The number of real solutions of the equation ${e^{4x}} + 4{e^{3x}} - 58{e^{2x}} + 4{e^x} + 1 = 0$ is ___________.
Answer (integer)
2
Solution
<p>Dividing by e<sup>2x</sup></p>
<p>${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$</p>
<p>$\Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$</p>
<p>Let ${e^x} + {e^{ - x}} = t \in [2,\infty )$</p>
<p>$\Rightarrow {t^2} + 4t - 60 = 0$</p>
<p>$\Rightarrow t = 6$ is only possible solution</p>
<p>${e^x} + {e^{ - x}} = 6 \Rightarrow {e^{2x}} - 6{e^x} + 1 = 0$</p>
<p>Let ${e^x} = p$,</p>
<p>${p^2} - 6p + 1 = 0$</p>
<p>$\Rightarrow p = {{3 + \sqrt 5 } \over 2}$ or ${{3 - \sqrt 5 } \over 2}$</p>
<p>So $x = \ln \left( {{{3 + \sqrt 5 } \over 2}} \right)$ or $\ln \left( {{{3 - \sqrt 5 } \over 2}} \right)$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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