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A light ray emits from the origin making an angle \( 30^{\circ} \) with the positive \( x \)-axis. After getting reflected by the line \( x+y=1 \), if this ray intersects \( x \)-axis at \( Q \), then the abscissa of \( Q \) is 

1. \( \frac{2}{3+\sqrt{3}} \) 

2. \( \frac{2}{3-\sqrt{3}} \) 

3. \( \frac{\sqrt{3}}{2(\sqrt{3}+1)} \) 

4. \( \frac{2}{(\sqrt{3}-1)} \)

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Correct option is 1. \(\frac{2}{3+\sqrt{3}}\)

Let Q be (h,o)

The equation of emitted ray is y = x/√3  (∵ tanθ = 1/√3)

∵ OP is reflected by line x+y = 1

∴ Image of Q(h,o) lies on line y = x/√3

\(∴\frac{x-h}{1}=\frac{y-o}{1}=\frac{-2(h-1)}{2}\)

∴ x = 1 , y = 1-h

It lies on y = x/√3.

∴ 1 - h = 1/√3

\(∴h=1-\frac{1}{\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}}=\frac{3-1}{\sqrt{3}(\sqrt{3}+1)}\)

\(=\frac{2}{\sqrt{3}(\sqrt{3}+1)}=\frac{2}{3+\sqrt{3}}\)

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