A) \[\sqrt{3}\]
B) \[\frac{\sqrt{3}}{2}\]
C) \[2+\sqrt{3}\]
D) \[2-\sqrt{3}\]
Correct Answer: A
Solution :
| \[a=\frac{\sqrt{3}}{2}\] |
| \[\therefore \]\[\sqrt{1+a}+\sqrt{1-a}=\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}\] |
| \[=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}\] |
| \[=\sqrt{\frac{4+2\sqrt{3}}{\sqrt{2}\times \sqrt{2}}}+\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}\times \sqrt{2}}\] |
| \[=\frac{\sqrt{{{(\sqrt{3}+1)}^{2}}}}{2}+\frac{\sqrt{{{(\sqrt{3}-1)}^{2}}}}{2}=\frac{\sqrt{3}+1}{2}+\frac{\sqrt{3}-1}{2}\] \[=\frac{\sqrt{3}+1+\sqrt{3}-1}{2}=\frac{2\sqrt{3}}{2}=\sqrt{3}\] |
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