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1.Relation and Function
medium
If ${e^x} = y + \sqrt {1 + {y^2}} $, then $y =$
A
$\frac{{{e^x} + {e^{ - x}}}}{2}$
B
$\frac{{{e^x} - {e^{ - x}}}}{2}$
C
${e^x} + {e^{ - x}}$
D
${e^x} - {e^{ - x}}$
Solution
(b) $\because \;{e^x} = y + \sqrt {1 + {y^2}} $
$\therefore$ ${e^x} – y = \sqrt {1 + {y^2}} $
Squaring both the sides, ${({e^x} – y)^2} = (1 + {y^2})$
${e^{2x}} + {y^2} – 2y{e^x} = 1 + {y^2} \Rightarrow {e^{2x}} – 1 = 2y{e^x}$
==> $2y = \frac{{{e^{2x}} – 1}}{{{e^x}}} \Rightarrow 2y = {e^x} – {e^{ – x}}$
Hence, $y = \frac{{{e^x} – {e^{ – x}}}}{2}$.
Standard 12
Mathematics
