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 $= 1$