MCQ
જો ${e^x} = y + \sqrt {1 + {y^2}} $, તો $y =$
- A$\frac{{{e^x} + {e^{ - x}}}}{2}$
- ✓$\frac{{{e^x} - {e^{ - x}}}}{2}$
- C${e^x} + {e^{ - x}}$
- D${e^x} - {e^{ - x}}$
$\therefore$ ${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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તો $\left[ {\vec p \,\,\,\vec q \,\,\, \vec r } \right]\, = ...$