A) \[0\]
B) \[1\]
C) \[\frac{\sqrt{3}}{2}\]
D) \[\sqrt{3}\]
Correct Answer: B
Solution :
| [b] On putting \[x=\frac{\sqrt{3}}{2},\]we get \[\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\] \[=\frac{2+\sqrt{3}}{\sqrt{2}\,(\sqrt{2}+\sqrt{3})}+\frac{2-\sqrt{3}}{\sqrt{2}\,(\sqrt{2}+\sqrt{2+\sqrt{3}})}\] \[=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4+2\sqrt{3}}}\] \[=\frac{2+\sqrt{3}}{2+(1+\sqrt{3})}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}\] \[=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\] \[=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\] \[=1\] |
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