MCQ
If ${e^x} = y + \sqrt {1 + {y^2}} $, then $y =$
  • A
    $\frac{{{e^x} + {e^{ - x}}}}{2}$
  • $\frac{{{e^x} - {e^{ - x}}}}{2}$
  • C
    ${e^x} + {e^{ - x}}$
  • D
    ${e^x} - {e^{ - x}}$

Answer

Correct option: B.
$\frac{{{e^x} - {e^{ - x}}}}{2}$
b
(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, $y = \frac{{{e^x} - {e^{ - x}}}}{2}$.

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