
e
x
,
y
=
|
x
−
1
|
,
x
=
2
→
e
x
=
|
x
−
1
|
→
e
x
−
(
1
−
x
)
=
0
intervals:
|
x
−
1
|
=
0
⇒
x
=
1
x
=
2
Area is given by,
A
=
∫
1
0
(
e
x
−
(
1
−
x
)
)
d
x
+
∫
2
1
(
e
x
−
(
1
−
x
)
)
d
x
=
∫
1
0
(
e
x
+
x
−
1
)
d
x
+
∫
2
1
(
e
x
+
x
−
1
)
d
x
=
[
e
x
+
x
2
2
−
x
]
1
0
+
[
e
x
+
x
2
2
−
x
]
2
1
=
e
+
1
2
−
1
+
(
e
2
−
2
+
2
)
−
(
e
−
1
2
+
1
)
=
e
2
−
2
Area bounded by the curve
y
2
(
2
a
−
x
)
=
x
3
and the line
x
=
2
a
, is
Report Question
0%
3
π
a
2
0%
3
π
a
2
2
0%
3
π
a
2
4
0%
π
a
2
4
Explanation
Given,
y
2
(
2
a
−
x
)
=
x
3
y
=
√
x
3
2
a
−
x
let,
x
=
2
a
sin
2
θ
d
x
=
4
a
sin
θ
cos
θ
d
θ
⇒
A
=
∫
2
a
0
√
x
3
2
a
−
x
d
x
A
=
∫
π
2
0
√
8
a
3
sin
6
θ
2
a
cos
2
θ
4
a
sin
θ
cos
θ
d
θ
=
8
a
2
∫
π
2
0
sin
4
θ
d
θ
=
8
a
2
∫
π
2
0
sin
2
θ
(
1
−
cos
2
θ
)
=
8
a
2
∫
π
2
0
(
1
−
cos
2
θ
2
)
−
1
4
sin
2
2
θ
d
θ
=
8
a
2
[
π
4
−
0
−
1
4
(
π
4
−
0
)
]
=
8
a
2
[
π
4
−
π
16
]
=
3
2
π
a
2
Area of the region bounded by
y
=
sin
−
1
|
sin
x
|
and
y
=
−
cos
−
1
|
cos
x
|
in the interval
[
0
,
2
π
]
is equal to
Report Question
0%
π
2
2
0%
π
2
0%
π
2
4
0%
2
π
2
The area bounded by the circles
x
2
+
y
2
=
r
2
,
r
=
1
,
2
and the rays given by
2
x
2
−
3
x
y
−
2
y
2
=
0
,
y
>
0
is
Report Question
0%
π
4
s
q
.
u
n
i
t
s
0%
π
2
s
q
.
u
n
i
t
s
0%
3
π
4
s
q
.
u
n
i
t
s
0%
π
s
q
.
u
n
i
t
s
The area bounding 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
The area enclosed between the curves
y
=
a
x
2
and
x
=
a
y
2
(
a
>
0
)
is
1
sq.unit, then the value of
a
is
Report Question
0%
1
/
√
3
0%
1
/
2
0%
1
0%
1
/
3
The area inside the parabola
5
x
2
−
y
=
0
but outside the parabola
2
x
2
−
y
+
9
=
0
, is
Report Question
0%
12
√
3
0%
6
√
3
0%
8
√
3
0%
4
√
3
Explanation
From given, we have,
5
x
2
−
y
=
0
2
x
2
−
y
+
9
=
0
we have boundaries,
5
x
2
=
2
x
2
+
9
3
x
2
=
9
x
=
±
√
3
A
=
∫
√
3
−
√
3
5
x
2
−
(
2
x
2
+
9
)
∫
√
3
−
√
3
5
x
2
−
(
2
x
2
+
9
)
=
∫
√
3
−
√
3
3
x
2
d
x
−
∫
√
3
−
√
3
9
d
x
=
6
√
3
−
18
√
3
=
−
12
√
3
∴
A
r
e
a
=
|
−
12
√
3
|
=
12
√
3
Area bounded by the curve
y
=
s
i
n
−
1
x
,
y
−
a
x
i
s
and
y
=
c
o
s
−
1
x
is equal to
Report Question
0%
(
2
+
√
2
)
0%
(
2
−
√
2
)
0%
(
1
+
√
2
)
0%
(
√
2
−
1
)
Explanation
Given,
y
=
sin
−
1
x
,
y
=
cos
−
1
x
,
y
−
a
x
i
s
⇒
sin
y
=
x
,
cos
y
=
x
⇒
sin
y
=
cos
y
⇒
tan
y
=
1
∴
y
=
tan
−
1
1
=
π
4
A
r
e
a
=
∫
π
4
0
cos
(
y
)
−
sin
(
y
)
d
y
=
∫
π
4
0
cos
(
y
)
d
y
−
∫
π
4
0
sin
(
y
)
d
y
=
1
√
2
−
(
−
1
√
2
+
1
)
=
√
2
−
1
The area of the figure bounded by the curves
y
=
ln
n
x
&
(
ln
n
x
)
2
is
Report Question
0%
e
+
1
0%
e
−
1
0%
3
−
e
0%
1
The area of the region bounded by x = 0, y = 0, x = 2, y = 2,
y
≤
e
x
and
y
≥
ℓ
n
x, is
Report Question
0%
6 - 4
ℓ
n
2
0%
4
ℓ
n
2 - 2
0%
2
ℓ
n
2 - 4
0%
6 - 2
ℓ
n
2
The area is bounded by
x
+
x
1
,
y
=
y
1
and
y
=
−
(
x
+
1
)
2
. Where
x
1
,
y
1
are the values of
x
,
y
satisfying the equation
s
i
n
−
1
x
+
s
i
n
−
1
y
=
−
π
will be (nearer to origin)
Report Question
0%
1
/
3
0%
3
/
2
0%
1
0%
2
/
3
Explanation
( Wrong question,
x
=
x
, instead of
x
+
x
1
)
Given that:
x
=
x
1
y
=
y
1
y
=
−
(
x
+
1
)
2
&
x
1
,
y
1
satisfy
sin
−
1
x
+
sin
−
1
y
=
z
We know:
−
z
2
≤
sin
−
1
x
≤
z
2
−
z
2
≤
sin
−
1
y
≤
z
2
∴
−
z
≤
sin
−
1
x
+
sin
−
1
y
≤
z
So, for
sin
−
1
x
+
sin
−
1
y
=
−
z
⇒
sin
−
1
x
=
−
π
2
&
sin
−
1
y
=
−
z
/
2
x
=
−
1
and
y
=
−
1
area required
=
1
−
|
∫
0
1
−
(
x
+
1
)
2
d
x
∣
=
1
−
|
∫
0
1
(
x
2
+
1
+
2
x
)
d
x
|
=
1
−
|
−
(
x
3
+
x
+
x
2
)
|
0
−
1
3
=
1
−
|
(
−
1
3
−
1
+
1
)
|
=
1
−
1
3
=
2
/
3
s
q
. Units
The area bounded by curves
y
=
|
x
|
−
1
and
y
=
−
|
x
|
+
1
is
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
First we form a table for different values of
x
and
y
for both equations as shown .
Here length of
A
C
=
2
units and distance(height) between vertices
D
and line
A
C
=
1
units and between vertices
B
and line
A
C
=
1
unit
∵
Area of triangle
=
1
2
×
b
×
h
∴
Area of
A
B
C
D
=
2
×
A
r
e
a
o
f
△
A
B
C
=
2
×
1
2
×
2
×
1
=
2
sq.units.
The area of the triangle formed by the lines joining the vertex of the parabola
x
2
=
12
y
to the ends of its latus rectum is-
Report Question
0%
16
sq. units
0%
12
sq. units
0%
18
sq. units
0%
24
sq. units
Explanation
Step 1: Make an appropriate diagram
Above figure shows a parabola of the form
x
2
=
4
a
y
Having vertex at
V
(
0
,
0
)
Focus at
S
(
0
,
a
)
Latus rectum is a line
A
B
passing through focus
S
and perpendicular to the axis of the parabola
Thus coordinates of end points of latus rectum is
A
(
−
2
a
,
a
)
and
B
(
2
a
,
a
)
Step 2: Find coordinates of end points of the triangle for given parabola
Given parabola is
x
2
=
12
y
⇒
4
a
=
12
∴
a
=
3
Therefore the coordinates of triangle are,
A
(
−
6
,
3
)
,
V
(
0
,
0
)
,
B
(
6
,
3
)
Area of
△
V
A
B
=
1
2
|
x
1
(
y
2
−
y
3
)
+
x
2
(
y
3
−
y
1
)
+
x
3
(
y
1
−
y
2
)
|
=
1
2
|
0
(
3
−
3
)
+
(
−
6
)
(
3
−
0
)
+
6
(
0
−
3
)
|
=
1
2
|
−
36
|
=
18
s
q
.
u
n
i
t
s
\
Hence, Option 'C' is correct
The area enclosed between the curves
y
=
l
o
g
e
(
x
+
e
)
,
x
=
l
o
g
e
(
1
y
)
and the x-axis is?
Report Question
0%
2
e
0%
e
0%
4
e
0%
None of these
The area (in sq. units) bounded by the parabola
y
=
x
2
−
1
, the tangent at the point (2,3) to it and the y-axis is:
Report Question
0%
14
3
0%
56
3
0%
8
3
0%
32
3
Explanation
Equation of tangent at
(
2
,
3
)
on
y
=
x
2
−
1
, is
y
=
(
4
x
−
5
)
.
.
.
.
.
.
.
(
i
)
∴
Required shaded area
=
a
r
(
△
A
B
C
)
−
3
∫
−
1
√
y
+
1
d
y
=
1
2
.
(
8
)
.
(
2
)
−
2
3
(
(
y
+
1
)
3
/
2
)
3
−
1
=
8
−
16
3
=
8
3
The area bounded by the curve
y
≤
x
2
+
3
x
,
0
≤
y
≤
4
,
0
≤
x
≤
3
, is
Report Question
0%
\dfrac{59}{6}
0%
\dfrac{57}{4}
0%
\dfrac{59}{3}
0%
\dfrac{57}{6}
Explanation
y = x^2 + 3x = 4 \Rightarrow x^2 + 3x - 4 = 0
(x + 4) (x - 1) = 0
x = 1
area
= \displaystyle \int_0^1 (x^2 + 3x ) dx + 2(4) = \dfrac{1}{3} + \dfrac{3}{2} + 8 = \dfrac{11}{6} + 8 = \dfrac{59}{6}
The area of the region bounded by the curve
y=\phi(x),y=0
and
x=10
is
Report Question
0%
\dfrac {81}{4}
0%
\dfrac {79}{4}
0%
\dfrac {73}{4}
0%
19
The area bounded by the curve
y=x
X-
axis and the lines
x=-1
and
x=1
is
Report Question
0%
0
0%
\dfrac {1}{3}
0%
\dfrac {2}{3}
0%
\dfrac {4}{3}
If the area (in sq. units) of the region
\{(x, y): y^2\le 4x, x+y\le 1, x\geq 0, y\ge 0 \}
is
a\sqrt{2}+b
, then
a-b
is equal to?
Report Question
0%
\dfrac{8}{3}
0%
\dfrac{10}{3}
0%
6
0%
-\dfrac{2}{3}
Explanation
\{(x, y): y^2\leq 4x, x+y\leq 1, x\geq 0, y\geq 0\}
A\displaystyle\int^{3-2\sqrt{2}}_02\sqrt{x}dx+\dfrac{1}{2}(1-(3-2\sqrt{2}))(1-(3-2\sqrt{2}))
=\dfrac{2[x^{3/2}]_0^{3-2\sqrt{2}}}{3/2}+\dfrac{1}{2}(2\sqrt{2}-2)(2\sqrt{2}-2)
=\dfrac{8\sqrt{2}}{3}+\left(-\dfrac{10}{3}\right)
a=\dfrac{8}{3}
,
b=-\dfrac{10}{3}
a-b=6
.
Let
f(x, y)=\{(x, y): y^2 \leq 4x, 0\leq x\leq \lambda\}
and
s(\lambda)
is area such that
\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5}
. Find the value of
\lambda
.
Report Question
0%
4\left(\dfrac{4}{25}\right)^{1/3}
0%
4\left(\dfrac{2}{25}\right)^{1/3}
0%
2\left(\dfrac{4}{25}\right)^{1/3}
0%
2\left(\dfrac{2}{25}\right)^{1/3}
Explanation
y^2=4x
S(I)=\left.2\displaystyle\int^{\lambda}_{0}2\sqrt{x}dx=\dfrac{4x^{3/2}}{3/2}\right]^{\lambda}_{0}=\dfrac{8}{3}\lambda^{3/2}
\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5}
\Rightarrow \dfrac{\lambda^{3/2}}{4^{3/2}}=\dfrac{2}{5}
\lambda =4\left(\dfrac{2}{5}\right)^{2/3}=4\left(\dfrac{4}{25}\right)^{1/3}
.
Region formed by
|x-y| \le 2
and
|x + y| \le 2
is
Report Question
0%
Rhombus of side is
2
0%
Square of area is
6
0%
Rhombus of area is
8\sqrt{2}
0%
Square of side is
2\sqrt{2}
Explanation
\dfrac{\text{ABCD is a square}}{Area = 4 \times \dfrac{1}{2}\times 2 \times 2 = 8}
side
= 2 \sqrt{2}
The region represented by
|x - y| \le 2
and
|x + y| \le 2
is bounded by a:
Report Question
0%
Square of side length
2\sqrt{2}
units
0%
Rhombus of side length
2
units
0%
Square of area
16 \,sq
units
0%
Rhombus of area
8\sqrt{2} sq.
units
Explanation
|x - y| \le 2
and
|x + y| \le 2
\Rightarrow x-y\le 2
or
x-y \ge 2
and
x+y \le 2
and
x+y \ge-2
After plotting them on the graph the common region is
Square whose side is
2\sqrt{2}
Let
S(\alpha) = \{ (x, y) : y^2 \le x, 0 \le x \le \alpha\}
and
A(\alpha)
is area of the region
S(\alpha)
. If for a
\lambda , 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5
, then
\lambda
equals
Report Question
0%
2\left(\dfrac{4}{25}\right)^{\frac{1}{3}}
0%
4\left(\dfrac{4}{25}\right)^{\frac{1}{3}}
0%
2\left(\dfrac{2}{5}\right)^{\frac{1}{3}}
0%
4\left(\dfrac{2}{5}\right)^{\frac{1}{3}}
Explanation
\displaystyle S(\alpha) =\{(x, y) : y^2 \le x, 0 \le x \le \alpha\}
\displaystyle A(\alpha)= 2 \int^{\alpha}_{0} \sqrt{x}dx = 2\alpha^{\frac{3}{2}}
A(4) = 2 \times 4^{3/2} = 16
A(\lambda) = 2\times \lambda^{3/2}
\dfrac{A(\lambda)}{A(4)} = \dfrac{2}{5} \Rightarrow \lambda = 4. \left(\dfrac{4}{25}\right)^{1/3}
The area (in sq. units) of the region
A=\{(x, y):x^2\leq y\leq x+2\}
is?
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0%
\dfrac{10}{3}
0%
\dfrac{9}{2}
0%
\dfrac{31}{6}
0%
\dfrac{13}{6}
Explanation
x^2\leq y\leq x+2
x^2=y; y=x+2
x^2=x+2
x^2-x-2=0
(x-2)(x-1)=0
x=2, -1
Area
=\displaystyle\int^2_{-1}(x+2)-x^2dx=\left[\dfrac {x^2}{2}+2x-\dfrac {x^3}{3}\right]^2_{-1}
=2+4-\dfrac 83-\dfrac 12 +2 -\dfrac 13=\dfrac{9}{2}
.
Area of the region bounded by
y^2\leq 4x, x+y\leq 1, x\geq 0, y\geq 0
is
a\sqrt{2}+b
, then value of
a-b
is?
Report Question
0%
4
0%
6
0%
8
0%
12
Explanation
Let P be the point common to
x+y=1
&
y^2=4x
So
y^2=4(1-y\Rightarrow y^2+4y-4=0
\Rightarrow y=\dfrac{-4\pm \sqrt{16+16}}{2}
\Rightarrow =-2+2\sqrt{2}
Hence
P(3, -2\sqrt{2}, -2+2\sqrt{2})
Hence started area
=
Area of region (OPN)
+
Area of (
\Delta
OPQ)
=\displaystyle\int^{3-2\sqrt{2}}_02\sqrt{x}dx+\dfrac{1}{2}[-1-(3-2\sqrt{2})]^2
=\dfrac{2}{3}\cdot 2(\sqrt{2}-1)(3-2\sqrt{2})+\dfrac{1}{2}[2(\sqrt{2}-1)]^2
=\dfrac{4}{3}\left\{-7+5\sqrt{2}\right\}+2(3-2\sqrt{2})=\left(\dfrac{20}{3}-4\right)\sqrt{2}+6-\dfrac{28}{3}=\dfrac{8}{3}\sqrt{2}-\dfrac{10}{3}
Hence
a=\dfrac{8}{3}, b=\dfrac{-10}{3}
So
a-b=6
.
If the area enclosed by the curves
{ y }^{ 2 }=4\lambda x
and
y=\lambda x
is
\cfrac { 1 }{ 9 }
square units then value of
\lambda
is
Report Question
0%
24
0%
37
0%
48
0%
38
Explanation
{ y }^{ 2 }=4\lambda x\quad \quad ;y=\lambda x
If
\lambda> 0
then
Hence
\int _{ 0 }^{ 4/\lambda }{ \left( 2\sqrt { \lambda } \sqrt { x } -\lambda x \right) } dx=\cfrac { 1 }{ 9 }
{ \left( \cfrac { 2\sqrt { \lambda } { x }^{ 3/2 } }{ 3/2 } -\cfrac { \lambda { x }^{ 2 } }{ 2 } \right) }^{ \cfrac { 4 }{ \lambda } }=\cfrac { 1 }{ 9 } \Rightarrow \cfrac { 4 }{ 3 } \sqrt { \lambda } \cfrac { 8 }{ { \lambda }^{ 3/2 } } -\lambda \cfrac { 8 }{ { \lambda }^{ 2 } } =\cfrac { 1 }{ 9 } \Rightarrow \cfrac { 32 }{ 3\lambda } =\cfrac { 1 }{ 9 } \Rightarrow \lambda =24
The area (in sq. units) of the region bounded by the curves
y={2}^{x}
and
y=\left| x+1 \right|
, in the first quadrant is:
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0%
\cfrac { 3 }{ 2 } -\cfrac { 1 }{ \log _{ e }{ 2 } }
0%
\cfrac { 1 }{ 2 }
0%
\log _{ e }{ 2 } +\cfrac { 3 }{ 2 }
0%
\cfrac { 3 }{ 2 }
Explanation
Required Area
\int _{ 0 }^{ 1 }{ \left( \left( x+1 \right) -{ 2 }^{ x } \right) } dx={ \left( \cfrac { { x }^{ 2 } }{ 2 } +x-\cfrac { { 2 }^{ x } }{ \ln { 2 } } \right) }_{ 0 }^{ 1 }=\left( \cfrac { 1 }{ 2 } +1-\cfrac { { 2 }^{ } }{ \ln { 2 } } \right) -\left( 0+0-\cfrac { 1 }{ \ln { 2 } } \right) =\cfrac { 3 }{ 2 } -\cfrac { 1 }{ \ln { 2 } }
If the area (in sq. units) bounded by the parabola
y^{2} = 4\lambda x
and the line
y = \lambda x, \lambda > 0
, is
\dfrac {1}{9}
, then
\lambda
is equal to
Report Question
0%
24
0%
48
0%
4\sqrt {3}
0%
2\sqrt {6}
Explanation
Area = \dfrac {1}{9} = \int_{0}^{\dfrac {4}{\lambda}}(\sqrt {4\lambda x} - \lambda x)dx
\Rightarrow \lambda = 24
.
The area bounded by the line
y=x
, x-axis and ordinates
x=-1
and
x=2
is?
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0%
\dfrac{3}{2}
0%
\dfrac{5}{2}
0%
2
0%
3
The area of the region
\left \{(x, y) : xy \leq 8, 1 \leq y\leq x^{2}\right \}
is
Report Question
0%
16\log_{6} 2 - 6
0%
8\log_{6} 2 - \dfrac {7}{3}
0%
16\log_{6} 2 - \dfrac {14}{3}
0%
8\log_{6} 2 - \dfrac {14}{3}
Explanation
xy \leq 8
1\leq y\leq x^{2}
x^{2} . x = 8
x = 2
Require Area
= \int_{1}^{4} \left (\dfrac {8}{y} - \sqrt {y}\right )dy = \left [8ln y - \dfrac {y^{3/2}}{3/2}\right ]_{1}^{4} = 8\ ln 4 - \dfrac {2}{3} . 8 - 0 + \dfrac {2}{3} = 16 ln2 - \dfrac {14}{3}
.
The area bounded by curve
y=\sin { 2x } \left( x=0\quad to\quad x=\pi \right)
and X-axis is ______
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0%
4
0%
2
0%
1
0%
\cfrac { 3 }{ 2 }
Explanation
As the crest and trugh are same we can write
A=2\int _{ 0 }^{ \pi /2 }{ \sin { 2x } } dx
=2{ \left[ \cfrac { -\cos { 2x } }{ 2 } \right] }_{ 0 }^{ \pi /2 }
=(1+1)
A=2
Area of the region bounded by the curve
y = \cos x
between
x = 0
and
x = \pi
is
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0%
1
sq. units
0%
4
sq. units
0%
2
sq. units
0%
3
sq. units
Explanation
Required area enclosed by the curve
y = \cos x, x = 0
and
x = \pi
A = \int_{0}^{\pi/2} \cos x dx + \left |\int_{\pi/2}^{\pi} \cos x dx\right |
=[sinx]_0^{\pi/2}+[sinx]_{\pi/2}^{\pi}
= \left [\sin \dfrac {\pi}{2} - \sin 0\right ] + \left |\sin \dfrac {\pi}{2} - \sin \pi \right |
= 1 + 1 = 2\ sq\ units
.
The area bounded by the curves
y = -x^2 + 3
and
y = 0
Report Question
0%
\sqrt{3} + 1
0%
\sqrt{3}
0%
4\sqrt{3}
0%
5\sqrt{3}
Explanation
\textbf{Step -1: Finding the points of intersection.}
\text{The two curves are }y=-x^2+3\text{ and }y=0
\text{Substituting }y=0\text{ in }y=-x^2+3
\Rightarrow 0=-x^2+3
\Rightarrow x^2=3
\Rightarrow x=\pm\sqrt3
\text{The points of intersection are }(-\sqrt3,0)\text{ and }(\sqrt3,0)
\textbf{Step -2: Finding the area between the curves.}
\text{Area between the curves}=\int_{-\sqrt3}^{\sqrt3}(-x^2+3)dx
=2\times\int_0^{\sqrt3}(-x^2+3)dx
=2\times[\dfrac{-x^3}{3}+3x]_0^{\sqrt3}
= 2[\dfrac{-3\sqrt3}{3}+3\sqrt3]
=4\sqrt3\text{ sq. units}
\textbf{Hence , the area bounded by the curves }\mathbf{y=-x^2+3}\textbf{ and }\mathbf{y=0}\textbf{ is }\mathbf{4\sqrt3}\textbf{ sq. units.}
The area (in sq. units) of the region
\left\{ \left( x,y \right) \in { R }^{ 2 }:{ x }^{ 2 }\le y\le 3-2x \right\}
, is:
Report Question
0%
\cfrac { 29 }{ 3 }
0%
\cfrac { 34 }{ 3 }
0%
\cfrac { 31 }{ 3 }
0%
\cfrac { 32 }{ 3 }
Explanation
From given data in question,
Point of intersection of given two curve
y = x^2
and
y = -2 x + 3
x^2 = -2 x + 3
x = -3 , 1
Area of shaded region
= \displaystyle \int_{-3}^1 ((-2x + 3) - x^2 ) dx
= \left[-x^2 + 3x - \dfrac{x^3}{3} \right]_{-3}^1
= - \left({1^2 - 3^2} \right) +3 \times (1-(-3)) - \dfrac{1^3 + 3^3}{3}
= 12 + 8 - \dfrac{28}{3} = \dfrac{32}{3}
.
The area bounded by
y = \sin^2 x , x = \dfrac{\pi}{2}
and
x = \pi
is
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0%
\dfrac{\pi}{2}
0%
\dfrac{pi}{4}
0%
\dfrac{\pi}{8}
0%
\dfrac{\pi}{16}
0%
2\pi
Explanation
Required area
= \displaystyle \int_{\frac{\pi}{2}}^{\pi} \sin^2 x dx
= \displaystyle \int^{\pi}_{\tfrac{\pi}{2}} \left[\dfrac{1 - \cos 2x}{2}\right] dx
= \dfrac{1}{2} \displaystyle \int^{\pi}_{\tfrac{\pi}{2}} (1 - \cos 2x) dx
= \dfrac{1}{2} \left[x - \dfrac{\sin 2x}{2} \right]_{\tfrac{\pi}{2}}^{\pi}
= \dfrac{1}{2} \left[(\pi - 0) - \left(\dfrac{\pi}{2} - 0\right) \right]
= \dfrac{1}{2} \left[\dfrac{\pi}{2} \right] = \dfrac{\pi}{4}
The area of the region bounded by the curve
y=2x-x^2
and the line
y=x
is ________ square units.
Report Question
0%
\dfrac{1}{6}
0%
\dfrac{1}{2}
0%
\dfrac{1}{3}
0%
\dfrac{7}{6}
Explanation
We note that the region bounded by these curves is in the region
x \in [0,1 ]
. In this region, the curve
y = x
lies below the curve
y = 2x-x^2
So, to calculate the area of said region, we evaluatie the following integral:
\int_0^1 2x-x^2 - x dx
= \int_0^1 x-x^2 dx
= [\frac{x^2}{2} - \frac{x^3}{3}]_{x=0}^{x=1}
= \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.
Given
f(x)=\begin{cases} x,0\le x<\dfrac { 1 }{ 2 } \\ \dfrac { 1 }{ 2 } ,x=\dfrac { 1 }{ 2 } \\ 1-x,\dfrac { 1 }{ 2 } <x\le 1 \end{cases}
and
g(x)=\left(x-\dfrac{1}{2}\right)^{2},x\epsilon R
, Then the area (in sq.units) of the region bounded by the curves
y=f(x)
and
y=g(x)
between the lines
2x=1
and
2x=\sqrt{3}
, is:
Report Question
0%
\dfrac{1}{3}+\dfrac{\sqrt{3}}{4}
0%
\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}
0%
\dfrac{1}{2}+\dfrac{\sqrt{3}}{4}
0%
\dfrac{\sqrt{3}}{4}-\dfrac{1}{3}
Explanation
\left\{\begin{matrix}x &,0<n<\dfrac{1}{2} \\\dfrac{1}{2} &,n=\dfrac{1}{2} \\ 1-n& ,\dfrac{1}{2}<n<1\end{matrix}\right.
and
g(x)=(x-\dfrac{1}{2})^2
graph of the current is as
refer above image.
Required image =are of trapezium
\displaystyle ABCD-\int^{\dfrac{\sqrt{3}}{2}}_{\dfrac{1}{2}}\left(x-\dfrac{1}{2}\right)^2dx
=\dfrac{1}{2}\left(\dfrac{\sqrt{3}-1}{2}\right)\left(\dfrac{1}{2}+1-\dfrac{\sqrt{3}}{2}\right)-\dfrac{1}{3}\left(\left(x-\dfrac{1}{2}\right)^3\right)^{\dfrac{\sqrt{3}}{2}}_{\dfrac{-1}{2}}
=\dfrac{\sqrt{3}}{4}-\dfrac{1}{3}
The area of the region, enclosed by the circle
x^2+y^2=2
which is not common to the region bounded by the parabola
y^2=x
and the straight line
y=x
, is:
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0%
\dfrac{1}{3}(15\pi-1)
0%
\dfrac{1}{6}(24\pi-1)
0%
\dfrac{1}{6}(12\pi-1)
0%
\dfrac{1}{3}(6\pi-1)
Explanation
\textbf{Step-1: Find the radius of the circle}
\text{The given equation of the circle is}
{x^2+y^2=2}
\text{Comparing with the general form of equation of a circle,}
{x^2+y^2=r^2}
,
\text{We get, }
r^2=2
r=\sqrt{2}
\textbf{Step-2: Find the area of required region}
\text{Here, Required Area= Area of circle-Area of shaded region}
{=\pi(\sqrt{2})^2-\int_{0}^{1}(\sqrt{x}-x)dx}
=2\pi-(\dfrac{2}{3}-\dfrac{1}{2} )
=2\pi-\dfrac{1}{6}
=\dfrac{12\pi-1}{6}
\textbf{Hence, Area of the required region is}
\mathbf{\dfrac{12\pi-1}{6}}
The area (in sq. units) of the region
\{(x, y) \in R^2 |4x^2 \le y \le 8x + 12\}
is :
Report Question
0%
\dfrac{128}{3}
0%
\dfrac{127}{3}
0%
\dfrac{125}{3}
0%
\dfrac{124}{3}
Explanation
Shaded are is the area between curve
Y = 4x^2
__(I)
and
Y = 8x + 12
__(II)
on solving equation (I) & (II) (to find intersection point of the curves).
4x^2 - 8x - 12 = 0
4(x +1 )(x - 3) = 0
x = -1
or
x = 3
\therefore
shaded Area
= \displaystyle \int_{-1}^3 [(8x + 12) - (4x^2)] dx
= \displaystyle \int_{-1}^3 8x dx + \int_{-1}^3 12 dx - 4 \int_{-1}^3 x^2 dx
= \left[4x^2 + 12 x - \dfrac{4}{3} x^3 \right]_{-1}^3
=4\cdot 9+12\cdot 3-\dfrac 43\cdot 27-4+12-\dfrac 43
= \dfrac{128}{3}
The area bounded by the curve
y =x^2 +2x +1
and tangent at
( 1 , 4)
and y -axis and
Report Question
0%
\frac {2}{3}
sq units
0%
\frac {1}{3}
sq units
0%
2 sq units
0%
None of these
If
A_n
is the area bounded by
y = ( 1 -x^2)^n
and coordinates axes ,
n \epsilon N
, then
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0%
A_n = A_{n-1}
0%
A_n < A_{n-1}
0%
A_n > A_{n-1}
0%
A_n =2 A_{n-1}
The area enclosed between the curves
y=log_{e}(x+e)\, , \,x=log_{e}\left ( \dfrac{1}{y} \right )
and the
x-axis
is
Report Question
0%
2
sq.units
0%
1
sq.units
0%
4
sq.units
0%
None of these
Explanation
y=log_{e}(x+e)\, , \, x=log_{e}\left ( \dfrac{1}{y} \right ) \Rightarrow y=e^{-x}
for
y=log_{e}(x+e)
shift the graph os
log_{e}x\, , e
units left hand side.
Required area =
\displaystyle \int_{1-e}^{0}log_e(x+e)dx\,+ \displaystyle \int_{0}^{\infty }e^{-x}dx
= |xlog_{e}(x+e)|_{1-e}^{0}-\displaystyle \int_{1-e}^{0}\dfrac{x}{x+e}dx-|e^{x}|_{0}^{\infty }
=\displaystyle \int_{0}^{1-e}\left ( 1-\dfrac{e}{x+e} \right )dx-e^{-\infty}+e^{0}
= |x-elog(x+e)|_{0}^{1-e}-0+1
=1-e +e\, loge+1 = 2sq.units
If
\displaystyle \left ( \alpha ^2,\alpha - 2 \right )
be a point interior to the region of the parabola
\displaystyle y^2 = 2x
bounded by the chord joining the points
\displaystyle \left ( 2,2 \right ) and \left ( 8,-4 \right )
then
\alpha
belongs to the interval
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0%
\displaystyle -2 + 2\sqrt {2} < \alpha <2
0%
\displaystyle \alpha > -2 + 2\sqrt {2}
0%
\displaystyle \alpha > -2 - 2\sqrt {2}
0%
None of these
The area of the region enclosed by the curves
y=x\log x
and
y=2x-2x^2
is
Report Question
0%
\dfrac{7}{12}\,sq.units
0%
\dfrac{1}{12}\,sq.units
0%
\dfrac{5}{12}\,sq.units
0%
None of these
Explanation
Curve taking :
y = x \log_ex
Clearly,
x > 0
,
For
0 < x < 1, x \log_ex < 0
, and for
x > 1, x \log_ex > 0
Also
x \log_ex = 0 \Rightarrow x = 1
.
Further,
\dfrac{dy}{dx} = 0 \Rightarrow 1 + \log_e x = 0 \Rightarrow x = 1/e
, which is a point of minima.
Required area
= \displaystyle \int_0^1 (2x - 2x^2)dx - \int_0^1 x \log x \ dx
= \left[x^2 - \dfrac{2x^3}{3}\right]_0^1 - \left[\dfrac{x^2}{2} \log x - \dfrac{x^2}{4}\right]_0^1
= \left( 1 - \dfrac{2}{3} \right) - \left[ 0 - \dfrac{1}{4} - \dfrac{1}{2} \underset{x \to 0}{\lim} x^2 \log x\right] = \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{7}{12}
.
Area of the region bounded by the curve
y =e^x, y=e^{-x}
and the straight line x= 1 given by
Report Question
0%
e-e^{-1} +2
0%
e-e^{-1} - 2
0%
e+ e^{-1} -2
0%
None of these
The area bounded by the curve
y = (x)
the x-axis and the ordinate
x= 1
and
x = b
is
(b- 1)
cos ( 3b + 4)
, then
f(x)
is given by
Report Question
0%
(x -1 ) sin (3x +4)
0%
(x-1) sin (3x-4)
0%
-3(x -1) sin ( 3x+ 4) + cos (3x+ 4)
0%
None of these
The area of the closed figure bounded by
y=\dfrac{x^{2}}{2}-2x+2
and the tangents to it at
(1,\dfrac{1}{2})
and
(4,2)
is
Report Question
0%
\dfrac{9}{8} \, sq.units
0%
\dfrac{3}{8} \, sq.units
0%
\dfrac{3}{2} \, sq.units
0%
\dfrac{9}{4} \, sq.units
Explanation
y= \dfrac{x^{2}}{2} -2x +2 = \dfrac{(x-2)^2}{2},
\dfrac{dy}{dx} = x-2 \, , x-2 \,\, , \left ( \dfrac{dy}{dx} \right )_{x=1}=-1\,\, , \left ( \dfrac{dy}{dx} \right )_{x=4}=2
\Rightarrow
Tangent at
(1, \dfrac{1}{2}
is
y- \dfrac{1}{2}=-1(x-1) \, or \, 2x+2y-3=0
Tangent at
(4,2)
is
y-2=2(x-4) \, or\, 2x-y-6=0
Hence, A=
\displaystyle \int_{1}^{\frac{5}{2}} \left ( \dfrac{x^{2}}{2}-2x+2-\dfrac{3-2x}{2} \right )dx+\displaystyle \int_{\frac{5}{2}}^{4}\left ( \dfrac{x^2}{2}-2x+2-(2x-6) \right )dx
=\displaystyle \int_{1}^{4}\left ( \dfrac{x^{2}}{2}-2x+2 \right )dx- \displaystyle \int_{1}^{\frac{5}{2}}\left ( \dfrac{3-2x}{2} \right )dx- \displaystyle \int_{\frac{5}{2}}^{4}(2x-6)dx
= \left ( \dfrac{x^{3}}{6}-x^{2}+2x \right )_{1}^{4}-\dfrac{1}{2}(3x-x^{2})_{1}^{\frac{5}{2}}-(x^{2}-6x)_{\frac{5}{2}}^{4}
= =\left ( \dfrac{63}{6}-15+6 \right )-\dfrac{1}{2}\left ( 3 \times \dfrac{3}{2}-\left ( \dfrac{25}{4}-1 \right ) \right )-\left ( \left ( 16-\dfrac{25}{4} \right )-6\left ( 4-\dfrac{5}{2} \right ) \right )
= \dfrac{3}{2}- -\dfrac{1}{2}\left ( \dfrac{9}{2}-\dfrac{21}{4} \right )-\left ( \dfrac{39}{4}-6\left ( \dfrac{3}{2} \right ) \right )
\dfrac{9}{8} \, sq.units
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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