If $$x = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}$$ and $$y = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}},$$ then the value of $$\left( {{x^2} + {y^2}} \right)$$ is?
A. 10
B. 13
C. 14
D. 15
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & x = \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 - 1} }} \times \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 + 1} }} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{ {3 - 1} }} \cr & \,\,\,\,\,\, = \frac{{3 + 1 + 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 + \sqrt 3 \cr & y = \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 + 1} }} \times \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 - 1} }} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{ {3 - 1} }} \cr & \,\,\,\,\,\, = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 - \sqrt 3 \cr & \therefore {x^2} + {y^2} = {\left( {2 + \sqrt 3 } \right)^2} + {\left( {2 - \sqrt 3 } \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {4 + 3} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$Related Questions on Square Root and Cube Root
The least perfect square, which is divisible by each of 21, 36 and 66 is:
A. 213444
B. 214344
C. 214434
D. 231444
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