QJEEM22Ma8e02: JEE Main 2022 Mathematics - Class 11 JEEM22Ma8e02 Solution: The number of real solutions of the equation is ___________.

Dividing by e^2x

${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$

$\Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$

Let ${e^x} + {e^{ - x}} = t \in [2,\infty )$

$\Rightarrow {t^2} + 4t - 60 = 0$

$\Rightarrow t = 6$ is only possible solution

${e^x} + {e^{ - x}} = 6 \Rightarrow {e^{2x}} - 6{e^x} + 1 = 0$

Let ${e^x} = p$ ,

${p^2} - 6p + 1 = 0$

$\Rightarrow p = {{3 + \sqrt 5 } \over 2}$ or ${{3 - \sqrt 5 } \over 2}$

So $x = \ln \left( {{{3 + \sqrt 5 } \over 2}} \right)$ or $\ln \left( {{{3 - \sqrt 5 } \over 2}} \right)$