Question

The number of real solutions of the equation $e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$ is

• A. 2.0
• B. 2.00
• C. 2

Answer 1

Answer: (A), (B), (C)

Dividing by $e^{2x}$
$e^{2x}+4e^x-58+4e^{-x}+e^{-2x}=0$
$\Rightarrow (e^x+e^{-x}{)}^2+4(e^x+e^{-x})-60=0$
Let $e^x+e^{-x}=t\in [2,\infty )$
$\Rightarrow t^2+4t-60=0$
$\Rightarrow t=6$ is only possible solution
$e^x+e^{-x}=6\Rightarrow e^{2x}-6e^x+1=0$
Let $e^x=p$,
$p^2-6p+1=0$
$\Rightarrow p=\frac{3+\sqrt{5}}{2}\text{OR}\frac{3-\sqrt{5}}{2}$
$\Rightarrow x=\ln \left(\frac{3+\sqrt{5}}{2}\right)\text{OR}\ln \left(\frac{3-\sqrt{5}}{2}\right)$
So, number of real solution is $2$