A) \[\frac{3}{2},0\]
B) \[0,\frac{3}{2}\]
C) \[\frac{2}{3},-1\]
D) \[1,\frac{2}{3}\]
Correct Answer: A
Solution :
[a] Given, \[{{2}^{x+y}}={{2}^{x-y}}=\sqrt{8}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,{{2}^{x+y}}={{2}^{x-y}}={{(8)}^{1/2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,{{2}^{x+y}}={{2}^{x-y}}={{2}^{3/2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,x+y=\frac{3}{2}\] and \[x-y=\frac{3}{2}\] |
Adding both equations, we get |
\[2x=3\Rightarrow x=\frac{3}{2}\] |
and \[y=\frac{3}{2}-\frac{3}{2}=0\] |
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