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CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Application Of Integrals
Quiz 7
If
∫
1
0
(
4
x
3
=
f
(
x
)
)
f
(
x
)
d
x
=
4
7
, then the area of region bounded by
y
=
f
(
x
)
,
x
−
axis and the line
x
=
and
x
=
2
is
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0%
11
2
0%
13
2
0%
15
2
0%
17
2
The are boundede by the curve
y
=
x
2
,
y
=
−
x
and
y
2
=
4
x
−
3
is
k
, them the value of
9
k
is
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0%
2
0%
3
0%
0
0%
4
If area bounded by to curves
y
2
=
4
a
x
and y=mx is
a
2
3
, then the value of m is
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0%
2
0%
−
1
0%
1
2
0%
none of these
Explanation
Given,
y
2
=
4
a
x
,
y
=
m
x
curves intersect at
(
4
a
m
2
,
a
m
)
Area enclosed is,
A
=
∫
4
a
m
2
0
(
√
4
a
x
−
m
x
)
d
x
∴
.... given
\left[2\sqrt a \ \dfrac{x^{\tfrac 32 }}{\tfrac 32 }- m \dfrac {x^2} 2 \right]^{4\frac{a}{m^2}}_0=\dfrac{a^2}{3}
\dfrac {4\sqrt a} 3 {({4\frac{a}{m^2}})^{\tfrac 3 2}}-\dfrac m 2 {(4\frac{a}{m^2})}^2-0-0=\dfrac{a^2}{3}
\dfrac{32 a ^2} {3m^3}-\dfrac{8 a ^2} {m^3}=\dfrac{a^2}{3}
\dfrac{8}{3}\dfrac{a^2}{m^3}=\dfrac{a^2}{3}
m^3=8
\therefore m=2
The area bounded by curves
3 x^2 + 5 y= 32
and
y = |x-2|
is
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0%
25
0%
33/2
0%
17/2
0%
33
The area of the plane region bounded by the curves
x+{ 2y }^{ 2 }=0
and
x+{ 3y }^{ 2 }=1
is equal to
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0%
\cfrac { 5 }{ 3 } sq.unit
0%
\cfrac { 1 }{ 3 } sq.unit
0%
\cfrac { 2 }{ 3 } sq.unit
0%
\cfrac { 4 }{ 3 } sq.unit
Explanation
The given curves are
x+2y^2=0
&
x+3y^2=1
1
st curve
\Rightarrow x^2+2y^2=0
\Rightarrow x^2=-2y^2
\therefore y^2=-x/2
(parabola)
2nd
curve
\Rightarrow x+3y^2=1
\Rightarrow 3y^2=1-x
y^2=(1-x)/3
(parabola)
So, area between parabolas is required
Solving
-\dfrac{x}{2}=\dfrac{(1-x)}{3}
\Rightarrow -3x=2-2x
\Rightarrow -3x+2x=2
\Rightarrow -x=2
=x=-2
\therefore y=1, -1
(-2, -1)
&
(-2, 1)
are points of intersection.
Required area
=2|\displaystyle\int^1_0(-2y^2-1+3y^2)dy|
=2|\displaystyle\int ^1_0(y^2-1)dy|=2\left[y-\dfrac{y^3}{3}\right]^1_0
=2|\left(\dfrac{1}{3}-1\right)|=2\times 2/3=4/3
square units.
The area of the figure formed by
|x| + |y| = 2
is (in sq. units)
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0%
2
0%
4
0%
6
0%
8
The area bounded by the curve
y=\ln (x)
and the lines
y=\ln (3),y=0
and
x=0
is equal
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0%
3
0%
3\ln (3)-2
0%
3\ln (3)+2
0%
2
The area (in sq.units) of the region described by
\left\{(x,y):y^{2}\le 2x\ and\ y\ge 4x-1\right\}
is
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0%
\dfrac{15}{64}
0%
\dfrac{9}{32}
0%
\dfrac{7}{32}
0%
\dfrac{5}{64}
The area bounded by the curves
y = \log _ { e } x
and
y = \left( \log _ { e } x \right) ^ { 2 }
is
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0%
3 - e
0%
e-3
0%
\frac { 1 } { 2 } ( 3 - e )
0%
\frac { 1 } { 2 } ( e - 3 )
Explanation
\begin{array}{l} y=\left( { { { \log }_{ e } }x } \right) \, \, \, \, \, \, \, ----\left( 1 \right) \\ y={ \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, \, \, \, -----\left( 2 \right) \\ For\, the\, po{ { int } }\, of\, { { int } }er\sec tion \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, =\left( { { { \log }_{ e } }x } \right) \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, -\left( { { { \log }_{ e } }x } \right) =0 \\ \left( { { { \log }_{ e } }x } \right) \left[ { \left( { { { \log }_{ e } }x } \right) \, -1 } \right] \, \\ \left( { { { \log }_{ e } }x } \right) =0\, \, and\, \left( { { { \log }_{ e } }x } \right) =1 \\ x=1\, \, and\, \, x=e \\ area\, bounded \\ =\left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ consider \\ I=\int { \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } \\ { \log _{ e } }x=t \\ x={ e^{ t } } \\ dx={ e^{ t.dt } } \\ Now, \\ I=\int { { e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} dt } \\ ={ e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} -\left\{ { \left( { 2t-1 } \right) { e^{ t } }-2{ e^{ t } } } \right\} \, \, \, \, \left[ { by\, { { int } }ergration\, by\, parts } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-t-2t+1+2 } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-3t+3 } \right] \\ =x\left[ { \log { { \left( x \right) }^{ 2 } } -3\log \left( x \right) +3 } \right] \\ Now, \\ \left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ =\left| { \left[ { x\left\{ { loh{ { \left( x \right) }^{ 2 } } } \right\} -3\log \left( x \right) +3 } \right] _{ 1 }^{ e } } \right| \\ =\left| { e\left( { 1-3+3 } \right) -1\left( { 0-0+3 } \right) } \right| \\ =\left| { e-3 } \right| \\ =3-e \end{array}
Hence, the option
B
is the correct answer.
The area common to the parabola
y=2{ x }^{ 2 }\quad
and
\quad y={ x }^{ 2 }+4
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0%
\dfrac { 2 }{ 3 } sq.\quad units\quad
0%
\dfrac { 3 }{ 2 } sq.\quad units\quad
0%
\dfrac { 32 }{ 3 } sq.\quad units\quad
0%
none of these.
Explanation
Find the P.O.I of the parabolas equate the equations
y = 2x^2
and
y = x^2 + y
we get
2x^2 = x^2 + 4
x^2 = 4
x = \pm 2
y = 8
\therefore
POI are
A(-2, 8)
&
C(2, 8)
\therefore
the required are ABCD
A = \displaystyle \int_{-2}^2 (y_1 - y_2) dx
(where
y_1 = x^2 + 4 \, \& \, y_2 = 2x^2
)
= \displaystyle \int_{-2}^2 (x^2 + 4 - 2x^2) dx = \int_{-2}^2 (4 - x^2) dx = \left(4x - \dfrac{x^3}{3} \right)^2_{-2}
\left[4(2) - \dfrac{(2)^3}{3} \right] - \left[4(-2) - \dfrac{(-2)^3}{3} \right] = \left[8 - \dfrac{8}{3} \right] - \left[8 + \dfrac{8}{3} \right]
= \dfrac{32}{3}
sq. units
The area bounded by a the curves y=x(1-/nX) and positive X-axis between
X={ e }^{ -1 }
amd X=e is:-
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0%
\left( \frac { { e }^{ 2 }-{ 4e }^{ -2 } }{ 5 } \right)
0%
\left( \frac { { e }^{ 2 }-{ 5e }^{ -2 } }{ 4 } \right)
0%
\left( \frac { { 4e }^{ 2 }-{ e }^{ -2 } }{ 5 } \right)
0%
\left( \frac { { 5e }^{ 2 }-{ e }^{ -2 } }{ 4 } \right)
The area bounded by the curves
y=x(x-3)^{2}
and
y=x
is (in
sq.units
) is
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0%
28
0%
32
0%
4
0%
8
The area bounded by the curve
y={ x }^{ 2 }
, X=axis and the ordinates z=1, z=3 is ____________.
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0%
\dfrac { 26 }{ 3 } sq.units
0%
\dfrac { 28 }{ 3 } sq.unit
0%
\dfrac { 1 }{ 3 } sq.units
0%
9\quad sq.units
The area bounded by the curves
y=x(x-3)^{2}
and
y=x
is (in sq.units) is
Report Question
0%
28
0%
32
0%
4
0%
8
The area bounded by the curve y = log x, X-axis and the ordinates x =1, x =2 is
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0%
log 4 sq. units
0%
log 2 sq units
0%
(log 4 - 1) sq.units
0%
(log 4 + 1)sq. units
The area enclosed by the curves
xy^{2}=a^{2}(a-x)
and
(a-x)y^{2}=a^{2}x
is
Report Question
0%
(\pi-2)a^{2}\ sq.units
0%
(4-\pi)a^{2}\ sq.units
0%
(\pi a^{2}/3\ sq.units
0%
\dfrac{\pi+a^{2}}{4}\ sq.units
The area bounded by the curved
{ y }^{ 2 }=16x
and the line x=4 is ___________________________.
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0%
\frac { 128 }{ 3 } sq-units
0%
\frac { 64 }{ 3 } squnits
0%
\frac { 32 }{ 3 } sq-units
0%
\frac { 16 }{ 3 } sq-units
If
A_{m}
represents the area bounded by the curve
y=\ln x^{m}
., the
x-
axis and the lines
x=1
and
x=2
, then
A_{m}+m\ A_{m-1}
is
Report Question
0%
m
0%
m^{2}
0%
m^{2}/2
0%
m^{2}-1
The area of the region bounded by the curve
y=x^{2}-3x
with
y \le 0
is
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0%
3
0%
\dfrac {9}{2}
0%
\dfrac {5}{2}
0%
none\ of\ these
Explanation
REF.Image
when x=0
\Rightarrow
y0
when y=0
\Rightarrow
x(x-3)=0
x=0 and x=3
y=x^{2}-3x
\frac{dy}{dx}=2x-3\Rightarrow x=1.5
min value at x=1.5
y=\frac{3}{2}(\frac{3}{2}-3)
=\frac{3}{2}(\frac{-3}{2})=\frac{-9}{4}=-2.25
Area of shaded region =
\int y.dx
=\int_{0}^{3}(x^{2}-3x)dx
=(\frac{x^{3}}{3}-\frac{3x^{2}}{2})^{3}_{0}
=\frac{3^{3}}{3}-3\times \frac{3^{2}}{2}=9-\frac{9\times 3}{2}=-4.5
considering magnitude = 4.5
If a curve
y = a \sqrt { x } +
bx passes through the point
( 1,2 )
and the area
bounded by the curve, line
x = 4
and
x
axis is
8
square units, then
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0%
a = 3 , b = - 1
0%
a = 3 , b = 1
0%
a = - 3 , b = 1
0%
a = - 3 , b = - 1
The area bounded by the circle
x^{2}+y^{2}=8
, the parabola
x^{2}=2y
and the line
y=x
in first quadrant is
\dfrac{2}{3}+k\pi
, where
k=
Report Question
0%
\dfrac{5}{7}
0%
2
0%
\dfrac{3}{5}
0%
3
The area enclosed between the curve
y=\log_{e}\left(x+e\right)
and the coordinate axes is
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
A]
A=\int_{1-e}^{0} ln(x+e)dx
=\int_{1-e}^{0} x^{0} ln (x+e)dx
=
(x ln(x+e))-\int_{1-e}^{0}\frac{x}{x+e}dx
=
-\int_{1-e}^{0}\frac{x+e-e}{x+e}dx
=\int_{1-e}^{0}(\frac{e}{x+e})dx
=[e ln(x+e)-x]_{1-e}^{0}=1
The area of the region formed by
{ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12\le 0,y\le x\quad and\quad x\quad \le \quad \dfrac { 5 }{ 2 } is
Report Question
0%
\frac { \pi }{ 6 } -\frac { \sqrt { 3}+1 }{ 8 }
0%
\frac { \pi }{ 6 } +\frac { \sqrt { 3}+1 }{ 8 }
0%
\frac { \pi }{ 6 } -\frac { \sqrt { 3}-1 }{ 8 }
0%
none of these
Area bounded by
y=2x^{2}
and
y=\dfrac{4}{(1+x^{2})}
will be (in sq units)
Report Question
0%
(2\pi+4/3)
0%
(2\pi-4/3)
0%
4/3-2\tan^{-1}2+\pi/2
0%
4/3-8\tan^{-1}2+2\pi
Explanation
\begin{array}{l} Po{ { int } }\, of\, intersection\, : \\ 2{ x^{ 2 } }=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ \Rightarrow { x^{ 4 } }+{ x^{ 2 } }-2=0 \\ \Rightarrow { x^{ 4 } }+2{ x^{ 2 } }-{ x^{ 2 } }-2=0 \\ \Rightarrow \left( { { x^{ 2 } }-1 } \right) \left( { { x^{ 2 } }+2 } \right) =0 \\ Then \\ Area\, bounded\, by\, y=2{ x^{ 2 } }\, and\, y=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ 2\int _{ 0 }^{ 1 }{ \left( { \dfrac { 4 }{ { 1+{ x^{ 2 } } } } -2{ x^{ 2 } } } \right) }dx=2\left[ { 4{ { \tan }^{ -1 } }x-\dfrac { { 2{ x^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 } \\ =2\left[ { 4\times \dfrac { \pi }{ 4 } -\dfrac { 2 }{ 3 } } \right] \\ =2\left[ { \pi -\dfrac { 2 }{ 3 } } \right] =2\pi -\dfrac { 4 }{ 3 } \\ Hence,\, option\, \, \, B\, \, is\, the\, correct\, answer. \end{array}
Letf(x)={ sin }^{ -1 }(sin\quad x)+{ cos }^{ -1 }(\quad cos\quad x),\quad g(x)=mx\quad and\quad h(x)=x\quad
are three functions. Now g(x) is divided area between f(x),x=
\pi
and y=0 into two equal parts.
The area bounded by the curve y=f(x), x=
\pi
and y=0 is:
Report Question
0%
\dfrac { { \pi }^{ 2 } }{ 4 } sq.\quad units
0%
{ \pi }^{ 2 }sq.units
0%
\dfrac { { \pi }^{ 2 } }{ 8 } sq.\quad units
0%
2{ \pi }^{ 2 }sq.units
Explanation
Given,
f(x)=\sin^{-1}\sin x+\cos^{-1}\cos x=x+x=2x
x=\pi
y=0
The above 3 forms a right angled triagle, with center as one vertex.
Area=\dfrac{1}{2} \times \pi times 2\pi
=\pi ^2sq.units
The area of the region bounded by the curves
1-y^{2}= \left | x \right | and \left | x \right |+\left | y \right |= 1
is
Report Question
0%
\frac{1}{3}sq. unit
0%
\frac{2}{3}sq. unit
0%
\frac{4}{3}sq. unit
0%
1 sq.unit
Find the area of the region enclosed by the curves
y=x\ \log x
and
y=2x-2x^{2}
.
Report Question
0%
1/12
0%
1/4
0%
2/12
0%
7/12
Area of the region bounded by
x^{2}+y^{2}-6y\leq 0
and
3y\leq x^{2}
is
Report Question
0%
\frac{9\pi }{2}-12
0%
\frac{9\pi }{4}-6
0%
9\pi
-24
0%
\frac{9\pi }{2}+6
The area enclosed by the curves y = cosx - sin x and y = [socx - sin x] and between x = 0 and
x =\dfrac{\pi}{2}
is
Report Question
0%
2(\sqrt{2} + 1)
sq. units
0%
2(\sqrt{2} - 1)
sq. units
0%
(\sqrt{2} - 1)
sq. units
0%
(\sqrt{2} + 1)
sq. units
The area bounded by the parabola
{{\text{y}}^{\text{2}}}{\text{ = 4}}\;{\text{ax}}\;{\text{and}}\;{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{4ay}}\;
is
Report Question
0%
\dfrac{{8{a^2}}}{3}
0%
\dfrac{{16{a^2}}}{3}
0%
\dfrac{{32{a^2}}}{3}
0%
\dfrac{{64{a^2}}}{3}
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%
\dfrac {14}{3}
0%
\dfrac {56}{3}
0%
\dfrac {8}{3}
0%
\dfrac {32}{3}
Explanation
Here,
y = x^{2} - 1
So, Equation of tangent at
(2, 3)
on
\Rightarrow
y =x^{2} - 1
, is
y = (4x - 5) ......(i)
\therefore
Required shaded area
\displaystyle = area (\triangle ABC) - \int_{-1}^{3} \sqrt {y + 1} dy
= \dfrac {1}{2} \cdot (8)\cdot (2) - \dfrac {2}{3} \left ((y + 1)^{3/2}\right )_{-1}^{3}
= 8 - \dfrac {16}{3} = \dfrac {8}{3}
(square units).
The area ehclosed by the curves y = f(x) and y =g(x), where f9x) = max
{x , x^2}
and g(x) = min
{x, x^2}
opver the interval [0,1] is
Report Question
0%
\dfrac{1}{6}
0%
\dfrac{1}{3}
0%
\dfrac{1}{2}
0%
1
The region in the
xy
- plane is bounded by curve
y=\sqrt {(25-x^2)}
and the line
y=0
. If the point
(a,a+1)
lies in the interior of the region, then
Report Question
0%
a \in \left( { - 4,3} \right)
0%
a \in (- \infty, -1) \cup (3, \infty)
0%
a \in (-1,3)
0%
None of these
The area of the region bounded by the parabolas
y^2= and x^2 = y, is
Report Question
0%
\dfrac{1}{3}
q.units
0%
\dfrac{8}{3}
q.units
0%
\dfrac{16}{3}
q.units
0%
\dfrac{4}{3}
q.units
If the area of the region bounded by the curves,
y=x^{2},y=\frac{1}{x}
and the lines y=0 and x=t (t > 1) is 1 sq. unit, then t is equal to:
Report Question
0%
{\dfrac{10}{3}}
0%
\dfrac{4}{3}
0%
\dfrac{7}{3}
0%
\dfrac{11}{3}
The area (in sq. units) in the first quadrant bounded by the parabola,
y = x^2 + 1
, the tangent to it at the point
(2, 5)
and the coordinate axes is:-
Report Question
0%
\dfrac{14}{3}
0%
\dfrac{187}{24}
0%
\dfrac{37}{24}
0%
\dfrac{8}{3}
The slope of the tangent to the curve y =f(x) at a point (x, Y) is 2x + 1 and the curve passes through (1, 2) The area of the region bounded by the curve, the x-axis and the line x= 1 is -
Report Question
0%
5/3 units
0%
5/6 units
0%
6/5 units
0%
6 units
The area (in sq. units) of the region
\{ { x },{ y }):{ y }^{ { 2 } }\geq 2{ x }
and
x ^ { 2 } + y ^ { 2 } \leq 4 x , x \geq 0 , y \geq 0 \}
is :
Report Question
0%
\pi - \dfrac { 4 \sqrt { 2 } } { 3 }
0%
\dfrac { \pi } { 2 } - \dfrac { 2 \sqrt { 2 } } { 3 }
0%
\pi - \dfrac { 4 } { 3 }
0%
\pi - \dfrac { 8 } { 3 }
The area of the region
A=[(x,y):0\le y\le x|x|+1
and
-1\le x\le 1]
in sq . units is :
Report Question
0%
\dfrac{2}{3}
0%
\dfrac{1}{3}
0%
2
0%
\dfrac{4}{3}
Explanation
The graph is as follows:
\displaystyle \int_{-1}^0(-x^2+1)dx+\displaystyle \int_0^1(x^2+1)dx=2
The area (in sq. units) of the region bounded
by the parabola,
y = x ^ { 2 } + 2
and the lines,
y = x + 1 , x = 0
and
x = 3 ,
is :
Report Question
0%
\dfrac { 15 } { 4 }
0%
\dfrac { 15 } { 2 }
0%
\dfrac { 21 } { 2 }
0%
\dfrac { 17 } { 4 }
Explanation
Req. area
= \int _ { 0 } ^ { 3 } \left( x ^ { 2 } + 2 \right) d x - \dfrac { 1 } { 2 } \cdot 5.3 = 9 + 6 - \dfrac { 15 } { 2 } = \dfrac { 15 } { 2 }
If the area enclosed between the curves
y=kx^2
and
x=ky^2
,
(k > 0)
, is
1
square unit. Then
k
is?
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0%
\dfrac{1}{\sqrt{3}}
0%
\dfrac{2}{\sqrt{3}}
0%
\dfrac{\sqrt{3}}{2}
0%
\sqrt{3}
Explanation
Area bounded by
y^2=4ax
&
x^2=4by
, a, b
\neq 0
is
\left|\dfrac{16ab}{3}\right|
by using formula:
4a=\dfrac{1}{k}=4b, k > 0
Area
=\left|\dfrac{16\cdot \dfrac{1}{4k}\cdot \dfrac{1}{4k}}{3}\right|=1
\Rightarrow k^2=\dfrac{1}{3}
\Rightarrow k=\dfrac{1}{\sqrt{3}}
.
The area of the region bounded by
y=\left | x-1 \right | and \,\,y=1
is
Report Question
0%
1
0%
2
0%
1/2
0%
None of these
The area of the region
\left\{ ( x , y ) : x ^ { 2 } + y ^ { 2 } \leq 1 \leq x + y \right\}
is
Report Question
0%
\dfrac { \pi ^ { 2 } } { 5 } \text { unit } ^ { 2 }
0%
\dfrac { \pi ^ { 2 } } { 2 } \text { unit } ^ { 2 }
0%
\dfrac { \pi ^ { 2 } } { 3 } \text { unit } ^ { 2 }
0%
\left( \dfrac { \pi } { 4 } - \dfrac { 1 } { 2 } \right)\text { unit } ^ { 2 }
The area of the region bounded by the parabola y =
x^2
3x with y 0 is
Report Question
0%
3
0%
\dfrac{3}{2}
0%
\dfrac{9}{2}
0%
{9}{}
The area of the quadrilateral formed by the tangents at the endpoints of the latus recta to the ellipse,
\dfrac{x^{2}}{9}+\dfrac{y^{2}}{5}=1
is
Report Question
0%
\dfrac{27}{4}
0%
18
0%
\dfrac{27}{2}
0%
27
Explanation
Given equation of ellipse is
\dfrac{{x}^{2}}{9}+\dfrac{{y}^{2}}{5}=1
......
(1)
\therefore {a}^{2}=9,{b}^{2}=5
\Rightarrow a=3,b=\sqrt{5}
Now,
e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{5}{9}}=\dfrac{2}{3}
Foci
=\left(\pm ae,0\right)=\left(\pm 2,0\right)
and
\dfrac{{b}^{2}}{a}=\dfrac{5}{3}
\therefore
Extremities of one of latus rectum are
\left(2,\dfrac{5}{3}\right)
and
\left(2,\dfrac{-5}{3}\right)
\therefore
Equation of tangent at
\left(2,\dfrac{5}{3}\right)
is
\dfrac{x\left(2\right)}{9}+\dfrac{y\left(\dfrac{5}{3}\right)}{5}=1
or
2x+3y=9
.......
(2)
Eqn
(2)
intersects
X
and
Y
axes at
\left(\dfrac{9}{2},0\right)
and
\left(0,3\right)
respectively.
\therefore
Area of quadrilateral
4\times
Area of
\triangle{POQ}
=4\times\dfrac{1}{2}\times\dfrac{9}{2}\times 3=27
sq.units.
The area bounded by curve
y=x^{2}-1
and tangents to it at
(2,3)
and
y-
axis is
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0%
8/3
0%
2/3
0%
4/3
0%
1/3
Explanation
x-axis:(-1,0)
Area
=\int_{-1}^{0}(x^2-1)dx
=\left [ \dfrac{x^3}{3}-x \right ]_{-1}^0
=\left [ \dfrac{-1}{3}-(-1) \right ]-[0]
=-\dfrac{1}{3}+1
=\dfrac{2}{3}
sq. units
The area bounded by the curves
x+2|y|=1
and
x=0
is?
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0%
\dfrac{1}{4}
0%
\dfrac{1}{2}
0%
1
0%
2
Explanation
x+2\left| y \right| =1
and
x=0
y=+ve
x+2y=1
y=-ve
x-2y=1
Area of triangle
=\cfrac{1}{2}\times base\times height\\=\cfrac{1}{2}\times 1\times 1\\=\cfrac{1}{2}\text{ }unit
Area included between
{ y }=\dfrac { { x }^{ { 2 } } }{ 4{ a } }
and
y = \dfrac { 8 a ^ { 3 } } { x ^ { 2 } + 4 a ^ { 2 } }
is
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\dfrac { a ^ { 2 } } { 3 } ( 6 \pi - 4 )
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\dfrac { a ^ { 2 } } { 3 } ( 4 \pi + 3 )
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\dfrac { a ^ { 2 } } { 3 } ( 8 \pi + 3 )
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None of these
The area of the figure formed by
a|x|+b|y|+c=0
, is
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\dfrac{c^{2}}{|ab|}
0%
\dfrac{2c^{2}}{|ab|}
0%
\dfrac{c^{2}}{2|ab|}
0%
None\ of\ these
The area of the region bounded by the curves
y = \sin x
and
y = \cos x ,
and lying between the lines
x = \dfrac { \pi } { 4 }
and
x = \dfrac { 5 \pi } { 4 } ,
is
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2 + \sqrt { 2 }
0%
2
0%
2 \sqrt { 2 }
0%
2 - \sqrt { 2 }
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