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CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Applications Of Integrals
Quiz 11
The area bounded by the curves $$y=\log x$$, $$y=\log \left | x \right |$$, $$y=\left | \log x \right |$$ and $$y=\left | \log \left | x \right | \right |$$
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4 sq. units
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6 sq. units
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10 sq. units
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None of these
Explanation
Required Area $$=2\int_{0}^{1}\left | \log \left | x \right | \right |dx$$
$$\displaystyle =2\left [ \left ( x\left | \log \left | x \right | \right | \right ) \right ]_{0}^{1}-\int_{0}^{1}\left ( -\frac{1}{x} \right )xdx$$
$$=2\left [ \left ( 1-0 \right )+\left ( x \right )_{0}^{1} \right ]=4$$ units
Find the area bounded by $$\displaystyle y=\sqrt{x}$$ and $$y = x$$.
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$$\displaystyle \frac{1}{8}sq.units$$
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$$\displaystyle \frac{1}{4}sq.units$$
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$$\displaystyle \frac{1}{12}sq.units$$
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$$\displaystyle \frac{1}{6}sq.units$$
The area of the figure bounded by $$y^{2}= 2x+1$$ and $$x-y-1= 0$$ is:
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$$2/3$$
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$$4/3$$
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$$8/3$$
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$$11/3$$
Explanation
From figure required area
$$\displaystyle -\int _{ 0 }^{ -\frac { 1 }{ 2 } }{ \left( \sqrt { 2x+1 } -\left( 1-x \right) \right) dx } +\int _{ -\frac { 1 }{ 2 } }^{ 4 }{ \left( \sqrt { 2x+1 } -\left( 1-x \right) \right) dx } $$
$$\displaystyle ={ \left[ \frac { 1 }{ 2 } \frac { { \left( 2x+1 \right) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -x+\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ -\frac { 1 }{ 2 } }+{ \left[ \frac { 1 }{ 2 } \frac { { \left( 2x+1 \right) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -x+\frac { { x }^{ 2 } }{ 2 } \right] }_{ -\frac { 1 }{ 2 } }^{ 4 }=\frac { 11 }{ 3 } $$
The parabolas $$y^{2}=4x$$ and $$x^{2}=4y$$ divide the square region bounded by the lines $$x=4, y=4$$ and the coordinate axes. If $$S_{1}$$, $$S_{2}$$, $$S_{3}$$ are respectively the areas of these parts numbered from top to bottom; $$S_{1}: S_{2}: S_{3}$$ is
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$$1: 2: 3$$
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$$1: 2: 1$$
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$$1: 1: 1$$
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$$2: 1: 2$$
Explanation
Total area $$=4\times 4=16$$ sq. units
Area of $$\displaystyle S_{3}=\int_{0}^{4}\frac{x^{2}}{4}dx=\frac{16}{3}=S_{1}$$
$$\therefore $$ $$\displaystyle S_{2}=16-\frac{16}{3}\times 2=\frac{16}{3}$$
$$\therefore $$ $$S_{1}:S_{2}:S_{3}$$ is $$1:1:1$$
Find the area of the region bounded by the curves $$\displaystyle x=\frac{1}{2},x=2,y=logx$$ and $$y=2^{x}$$
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$$\displaystyle \frac{4-\sqrt{2}}{\log2}-\frac{5}{2} \log2+\frac{3}{2}sq.units$$
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$$\displaystyle \frac{4+\sqrt{2}}{\log2}-\frac{3}{2} \log2+\frac{5}{2}sq.units$$
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$$\displaystyle \frac{4-\sqrt{2}}{\log2}-\frac{3}{2} \log2+\frac{5}{2}sq.units$$
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$$\displaystyle \frac{4+\sqrt{2}}{\log2}-\frac{5}{2} \log2+\frac{3}{2}sq.units$$
Explanation
Therefore the required area
$$\displaystyle =\int _{ \dfrac { 1 }{ 2 } }^{ 2 }{ \left( { 2 }^{ x }-\log { x } \right) dx } $$
$$\displaystyle ={ \left[ \dfrac { { 2 }^{ x } }{ \log { 2 } } -\left( x\log { x } -x \right) \right] }_{ \dfrac { 1 }{ 2 } }^{ 2 }$$
$$\displaystyle =\frac { 4-\sqrt { 2 } }{ \log 2 } -\frac { 5 }{ 2 } \log 2+\frac { 3 }{ 2 } $$
The curve $$f(x) = \displaystyle Ax^{2}+Bx+C$$ passes through the point (1, 3) and line $$4x + y = 8$$ is tangent to it at the point (2, 0). The area enclosed by $$y = f(x),$$ the tangent line and the y-axis is
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$$\dfrac{4}{3}$$
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$$\dfrac{8}{3}$$
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$$\dfrac{16}{3}$$
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$$\dfrac{32}{3}$$
Explanation
Given curve is $$\displaystyle y=f(x)=Ax^{2}+Bx+C.........(i)$$
It passes through $$(1, 3)$$
$$\displaystyle \therefore 3=A+B+C..........(ii)$$
Point $$(2, 0)$$ also lie on the curve $$(i)$$
$$\displaystyle \therefore 0=4A+2B+C.........(iii)$$
$$\displaystyle -C=4A+2B$$
from eq (ii)
$$\displaystyle 3=A+B-4A-2B$$
$$\displaystyle B=3(1-A)........(v)$$
Slope of tangent is $$-4$$
$$\displaystyle \therefore -4=4A+B.........(iv)$$
$$\displaystyle \therefore $$ from $$(ii), (iii)$$ & $$(iv)$$ and $$(v)$$ we get
$$A = -1, B = 0, C = 4$$
Thus, the required curve is $$\displaystyle y=-x^{2}+4$$
Hence required area $$=$$ area of $$\displaystyle \Delta OAB-\int_{0}^{2}\left ( -x^{2}+4 \right )dx=8-\left(-\frac{8}{3}+8\right)=\frac{8}{3}$$
For which of the following values of m is the area of the region bounded by the curve $$\displaystyle y=x-x^{2}$$ and the line $$y = mx$$ equals to 9/2 ?
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- 4
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- 2
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2
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4
Explanation
The two curves meet at
$$\displaystyle mx=x-x^{2}\:or\:x^{2}=x\left ( 1-m \right )\therefore x=0,1-m$$
$$\displaystyle \int_{0}^{1-m}\left ( y_{1}-y_{2} \right )dx=\int_{0}^{1-m}\left ( x-x^{2}-mx \right )dx=\left [ \left ( 1-m \right )\frac{x^{2}}{2}-\frac{x^{3}}{3} \right ]^{1-m}_{0}$$
$$\displaystyle \frac{9}{2}$$ if m < 1or $$\displaystyle \left ( 1-m \right )^{3}\left [ \frac{1}{2}-\frac{1}{3} \right ]=\frac{9}{2}$$ or $$\displaystyle \left ( 1-m \right )^{3}=27$$
$$\displaystyle \therefore m =-2$$
But if m > 1 then 1 - m is negative then
$$\displaystyle \left [ \left ( 1-m \right )\frac{x^{2}}{2}-\frac{x^{}3}{3} \right ]^{0}_{1-m}=\frac{9}{2}-\left ( 1-m \right )^{3}\left ( \frac{1}{2}-\frac{1}{3} \right )=\frac{9}{2}$$
$$\displaystyle \therefore -\left ( 1-m \right )^{3}=-27$$ or $$\displaystyle 1-m=-3\therefore m=4$$
The area bounded between the curve $$\displaystyle y = \tan x$$;
tangent drawn to it at $$\displaystyle x=\frac { \pi }{ 4 } $$ and $$\displaystyle y\ge 0$$ is
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$$\displaystyle \frac { 1 }{ 4 } \left( { \log }_{ e }4-1 \right) $$
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$$\displaystyle \frac { 1 }{ 2 } \left( {\log }_{ e }4-1 \right) $$
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$$\displaystyle \frac { 1 }{ 2 } \left( { \log }_{ e }4+1 \right) $$
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$$\displaystyle \frac { 1 }{ 4 } \left( {\log }_{ e }4+1 \right) $$
Explanation
$$ y=\tan X$$
now slope(m) of the line is $$\dfrac{dy}{dx}_{(x=\dfrac{\pi}{4})}=\sec^{2}X=2$$
now in graph of $$ \tan X$$ when $$x=\dfrac{\pi}{4},y=1$$
so equation of line is $$(y-y_1)=m(x-x_1)$$
so equation of line is $$ y=2x+1-\dfrac{\pi}{2}$$
now at $$ y=0, x=\dfrac{\pi}{4}-\dfrac{1}{2}$$
now $$A=\int_{0}^{\tfrac{\pi}{4}}(\tan X)-\int_{(\tfrac{\pi}{4}-\tfrac{1}{2})}^{\tfrac{\pi}{4}}(y=2x+1-\dfrac{\pi}{2})$$
$$A=\ln(\sec X)-[x^2-\dfrac{\pi x}{2}+x]$$
now putting upper and lower limt,and on simplified the term, we will get
$$A=\dfrac{(\ln 4-1)}{4}$$
The area bounded by $${ y }^{ 2 }+8x=16$$ and $${ y }^{ 2 }-24x=48$$ is $$\displaystyle \frac { a\sqrt { 6 } }{ c } $$, then $$a+c=$$
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$$30$$
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$$32$$
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$$35$$
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None
Explanation
The curve $${ y }^{ 2 }+8x=16$$ and $${ y }^{ 2 }-24x=48$$ cuts at $$x=-1$$
$${ y }^{ 2 }=24\Rightarrow y=\pm \sqrt { 24 } $$
Required area
$$\displaystyle =\int _{ -\sqrt { 24 } }^{ \sqrt { 24 } }{ \left( { x }_{ 1 }-{ x }_{ 2 } \right) dy } =\int _{ -\sqrt { 24 } }^{ \sqrt { 24 } }{ { \left( \left( 2-\frac { { y }^{ 2 } }{ 8 } \right) -\left( \frac { { y }^{ 2 } }{ 24 } -2 \right) \right)}dy } $$
$$\displaystyle =4{ \left[ y \right] }_{ -\sqrt { 24 } }^{ \sqrt { 24 } }-\frac { 1 }{ 18 } { \left[ { y }^{ 3 } \right] }_{ -\sqrt { 24 } }^{ \sqrt { 24 } }$$
$$=8\sqrt { 24 } -\dfrac { 2 }{ 18 } (24\sqrt { 24 }) $$
$$\displaystyle =\frac { 16\sqrt { 24 } }{ 3 } =\frac { 32\sqrt { 6 } }{ 3 } $$
Area is given as $$\dfrac{a\sqrt{6}}{c}$$
$$\Rightarrow a=32, c=3$$
$$\therefore a+c=32+3=35$$
Suppose $$g(x) = 2x+1$$ and $$h(x) = \displaystyle 4x^{2}+4x+5$$ and $$h(x) = (fog)(x)$$ The area enclosed by the graph of the function $$y = f(x)$$ and the pair of tangents drawn to it from the origin is
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8/3
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16/3
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32/3
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none
Explanation
$$g(x) = 2x+1$$ and
$$h(x) = \displaystyle 4x^{2}+4x+5 = (2x+1)^2 + 4 = g(x)^2 + 4$$
$$h(x) = (fog)(x)$$
$$f(x) = x^2+4$$
Let the point where tangent touches the parabola be (x,y)= $$(x, x^2+4)$$
Slope = $$ 2x$$
Equation of tangent
$$\cfrac{x^2+4}{x} = \cfrac{2x}{1}$$
$$ x = \pm 2$$
$$ y = 8$$
Area enclosed
$$ = 2(\cfrac{1}{2} 4 \times 8 - \int _4 ^8 \sqrt{y-4}dy)$$
$$ = \cfrac{16}{3} $$
Consider two curves $$\displaystyle C_{1}:y=\frac{1}{x}$$ and $$\displaystyle C_{2}$$ : $$y = \displaystyle lnx$$ on the xy plane Let $$\displaystyle D_{1}$$ denotes the region surrounded by $$\displaystyle C_{1}$$, $$\displaystyle C_{2}$$ and the line $$x = 1$$ and $$\displaystyle D_{2}$$ denotes the region surrounded by $$\displaystyle C_{1}$$, $$\displaystyle C_{2}$$ and the line $$x = a$$ If $$\displaystyle D_{1}$$=$$\displaystyle D_{2}$$ then the value of 'a':
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$$\displaystyle \frac{e}{2}$$
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$$e$$
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$$e-1$$
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$$2(e-1)$$
Explanation
Let the x point of intersection of the two curves be (k)
$$D1 = \int _1 ^k (\cfrac{1}{x} - \ln \ \ x)dx$$
$$D2 = \int _k ^a (- \cfrac{1}{x} \ln \ \ x)dx$$
$$ D1 +D2 =0$$
$$\int _1 ^a (\cfrac{1}{x}dx = \int _1 ^a (\ln \ \ x)dx$$
$$ \ln \ \ a = a \ \ \ln \ \ a - a +1$$
$$ (\ln \ \ a -1)(1-a) = 0$$
$$ a =e$$
Suppose $$y = f(x)$$ and $$y = g(x)$$ are two functions whose graphs intersect at three points $$(0, 4), (2, 2)$$ and $$(4, 0)$$ with $$f(x) > g(x)$$ for $$0 < x < 2$$ and $$f(x) < g(x)$$ for $$2 < x < 4 $$. if $$\displaystyle \int_{0}^{4}\left ( f(x)-g(x) \right )dx=10$$ and $$\displaystyle \int_{2}^{4}\left ( g(x)-f(x) \right )dx=5$$, the area between two curves for $$0 < x < 2$$, is:
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5
0%
10
0%
15
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20
Explanation
$$f(x) > g(x)$$ for $$0 < x < 2$$ and $$f(x) < g(x)$$ for $$2 < x < 4 $$.
if $$\displaystyle \int_{0}^{4}\left [ f(x)-g(x) \right ]dx=10$$ and $$\displaystyle \int_{2}^{4}\left [ g(x)-f(x)) \right ]dx=5$$,
$$\displaystyle \int_{0}^{4}\left [ f(x)-g(x) \right ]dx=10$$
$$\displaystyle \int_{0}^{2} f(x)-g(x) dx + \int_{2}^{4} f(x)-g(x) dx =10$$
$$\displaystyle \int_{0}^{2} ( f(x)-g(x)) dx + \int_{2}^{4} ( g(x)-f(x)) dx =10$$
$$\displaystyle \int_{0}^{2} ( f(x)-g(x)) dx =15$$
The area bounded by the curves $$\displaystyle y=-\sqrt{-x}$$ and $$\displaystyle x=-\sqrt{-y}$$ were $$\displaystyle x,y\leq 0$$
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Can not be determined
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is 1/3
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is 2/3
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is same as that of the figure by the curves $$\displaystyle y=\sqrt{-x};x\leq 0$$ and $$\displaystyle x=\sqrt{-y};y\leq 0$$
Explanation
$$y_1=-\sqrt{-x}$$
$$y_2= -x^2$$
Area bound is given by,
$$\displaystyle = \int _{-1} ^0 (y_1+y_2)dx $$
$$\displaystyle = \int _{-1} ^0 (-\sqrt{-x}+x^2)dx $$
$$\displaystyle = \left[\frac{-x^{3/2}}{3/2}+\dfrac {x^3}{3}\right]_{-1}^0$$
$$=(0-0)-\left(\dfrac{-2}{3}+\dfrac{1}{3}\right)$$
$$ = \cfrac{1}{3}$$
The area of the figure bounded by $${ y }^{ 2 }=2x+1$$ and $$x-y-1=0$$ is
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$$\displaystyle \frac { 16 }{ 3 } $$
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$$\displaystyle \frac { 8 }{ 3 } $$
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$$\displaystyle \frac { 4 }{ 3 } $$
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None of these
Explanation
Solving $${ y }^{ 2 }=2x+1$$ and $$x-y-1=0$$
we get $$\displaystyle \frac { { y }^{ 2 }-1 }{ 2 } =y+1\Rightarrow { y }^{ 2 }-2y-3=0$$
$$\Rightarrow \left( y-3 \right) \left( y+1 \right) =0\Rightarrow y=3,-1$$
Area of strip is $$\displaystyle =\int _{ -1 }^{ 3 }{ \left( y+1-\frac { { y }^{ 2 }-1 }{ 2 } \right) dy } ={ \left[ \frac { { y }^{ 2 } }{ 2 } -\frac { { y }^{ 3 } }{ 6 } +\frac { 3 }{ 2 } y \right] }_{ -1 }^{ 3 }$$
$$\displaystyle =\left( \frac { 9 }{ 2 } -\frac { 1 }{ 2 } \right) -\left( \frac { 27 }{ 6 } -\frac { -1 }{ 6 } \right) +\frac { 3 }{ 2 } \left[ 3-\left( -1 \right) \right] =\frac { 16 }{ 3 } \quad $$
Area of the region enclosed between the curves $$\displaystyle x=y^{2}-1$$ and $$\displaystyle x = \left | y \right |\sqrt{1-y^{2}}$$ is
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$$1$$
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$$4/3$$
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$$2/3$$
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$$2$$
Explanation
Required area is,
$$\displaystyle A=2\int_{0}^{1}\left [ y\sqrt{1-y^{2}}-\left ( y^{2}-1 \right ) \right ]dy$$
$$\displaystyle =\left(\frac{-2}{3}\left ( 1-y^{2} \right )^{3/2}\right)_{0}^{1}-\left ( \frac{2y^{3}}{3}-2y \right )^{1}_{0}=2$$
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 line $$3x + 2y =13$$ divides the area enclosed by the curve $$\displaystyle 9x^{2}+4y^{2}-18x-16y-11=0$$ in two parts Find the ratio of the larger area to the smaller area
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$$\displaystyle \frac{3\pi+2}{\pi -2}$$
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$$\displaystyle \frac{3\pi-2}{\pi +2}$$
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$$\displaystyle \frac{\pi+2}{\pi -2}$$
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$$\displaystyle \frac{\pi-2}{\pi +2}$$
Explanation
$$9{ x }^{ 2 }+4{ y }^{ 2 }-18x-16y-11=0\Rightarrow 9\left( { x }^{ 2 }-2x \right) +4\left( { y }^{ 2 }-4y \right) =11\\ \Rightarrow 9\left( { \left( x-1 \right) }^{ 2 }-1 \right) +4\left( { \left( y-2 \right) }^{ 2 }-4 \right) =11\Rightarrow 9{ \left( x-1 \right) }^{ 2 }+4{ \left( y-2 \right) }^{ 2 }=36$$
$$\displaystyle \\ \Rightarrow \dfrac { { \left( x-1 \right) }^{ 2 } }{ 4 } +\dfrac { { \left( y-2 \right) }^{ 2 } }{ 9 } =1$$
Let $$x-1=X$$ and $$y-2=Y$$
$$\displaystyle \Rightarrow \frac { { X }^{ 2 } }{ 4 } +\frac { { Y }^{ 2 } }{ 9 } =1$$
Hence $$3x+2y=13$$
$$\displaystyle \Rightarrow 3\left( X+1 \right) +2\left( Y-2 \right) =13\Rightarrow 3X+2Y=6\Rightarrow \frac { X }{ 2 } +\frac { Y }{ 3 } =1$$
Therefore area of triangle $$\displaystyle OPQ=\frac { 1 }{ 2 } \times 2\times 3=3$$
Also area of ellipse $$=\pi $$(semimajor axes)(semi minor axis) $$=\pi .3.2=6\pi $$
$$\displaystyle { A }_{ 1 }=\frac { 6\pi }{ 4 } -$$ area of $$\displaystyle \triangle OPQ=\frac { 3\pi }{ 2 } -3$$
$$\displaystyle { A }_{ 2 }=3\left( \frac { 6\pi }{ 4 } \right) +$$ area of $$\displaystyle \triangle OPQ=\frac { 9\pi }{ 2 } +3$$
Hence $$\displaystyle \dfrac { { A }_{ 2 } }{ { A }_{ 1 } } =\dfrac { \dfrac { 9\pi }{ 2 } +3 }{ \dfrac { 3\pi }{ 2 } -3 } =\dfrac { 3\pi +2 }{ \pi -2 } $$
If the area enclosed by the parabolas $$\displaystyle y=a-x^{2}$$ and $$\displaystyle y=x^{2}$$ is $$\displaystyle 18\sqrt {2}$$ sq. units Find the value of 'a'
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$$a = -9$$
0%
$$a= 6$$
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$$a =9$$
0%
$$a=-6$$
Explanation
For the point of intersection
$$x^{2}=a-x^{2}$$
$$2x^{2}=a$$
$$x=\pm\sqrt{\dfrac{a}{2}}$$
Hence the required area will be
$$=\int_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}} x^{2}-(a-x^{2}).dx$$
$$=\int_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}} 2x^{2}-a.dx$$
$$=2\int_{0} ^{\sqrt{\frac{a}{2}}} 2x^{2}-a.dx$$
$$=2[\dfrac{2x^{3}}{3}-ax]_{0} ^{\sqrt{\frac{a}{2}}}$$
$$=2[\dfrac{2.a\sqrt{a}}{6.\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}]$$
$$=2[\dfrac{a\sqrt{a}}{3\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}]$$
$$=18\sqrt{2}$$
$$|\dfrac{a\sqrt{a}}{3\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}|=9\sqrt{2}$$
$$\dfrac{2a\sqrt{a}}{3\sqrt{2}}=9\sqrt{2}$$
$$a\sqrt{a}=27$$
$$a^{\frac{3}{2}}=3^{3}$$
$$a=3^{2}$$
Hence $$a=9$$.
The area enclosed between the curves $$y=x^3$$ and $$y=\sqrt{x}$$ is (in square units)
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$$\displaystyle\frac{5}{3}$$
0%
$$\displaystyle\frac{5}{4}$$
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$$\displaystyle\frac{5}{12}$$
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$$\displaystyle\frac{12}{5}$$
Explanation
$$y=\sqrt { x } $$ or $${ y }^{ 2 }=x\left( y\ge 0 \right) $$ and $$y={ x }^{ 3 }.$$
We get points of intersection $$(0,0)$$ and $$(1,1)$$
$$\therefore$$ Required area $$\displaystyle =\int _{ 0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 3 } \right) } dx$$
$$\displaystyle =\left[ \dfrac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -\dfrac { { x }^{ 4 } }{ 4 } \right] _{ 0 }^{ 1 }=\dfrac{2}{3}-\dfrac{1}{4}=\dfrac { 5 }{ 12 } $$ sq. unit.
Find the area bounded by $$\displaystyle y = \cos ^{-1}x,y=\sin ^{-1}x$$ and $$y-$$axis
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$$\displaystyle \left ( 2-\sqrt{2} \right )$$ sq. units
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$$\displaystyle \left ( \sqrt{2}-{2} \right )$$ sq. units
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$$2 \sqrt{2}$$ sq. units
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$$\sqrt {2}$$ sq. units
Explanation
$$y=\cos ^{ -1 }{ x } $$ and $$y=\sin ^{ -1 }{ x } $$ intersect at $$P\equiv\displaystyle \left( \frac { 1 }{ \sqrt { 2 } } ,\frac { \pi }{ 4 } \right) $$
Hence, required area is,
$$\displaystyle \int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \left( \cos ^{ -1 }{ x } -\sin ^{ -1 }{ x } \right) dx } =\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \cos ^{ -1 }{ x } dx } -\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \sin ^{ -1 }{ x } dx } $$
$$\displaystyle ={ \left[ x\cos ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }-\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } -{ \left[ x\sin ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx }$$
$$\displaystyle ={ \left[ \dfrac1{\sqrt2}\cos ^{ -1 }{\dfrac1{\sqrt2} } \right] } -{ \left[ \dfrac1{\sqrt2}\sin ^{ -1 }{ \dfrac1{\sqrt2} } \right] }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ 2x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx }$$
$$\displaystyle =\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ 2x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx }$$
Assuming $$1-x^2 = t^2$$
Differentiating both sides, we get
$$-2xdx = 2tdt$$
Substituting in the integral, we get
$$\displaystyle =\int _{ \frac { 1 }{ \sqrt { 2 } } }^12\ dt$$
$$ =2-\sqrt { 2 } $$
Find the area enclosed between the curves $$\displaystyle y=\log_{e}\left ( x+e \right ), x=\log_{e}\left ( 1/y \right )$$ & the x-axis
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1 sq. units
0%
2 sq. units
0%
3 sq. units
0%
4 sq. units
Explanation
Area enclosed
$$ = \int _{1-e} ^0 ln(x+e) dx + \int _{0} ^{\infty} e^{-x} dx$$
$$ = (xln(x+e) -x)| _{1-e} ^0 -e^{-x} | _{0} ^{\infty}$$
$$ =2$$
Find the value(s) of the parameter 'a' (a > 0) for each of which the area of the figure bounded by the straight line $$\displaystyle y=\frac{a^{2}-ax}{1+a^{4}}$$ & the parabola $$\displaystyle y=\frac{x^{2}+2ax+3a^{2}}{1+a^{4}}$$ is the greatest
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$$\displaystyle a=2^{1/4}$$
0%
$$\displaystyle a=5^{1/4}$$
0%
$$\displaystyle a=7^{1/4}$$
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$$\displaystyle a=3^{1/4}$$
Explanation
$$\displaystyle A=\int_{-a}^{-2a}\frac{a^{2}-ax-\left ( x^{2}+2ax+3a^{2} \right )}{1+a^{4}}dx$$
$$\displaystyle =\frac{3}{2}\frac{a^{3}}{1+a^{4}}$$ Now f(a) $$\displaystyle =\frac{3}{2}\frac{a^{3}}{1+a^{4}}\Rightarrow f'(a)=0\Rightarrow \left ( 1+a^{4} \right )3a^{2}-a^{3}4a^{3}=0$$
$$\displaystyle \Rightarrow a_{min}=0,\:\:a_{min}=3^{1/4}$$
Find the positive value of 'a' for the which the parabola $$\displaystyle y=x^{2}+1$$ bisects the area of the rectangle with vertices $$(0, 0), (a, 0), (0, \displaystyle a^{2}+1)$$ and $$(a, \displaystyle a^{2}+1)$$
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$$\displaystyle \sqrt 2$$
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$$\displaystyle \sqrt 3$$
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$$\displaystyle \sqrt 5$$
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$$\displaystyle \sqrt 7$$
Explanation
Area of the rectangle outside the parabola
$$= \int _0 ^{a} (x^2+1)dx$$
$$ = \cfrac{a^3}{3} + a$$
Given it is equal to half the area of rectangle
$$ \cfrac{a(a^2+1)}{2} = \cfrac{a^3}{3} + a$$
On solving
$$a = \sqrt{3}$$
A figure is bounded by the curves $$\displaystyle y=\left | \sqrt{2}\sin \frac{\pi x}{4} \right |$$ $$y = 0, x = 2$$ & $$x = 4$$. At what angles to the positive $$x$$-axis straight lines must be drawn through $$(4, 0)$$, so that these lines divide the figure into three parts of the same size
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$$\displaystyle \pi -\tan ^{-1}\frac{2\sqrt{2}}{3\pi },\pi +\tan^{-1} \frac{4\sqrt{2}}{3\pi }$$
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$$\displaystyle \pi -\tan ^{-1}\frac{2\sqrt{2}}{2\pi },\pi -\tan ^{-1}\frac{4\sqrt{2}}{2\pi }$$
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$$\displaystyle \pi -\tan ^{-1}\frac{2\sqrt{2}}{3\pi },\pi -\tan ^{-1}\frac{4\sqrt{2}}{3\pi }$$
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$$\displaystyle \pi +\tan ^{-1}\frac{2\sqrt{2}}{2\pi },\pi -\tan ^{-1}\frac{4\sqrt{2}}{2\pi }$$
Explanation
Let equation of line is $$y = mx - 4m$$
$$\displaystyle A=\int_{2}^{4}\sqrt{2}\sin \frac{\pi }{4}x\ dx=\left [ -\sqrt{2}\frac{4}{\pi }\cos \frac{\pi x}{4} \right ]^{4}_{2}=\frac{4\sqrt{2}}{\pi }....(i)$$
Also area of $$\displaystyle \Delta ABC=\frac{1}{2}.2.\left ( -2m_{1} \right )=-2m_{1}...(ii)$$ from (i) and (ii)
$$\displaystyle -2m_{1}=\frac{4\sqrt{2}}{3\pi }\Rightarrow m_{1}=\frac{-2\sqrt{2}}{3\pi }$$
$$\displaystyle \Rightarrow \tan \left ( \pi -\theta _{1} \right )=\frac{-2\sqrt{2}}{3\pi }\Rightarrow \pi -\theta _{1}=\tan ^{-1}\frac{2\sqrt{2}}{3\pi }$$
$$\displaystyle \Rightarrow \theta _{1}=\pi -\tan ^{-1}\frac{2\sqrt{2}}{3\pi }$$ or
$$\displaystyle \frac{1}{2}.(2)\left ( -2m_{2} \right )=\frac{8\sqrt{2}}{3\pi }$$
$$\displaystyle \Rightarrow m_{2}=\frac{-4\sqrt{2}}{3\pi }\Rightarrow \tan \left ( \pi -\theta _{2} \right )=\frac{-4\sqrt{2}}{3\pi }$$
$$\displaystyle \Rightarrow \theta _{2}=\pi -\tan ^{-1}\frac{4\sqrt{2}}{3\pi }$$
In what ratio does the x-axis divide the area of the region bounded by the parabolas $$\displaystyle y=4x-x^{2}$$ and $$\displaystyle y=x^{2}-x$$?
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4 : 121
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4 : 144
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4 : 169
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4 : 100
For what value of 'a' is the area of the figure bounded by $$\displaystyle y=\frac{1}{x}, y=\frac{1}{2x-1}$$ $$x = 2$$ & $$x = a$$ equal to $$\displaystyle ln\frac{4}{\sqrt{5}}$$?
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$$\displaystyle a=4\:$$
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$$\displaystyle a=8\:$$
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$$\displaystyle a=4\: or \frac{2}{5}\left ( 6-\sqrt{21} \right )$$
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none of these
Explanation
The required area is
$$=\int_{2} ^{a} \dfrac{1}{x}-\dfrac{1}{2x-1}.dx$$
$$=[lnx-\dfrac{ln(2x-1)}{2}]_{2} ^{a}$$
$$=[ln\dfrac{x}{\sqrt{2x-1}}]_{2} ^{a}$$
$$=-ln\dfrac{(2)}{\sqrt{3}}+ln\dfrac{a}{\sqrt{2a-1}}$$
$$=ln\dfrac{4}{\sqrt{5}}$$
Hence
$$ln(\dfrac{a}{\sqrt{2a-1}})=ln\dfrac{4}{\sqrt{5}}+ln\dfrac{(2)}{\sqrt{3}}$$
$$\Rightarrow ln(\dfrac{a}{\sqrt{2a-1}})=ln\dfrac{8}{\sqrt{15}}$$
Hence
$$2a-1=15$$
Hence
$$a=8$$.
Therefore a=8 is one solution.
Now
$$\dfrac{a^{2}}{2a-1}=\dfrac{64}{15}$$
$$\Rightarrow 15a^{2}=128a-64$$
$$\Rightarrow 15a^{2}-128a+64=0$$
Hence $$a=8$$ and $$a=\dfrac{8}{\sqrt{15}}$$
A polynomial function f(x) satisfies the condition $$f(x + 1) = f(x) + 2x + 1$$ Find f(x) if $$f(0) = 1$$ Find also the equations of the pair of tangents from the origin on the curve $$y = f(x)$$ and compute the area enclosed by the curve and the pair of tangents
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$$\displaystyle f(x)=x^{2}+1;y=\pm x; A=\frac{2}{3}sq. units$$
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$$\displaystyle f(x)=x^{2}+1;y=\pm x; A=\frac{4}{3}sq. units$$
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$$\displaystyle f(x)=x^{2}+1;y=\pm 2x; A=\frac{2}{3}sq. units$$
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$$\displaystyle f(x)=x^{2}+1;y=\pm 2x; A=\frac{4}{3}sq. units$$
Explanation
$$\displaystyle f\left ( x+1 \right )=f(x)+2x+1$$
$$\displaystyle \Rightarrow f''\left ( x+1 \right )=f''(x)\:\:\:\:\forall \times \in R$$ Let $$f"(x) = a \displaystyle \Rightarrow$$ $$ f'(x) = ax + b \displaystyle \Rightarrow f(x)=\frac{ax^{2}}{2}+bx+c\Rightarrow c=1$$ $$\displaystyle \left [ \because f(0)=1 \right ]$$
Now $$f(x + 1) - f(x) = 2x + 1 \displaystyle \Rightarrow \left [ \frac{a}{2}\left ( x+1 \right )^{2}+b\left ( x+1 \right )+c \right ]-\left [ \frac{ax^{2}}{2}+bx+c \right ]
=2x+1$$ $$\displaystyle \Rightarrow ax+\frac{a}{2}+b=2x+1$$ on comparing we get a = 2
or $$\displaystyle \frac{a}{2}+b=1\Rightarrow b=0$$
$$\displaystyle \therefore f(x)=x^{2}+1.........(i)$$
Now let equation of tangent be $$y = mx ....... (ii) $$
From (i) and (ii), we get
$$\displaystyle x^{2}-mx+1=0\Rightarrow m=\pm 2$$
$$\displaystyle \therefore tangents\:\:\: are\:\:\:y=2x \:or\:\:y=-2x $$ $$\displaystyle A=2\int_{0}^{1}\left ( x^{2}+1-2x \right )dx=\frac{2}{3}$$
Area enclosed between the curves $${ y }^{ 2 }=x$$ and $${ x }^{ 2 }=y$$ is equal to
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$$\displaystyle 2\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) dx } $$
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$$\displaystyle \frac { 1 }{ 3 } $$
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area of region $$\left\{ \left( x,y \right) :{ x }^{ 2 }\le y\le \left| x \right| \right\} $$
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$$\displaystyle \frac { 2 }{ 3 } $$
Explanation
The $$\displaystyle { y }^{ 2 }=x$$ and $${ x }^{ 2 }=y$$ intersect each other at point $$(0,0)$$ and $$(1,1)$$ as shown below.
$$\therefore$$ required area is given by $$\displaystyle A=\int _{ 0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 2 } \right) } dx=2{ A }_{ 2 }$$
Where $${ A }_{ 2 }=$$area between $$y=x$$ and $$y={x}^{2}$$ between
$$x=0$$ and $$x=1=\left\{ \left( x,y \right) :{ x }^{ 2 }\le y\le \left| x \right| \right\} $$
$$\displaystyle =2\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) } dx=2\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }$$
$$\displaystyle =2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =\frac { 1 }{ 3 } $$ square units.
Let $$f(x)$$ be a continuous function given by $$\displaystyle f\left ( x \right )=2x$$ for $$\displaystyle \left | x \right |\leq 1$$ for $$\displaystyle f\left ( x \right )=x^{2}+ax+b$$ for $$\displaystyle \left | x \right |> 1$$. Find the area of the region in the third quadrant bounded by the curves $$\displaystyle x=-2y^{2}$$ and $$y = f(x)$$ lying on the left of the line $$8x + 1 = 0$$
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$$\dfrac{235}{192} ; a = 2 ; b = - 1$$
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$$\dfrac{235}{192} ; a = -1 ; b = 2$$
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$$\dfrac{257}{192} ; a = -1 ; b = 2$$
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$$\dfrac{257}{192} ; a = 2 ; b = - 1$$
Let $$\displaystyle C_{1}$$ & $$\displaystyle C_{2}$$ be two curves passing through the origin as shown in the figure A curve C is said to "bisect the area" the region between $$\displaystyle C_{1}$$ & $$\displaystyle C_{1}$$ if for each point P of C the two shaded regions A & B shown in the figure have equal areas Determine the upper curve $$\displaystyle C_{2}$$ given that the bisecting curve C has the equation $$\displaystyle y=x^{2}$$ & that the lower curve $$\displaystyle C_{1}$$ has the equation $$\displaystyle y=x^{2}/2$$
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$$\displaystyle \left ( 16/9 \right )x^{2}$$
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$$\displaystyle \left ( 25/9 \right )x^{2}$$
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$$\displaystyle \left ( 25/16 \right )x^{2}$$
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$$\displaystyle \left ( 9/25 \right )x^{2}$$
Explanation
According to question
$$\displaystyle \int_{0}^{a^{2}}\left ( -f^{-1}\left ( y \right )+\sqrt{y} \right )\:\:dy=\int_{0}^{a}\left ( x^{2}-\frac{x^{2}}{2} \right )dx$$
$$\displaystyle \Rightarrow \left [ f^{-1}(a^{2})-a \right ]2a=-\frac{a^{2}}{2}\Rightarrow f^{-1}(a^{2})=\frac{3a}{4}\Rightarrow f\left ( \frac{3a}{4} \right )-a^{2}$$
or $$\displaystyle f(x)=\frac {16}{9}x^{2}$$
Find the area bounded by $$y = x + sinx$$ and its inverse between $$x = 0$$ and $$x = \displaystyle 2\pi$$
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$$2$$
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$$4$$
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$$6$$
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$$8$$
Explanation
graph of $$x+\sin x $$ and its inverse is shown in the figure which is symmetric with $$y=x$$
let $$f(x)=x+\sin x$$
from the figure it is clear that $$A=\int _{ 0 }^{ 2\pi }{ \left( f\left( x \right) -f^{ -1 }\left( x \right) \right) dx } =4\int _{ 0 }^{ \pi }{ \left( f\left( x \right) -x \right) dx } =4\int _{ 0 }^{ \pi }{ \left( x+\sin { x } -x \right) dx } =8$$
The area of the region bounded by the parabola $$y={x}^{2}-4x+5$$ and the straight line $$y=x+1$$ is
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$$\cfrac{1}{2}$$
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$$2$$
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$$3$$
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$$\cfrac{9}{2}$$
Explanation
Given equation of parabola is
$$y={x}^{2}-4x+5$$
$$\Rightarrow$$ $$y={(x-2)}^{2}+1$$
$$\Rightarrow$$ $${(x-2)}^{2}=(y-1)....(i)$$
and equation of line is
$$y=x+1$$
$$\Rightarrow$$ $$x-y=-1.....)(ii)$$
On putting the value of $$(y-1)$$ from eqs $$(ii)$$ in $$(i)$$ we get
$${(x-2)}^{2}=x$$ $$\Rightarrow$$ $${x}^{2}+4-4x=x$$
$$\Rightarrow$$ $${x}^{2}-5x+4=0$$
$$\Rightarrow$$ $${x}^{2}-4x-x+4=0$$
$$\Rightarrow$$ $$x(x-4)-1(x-4)=0$$
$$\Rightarrow$$ $$(x-1)(x-4)=0$$ or $$x=1,4$$
then from is $$(ii)$$ $$y=2,5$$
$$\therefore$$ Required area $$=\displaystyle \int _{ 1 }^{ 4 }{ \left( -{ x }^{ 2 }+5x-4 \right) dx } ={ \left[ \cfrac { -{ x }^{ 3 } }{ 3 } +\cfrac { 5{ x }^{ 2 } }{ 2 } -4x \right] }_{ 1 }^{ 4 }$$
$$=\left[ -\cfrac { 64 }{ 3 } +40-16+\cfrac { 1 }{ 3 } -\cfrac { 5 }{ 2 } +4 \right] $$
$$=\left(-21-\cfrac { 5 }{ 2 } +28\right)=\cfrac { 9 }{ 2 } $$
Area bounded by $$\displaystyle y=1-\ell { n }^{ 2 }x,x-$$axis which is common with $$\displaystyle x\ge 1$$, is -
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$$\displaystyle e-\frac { 5 }{ e } $$
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$$\displaystyle e-2$$
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$$\displaystyle 3$$
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$$\displaystyle 1$$
Explanation
$$\displaystyle A=\int _{ 1 }^{ e }{ \left( 1-\ell { n }^{ 2 }x \right)
dx } =\left( e-1 \right) -\int _{ 1 }^{ e }{ 1.\ell { n }^{ 2 }xdx } $$
$$\displaystyle =e-1-\left[ [x\ell { n }^{ 2 }x{ ] }_{ 1 }^{ e }-\int _{ 1 }^{ e }{ x.2\ell nx.\frac { 1 }{ x } dx } \right] $$
$$\displaystyle =e-1-\left[ e-0-2.[x\ell nx-x{ ] }_{ 1 }^{ e } \right] $$
$$\displaystyle =e-1-e+2\left[ e-e-0+1 \right]$$
$$ =1$$
Area bounded by $$\displaystyle y=2x-{ x }^{ 2 }$$ & $$\displaystyle (x-1{ ) }^{ 2 }+{ y }^{ 2 }=1$$ in first quadrant, is:
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$$\displaystyle \frac { \pi }{ 2 } -\frac { 4 }{ 3 } $$
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$$\displaystyle \frac { \pi }{ 2 } -\frac { 2 }{ 3 } $$
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$$\displaystyle \frac { \pi }{ 2 } +\frac { 4 }{ 3 } $$
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$$\displaystyle \frac { \pi }{ 2 } +\frac { 2 }{ 3 } $$
Explanation
Area of circle is $$\pi r^2$$
But here only half circle is in $$1^{st}$$ quadrant
Then, $$A^*=\dfrac{\pi r^2}{2}$$ .......$$( r=1)$$
$$\Rightarrow A^{*}=\dfrac{\pi}{2} ......(1)$$
Now area bounded by parabola $$y=2x-x^2$$ is
$$A^@=\displaystyle \int_{0}^{2}(2x-x^2)$$
$$A=\dfrac{4}{3} ......(2)$$
Now area bounded by both curves is
$$A=A^*-A^@ $$
$$\Rightarrow A=\dfrac{\pi}{2}-\dfrac{4}{3}$$
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 $$=\displaystyle \int_{0}^{1} (y_{1} - y_{2})dx$$
$$=\displaystyle \int_{0}^{1} [\left \{\sin^{-1} x + x(1 - x)\right \} - \left \{\sin^{-1} x - x(1 - x) \right \}]dx$$
$$= 2\displaystyle \int_{0}^{1} (x - x^{2}) dx$$
$$= 2\left [\dfrac {x^{2}}{2} - \dfrac {x^{3}}{3}\right ]_{0}^{1} = 2\left (\dfrac {1}{2} - \dfrac {1}{3}\right ) = \dfrac {1}{3}$$
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$$=area\,of\,OBA{B}^{\prime}O=2\,area\,of \,OBAO$$
$$=2\left[area\,of\,OBDO+area\,of\,DBAD\right]$$
$$=2\left[\displaystyle\int_{0}^{2}{\sqrt{6x}dx}+\displaystyle\int_{2}^{4}{\sqrt{16-{x}^{2}}dx}\right]$$
$$=2\left[\sqrt{6}\left[\dfrac{{x}^{\frac{3}{2}}}{\dfrac{3}{2}}\right]_{0}^{2}+\left[\dfrac{1}{2}x\sqrt{16-{x}^{2}}+\dfrac{1}{2}\times 16{\sin}^{-1}{\dfrac{x}{4}}\right]_{2}^{4}\right]$$
$$=2\left[\left[\sqrt{6}\times\dfrac{2}{3}\times{2}^{\frac{3}{2}}-0\right]+\left[\dfrac{1}{2}\times\,4\sqrt{16-16}+\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{4}{4}}-\dfrac{1}{2}\times\,2\sqrt{16-4}-\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{1}{2}}\right]\right]$$
$$=2\left[\dfrac{8\sqrt{3}}{3}+\dfrac{8\pi}{2}-2\sqrt{3}-\dfrac{8\pi}{6}\right]$$
$$=2\left[\dfrac{8\sqrt{3}-6\sqrt{3}}{3}+8\left(\dfrac{\pi}{2}-\dfrac{\pi}{6}\right)\right]$$
$$=2\left[\dfrac{2\sqrt{3}}{3}+8\times\dfrac{2\pi}{6}\right]$$
$$=\dfrac{4\sqrt{3}}{3}+\dfrac{16\pi}{3}$$
The area of the region enclosed between parabola $${y}^{2}=x$$ and the line $$y=mx$$ is $$\cfrac{1}{48}$$. Then, the value of $$m$$ is
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$$-2$$
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$$-1$$
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$$1$$
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$$2$$
Explanation
Equation of parabola is $${y}^{2}=x$$ and line $$y=mx$$
For intersection point of both curves put $$x={y}^{2}$$,
we get
$$y=m{y}^{2}\Rightarrow$$ $$y(my-1)=0$$
$$\Rightarrow$$ $$y=0$$ or $$y=\cfrac{1}{m}$$
Then, $$x=0$$ or $$x=\cfrac{1}{{m}^{2}}$$
$$\therefore$$ Intersection points are $$(0,0)$$ and $$P(\cfrac{1}{{m}^{2}},\cfrac{1}{m})$$
$$\therefore$$ Required area
$$=\displaystyle \int _{ 0 }^{ 1/m }{ \left| \left( \cfrac { y }{ m } -{ y }^{ 2 } \right) \right| dy } =\left| { \left[ \cfrac { { y }^{ 2 } }{ 2m } -\cfrac { { y }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1/m } \right| \quad $$
$$=\left| \cfrac { 1 }{ 2{ m }^{ 3 } } -\cfrac { 1 }{ 3{ m }^{ 3 } } \right| =\left| \cfrac { 1 }{ 6{ m }^{ 3 } } \right| =\cfrac { 1 }{ 48 } $$
$$\Rightarrow \cfrac { 1 }{ 6{ m }^{ 3 } } =\pm \cfrac { 1 }{ 48 } \Rightarrow { m }^{ 3 }=\pm 8$$
Now if $${m}^{3}=8$$
$${ m }^{ 3 }=8\quad \Rightarrow { m }^{ 3 }={ \left( 2 \right) }^{ 3 }\Rightarrow m=2$$
If $${m}^{3}=-8$$
$$\Rightarrow { m }^{ 3 }=-8$$
$$\Rightarrow { m }^{ 3 }={ -\left( 2 \right) }^{ 3 }$$
$$\Rightarrow m=-2$$
The area of the region bounded by the curves, $$y^{2} = 8x$$ and $$y = x$$ is
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$$\dfrac {64}{3}$$
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$$\dfrac {32}{3}$$
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$$\dfrac {16}{3}$$
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$$\dfrac {8}{3}$$
Explanation
Given curves,
$$y^2=8x \rightarrow$$ (i)
and $$y=x \rightarrow$$ (ii)
On solving Eqs. (i) and (ii), we get
$$x^2-8x=0$$
$$x(x-8)=0$$
and $$y=0,8$$
$$\therefore$$ Required area (OPA) $$=\displaystyle \int_0^8(\sqrt{8x}-x)dx$$
$$=\left[2\sqrt2. \dfrac23.x^{3/2}-\dfrac{x^2}2\right]_0^8$$
$$=\dfrac{4\sqrt2}3.(8)^{3/2}-\dfrac{8^2}2$$
$$=\dfrac{32}3$$
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
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3
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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 (in square units) of the region bounded by $$x=-1, x=2, y=x^2+1$$ and $$y=2x-2$$ is
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10
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7
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8
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9
Explanation
$$A_1=ar (ABC)$$
$$=\frac{1}{2}\times |2||4|$$
$$=4$$ units
$$A_2=ar(CDE)$$
$$=\frac{1}{2}\times 1\times 2$$
$$A_2=1$$ units
Area of region between parabola and x-axis
$$A_3=\int_{-1}^2(x^2+1)dx=\left[\dfrac{x^3}{3}+x\right]_{-1}^{2}$$
Area required $$=A_3+A_1-A_2$$
$$=6+4-1$$
$$=9$$ units
Area of region $$\displaystyle \left\{ \left( x,y \right) \in { R }^{ 2 }:y\ge \sqrt { \left| x+3 \right| } ,5y\le x+9\le 15 \right\} $$ is equal to
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$$\displaystyle \frac { 1 }{ 6 } $$
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$$\displaystyle \frac { 4 }{ 3 } $$
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$$\displaystyle \frac { 3 }{ 2 } $$
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$$\displaystyle \frac { 5 }{ 3 } $$
Explanation
Area $$ABE$$ (under parabola) $$=\displaystyle \int^{-3}_{-4}\sqrt{-x-3}dx=\frac{2}{3}$$
Area $$BCD$$ (under parabola) $$=\displaystyle \int^{1}_{-3}\sqrt{x+3}dx=\frac{16}{3}$$
Area of trapezium $$ACDE$$ $$=\dfrac{1}{2}(1+2)5=\dfrac{15}{2}$$
Required area $$= \dfrac{15}{2}-\dfrac{16}{3}-\dfrac{2}{3}=\dfrac{3}{2}$$
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 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 $$(a>0)$$
The ratio of the areas of two regions of the curve $$C_1 : 4x^2 + \pi^2y^2 = 4\pi^2$$ divided by the curve $$C_2 : y = -sgn \left(x - \dfrac{\pi}{2}\right) \cos x$$ (where sgn(x) denotes signum function) is -
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$$\dfrac{\pi^2 +4}{\pi^2-2\sqrt{2}}$$
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$$\dfrac{\pi^2-2}{\pi^2+2}$$
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$$\dfrac{\pi^2+6}{\pi^2+3\sqrt{3}}$$
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$$\dfrac{\pi^2+1}{\pi^2-\sqrt{2}}$$
If $$z$$ is not purely real then area bounded by curves $$lm\left(z+\dfrac{1}{z}\right) = 0$$ and $$|z-1| = 2$$ is (in square units)-
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$$4\pi$$
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$$3\pi$$
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$$2\pi$$
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$$\pi$$
The graphs of $$f(x)=x^{2}$$ and $$g(x)=cx^{3}(c>0)$$ intersect at the points $$(0, 0)$$ and $$(\dfrac{1}{c}, \dfrac{1}{c^{2}})$$. If the region which lies between these graphs and over the interval $$[0, \dfrac{1}{c}]$$ has the area equal to $$(\dfrac{2}{3})sq.\ units$$, then the value of $$c$$ is :
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{2}$$
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$$1$$
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$$2$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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