Question

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

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

Answer: (C)

$e^{4x}+e^{3x}-4e^x+e^x+1=0$
Divide by $e^{2x}$
$\Rightarrow e^{2x}+e^x-4+\frac{1}{e^x}+\frac{1}{e^{2x}}=0$
$\Rightarrow \left(e^{2x}+\frac{1}{e^{2x}}\right)+\left(e^x+\frac{1}{e^x}\right)-4=0$
$\Rightarrow {\left(e^x+\frac{1}{e^x}\right)}^2-2+\left(e^x+\frac{1}{e^x}\right)-4=0$
Let $e^x+\frac{1}{e^x}=t\Rightarrow {\left(e^x-1\right)}^2=0\Rightarrow x=0.$
$\therefore$ Number of real roots $=1$