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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 $${\int}_{0}^{1}\left(4x^{3}=f(x)\right)f(x)dx=\dfrac{4}{7}$$, 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%
$$\dfrac{11}{2}$$
0%
$$\dfrac{13}{2}$$
0%
$$\dfrac{15}{2}$$
0%
$$\dfrac{17}{2}$$
The are boundede by the curve $$y=x^{2},y=-x$$ and $$y^{2}=4x-3$$ is $$k$$, them the value of $$9k$$ is
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$$2$$
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$$3$$
0%
$$0$$
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$$4$$
If area bounded by to curves $$y^2 = 4ax$$ and y=mx is $$\dfrac{a^2}{3}$$, then the value of m is
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$$2$$
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$$-1$$
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$$\dfrac{1}{2}$$
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none of these
Explanation
Given,
$$y^2=4ax,y=mx$$ curves intersect at $$\left ( 4\dfrac{a}{m^2},\dfrac{a}{m} \right )$$
Area enclosed is,
$$A=\displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx$$
$$\therefore \displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx=\dfrac{a^2}{3}$$ .... 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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$$\cfrac { 5 }{ 3 } sq.unit$$
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$$\cfrac { 1 }{ 3 } sq.unit$$
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$$\cfrac { 2 }{ 3 } sq.unit$$
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$$\cfrac { 4 }{ 3 } sq.unit$$
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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$$2$$
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$$4$$
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$$6$$
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$$8$$
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%
$$3$$
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$$3\ln (3)-2$$
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$$3\ln (3)+2$$
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$$2$$
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%
$$\dfrac{15}{64}$$
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$$\dfrac{9}{32}$$
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$$\dfrac{7}{32}$$
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$$\dfrac{5}{64}$$
The area bounded by the curves $$y = \log _ { e } x$$ and $$y = \left( \log _ { e } x \right) ^ { 2 }$$ is
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$$3 - e$$
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$$e-3$$
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$$\frac { 1 } { 2 } ( 3 - e )$$
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$$\frac { 1 } { 2 } ( e - 3 )$$
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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$$\dfrac { 2 }{ 3 } sq.\quad units\quad $$
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$$\dfrac { 3 }{ 2 } sq.\quad units\quad $$
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$$\dfrac { 32 }{ 3 } sq.\quad units\quad $$
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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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$$\left( \frac { { e }^{ 2 }-{ 4e }^{ -2 } }{ 5 } \right) $$
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$$\left( \frac { { e }^{ 2 }-{ 5e }^{ -2 } }{ 4 } \right) $$
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$$\left( \frac { { 4e }^{ 2 }-{ e }^{ -2 } }{ 5 } \right) $$
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$$\left( \frac { { 5e }^{ 2 }-{ e }^{ -2 } }{ 4 } \right) $$
The area bounded by the curves $$y=x(x-3)^{2}$$ and $$y=x$$ is (in $$sq.units$$) is
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0%
$$28$$
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$$32$$
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$$4$$
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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$$\dfrac { 26 }{ 3 } sq.units$$
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$$\dfrac { 28 }{ 3 } sq.unit$$
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$$\dfrac { 1 }{ 3 } sq.units$$
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$$9\quad sq.units$$
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 = 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
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(log 4 - 1) sq.units
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(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
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0%
$$(\pi-2)a^{2}\ sq.units$$
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$$(4-\pi)a^{2}\ sq.units$$
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$$(\pi a^{2}/3\ sq.units$$
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$$\dfrac{\pi+a^{2}}{4}\ sq.units$$
The area bounded by the curved $${ y }^{ 2 }=16x$$ and the line x=4 is ___________________________.
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$$\frac { 128 }{ 3 } sq-units$$
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$$\frac { 64 }{ 3 } squnits$$
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$$\frac { 32 }{ 3 } sq-units$$
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$$\frac { 16 }{ 3 } sq-units$$
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
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$$m$$
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$$m^{2}$$
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$$m^{2}/2$$
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$$m^{2}-1$$
The area of the region bounded by the curve $$y=x^{2}-3x$$ with $$y \le 0$$ is
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$$3$$
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$$\dfrac {9}{2}$$
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$$\dfrac {5}{2}$$
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$$none\ of\ these$$
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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$$a = 3 , b = - 1$$
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$$a = 3 , b = 1$$
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$$a = - 3 , b = 1$$
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$$a = - 3 , b = - 1$$
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 $$k=$$
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$$\dfrac{5}{7}$$
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$$2$$
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$$\dfrac{3}{5}$$
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$$3$$
The area enclosed between the curve $$y=\log_{e}\left(x+e\right)$$ and the coordinate axes is
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$$1$$
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$$2$$
0%
$$3$$
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$$4$$
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 $${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12\le 0,y\le x\quad and\quad x\quad \le \quad \dfrac { 5 }{ 2 } is$$
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$$\frac { \pi }{ 6 } -\frac { \sqrt { 3}+1 }{ 8 } $$
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$$\frac { \pi }{ 6 } +\frac { \sqrt { 3}+1 }{ 8 } $$
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$$\frac { \pi }{ 6 } -\frac { \sqrt { 3}-1 }{ 8 } $$
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none of these
Area bounded by $$y=2x^{2}$$ and $$y=\dfrac{4}{(1+x^{2})}$$ will be (in sq units)
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$$(2\pi+4/3)$$
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$$(2\pi-4/3)$$
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$$4/3-2\tan^{-1}2+\pi/2$$
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$$4/3-8\tan^{-1}2+2\pi$$
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:
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$$\dfrac { { \pi }^{ 2 } }{ 4 } sq.\quad units$$
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$${ \pi }^{ 2 }sq.units$$
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$$\dfrac { { \pi }^{ 2 } }{ 8 } sq.\quad units$$
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$$2{ \pi }^{ 2 }sq.units$$
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
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$$\frac{1}{3}sq. unit $$
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$$\frac{2}{3}sq. unit $$
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$$\frac{4}{3}sq. unit $$
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$$1 sq.unit$$
Find the area of the region enclosed by the curves $$y=x\ \log x$$ and $$y=2x-2x^{2}$$.
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$$1/12$$
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$$1/4$$
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$$2/12$$
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$$7/12$$
Area of the region bounded by $$x^{2}+y^{2}-6y\leq 0$$ and $$3y\leq x^{2}$$ is
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$$\frac{9\pi }{2}-12$$
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$$\frac{9\pi }{4}-6$$
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$$9\pi$$-24
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$$\frac{9\pi }{2}+6$$
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
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$$2(\sqrt{2} + 1)$$ sq. units
0%
$$2(\sqrt{2} - 1)$$ sq. units
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$$(\sqrt{2} - 1)$$ sq. units
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$$(\sqrt{2} + 1)$$ 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
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0%
$$\dfrac{{8{a^2}}}{3}$$
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$$\dfrac{{16{a^2}}}{3}$$
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$$\dfrac{{32{a^2}}}{3}$$
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
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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
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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
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0%
$$a \in \left( { - 4,3} \right)$$
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$$a \in (- \infty, -1) \cup (3, \infty)$$
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$$a \in (-1,3) $$
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None of these
The area of the region bounded by the parabolas $$y^2= and x^2 = y, is$$
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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:
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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:-
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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 -
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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 :
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$$\pi - \dfrac { 4 \sqrt { 2 } } { 3 }$$
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$$\dfrac { \pi } { 2 } - \dfrac { 2 \sqrt { 2 } } { 3 }$$
0%
$$\pi - \dfrac { 4 } { 3 }$$
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$$\pi - \dfrac { 8 } { 3 }$$
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 :
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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 :
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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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$$\dfrac{1}{\sqrt{3}}$$
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
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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
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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
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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
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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%
$$2/3$$
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$$4/3$$
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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0%
$$\dfrac{c^{2}}{|ab|}$$
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$$\dfrac{2c^{2}}{|ab|}$$
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$$\dfrac{c^{2}}{2|ab|}$$
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$$None\ of\ these$$
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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0%
$$2 + \sqrt { 2 }$$
0%
$$2$$
0%
$$2 \sqrt { 2 }$$
0%
$$2 - \sqrt { 2 }$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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