Processing math: 7%
MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
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
= \int_{0}^{2}[2^{x} - (2x - x^{2})]dx
= \int_{0}^{2} (2^{x} -2x + x^{2})dx
= \left [\dfrac {2^{x}}{\log 2} - x^{2} + \dfrac {x^{3}}{3}\right ]_{0}^{2}
= \dfrac {4}{\log 2} - 4 + \dfrac {8}{3} - \dfrac {1}{\log 2}
= \dfrac {3}{\log 2} - \dfrac {4}{3}
.
The area of the portion of the circle
{ x }^{ 2 }+{ y }^{ 2 }=64
which is exterior to the parabola
{ y }^{ 2 }=12x
, is
Report Question
0%
\left( 8\pi -\sqrt { 3 } \right)
sq units
0%
\dfrac { 16 }{ 3 } \left( 8-\sqrt { 3 } \right)
sq units
0%
\dfrac { 16 }{ 3 } \left( 8\pi -\sqrt { 3 } \right)
sq units
0%
None of the above
Explanation
Required shaded area
=
=Area\, of \, circle -2\left[ \displaystyle\int _{ 0 }^{ 4 }{ 2\sqrt { 3 } \sqrt { x } dx } +\displaystyle\int _{ 4 }^{ 8 }{ \sqrt { 64-{ x }^{ 2 } } dx } \right]
=64\pi -2\left[ { \left( 2\sqrt { 3 } { x }^{ { 3 }/{ 2 } }\times \dfrac { 2 }{ 3 } \right) }_{ 0 }^{ 4 }+{ \left( \dfrac { x }{ 2 } \sqrt { 64-{ x }^{ 2 } } +\dfrac { 64 }{ 2 } \sin ^{ -1 }{ \dfrac { x }{ 8 } } \right) }_{ 4 }^{ 8 } \right]
=64\pi -\dfrac { 64 }{ \sqrt { 3 } } -32\pi +16\sqrt { 3 } +\dfrac { 32\pi }{ 3 }
=\dfrac { 128\pi }{ 3 } -\dfrac { 16\sqrt { 3 } }{ 3 }
=\dfrac { 16 }{ 3 } \left( 8\pi -\sqrt { 3 } \right)
sq units.
The area bounded byu the curve y = cos x, the line joining
(- \pi / 4 , \cos (- \pi / 4 ))
and (0, 2) and the line joining
( \pi / 4 , \cos ( \pi / 4 ))
and (0, 2) is :
Report Question
0%
\frac{4 + \sqrt{2}}{8} \pi - \sqrt{2}
0%
\frac{4 + \sqrt{2}}{8} \pi + \sqrt{2}
0%
\frac{4 + \sqrt{2}}{4} \pi - \sqrt{2}
0%
\frac{4 + \sqrt{2}}{4} \pi + \sqrt{2}
Explanation
Area between the curves
f(x)
and
g(x)
is given by
\int (f(x)-g(x))dx
Thus, the area between the given curves is
A = \int_{-\dfrac{\pi}{4}}^0 ((2+\dfrac{2-\dfrac{1}{\sqrt{2}}}{\dfrac{\pi}{4}}x)-\cos x)dx + \int_0^{\dfrac{\pi}{4}} ((2-\dfrac{2-\dfrac{1}{\sqrt{2}}}{\dfrac{\pi}{4}}x)-\cos x)dx
\Rightarrow A = \dfrac{2+\dfrac{1}{\sqrt{2}}}{2} \dfrac{\pi}{2}-\sqrt{2} = \drac{4+\sqrt{2}}{8}\pi - \sqrt{2}
Area common to the curves
y^{2} = ax
and
x^{2} + y^{2} = 4ax
is equal to
Report Question
0%
(9\sqrt {3} + 4\pi) \dfrac {a^{2}}{3}
0%
(9\sqrt {3} + 4\pi)a^{2}
0%
(9\sqrt {3} - 4\pi) \dfrac {a^{2}}{3}
0%
None of these
Explanation
Area common to the curves
{ y }^{ 2 }=a_{ x }\quad { x }^{ 2 }+{ y }^{ 2 }=4ax
is equal to :
{ x }^{ 2 }+{ y }^{ 2 }=4ax\\ { y }^{ 2 }=a_{ x }\\ x=0,3a\\ y=0,\pm \sqrt { 3 } a
Shaded region
\int _{ 0 }^{ 3 }{ \sqrt { ax } dx } +\int _{ 3a }^{ 4a }{ \sqrt { 4ax-{ x }^{ 2 } } dx } \\ =\left[ \cfrac { \sqrt { a } { x }^{ \cfrac { 3 }{ 2 } } }{ \cfrac { 3 }{ 2 } } \right] _{ 0 }^{ 3a }+\int _{ 3a }^{ 4a }{ \sqrt { (2a)^{ 2 }-(x-2a)^{ 2 } } dx } \\ =\cfrac { 2.\sqrt { a } .3\sqrt { 3 } .a\sqrt { a } }{ 3 } +\left[ \cfrac { x-2a }{ 2 } \sqrt { 4a_{ x }-{ x }^{ 2 } } +\cfrac { 4{ a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \cfrac { x-2a }{ 2a } } \right] _{ 3a }^{ 4a }\\ =2\sqrt { 3 } { a }^{ 3 }+\left[ 0+2{ a }^{ 2 }\left( \cfrac { \pi }{ 2 } \right) -\cfrac { \sqrt { 3 } { a }^{ 2 } }{ 2 } -2{ a }^{ 2 }\left( \cfrac { \pi }{ b } \right) \right] \\ =2\sqrt { 3 } { a }^{ 3 }-\cfrac { \sqrt { 3 } { a }^{ 2 } }{ 2 } +\pi { a }^{ 2 }-\cfrac { \pi }{ 3 } { a }^{ 2 }\\ =\cfrac { 3\sqrt { 3 } { a }^{ 2 } }{ 2 } +\cfrac { 2\pi { a }^{ 2 } }{ 3 } =\left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 }
Required area is :
=2(shaded\quad area)
=2\left( \left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 } \right) \\ =\left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 } sq.units
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,since, 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=
[\frac{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
2x^2-3xy-2y^2=0
(
y > 0
) is?
Report Question
0%
\pi
0%
\dfrac{3\pi}{4}
0%
\dfrac{\pi}{2}
0%
\dfrac{\pi}{4}
On the real line R, we define two functions f and g as follows:
f(x) = min [x - [x], 1 - x + [x]]
,
g(x) = max [x - [x], 1 - x + [x]]
,
where [x] denotes the largest integer not exceeding x.
The positive integer n for which
\displaystyle \int_{0}^{n}{(g(x) - f(x) ) dx = 100}
is?
Report Question
0%
100
0%
193
0%
200
0%
202
Explanation
Given,
f(x) = min [x - [x], 1 - x + [x]]=min[f,1-f]
and
g(x) = max [x - [x], 1 - x + [x]]=max[f,1-f]
where
f
is the fractional part of the function.
and
f
is always
0\le f <1
.
\therefore
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
\displaystyle \int_0^n(g(x)-f(x))dx=
\displaystyle n\int_0^1(g(x)-f(x))dx
=\displaystyle n\int_0^{0.5}(1-2x)dx+
\displaystyle n\int_{0.5}^1(2x-1)dx=n(0.5-0.25)+n(0.75-0.5)
0.5 n=100\implies n=200
The parabola
y^2=4x+1
divides the disc
x^2+y^2\leq 1
into two regions with areas
A_1
and
A_2
. Then
|A_1-A_2|
equals.
Report Question
0%
\displaystyle\frac{1}{3}
0%
\displaystyle\frac{2}{3}
0%
\displaystyle\frac{\pi}{4}
0%
\displaystyle\frac{\pi}{3}
Explanation
\Rightarrow
Area of semi-circle
=\pi/2
suppose the area of the parabola between
-1/4
to
0=P
.
Thus,
A_1=\pi/2-P
and
A_2=\pi/2+P
\Rightarrow A_2-A_1=2P=\displaystyle2\times 2\int_{-1/4}^0(\sqrt{4x+1})dx
\Rightarrow A_2-A_1=\dfrac{4\times2(4x+1)^{3/2}}{3\times 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 = \pi /2
is
Report Question
0%
2 + \sqrt {2}
0%
\sqrt {2}
0%
2
0%
2 - \sqrt {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
\to
represents
max (|x|, |y|) = 4
\to
represents
min (|x|, |y|)=2
To solve this question we first need to understand the meaning of
min(|x|,|y|)=2
and
max(|x|,|y|)=4
min(|x|,|y|)=2
means that either
|x|=2
and
|y|\geq 2
or
|y|=2
and
|x|\geq 2
Similarly,
max(|x|,|y|)=4
means that either
|x|=4
and
|y|\leq 4
or
|y|=4
and
|x|\leq 4
As can be seen from the image of the graph plotted for this, we can see 4 squares each of size
2\times2
So, total area=
4\times(2\times2)=16
Hence, correct answer is option
B
Area bounded by the curves
\displaystyle y = \left[ \frac{x^2}{64} + 2 \right]
([.] 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 = \dfrac {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
\frac{x^2}{4} - 1 = 7-x
\implies x^2-4 = 28 - 4x
\implies x^2 +4x -32 = 0 \implies (x+8)(x-4) = 0
\implies x = 4
(since
x>0
)
\implies y = 7-x= 7-4 = 3
So,
A \equiv (4,3)
Area bounded by the straight line and the x-axis
=\int _{ 0 }^{ 4 } (7-x)dx= \left( 7x - \frac { x^{ 2 } }{ 2 } \right) _{ 0 }^{ 4 }=28-\frac { 4^{ 2 } }{ 2 } =20
Area bounded by the parabola
=\int _{ 0 }^{ 2 }{ \left( 1-\frac { x^{ 2 } }{ 4 } \right) } dx+\int _{ 2 }^{ 4 }{ \left( \frac { x^{ 2 } }{ 4 } -1 \right) dx } ={ \left( x-\frac { x^{ 3 } }{ 12 } \right) }_{ 0 }^{ 2 }+{ \left( \frac { x^{ 3 } }{ 12 } -x \right) }_{ 2 }^4=4
So, the shaded area on the right hand side becomes
(20-4)=16
\therefore
The total area
=2 \times 16 = 32
Consider two curves
C_1 : (y - \sqrt 3)^2 = 4 ( x - \sqrt2)
and
C_2 : x^2 + y^2 = ( 6 + 2 \sqrt2 ) x + 2 \sqrt{3y} - 6 ( 1 + \sqrt2)
then
Report Question
0%
C_1 and C-2
touch each other only at one point.
0%
C_1 and C-2
touch each other exactly at two points.
0%
C_1 and C-2
intersect (but do not touch) at exactly two points.
0%
C_1 and C-2
neither intersect nor touch each other.
The area bounded by
x^2+y^2-2x=0
&
y=\sin\displaystyle\frac{\pi x}{2}
in the upper half of the circle is?
Report Question
0%
\displaystyle\frac{\pi}{2}-\frac{4}{\pi}
0%
\displaystyle\frac{\pi}{4}-\frac{2}{\pi}
0%
\displaystyle \pi -\frac{8}{\pi}
0%
None
Explanation
The area bonded by
{ x }^{ 2 }+{ y }^{ 2 }-2x=0\quad y=\sin { \cfrac { \pi x }{ 2 } }
In the upper half of the circle is
Required Area area = shaded area
=\cfrac { 1 }{ 2 } \pi { r }^{ 2 }-\int _{ 0 }^{ 2 }{ ydx } \\ =\cfrac { 1 }{ 2 } \pi { (1) }^{ 2 }-\int _{ 0 }^{ 2 }{ \cfrac { \sin { \pi x } }{ 2 } dx } \\ =\cfrac { \pi }{ 2 } +\left[ \cfrac { \cos { \cfrac { \pi x }{ 2 } } }{ \cfrac { \pi }{ 2 } } \right] _{ 0 }^{ 2 }=\cfrac { \pi }{ 2 } +\cfrac { 2 }{ \pi } \left[ \cos { \cfrac { \pi x }{ 2 } } \right] _{ 0 }^{ 2 }\\ =\cfrac { \pi }{ 2 } +\cfrac { 2 }{ \pi } \left[ \cos { \pi } -\cos { 0 } \right] =\cfrac { \pi }{ 2 } +\cfrac { 2 }{ \pi } -2\\ =\cfrac { \pi }{ 2 } +\cfrac { 4 }{ \pi } sq.\quad units
Hence the correct answer is
\cfrac { \pi }{ 2 } +\cfrac { 4 }{ \pi } sq.\quad units
Consider two curves
{C_1}\,:\,y = \frac{1}{x}\,and\,{C_2}:\,y = \,\ell nx
on the
xy
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%
\frac{e}{2}
0%
e
0%
1
0%
2\left( {e - 1} \right)
What is the area of the region bounded by the parabola
{ y }^{ 2 }=6(x-1)
and
{ y }^{ 2 }=3x
Report Question
0%
\cfrac { \sqrt { 6 } }{ 3 }
0%
\cfrac { 2\sqrt { 6 } }{ 3 }
0%
\cfrac { 4\sqrt { 6 } }{ 3 }
0%
\cfrac { 5\sqrt { 6 } }{ 3 }
Explanation
Solving
y^2=6(x-1)
and
y^2=3x
we get
6(x-1)=3x
\Rightarrow x=2
Hence
y=\pm \sqrt{6}
y^2=6(x-1)\Rightarrow x=1+\dfrac{y^2}{6}
and
y^2=3x \Rightarrow x=\dfrac{y^2}{3}
Area
=\int_{-\sqrt{6}}^{\sqrt{6}}\left(1+\dfrac{y^2}{6}-\dfrac{y^2}{3}\right)dy
=2\int_{0}^{\sqrt{6}}\left(1-\dfrac{y^2}{6}\right)dy
=2\left[y-\dfrac{y^3}{18}\right]_{0}^{\sqrt{6}}
=2 \times \dfrac{2\sqrt{6}}{3}=\dfrac{4\sqrt{6}}{3}
Area
=\dfrac{4\sqrt{6}}{3}
Area common to the circle
x^{2}+y^{2}=64
and the parabola
y^{2}=4x
is
Report Question
0%
\dfrac{16}{3}(4\pi + \sqrt{3})
0%
\dfrac{16}{3}(8\pi + \sqrt{3})
0%
\dfrac{16}{3}(4\pi - \sqrt{3})
0%
none\ of\ these
The area bounded by the curves
x= a \cos^3t, y= a \sin^3 t
is
Report Question
0%
\dfrac{3\pi a^2}{8}
0%
\dfrac{3\pi a^2}{16}
0%
\dfrac{3\pi a^2}{32}
0%
None of the above
Explanation
x=a\cos^{3}t
,
y=a\sin^{2}t
x^{\dfrac{2}{3}}+y^{\dfrac{2}{3}}=a^{\dfrac{2}{3}}
A=\int_{0}^{2\pi}{x}dy
A=a^{2}\int_{0}^{2\pi}{\cos^{3}t\times3\sin^{2}t\times\cos t}dt
=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt
\rightarrow(1)
similarily
A=\int_{0}^{2\pi}{y}dx
=3a^{2}\int_{0}^{2\pi}{\sin^{2}t\times\cos^{2}t}dt
\rightarrow(2)
Adding (1) and (2)
$$2A=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt
A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{\sin^{2}2}dt
\rightarrow(3)
Similarily
A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{\cos^{2}t}dt
\rightarrow(4)
By putting
t=t+\dfrac{\pi}{4}
in (3)
Adding (3) and (4)
2A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{dt}
A=\dfrac{3a^{2}}{8}\pi
Let\quad f(x)=2-\left| x-1 \right| and\quad g(x)={ \left( x-1 \right) }^{ 2 },\quad then\quad
Report Question
0%
area bounded by
f(x)
and
g(x)
is
\cfrac { 7 }{ 6 }
0%
area bounded by
f(x)
and
g(x)
is
\cfrac { 7 }{ 3 }
0%
area bounded by
f(x)
g(x)
and
x-
axis is
\cfrac { 5 }{ 3 }
0%
area bounded by
f(x)
g(x)
and
x-
axis is
\cfrac { 5 }{ 6 }
Explanation
Area of parabola w.r.t x axis
=g(X)
\int_{0}^{2}{(x-1)^{2}dx}
\Rightarrow
\left[\dfrac{(x-1)^{3}}{3}\right]_{0}^{2}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{2}{3}
Area of
f(x)
=2\times1+\dfrac{1}{2}\times2\times1
\Rightarrow
2+1=3
Area bounded by
f(x)
and
g(x)
=3-\dfrac{2}{3}=\dfrac{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(\pi-2)
0%
50(\pi-2)
0%
25\pi
0%
50\pi
0%
40\pi (5-\sqrt{2})
The area bounded by the curves
y={ \left( x-1 \right) }^{ 2 },y={ \left( x+1 \right) }^{ 2 }
and
y=\dfrac { 1 }{ 4 }
is
Report Question
0%
\dfrac { 1 }{ 3 } sq\ unit
0%
\dfrac { 2 }{ 3 } sq\ unit
0%
\dfrac { 1 }{ 4 } sq\ unit
0%
\dfrac { 1 }{ 5 } sq\ unit
Find area curved by three circles
Report Question
0%
(5\pi-3\sqrt{3})units^{2}
0%
(5\pi+4\sqrt{3})units^{2}
0%
(5\pi+3\sqrt{3})units^{2}
0%
(5\pi+3\sqrt{2})units^{2}
Explanation
Simple area concept
If
f\left(x\right)=
max
\left\{\sin{x},\cos{x},\dfrac{1}{2}\right\}
, then the area of the region bounded by the curves
y=f\left(x\right),x-
axis
y-
axis and
x=2\pi
is
Report Question
0%
\left(\dfrac{5\pi}{12}+3\right)
.sq.unit
0%
\left(\dfrac{5\pi}{12}+\sqrt{2}\right)
.sq.unit
0%
\left(\dfrac{5\pi}{12}+\sqrt{3}\right)
.sq.unit
0%
\left(\dfrac{5\pi}{12}+\sqrt{2}+\sqrt{3}\right)
.sq.unit
Explanation
\because f\left(x\right)=
max
\left\{\sin{x},\cos{x},\dfrac{1}{2}\right\}
Interval value of
f\left(x\right)
For
0\le x<\dfrac{\pi}{4}, \cos{x}
For
\dfrac{\pi}{4}\le x<\dfrac{5\pi}{6}, \sin{x}
For
\dfrac{5\pi}{6}\le x<\dfrac{5\pi}{3}, \dfrac{1}{2}
For
\dfrac{5\pi}{3}\le x<2\pi, \cos{x}
Hence, required area
=\int_{0}^{\frac{\pi}{4}}{\cos{x}dx}+\int_{\frac{\pi}{4}}^{\frac{5\pi}{6}}{\sin{x}dx}+\int_{\frac{5\pi}{6}}^{\frac{5\pi}{3}}{\dfrac{1}{2}dx}+\int_{\frac{5\pi}{3}}^{2\pi}{\cos{x}dx}
=\left[\sin{x}\right]_{0}^{\frac{\pi}{4}}-\left[\cos{x}\right]_{\frac{\pi}{4}}^{\frac{5\pi}{6}}+\dfrac{1}{2}\left[x\right]_{\frac{5\pi}{6}}^{\frac{5\pi}{3}}+\left[\sin{x}\right]_{\frac{5\pi}{3}}^{2\pi}
=\left(\dfrac{1}{\sqrt{2}}-0\right)-\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\right)+\dfrac{1}{2}\left(\dfrac{5\pi}{3}-\dfrac{5\pi}{6}\right)+\left(0+\dfrac{\sqrt{3}}{2}\right)
On simplification, we get
=\left(\dfrac{5\pi}{12}+\sqrt{2}+\sqrt{3}\right)
.sq.unit.
The parabola
y=\dfrac{x^2}{2}
divides the circle
x^2+y^2=8
into two parts. Find the area of both parts.
Report Question
0%
6\pi+\dfrac{4}{3}
,
2\pi-\dfrac{4}{3}
0%
6\pi-\dfrac{4}{3}
,
2\pi+\dfrac{4}{3}
0%
6\pi+\dfrac{2}{3}
,
2\pi-\dfrac{2}{3}
0%
6\pi-\dfrac{2}{3}
,
2\pi+\dfrac{2}{3}
Explanation
Let us first find the x-values of the points of intersection using the given equations of parabola
y=\dfrac {x^2}{2}
and circle
x^2+y^2=8
as follows:
x^{ 2 }+y^{ 2 }=8\\ \Rightarrow x^{ 2 }+\left( \dfrac { x^{ 2 } }{ 2 } \right) ^{ 2 }=8\quad \quad \quad \quad \quad \left( \because \quad y=\dfrac { x^{ 2 } }{ 2 } \right) \\ \Rightarrow x^{ 2 }+\frac { x^{ 4 } }{ 4 } =8\\ \Rightarrow 4x^{ 2 }+x^{ 4 }=32
\Rightarrow x^{ 4 }+4x^{ 2 }-32=0\\ \Rightarrow x^{ 4 }+8x^{ 2 }-4x^{ 2 }-32=0\\ \Rightarrow x^{ 2 }(x^{ 2 }+8)-4(x^{ 2 }+8)=0\\ \Rightarrow (x^{ 2 }-4)(x^{ 2 }+8)=0\\ \Rightarrow (x^{ 2 }-4)=0,\quad (x^{ 2 }+8)=0\\ \Rightarrow x^{ 2 }=4,\quad x^{ 2 }=-8\\ \Rightarrow x=\pm \sqrt { 4 } \\ \Rightarrow x=\pm 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 }=\int _{ -2 }^{ 2 }{ \left( \sqrt { 8-{ x }^{ 2 } } -\dfrac { 1 }{ 2 } { x }^{ 2 } \right) } dx\\ =2\int _{ 0 }^{ 2 }{ \left( \sqrt { 8-{ x }^{ 2 } } -\dfrac { 1 }{ 2 } { x }^{ 2 } \right) } dx\\ =2\left[ \dfrac { 1 }{ 2 } \times 8\sin ^{ -1 }{ \left( \dfrac { 2 }{ \sqrt { 8 } } \right) } +\dfrac { 1 }{ 2 } \times 2\sqrt { 8-{ 2 }^{ 2 } } -\dfrac { 1 }{ 2 } { \left[ \dfrac { 1 }{ 3 } { x }^{ 3 } \right] }_{ 0 }^{ 2 } \right] \\ =8\sin ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) } +2\sqrt { 4 } -\dfrac { 8 }{ 3 }
=8\times \dfrac { \pi }{ 4 } +4-\dfrac { 8 }{ 3 } \\ =2\pi +\dfrac { 4 }{ 3 }
We know that the area of the circle is
\pi r^2
, therefore the area of the circle
x^2+y^2=8
with radius
r=\sqrt {8}
is:
\pi \left( \sqrt { 8 } \right) ^{ 2 }=8\pi
Thus, we have
A_{ 2 }=8\pi -\left( 2\pi +\dfrac { 4 }{ 3 } \right) =6\pi -\dfrac { 4 }{ 3 }
Hence, the area of parabola and circle is
2\pi +\dfrac { 4 }{ 3 }
and
6\pi -\dfrac { 4 }{ 3 }
respectively.
The area enclosed by the curve
y=\sqrt{(4-x^2)}, y\geq \sqrt{2}\sin\left(\dfrac{x\pi}{2\sqrt{2}}\right)
and x-axis is divided by y-axis in the ratio.
Report Question
0%
\dfrac{\pi^2-8}{\pi^2+8}
0%
\dfrac{\pi^2-4}{\pi^2+4}
0%
\dfrac{\pi -4}{\pi +4}
0%
\dfrac{2\pi^2}{\pi^2+2\pi -8}
Explanation
y=\sqrt{4-x^{2}} \ldots
(given)
\cdots(i)
\Rightarrow y^{2}-4 x^{2}
\Rightarrow x^{2}+y^{2}=4
y \geq \sqrt{2} \sin \left(\dfrac{\pi x}{2 \sqrt{2}}\right)
(given) ... (ii)
To find the period,
=\frac{2 \not \pi}{\not \pi} \times 2 \sqrt{2}
=4 \sqrt{2} \Rightarrow
period.
=\dfrac{4 \sqrt{2}}{2}=2 \sqrt{2}
\quad=2 \times 1.732
\quad[3.4]
To find the intersection point using (i) and (ii)
x=\sqrt{2}
Area of circle
=\pi r^{2}=\pi \times y=4 \pi
Area of circle
=\pi=A_{1}
.. (iii)
from (-2,0) to (0,0)
A_{2}=\int_{0}^{\sqrt{2}} \sqrt{4-x^{2}}-\sqrt{2} \sin \left(\dfrac{\pi}{2 \sqrt{2}} x\right) d x
\Rightarrow \int_{0}^{\sqrt{2}} \sqrt{4-x^{2}} d x-\sqrt{2} \int_{0}^{\sqrt{2}} \sin \left[\dfrac{\pi x}{2 \sqrt{2}}\right] d x
\left[\begin{array}{ll}\therefore & \sqrt{a^{2}-x^{2}} & d x=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}\end{array}\right]
=\left.\left[\dfrac{x}{2} \sqrt{4-x^{2}}+\dfrac{4}{2} \sin ^{-1} \dfrac{x}{2}\right]_{0}^{\sqrt{2}}-\dfrac{\sqrt{2}}{\left(\dfrac{\pi}{2 \sqrt{2}}\right)}[-\operatorname{cos}\left(\dfrac{\pi x}{2 \sqrt{2}}\right)\right]_{0}^{\sqrt{2}}
=\left[\dfrac{2}{2} \pi \sqrt{2}+\dfrac{4}{2} \sin ^{-1} \left[\dfrac{1}{\sqrt{2}}\right]\right]+\dfrac{2 \times 2}{\pi}\left|\cos \left(\dfrac{\pi x}{2 \sqrt{2}}\right)\right|_{0}
=\left[1+\dfrac{4}{2} \sin ^{-1}\left[\dfrac{1}{\sqrt{2}}\right]\right]+\dfrac{4}{\pi}\left[\cos \left(\dfrac{\pi}{2\sqrt 2} \times \sqrt2 )-\operatorname{cos} 0\right]\right.
\Rightarrow[1+\pi / 2-4 / \pi]
\Rightarrow \dfrac{2 \pi+\pi^{2}-8}{2 \pi}
.. (iv)
Using equation (iii) and (iv).
\dfrac{A_{1}}{A_{2}}=\dfrac{\pi \times 2 \pi}{2 \pi+\pi^{2}-8}=\dfrac{2 \pi^{2}}{2 \pi+\pi^{2}-8}
Answer (D)
If
k=2
then
f\left(x\right)
attains point of inflection at
Report Question
0%
0
0%
\sqrt{2}
0%
-\sqrt{2}
0%
None of these
Area bounded by
|x-1| \le 2
and
x^{2}-y^{2}=1
, is
Report Question
0%
6 \sqrt{2}+\dfrac{1}{2}
In
|3+2\sqrt{2}|
0%
6 \sqrt{2}+\dfrac{1}{2}
In
|3-2\sqrt{2}|
0%
6 \sqrt{2}-
In
|3+2\sqrt{2}|
0%
none\ of\ these
Explanation
Given :
|x-1|\leq 2
and
x^2-y^2=1
We have to find area bounded by both curve.
As we know
|x|\leq a
means
-a \leq x \leq a
So
|x-1|\leq 2
-2\leq x-1\leq 2
-2\leq x-1\leq 2
-2+1\leq x\leq 2+1
\boxed{-1 \leq x \leq 3}
Area of curve
|x-1|\leq 2
Area
=2 \int_{3}^{1} \sqrt {x^2-1}dx
\because \int \sqrt {x^2-a^2}dx=\dfrac{x}{2}=\dfrac{a^2}{2}ln(x+\sqrt{x^2-a^2})
=2\left[\dfrac{x}{2}\sqrt{x^2-1}-\dfrac{1}{2}in (x+\sqrt{x^2-1})\right]^3_1
2\left[\dfrac{3}{2}\sqrt8-\dfrac{1}{2}in (3+\sqrt 8)\right]-2\left[\dfrac{1}{2}\sqrt 0-\dfrac{1}{2}in (1+0)\right]
6\sqrt 2 in (3+2\sqrt 2)-0
A=6\sqrt 2-in (3+2\sqrt 2)
Consider the two curves
{ C }_{ 1 } :{ y }^{ 2 }=4x
{ C }_{ 2 } : { x }^{ 2 }+ { y }^{ 2 } - 6x + 1 = 0
Then, the area of region between these curves?
Report Question
0%
\dfrac{20}{3}-2\pi
0%
\dfrac{10}{3}-2\pi
0%
\dfrac{20}{3}-\pi
0%
\dfrac{10}{3}-\pi
The area enclosed between the curve
y=x^3
and
y=\sqrt{x}
is
Report Question
0%
\dfrac{5}{3}
0%
\dfrac{5}{4}
0%
\dfrac{5}{12}
0%
None of these
Explanation
Given two curves are
y=x^3
and
y=\sqrt x
Both curves intersect at point
(1,1)
Now equating the curves we get,
x^3=\sqrt x;
when
x=1
\therefore
The curve represented by
y=\sqrt x
is upward
to the curve represented by
y=x^3
\therefore
Area enclosed by the curves= Area of the shaded region
\Rightarrow
Area enclosed by the curves
=\int_{1}^{0}(\sqrt x-x^3)dx
=\\int_{1}^{0}x\dfrac{1}{2}dx-\int_{2}^{0}x^3dx
=\left[\dfrac{x^{3/2}}{\dfrac{3}{2}}\right]^1_0-\left[\dfrac{x^4}{4}\right]^1_0
=\left[\dfrac{(1)^{3/2}}{\dfrac{3}{2}}-0\right]-\left[\dfrac{(1)^4}{4}-0$\right]
=\dfrac{1}{\dfrac{3}{2}}-\dfrac{1}{4}
$$=\dfrac{2}{3}-\dfrac{1}{4}$
$
=\dfrac{8-3}{12}
\therefore
Area enclosed by the curves
=\dfrac{5}{12}
sq. units
The area between the curves y=tan x, cot x and axis in the interval
\left[0,\pi \right/2
]is ?
Report Question
0%
log
2
0%
log
3
0%
log
5
0%
none of these
Explanation
y=tanx,y=cotx
are symmetrical about
x=\frac { \pi }{ 4 }
,where they intersect
\frac { \pi }{ 4 } Req Area=2|∫tanxdx|
\frac { \pi }{ 4 } |2∫cotxdx|=2\times \left| \left( lnsec\frac { \pi }{ 4 } \right) -lnsec0 \right|
\frac { \pi }{ 22 } \times |ln\sqrt { 2 } -0|=2\times \frac { 1 }{ 2 } \times ln2=ln2
The area bounded by the curves
y = \sin \left( {x - \left[ x \right]} \right),\,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}}\left| {\cos \,x} \right|
and
y = {\left( {{{\cos }^{ - 1}}\left| {\cos \,x} \right|} \right)^2},x \in \left[ {0,\pi } \right]
is
Report Question
0%
\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)
0%
\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} + 1} \right)
0%
\frac{2}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)
0%
\frac{2}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} + 1} \right)
The area bounded by the curves
y = sin^{-1} |sin \, x|
and
y = (sin^{-1} | sin \, x|)^2 , \, 0 \le x \le 2 \pi
is
Report Question
0%
\left(\dfrac{\pi^3}{3} + \dfrac{4}{3} \right)
sq. unit
0%
\left(\dfrac{\pi^3}{6} - \dfrac{\pi^2}{2} + \dfrac{4}{3} \right)
sq. unit
0%
\left(\dfrac{\pi^2}{2} - \dfrac{4}{3} \right)
sq. unit
0%
\left(\dfrac{\pi^2}{6} - \dfrac{\pi}{4} + \dfrac{4}{3} \right)
sq. unit
Explanation
y=\sin ^{-1}|\sin n |
and
y=(\sin ^{-1}|\sin n|^2)^2
Required area
=4\int_{1}^{0}[x-(\sin^{-1}(\sin n))^2]dn+4\int_{\dfrac{\pi}{1}}^{1}[(\sin^{-1}(\sin n))^2-n]dn
=4\left(\dfrac{1}{2}\right)-4\int_{1}^{0}(\sin ^{-1}(\sin x))^2dn+4\int_{\dfrac{\pi}{2}}^{1}(\sin ^{-1}(\sin n))^2dn-\dfrac{4}{2}\left(\dfrac{\pi^2}{4}-1\right)^1
=4-\dfrac{\pi}{2}-4\int_{1}^{0}(\sin ^{-1}(\sin n))^2dn +4\int_{\dfrac{\pi}{2}}^{0}(\sin ^{-1}(\sin n))^2dn
=4-\dfrac{\pi^2}{2}-4\int_{1}^{0}n^2dn+4\int_{\dfrac{\pi}{2}}^{1}n^2dn
=4-\dfrac{\pi^2}{2}-\dfrac{4}{3}+\dfrac{4}{3}\left[\dfrac{\pi^3}{8}-1^1\right]
=4-\dfrac{\pi^2}{2}-\dfrac{8}{3}+\dfrac{\pi^3}{6}
=\boxed{\left(\dfrac{4}{3}-\dfrac{\pi^2}{2}+\dfrac{\pi^3}{6}\right)sq\, unit}
The area bounded by the curves
{y^2} = 4x
and
{x^2} = 4y
is :
Report Question
0%
\frac{{32}}{3}
0%
\frac{{16}}{3}
0%
\frac{8}{3}
0%
0
Explanation
x^2=4y
y^2=4x
x=2\sqrt{y}
=4\times 2\sqrt{y}
y^2=8\sqrt{y}
\Rightarrow y^4-8y=0
then
y=0, 4
x^2=4y
we get
x=4, 0
So, the points of intersection are
(0, 0)
&
(4, 4)
Area
=\displaystyle\int^4_0\displaystyle\int^{x^2/4}_{2\sqrt{x}}dydx=\displaystyle\int^4_0[y]^{x^2/4}_{2\sqrt{x}}dx
=\displaystyle\int^4_0\left[\dfrac{x^2}{4}-2\sqrt{x}\right]dx
=\left[\dfrac{x^3}{12}-\dfrac{4}{3}\cdot x^{3/2}\right]^4_0
=\left[\dfrac{4^3}{12}-\dfrac{4}{3}(4)^{3/2}\right]
=\dfrac{16}{3}
sq. units.
The area bounded by
y=2-\left| 2-x \right|
and
y=\frac { 3 }{ \left| x \right| }
is :
Report Question
0%
\frac { 4+3\ell n3 }{ 2 }
0%
\frac { 4-3\ell n3 }{ 2 }
0%
\frac { 3 }{ 2 } +\ell n3
0%
\frac { 1 }{ 2 } +\ell n3
Explanation
y=2-\left|2-x\right|
y=\dfrac{3}{\left|x\right|}
1
can be rewriiten as
y=x if x<+2
On solving these two equation, we get
x=\sqrt{3}
and
x=3
Area=\displaystyle \int_{\sqrt{3}}^{3}{\left({y}_{1}-{y}_{2}\right)}dx
=\displaystyle \int_{\sqrt{3}}^{3}{\left(2-\left|2-x\right|\right)}-\dfrac{3}{\left|x\right|}dx
=\displaystyle \int_{\sqrt{3}}^{2}{x-\dfrac{3}{\left|x\right|}+\int_{2}^{3}{4-x-\dfrac{3}{\left|x\right|}}}dx
{\left[\dfrac{{x}^{2}}{2}-3\log_{e}{x}\right]}_{\sqrt{3}}^{2}+{\left[4x-\dfrac{{x}^{2}}{2}-3\log_{e}{x}\right]}_{2}^{3}
=\dfrac{1}{2}+3\log_{e}{\dfrac{\sqrt{3}}{2}}+12-\dfrac{9}{2}-3\log_{e}{3}-8+2+3\log_{e}{2}
=\dfrac{1}{2}+3\ln{\dfrac{\sqrt{3}}{2}}+\dfrac{3}{2}3\ln{\dfrac{2}{3}}
2+3\ln{\dfrac{\sqrt{3}}{2}}=2-3\ln{\sqrt{3}}=\dfrac{4-3\ln{3}}{2}
Area of the region defined by
1\ \le |x|+|y|
and
x^{2}-2x+1 \le 1-y^{2}
is
k \pi
then
k=.....sq
units
Report Question
0%
\dfrac {3}{4}
0%
\dfrac {7}{6}
0%
\dfrac {128}{5}
0%
\dfrac {10}{3}
Area bounded by the curves
y=\log _{ e }{ x } \quad
and
y={ \left( \log _{ e }{ x } \right) }^{ 2 }
is ?
Report Question
0%
e-2
0%
3-e
0%
e
0%
e-1
The maximum area bounded by the curves
{y^2} = 4ax,\,\,\,\,y = ax\,\,a
and
y = \frac{x}{a}\,\,,1 \le a \le 2
is
Report Question
0%
44 sq. units
0%
74 sq. units
0%
84 sq.units
0%
114 sq. units
The area bounded by the curves
y=xe^{x},y=xe^{-x}
and the line
x=1
, is
Report Question
0%
\dfrac {2}{e}
0%
1-\dfrac {2}{e}
0%
\dfrac {1}{e}
0%
1-\dfrac {1}{e}
The area (in sq.units) of the region
\left\{ ( x , y ) :{ y} ^ { 2 } \ge 2 x\right.
and
{x} ^ { 2 } +{ y} ^ { 2 } \le 4 x , x \ge 0 , y \ge 0
is
Report Question
0%
\pi - \dfrac { 8 } { 3 }
0%
\pi - \dfrac { 4 \sqrt { 2 } } { 3 }
0%
\dfrac { \pi } { 2 } - \dfrac { 2 \sqrt { 2 } } { 3 }
0%
\pi - \dfrac { 4 } { 3 }
Explanation
Consider
y^2=2x
and
x^2+y^2=4x
, on solving,
we get
x^2+2x=4x
x^2=2x\Rightarrow x=0, 2
The integral lies between
0
and
2
y^2=2x
y^2+x^2=4x
\Rightarrow y=\sqrt{2x}
\Rightarrow y=\sqrt{4x-x^2}
Area
=\displaystyle\int^2_0\sqrt{4x-x^2}-\sqrt{2x}dx
=\displaystyle\int^2_0\sqrt{2^2-(2-x)^2}-\displaystyle\int^2_0\sqrt{2x}dx
=\left[\dfrac{x-2}{2}\sqrt{4x-x^2}-\dfrac{4}{2}\sin^{-1}\left(\dfrac{x-2}{2}\right)\right]^2_0-\left[\dfrac{\sqrt{2}}{3/2}x^{3/2}\right]^2_0
=\left|2\sin^{-1}(-1)\right|-\dfrac{2\sqrt{2}}{3}\cdot 2\sqrt{2}
=\pi -\dfrac{8}{3}
.
The area bounded by the curves
\sqrt{x}+\sqrt{y}=1
and
{x}+{y}=1
is ?
Report Question
0%
\dfrac{1}{3}
0%
\dfrac{1}{6}
0%
\dfrac{1}{2}
0%
\dfrac{5}{6}
0%
\dfrac{1}{4}
Explanation
\textbf{Step -1: Finding the point of intersection.}
\text{Find the points of intersection of the curves}
\implies \sqrt x=1-\sqrt y
\implies x=(1-\sqrt y)^2=1+y-2\sqrt y
\implies 1+y-2\sqrt y +y=1
\implies 2y=2\sqrt y
\implies y=0,1
\implies \text{Point of intersections are }(1,0)\text{ and }(0,1)
\textbf{Step -2: Calculating the area .}
\sqrt y=1-\sqrt x
\implies y=1+x-2\sqrt x
\implies f_1(x)=x-2\sqrt x+1
\implies f_2(x)=1-x
\implies \text{Area}=\int_0^1(f_1(x)-f_2(x))dx
\implies \int_0^1(2x-2\sqrt x)dx
\implies x^2-2\dfrac{x^{\frac32}}{\frac32}|_0^1
\implies |1-\dfrac43|=|-\dfrac13|=\dfrac13
\textbf{Hence, The required area is }\mathbf{\dfrac13}\textbf{ 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\pi\ sq.\ units
0%
16\pi\ sq.\ units
0%
32\pi\ sq.\ units
0%
None\ of\ these
Area bounded between the curves
y=\sqrt{4-x^2}
and
y^2=3|x|
is/are?
Report Question
0%
\dfrac{\pi -1}{\sqrt{3}}
0%
\dfrac{2\pi -1}{3\sqrt{3}}
0%
\dfrac{2\pi -\sqrt{3}}{3}
0%
\dfrac{2\pi -\sqrt{3}}{3\sqrt{3}}
Area bounded by the curves
y=\cos^{-1}(\sin x)
and
y=\sin^{-1}(\sin x)
in the interval
[0, \pi]
is
Report Question
0%
\dfrac{\pi^{2}}{16}
0%
\dfrac{\pi^{2}}{32}
0%
\dfrac{\pi^{2}}{4}
0%
\dfrac{\pi^{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=tanx\left( \dfrac { -\pi }{ 3 } \le x\le \dfrac { \pi }{ 3 } \right) and\quad y=cotx\left( \dfrac { \pi }{ 6 } \le x\le \dfrac { 3\pi }{ 2 } \right)
Report Question
0%
log\dfrac { 3 }{ 2 }
0%
log\sqrt { \dfrac { 3 }{ 2 } }
0%
2log\dfrac { 3 }{ 2 }
0%
log\left( \dfrac { 3 }{ \sqrt { 2 } } \right)
The area bounded by the curves is
\sqrt{\left|x\right|}+\sqrt{\left|y\right|}=\sqrt{a}
and
x^{2}+y^{2}=a^{2}
(where
a>0
) is
Report Question
0%
\left(\pi-\dfrac{2}{3}\right)a^{2}\ sq\ units
0%
\left(\pi+\dfrac{2}{3}\right)a^{2}\ sq\ units
0%
\left(\pi+\dfrac{2}{3}\right)a^{3}\ sq\ units
0%
\left(\pi-\dfrac{2}{3}\right)a^{3}\ sq\ units
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page