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CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Application Of Integrals
Quiz 11
The smaller area enclosed by $$y=f(x)$$, where $$f(x)$$ is polynomial of least degree satisfying $$\displaystyle{ \left[ \lim _{ x\rightarrow 0 }{ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } } \right] }^{ \tfrac { 1 }{ x } }=e$$ and the circle $$x^2+y^2=2$$ above the $$x-$$axis is
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$$\displaystyle\frac { \pi }{ 2 } +\frac { 3 }{ 5 } $$
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$$\displaystyle\frac { \pi }{ 2 } -\frac { 3 }{ 5 } $$
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$$\dfrac { \pi }{ 2 } -\dfrac { 6 }{ 5 } $$
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None of these
Explanation
Since $$\displaystyle\lim _{ x\rightarrow 0 }{ { \left[ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } \right] }^{ \tfrac { 1 }{ x } } } $$ exists, so $$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { f\left( x \right) }{ { x }^{ 3 } } } =0$$
$$\therefore f(x)={ a }_{ 4 }{ x }^{ 4 }+{ a }_{ 5 }{ x }^{ 5 }+...+{ a }_{ n }{ x }^{ n },a_n\neq 0,n\ge 4$$
Since, $$f(x)$$ is of least degree $$\Rightarrow f(x)={ a }_{ 4 }{ x }^{ 4 }$$
The graph of $$y=x^4$$ and $$x^2+y^2=2$$ are shown in the figure
$$\therefore$$ The required area $$\displaystyle=2\int _{ 0 }^{ 1 }{ \left( \sqrt { 2-{ x }^{ 2 } } -{ x }^{ 4 } \right) } dx=\frac { \pi }{ 2 } +\frac { 3 }{ 5 } $$
The area of the region described by $$\left \{(x, y)/ x^{2} + y^{2} \leq 1\ and\ y^{2} \leq 1 - x\right \}$$ is
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$$\dfrac {\pi}{2} - \dfrac {2}{3}$$
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$$\dfrac {\pi}{2} + \dfrac {2}{3}$$
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$$\dfrac {\pi}{2} + \dfrac {4}{3}$$
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$$\dfrac {\pi}{2} - \dfrac {4}{3}$$
The area (in square units) of the region bounded by the curves $$x=y^2$$ and $$x=3-2y^2$$ is
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$$\dfrac{3}{2}$$
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2
0%
3
0%
4
Explanation
Required area $$\displaystyle =2\int_0^1(3-2y^2-y^2)dy=6\int_0^1(1-y^2)dy=6\left(y-\dfrac{y^3}{3}\right)_0^1=4$$
Since given curves are symmetrical about x-axis
The area (in square units) bounded by the curves $$x\, =\, -2y^2$$ and $$x\, =\, 1-3y^2$$ is
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$$\displaystyle \frac{2}{3}$$
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$$1$$
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$$\displaystyle \frac{4}{3}$$
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$$\displaystyle \frac{5}{3}$$
Explanation
$$x=-2y^2$$
$$x=1-3y^2$$
$$-2y^2=1-3y^2\Rightarrow y^2=1\Rightarrow y=\pm 1$$
Area between the curves $$= A_1-A_2$$
$$\displaystyle =\int^1_{-1}(1-3y^2+2y^2)dy$$
$$=\int^1_{-1}(1-y^2)dy$$
$$=[y-\dfrac{y^3}{3}]^1_{-1}=\dfrac{4}{3}$$
The area of the region bounded by the curves $$x^{2} + y^{2} = 8$$ and $$y^{2} = 2x$$ is
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$$2\pi + \dfrac {1}{3}$$
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$$\pi + \dfrac {1}{3}$$
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$$2\pi + \dfrac {4}{3}$$
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$$\pi + \dfrac {4}{3}$$
Explanation
Given curves,
$$x^{2} + y^{2} = 8$$ .....(i)
and $$y^{2} = 2x$$ .....(ii)
On solving Eqs. (i) and (ii), we get
$$x^{2} + 2x - 8 = 0$$
$$x^{2} + 4x - 2x - 8 = 0$$
$$x (x - 4) - 2 (x + 4) = 0$$
$$(x- 2) (x +4) = 0$$
$$\therefore x = 2$$ and $$y = \pm 2$$
$$\therefore$$ Required area
$$= 2[\text {Area of OAP} + \text {Area of PAB}]$$
$$= 2\left [\displaystyle\int_{0}^{2} \sqrt {2x} dx + \int_{2}^{2\sqrt {2}} \sqrt {8 - x^{2}} dx \right ]$$
$$= 2\left [\sqrt {2} \left (x^{3/2} \cdot \dfrac {2}{3}\right )^{2}_{0} + \left (\dfrac {x}{2} \sqrt {8 - x^{2}} + \dfrac {8}{2} \sin^{-1} \dfrac {x}{2\sqrt {2}} \right )_{2}^{2\sqrt {2}} \right ]$$
$$= 2\left [\dfrac {2\sqrt {2}}{3} (2^{3/2}) + 4\times \dfrac {\pi}{2} - 2 - 4\times \dfrac {\pi}{4}\right ]$$
$$= 2\left [\dfrac {2\sqrt {2}}{3} \cdot 2\sqrt {2} + 2\pi - 2 - \pi \right ]$$
$$= 2\left [\dfrac {8}{3} - 2 + \pi \right ] + 2\left (\dfrac {2}{3} + \pi \right ) = 2\pi + \dfrac {4}{3}$$
Area common to the curves $$5x^2 = 0$$ and $$ 2x^2 + 9 = 0$$ is equal to
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$$12 \sqrt 3 $$
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$$ 6 \sqrt3$$
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$$36$$
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$$18$$
Explanation
Given curves
$$y=5x^2$$
$$\Rightarrow x^2=\dfrac{y}{5}$$ ....(1)
which is a parabola opening upward having vertex at (0,0)
Other curve is $$2x^2+9=y$$ .....(2)
$$\Rightarrow x^2=\dfrac{y-9}{2}$$
which is a parabola having vertex at (0,9).
Now, solving eqn (1) and (2), we get
$$5x^2=2x^2+9$$
$$\Rightarrow 3x^2=9$$
$$\Rightarrow x=\pm\sqrt{3}$$
$$\Rightarrow y=15$$
Point of intersection of the curves is $$(-\sqrt{3},15)$$ and $$(\sqrt{3},15)$$
Required area $$=2(ar OAB)$$
$$=2 \int ^\sqrt 3 _0 (y_1-y_2)dx$$
$$= 2 \int ^\sqrt 3 _0 \{ (2x^2 + 9) - 5 x^2\} dx$$
$$ 2 \int ^\sqrt 3 _ 0 (9-3x^2) dx$$
$$ = 2 \left( 9x - 3 \displaystyle \frac{x^3}{3}\right) ^\sqrt3 _0$$
$$ = 2(9 \sqrt 3 - 3 \sqrt 3 ) = 12 \sqrt 3 $$ sq. units
The area of the region, bounded by the curves $$y = \sin^{-1} x + x (1 - x)$$ and $$y = \sin^{-1} x - x (1 - x)$$ in the first quadrant, is
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$$1$$
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$$\dfrac {1}{2}$$
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$$\dfrac {1}{3}$$
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$$\dfrac {1}{4}$$
Explanation
$$\sin^{-1} x$$ is defined, if $$-1 \leq x \leq 1$$
In first quadrant $$0\leq x\leq 1$$ and $$x(1 - x) \geq 0$$
$$\therefore y = \sin^{-1} x + x (1 - x)$$ ...... (i)
Lies above $$y = \sin^{-1} x - x (1 - x)$$ ...... (ii)
On solving, we get
$$2x (1 - x) = 0$$
$$\Rightarrow x = 0, 1$$
$$\therefore$$ Required area $$=\displaystyle \int_{0}^{1} (y_{1} - y_{2})dx$$
$$=\displaystyle \int_{0}^{1} [\left \{\sin^{-1} x + x(1 - x)\right \} - \left \{\sin^{-1} x - x(1 - x) \right \}]dx$$
$$= 2\displaystyle \int_{0}^{1} (x - x^{2}) dx$$
$$= 2\left [\dfrac {x^{2}}{2} - \dfrac {x^{3}}{3}\right ]_{0}^{1} = 2\left (\dfrac {1}{2} - \dfrac {1}{3}\right ) = \dfrac {1}{3}$$
If the area bounded by the curves $$y=a{ x }^{ 2 }$$ and $$x=a{ y }^{ 2 }$$, $$(a>0)$$ is $$1$$ sq.units, then the value of $$a$$ is
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$$\cfrac { 2 }{ 3 } $$
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$$\cfrac { 1 }{ \sqrt 3 } $$
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$$1 $$
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$$4$$
Explanation
Points of intersection of $$y=ax^2$$ and $$x=ay^2$$ are $$(0,0)$$ and $$\left (\dfrac {1}{a},\dfrac {1}{a}\right)$$.
Hence, $$\displaystyle \int _0 ^{\tfrac1a} \left (\sqrt {\dfrac {x}{a}}-ax^2\right)dx=1$$
$$\Rightarrow \displaystyle\cfrac{2x^{\tfrac32}}{3\sqrt a} \big|_0^{\tfrac1a} - \cfrac{ax^3}3\bigr|_0^{\tfrac1a} =1$$
$$\Rightarrow \cfrac{2}{3a^2} - \cfrac1{3a^2} =1$$
$$\Rightarrow \cfrac{1}{3a^2} =1$$
$$\Rightarrow a=\dfrac {1}{\sqrt3}$$ ....As $$(a>0)$$
Area bounded by curve $$y=x^2$$ and $$y=2-x^2$$ is ?
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$$\dfrac{8}{3}$$ sq units
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$$\dfrac{3}{8}$$ sq units
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$$\dfrac{3}{2}$$ sq units
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None of these
Explanation
$$y=x^2$$ and $$y=2-x^2$$
now both he curve intersect each other ,
from both the equation
$$ x^2=2-x^2$$
$$ x=+1,-1$$
now area bounded by both curve is
$$A =\displaystyle \int_{-1}^{1}(2-x^2)-\displaystyle \int_{-1}{1}(x^2)$$
because both are even function hence
$$A =2[\displaystyle \int_{0}^{1}(2-x^2)-\displaystyle \int_{0}{1}(x^2)$$]
$$A=2[(2x-\dfrac{x^3}{3})-(\dfrac{x^3}{3})]$$
now on putting upper and lower value of limit ,we will get
$$A=\dfrac{8}{3}$$
If the line $$x = \alpha $$ divides the area of region $$R=\left\{ \left( x,y \right) \in { R }^{ 2 }:{ x }^{ 3 }\le y\le x,0\le x\le 1 \right\} $$ into two equal parts, then
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$$2{ \alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$$
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$${ \alpha }^{ 4 }+4{ \alpha }^{ 2 }-1=0$$
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$$0 < \alpha \le \dfrac { 1 }{ 2 } $$
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$$\dfrac { 1 }{ 2 } < \alpha < 1$$
Explanation
Area $$(x=0\to x=\alpha) =$$ Area of $$\triangle OAB - $$ Area of $$\triangle OCB$$
$$\displaystyle= \dfrac12\alpha\cdot\alpha - \int_0^\alpha x^2 dx$$
$$=\cfrac{\alpha^2}2 - \cfrac{\alpha^2}4$$
Area $$(0=\alpha \to x=1) = $$ Area of $$\Box BAEF - $$ Area of $$\Box BCEF$$
$$\cfrac12(\alpha+1)(1-\alpha) -\displaystyle \int_\alpha^1x^3\ dx$$
$$\cfrac{1-\alpha^2}{2} - \left(\cfrac14-\cfrac{\alpha^2}4\right)$$
According to the question:
$$\cfrac{\alpha^2}2-\cfrac{\alpha^4}4 = \cfrac12-\cfrac{\alpha^2}2 - \cfrac14 + \cfrac{\alpha^4}4$$
$$\Rightarrow 2\alpha^4-4\alpha^2+1=0$$ ... Option A
$$f(a) = 2\alpha^2 - 4\alpha^2 + 1 = 0$$
At $$\alpha = 0, f(a) = 1$$
At $$\alpha = 1, f(\alpha) = -1$$
At $$\alpha =\cfrac12, f(a) = \cfrac18 > 0$$
Therefore, Root lies in $$\alpha \in \left(\cfrac12,1\right)$$ .. Option D.
The area bounded by the parabolas $$y^2 = 4a(x + a)$$ and $$y^2 = - 4a (x - a)$$ is
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$$\dfrac{16}{3} a^2$$ sq units
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$$\dfrac{8}{3} $$ sq units
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$$\dfrac{4}{3} a^2$$ sq units
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None of these
Explanation
$$y^{2}=4a\left ( x+a \right )$$
$$y=\pm \sqrt{4ax+4a^{2}}$$
As the area bounded by this curve is in -ve $$x$$ quadrant . Therfore
$$y=-\sqrt{-4ax+4a^{2}}$$
Area bounded by this curve will be $$\int_{-a}^{0}\sqrt{4ax+4a^{2}dx} $$
Area bounded by this curve will be $$ \dfrac {8}{3} a^{2} $$
Similarly , The area bounded by $$y^{2}=-4a\left ( x-a \right )$$ curve will be
$$\int_{0}^{a}\sqrt{-4ax+4a^{2}dx}$$
Area =
$$ \dfrac {8}{3} a^{2} $$
thus , total bounded area will be
$$ \dfrac {16 }{3} a^{2} $$
The area of the region bounded by the curves $$y = 2^{x}, y = 2x - x^{2}$$ and $$x = 2$$ is
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$$\dfrac {3}{\log 2} - \dfrac {4}{3}$$
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$$\dfrac {3}{\log 2} - \dfrac {4}{9}$$
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$$\dfrac {3}{2} - \dfrac {\log 2}{9}$$
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None of these
Explanation
Given curves are $$y = 2^{x}, y = 2x - x^{2}$$ and $$x = 2$$
$$\therefore$$ 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
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$$\left( 8\pi -\sqrt { 3 } \right) $$ sq units
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$$\dfrac { 16 }{ 3 } \left( 8-\sqrt { 3 } \right) $$ sq units
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$$\dfrac { 16 }{ 3 } \left( 8\pi -\sqrt { 3 } \right) $$ sq units
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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.
Area common to the curves $$y^{2} = ax$$ and $$x^{2} + y^{2} = 4ax$$ is equal to
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$$(9\sqrt {3} + 4\pi) \dfrac {a^{2}}{3}$$
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$$(9\sqrt {3} + 4\pi)a^{2}$$
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$$(9\sqrt {3} - 4\pi) \dfrac {a^{2}}{3}$$
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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 : $$=2(shaded\quad area)$$
$$=2\left( \left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 } \right) \\ =\left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 } sq.units$$
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?
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$$\displaystyle\frac{\pi}{2}-\frac{4}{\pi}$$
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$$\displaystyle\frac{\pi}{4}-\frac{2}{\pi}$$
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$$\displaystyle \pi -\frac{8}{\pi}$$
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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 $$\cfrac { \pi }{ 2 } +\cfrac { 4 }{ \pi } sq.\quad units$$
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?
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$$100$$
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$$193$$
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$$200$$
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$$202$$
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.
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$$\displaystyle\frac{1}{3}$$
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$$\displaystyle\frac{2}{3}$$
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$$\displaystyle\frac{\pi}{4}$$
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$$\displaystyle\frac{\pi}{3}$$
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
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$$2 + \sqrt {2}$$
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$$\sqrt {2}$$
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$$2$$
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$$2 - \sqrt {2}$$
The area bounded by min (|x|, |y|) = 2 and max (|x|, |y|) = 4 is
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8 sq unit
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16 sq unit
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24 sq unit
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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 $$B$$
The area of the region $$\left\lfloor x \right\rfloor +\left\lfloor y \right\rfloor =1,-1\le x\le 1$$ and $$xy\le 1/2$$
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rational
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$$\cfrac { 1 }{ 2 } \left( \cfrac { 3 }{ 2 } +\ln { 2 } \right) $$
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$$\cfrac { 1 }{ 2 } \left( \cfrac { 5 }{ 2 } +\ln { 2 } \right) $$
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Irrational
Explanation
NOTE: $$xy\leq1/2$$ is equation of region below the parabola $$y\leq1/2x$$ and above the co-ordinate axis (shown by blue)
$$\bf{CASE-1:}$$ $$x\in[-1,0)$$
$$\left \lfloor{x}\right \rfloor=-1$$, so we want
$$\left \lfloor{y}\right \rfloor=2$$
$$\rightarrow$$we want $$y\in[2,3)$$
$$\rightarrow$$This region is depited by full red square in image
$$\rightarrow Area_1=1*1 = 1$$
$$\bf{CASE-2}$$ $$x\in[0,1)$$
$$\left \lfloor{x}\right \rfloor=0$$
, so we want
$$\left \lfloor{y}\right \rfloor=1$$
$$\rightarrow$$we want $$y\in[1,2)$$
$$\rightarrow$$ so this case depicts a square (with lower-left vertex at $$(0,1)$$ and upper-right vertex at $$(1,2)$$ ) on co-ordinate axis, but note that $$xy\leq1/2$$ cuts of this square as shown in image
We can find the shaded red region by doing,
$$Area_2= \int_{1}^{2} \dfrac{1}{2y}dy$$ ,treating x as a function of y, ($$x=1/2y$$)
$$\dfrac{1}{2}$$ $$ \int_{1}^{2}\dfrac{1}{y}dy=\dfrac{1}{2} (ln(2)-ln(1))=\dfrac{ln(2)}{2}$$
Total area=$$Area_1+Area_2=1+ln(2)/2=\dfrac{1}{2}(2+ln(2)) \rightarrow IRRATIONAL$$
Note: ln(n) -> IRRATIONAL for all integers n$$\geq$$2
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
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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
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$$18$$
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$$32$$
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$$36$$
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$$64$$
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
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$$C_1 and C-2$$ touch each other only at one point.
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$$C_1 and C-2$$ touch each other exactly at two points.
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$$C_1 and C-2$$ intersect (but do not touch) at exactly two points.
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$$C_1 and C-2$$ neither intersect nor touch each other.
What is the area of the region bounded by the parabola $${ y }^{ 2 }=6(x-1)$$ and $${ y }^{ 2 }=3x$$
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$$\cfrac { \sqrt { 6 } }{ 3 } $$
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$$\cfrac { 2\sqrt { 6 } }{ 3 } $$
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$$\cfrac { 4\sqrt { 6 } }{ 3 } $$
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$$\cfrac { 5\sqrt { 6 } }{ 3 } $$
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 $$=\dfrac{4\sqrt{6}}{3}$$
The area bounded by the curves $$x= a \cos^3t, y= a \sin^3 t$$ is
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$$\dfrac{3\pi a^2}{8}$$
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$$\dfrac{3\pi a^2}{16}$$
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$$\dfrac{3\pi a^2}{32}$$
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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$$
The area of the region lying between the line x-y+2=0 and the curve x=$$\sqrt y $$.
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9
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9/2
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10/3
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none of these
$$Let\quad f(x)=2-\left| x-1 \right| and\quad g(x)={ \left( x-1 \right) }^{ 2 },\quad then\quad $$
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area bounded by $$f(x)$$ and $$g(x)$$ is $$\cfrac { 7 }{ 6 } $$
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area bounded by $$f(x)$$ and $$g(x)$$ is $$\cfrac { 7 }{ 3 } $$
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area bounded by $$f(x)$$$$g(x)$$ and $$x-$$ axis is $$\cfrac { 5 }{ 3 } $$
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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?
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$$25(\pi-2)$$
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$$50(\pi-2)$$
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$$25\pi$$
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$$50\pi$$
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$$ 40\pi (5-\sqrt{2})$$
The area bounded by the curves $$y={ \left( x-1 \right) }^{ 2 },y={ \left( x+1 \right) }^{ 2 }$$ and $$y=\dfrac { 1 }{ 4 }$$ is
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$$\dfrac { 1 }{ 3 } sq\ unit$$
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$$\dfrac { 2 }{ 3 } sq\ unit$$
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$$\dfrac { 1 }{ 4 } sq\ unit$$
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$$\dfrac { 1 }{ 5 } sq\ unit$$
Area common to the circle $$x^{2}+y^{2}=64$$ and the parabola $$y^{2}=4x$$ is
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$$\dfrac{16}{3}(4\pi + \sqrt{3})$$
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$$\dfrac{16}{3}(8\pi + \sqrt{3})$$
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$$\dfrac{16}{3}(4\pi - \sqrt{3})$$
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$$none\ of\ these$$
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
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$$\left(\dfrac{5\pi}{12}+3\right)$$.sq.unit
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$$\left(\dfrac{5\pi}{12}+\sqrt{2}\right)$$.sq.unit
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$$\left(\dfrac{5\pi}{12}+\sqrt{3}\right)$$.sq.unit
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$$\left(\dfrac{5\pi}{12}+\sqrt{2}+\sqrt{3}\right)$$.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.
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$$6\pi+\dfrac{4}{3}$$ ,
$$2\pi-\dfrac{4}{3}$$
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$$6\pi-\dfrac{4}{3}$$ ,
$$2\pi+\dfrac{4}{3}$$
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$$6\pi+\dfrac{2}{3}$$ ,
$$2\pi-\dfrac{2}{3}$$
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$$6\pi-\dfrac{2}{3}$$ ,
$$2\pi+\dfrac{2}{3}$$
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.
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$$\dfrac{\pi^2-8}{\pi^2+8}$$
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$$\dfrac{\pi^2-4}{\pi^2+4}$$
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$$\dfrac{\pi -4}{\pi +4}$$
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$$\dfrac{2\pi^2}{\pi^2+2\pi -8}$$
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
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$$0$$
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$$\sqrt{2}$$
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$$-\sqrt{2}$$
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None of these
The area of the region enclosed between by the $${x^2} + {y^2} = 16$$ and the parabola $${y^2} = 6x$$.
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$$\dfrac{2}{3} (\sqrt 3 + 4\pi)$$ sq. units
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$$\dfrac{4}{3} (\sqrt 3 + 4\pi)$$ sq. units
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$$\dfrac{2}{3} (\sqrt 3 + 8\pi)$$ sq. units
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$$\dfrac{4}{3} (\sqrt 3 + 8\pi)$$ sq. units
Explanation
Point of intersection of the parabola and the circle is obtained by solving the equations:
$${x}^{2}+{y}^{2}=16$$ and $${y}^{2}=6x$$
$$\Rightarrow\,{x}^{2}+6x-16=0$$
$$\Rightarrow\,{x}^{2}+8x-2x-16=0$$
$$\Rightarrow\,x\left(x+8\right)-2\left(x+8\right)=0$$
$$\Rightarrow\,\left(x-2\right)\left(x+8\right)=0$$
$$\Rightarrow\,x=2,\,x=-8$$
$$\therefore\,x=2$$ is the only possible solution(from the fig.)
$$\therefore\,$$ when $$x=2,\,y=\pm\sqrt{6\times 2}=\pm\,2\sqrt{3}$$
$$\therefore\,B\left(2,2\sqrt{3}\right)$$ and $${B}^{\prime}\left(2,-2\sqrt{3}\right)$$ are the points of intersection of parabola and the circle.
Required area$$=area\,of\,OBA{B}^{\prime}O=2\,area\,of \,OBAO$$
$$=2\left[area\,of\,OBDO+area\,of\,DBAD\right]$$
$$=2\left[\displaystyle\int_{0}^{2}{\sqrt{6x}dx}+\displaystyle\int_{2}^{4}{\sqrt{16-{x}^{2}}dx}\right]$$
$$=2\left[\sqrt{6}\left[\dfrac{{x}^{\frac{3}{2}}}{\dfrac{3}{2}}\right]_{0}^{2}+\left[\dfrac{1}{2}x\sqrt{16-{x}^{2}}+\dfrac{1}{2}\times 16{\sin}^{-1}{\dfrac{x}{4}}\right]_{2}^{4}\right]$$
$$=2\left[\left[\sqrt{6}\times\dfrac{2}{3}\times{2}^{\frac{3}{2}}-0\right]+\left[\dfrac{1}{2}\times\,4\sqrt{16-16}+\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{4}{4}}-\dfrac{1}{2}\times\,2\sqrt{16-4}-\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{1}{2}}\right]\right]$$
$$=2\left[\dfrac{8\sqrt{3}}{3}+\dfrac{8\pi}{2}-2\sqrt{3}-\dfrac{8\pi}{6}\right]$$
$$=2\left[\dfrac{8\sqrt{3}-6\sqrt{3}}{3}+8\left(\dfrac{\pi}{2}-\dfrac{\pi}{6}\right)\right]$$
$$=2\left[\dfrac{2\sqrt{3}}{3}+8\times\dfrac{2\pi}{6}\right]$$
$$=\dfrac{4\sqrt{3}}{3}+\dfrac{16\pi}{3}$$
Area bounded by $$|x-1| \le 2$$ and $$x^{2}-y^{2}=1$$, is
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$$6 \sqrt{2}+\dfrac{1}{2}$$ In $$|3+2\sqrt{2}|$$
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$$6 \sqrt{2}+\dfrac{1}{2}$$ In $$|3-2\sqrt{2}|$$
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$$6 \sqrt{2}-$$ In $$|3+2\sqrt{2}|$$
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$$none\ of\ these$$
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 $$=2 \int_{3}^{1} \sqrt {x^2-1}dx$$
$$\because \int \sqrt {x^2-a^2}dx=\dfrac{x}{2}=\dfrac{a^2}{2}ln(x+\sqrt{x^2-a^2})$$
$$=2\left[\dfrac{x}{2}\sqrt{x^2-1}-\dfrac{1}{2}in (x+\sqrt{x^2-1})\right]^3_1$$
$$2\left[\dfrac{3}{2}\sqrt8-\dfrac{1}{2}in (3+\sqrt 8)\right]-2\left[\dfrac{1}{2}\sqrt 0-\dfrac{1}{2}in (1+0)\right]$$
$$6\sqrt 2 in (3+2\sqrt 2)-0$$
$$A=6\sqrt 2-in (3+2\sqrt 2)$$
Find area curved by three circles
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$$(5\pi-3\sqrt{3})units^{2}$$
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$$(5\pi+4\sqrt{3})units^{2}$$
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$$(5\pi+3\sqrt{3})units^{2}$$
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$$(5\pi+3\sqrt{2})units^{2}$$
Explanation
Simple area concept
The whole area of the curves $$x=a\cos^3t, y=b\sin^3t$$ is given by?
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$$\dfrac{3}{8}\pi ab$$
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$$\dfrac{5}{8}\pi ab$$
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$$\dfrac{1}{8}\pi ab$$
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None of these
Explanation
$$x=a{ \cos }^{ 3 }\theta $$
$$y=b{ \sin }^{ 3 }\theta $$
To get the area,
$$A=\int _{ 0 }^{ 2x }{ x } dy$$
$$=\int _{ 0 }^{ 2\pi }{ a } { \cos }^{ 3 }\theta 3b{ \sin }^{ 2 }\theta \cos\theta d\theta $$
$$=3ab\int _{ 0 }^{ 2\pi }{ { \sin }^{ 2 }\theta { \cos }^{ 4 }\theta } d\theta \quad \longrightarrow \left( 1 \right) $$
Substituting $$\theta \rightarrow \theta +\pi /2$$ in $$1$$,
$$A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }\theta { \sin }^{ 4 }\theta } d\theta \quad \longrightarrow \left( 2 \right) $$
Adding $$(1)$$ and $$(2)$$
$$2A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }\theta } { \sin }^{ 2 }\theta d\theta \quad \longrightarrow \left( 3 \right) $$
$$\left\{ { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1 \right\} $$
Multiplying $$(3)$$ by $$(4)$$
$$8A=3ab\int _{ 0 }^{ 2\pi }{ { \sin }^{ 2 }2\theta } d\theta \quad \longrightarrow \left( 4 \right) $$
Substituting $$\theta \rightarrow \theta +\pi /4$$ in $$(4)$$
$$8A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }2\theta } d\theta \quad \longrightarrow \left( 5 \right) $$
Adding $$(4)$$ and $$(5)$$
$$16A=3ab\int _{ 0 }^{ 2\pi }{ 1 } d\theta $$
$$=6ab\pi $$
$$A=\dfrac { 3\pi ab }{ 8 } $$
The triangle formed by the tangent to the parabola $$y^2=4x$$ at the point whose abscissa lies in the interval $$\left[a^2, 4a^2\right]$$, the ordinate and the x-axis, has the greatest area equal to?
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$$12a^3$$
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$$8a^3$$
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$$16a^3$$
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$$20a^3$$
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?
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$$\dfrac{20}{3}-2\pi$$
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$$\dfrac{10}{3}-2\pi$$
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$$\dfrac{20}{3}-\pi$$
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$$\dfrac{10}{3}-\pi$$
Area bounded by the curves $$y=\log _{ e }{ x } \quad$$ and $$y={ \left( \log _{ e }{ x } \right) }^{ 2 }$$ is ?
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$$e-2$$
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$$3-e$$
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$$e$$
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$$e-1$$
The area enclosed between the curve $$y=x^3$$ and $$y=\sqrt{x}$$ is
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$$\dfrac{5}{3}$$
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$$\dfrac{5}{4}$$
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$$\dfrac{5}{12}$$
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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$$]is ?
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log $$2$$
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log $$3$$
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log $$5$$
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none of these
Explanation
$$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$$ and the x-axis is
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$$\sin 1$$
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$$1 - \sin 1$$
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$$1 + \sin 1$$
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None of these
The area (in square units) bounded by the curves $$y = {\cos ^{ - 1}}\left| {\cos \,x} \right|$$ and $$y = {\left( {{{\cos }^{ - 1}}\left| {\cos \,x} \right|} \right)^2},x \in \left[ {0,\pi } \right]$$ is
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$$\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)$$
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$$\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} + 1} \right)$$
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$$\frac{2}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)$$
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$$\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$$ is
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$$\left(\dfrac{\pi^3}{3} + \dfrac{4}{3} \right)$$ sq. unit
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$$\left(\dfrac{\pi^3}{6} - \dfrac{\pi^2}{2} + \dfrac{4}{3} \right)$$ sq. unit
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$$\left(\dfrac{\pi^2}{2} - \dfrac{4}{3} \right)$$ sq. unit
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$$\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$$ is :
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$$\frac{{32}}{3}$$
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$$\frac{{16}}{3}$$
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$$\frac{8}{3}$$
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$$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| }$$ is :
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$$\frac { 4+3\ell n3 }{ 2 } $$
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$$\frac { 4-3\ell n3 }{ 2 } $$
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$$\frac { 3 }{ 2 } +\ell n3$$
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$$\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$$ units
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$$\dfrac {3}{4}$$
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$$\dfrac {7}{6}$$
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$$\dfrac {128}{5}$$
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$$\dfrac {10}{3}$$
The maximum area bounded by the curves $${y^2} = 4ax,\,\,\,\,y = ax\,\,a$$ and $$y = \frac{x}{a}\,\,,1 \le a \le 2$$ is
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0%
$$ 44 sq. units $$
0%
$$ 74 sq. units $$
0%
$$84 sq.units $$
0%
$$ 114 sq. units $$
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