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