| If \[x=\frac{\sqrt{3}}{2},\]then the value of \[\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}\] [SSC (CGL) 2012] |
A) \[0\]
B) \[1\]
C) \[\frac{\sqrt{3}}{2}\]
D) \[\sqrt{3}\]
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 |
| \[\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{\sqrt{1-\frac{\sqrt{3}}{2}}}\] |
| \[=\frac{2+\sqrt{3}}{\sqrt{2}(\sqrt{2}+\sqrt{2}+\sqrt{3})}+\frac{2-\sqrt{3}}{\sqrt{2}(\sqrt{2}-\sqrt{2}-\sqrt{3})}\] |
| \[=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\] |
| \[=\frac{2+\sqrt{3}}{2+\sqrt{{{(1)}^{2}}+{{(\sqrt{3})}^{2}}+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-{{(1)}^{2}}+{{(\sqrt{3})}^{2}}-2\sqrt{3}}\]\[=\frac{2+\sqrt{3}}{2+\sqrt{{{(1+\sqrt{3})}^{2}}}}+\frac{2-\sqrt{3}}{2-\sqrt{{{(\sqrt{3}-1)}^{2}}}}\] |
| \[=\frac{2+\sqrt{3}}{2+(1+\sqrt{3})}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\] |
| \[=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\] |
| \[=\frac{6+3\sqrt{3}-2\sqrt{3}-3+6-3\sqrt{3}+2\sqrt{3}-3}{(3+\sqrt{3})(3-\sqrt{3})}\] |
| \[=\frac{6}{9+3\sqrt{3}-3\sqrt{3}-3}=\frac{6}{6}=1\] |
You need to login to perform this action.
You will be redirected in
3 sec
