Consider the following regions in the plane R1- x- y -0- 1 and 0- y- 1 - and R2- x-y -x2-y2- 4-3 The area of the region R1- R2 can be expressed as a-3-b-9 where a and b are integers then

The correct options are A a = 3 D b = 1 A=1/√(3)+∫_1/√(3)^1√(4/3 x^2)dx =1/√(3)+ [ x/2√(4/3 x^2)+2/3sin ^ 1 ( x√(3)/2 ) ]^1_1/√(3) =1 ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Planes

Consider the ...

Question

Consider the following regions in the plane $R_{1} = {(x, y) : 0 \leq \times \leq 1 a n d 0 \leq y \leq 1} .$ and $R_{2} = {(x, y); x^{2} + y^{2} \leq 4 / 3}$
The area of the region $R_{1} \cap R_{2}$ can be expressed as $\frac{a \sqrt{3} + b π}{9}$ where a and b are integers then

a = 3

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a = 1

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b = 1

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b = 3

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Solution

The correct options are
A $a = 3$
D $b = 1$
$A = \frac{1}{\sqrt{3}} + \int_{1 / \sqrt{3}}^{1} \sqrt{\frac{4}{3} - x^{2}} d x$
$= \frac{1}{\sqrt{3}} + {[\frac{x}{2} \sqrt{\frac{4}{3} - x^{2}} + \frac{2}{3} {sin}^{- 1} (\frac{x \sqrt{3}}{2})]}_{1 / \sqrt{3}}^{- 1}$
$= \frac{1}{\sqrt{3}} + [(\frac{1}{2 \sqrt{3}} - \frac{1}{2 \sqrt{3}}) + \frac{2}{3} (\frac{π}{3} - \frac{π}{6})] = \frac{3 \sqrt{3} + π}{9}$

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Similar questions

Q. Consider the following regions in the

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R_{1} = {(x, y) : 0 \leq x \leq 1 and 0 \leq y \leq 1}

R_{2} = {(x, y) : x^{2} + y^{2} \leq \frac{4}{3}}

The area of the region

R_{1} \cap R_{2}

can be expressed as

\frac{a \sqrt{3} + b π}{9},

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and

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Consider the following regions in the plane R1={(x,y):0≤×≤1and0≤y≤1}. and R2={(x,y);x2+y2≤4/3}The area of the region R1∩R2 can be expressed as a√3+bπ9 where a and b are integers then

The correct options are A a=3 D b=1A=1√3+∫11/√3√43−x2dx=1√3+[x2√43−x2+23sin−1(x√32)]11/√3=1√3+[(12√3−12√3)+23(π3−π6)]=3√3+π9

Consider the following regions in the plane $R_{1} = {(x, y) : 0 \leq \times \leq 1 a n d 0 \leq y \leq 1} .$ and $R_{2} = {(x, y); x^{2} + y^{2} \leq 4 / 3}$
The area of the region $R_{1} \cap R_{2}$ can be expressed as $\frac{a \sqrt{3} + b π}{9}$ where a and b are integers then