Question: If $ x=\frac{\sqrt{3}}{2}, $ then the value of $ \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}} $
Options:
A) $ 0 $
B) $ 1 $
C) $ \frac{\sqrt{3}}{2} $
D) $ \sqrt{3} $
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Answer:
Correct Answer: B
Solution:
- Given expression, $ \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}} $
On putting $ x=\frac{\sqrt{3}}{2}, $ we get
$ \frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{\sqrt{1-\frac{\sqrt{3}}{2}}} $
$ =\frac{2+\sqrt{3}}{\sqrt{2}(\sqrt{2}+\sqrt{2}+\sqrt{3})}+\frac{2-\sqrt{3}}{\sqrt{2}(\sqrt{2}-\sqrt{2}-\sqrt{3})} $
$ =\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}} $
$ =\frac{2+\sqrt{3}}{2+\sqrt{{{(1)}^{2}}+{{(\sqrt{3})}^{2}}+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-{{(1)}^{2}}+{{(\sqrt{3})}^{2}}-2\sqrt{3}} $ $ =\frac{2+\sqrt{3}}{2+\sqrt{{{(1+\sqrt{3})}^{2}}}}+\frac{2-\sqrt{3}}{2-\sqrt{{{(\sqrt{3}-1)}^{2}}}} $
$ =\frac{2+\sqrt{3}}{2+(1+\sqrt{3})}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}} $
$ =\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})} $
$ =\frac{6+3\sqrt{3}-2\sqrt{3}-3+6-3\sqrt{3}+2\sqrt{3}-3}{(3+\sqrt{3})(3-\sqrt{3})} $
$ =\frac{6}{9+3\sqrt{3}-3\sqrt{3}-3}=\frac{6}{6}=1 $
BETA
