Functions Question 675
Question: If $ e^{x}=y+\sqrt{1+y^{2}} $ , then y =
[MNR 1990, UPSEAT 2000]
Options:
A) $ \frac{e^{x}+{e^{-x}}}{2} $
B) $ \frac{e^{x}-{e^{-x}}}{2} $
C) $ e^{x}+{e^{-x}} $
D) $ e^{x}-{e^{-x}} $
Show Answer
Answer:
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}-2ye^{x}=1+y^{2}\Rightarrow e^{2x}-1=2ye^{x} $
Þ $ 2y=\frac{e^{2x}-1}{e^{x}}\Rightarrow 2y=e^{x}-{e^{-x}} $ Hence, $ y=\frac{e^{x}-{e^{-x}}}{2} $ .
BETA
