A) \[\frac{{{e}^{x}}+{{e}^{-x}}}{2}\]
B) \[\frac{{{e}^{x}}-{{e}^{-x}}}{2}\]
C) \[{{e}^{x}}+{{e}^{-x}}\]
D) \[{{e}^{x}}-{{e}^{-x}}\]
Correct Answer: B
Solution :
\[\because \ {{e}^{x}}=y+\sqrt{1+{{y}^{2}}}\] \\[{{e}^{x}}-y=\sqrt{1+{{y}^{2}}}\] Squaring both the sides, \[{{({{e}^{x}}-y)}^{2}}=(1+{{y}^{2}})\] \[{{e}^{2x}}+{{y}^{2}}-2y{{e}^{x}}=1+{{y}^{2}}\Rightarrow {{e}^{2x}}-1=2y{{e}^{x}}\] Þ \[2y=\frac{{{e}^{2x}}-1}{{{e}^{x}}}\Rightarrow 2y={{e}^{x}}-{{e}^{-x}}\] Hence, \[y=\frac{{{e}^{x}}-{{e}^{-x}}}{2}\].
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