If x-2-3- xy-1- then find x-2-x-y-2-y-

Finding x/2 x+y/2 yGiven that x=2+√(3), xy=1 x/2 x+y/2 y=x/2 x+y/2 y0ex0ex=x/2 x+xy/2x 1 (∵ xy=1 y=1/x )0ex0ex=x/2 x+1/2x 10ex0ex=2+√(3)/2 (2+√(3))+1/2(2+√(3 ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity

If x=2+√3, xy...

Question

If $x = 2 + \sqrt{3}, x y = 1$ , then find $\frac{x}{2 - x} + \frac{y}{2 - y}$ .

$- \sqrt{3}$

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$\sqrt{3}$

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$- \sqrt{2}$

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$- 2$

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Solution

The correct option is D
$- 2$

Finding $\frac{x}{2 - x} + \frac{y}{2 - y}$
Given that $x = 2 + \sqrt{3}, x y = 1$
$\frac{x}{2 - x} + \frac{y}{2 - y}$
$= \frac{x}{2 - x} + \frac{y}{2 - y} = \frac{x}{2 - x} + \frac{x y}{2 x - 1} (∵ x y = 1 \Leftrightarrow y = \frac{1}{x}) = \frac{x}{2 - x} + \frac{1}{2 x - 1} = \frac{2 + \sqrt{3}}{2 - (2 + \sqrt{3})} + \frac{1}{2 (2 + \sqrt{3}) - 1} = \frac{2 + \sqrt{3}}{2 - 2 - \sqrt{3}} + \frac{1}{2 \times 2 + 2 \sqrt{3} - 1} = \frac{2 + \sqrt{3}}{- \sqrt{3}} + \frac{1}{3 + 2 \sqrt{3}} = \frac{- 6 - 7 \sqrt{3} - 6 + \sqrt{3}}{\sqrt{3} (3 + 2 \sqrt{3})} (addingboththefraction) = \frac{- (12 + 6 \sqrt{3})}{\sqrt{3} (3 + 2 \sqrt{3})} = \frac{- (12 + 6 \sqrt{3})}{3 \sqrt{3} + 6} = \frac{- 6 (2 + \sqrt{3)}}{3 (2 + \sqrt{3)}} = \frac{- 6}{3} = - 2$
Hence, the correct answer is (D) $- 2$

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If x=2+3,xy=1, then find x2-x+y2-y.

The correct option is D -2Finding x2-x+y2-yGiven that x=2+3,xy=1 x2-x+y2-y=x2-x+y2-y=x2-x+xy2x-1(∵xy=1⇔y=1x)=x2-x+12x-1=2+32-(2+3)+12(2+3)-1=2+32-2-3+12×2+23-1=2+3-3+13+23=-6-73-6+333+23(addingboththefraction)=-12+6333+23=-12+6333+6=-6(2+3)3(2+3)=-63=-2Hence, the correct answer is (D)-2

If $x = 2 + \sqrt{3}, x y = 1$ , then find $\frac{x}{2 - x} + \frac{y}{2 - y}$ .