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

3

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4

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1

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2

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Solution

The correct option is C $1$
$e^{4 x} + e^{3 x} - 4 e^{2 x} + e^{x} + 1 = 0$
$\Rightarrow e^{2 x} + e^{x} - 4 + \frac{1}{e^{x}} + \frac{1}{e^{2 x}} = 0$

$\Rightarrow (e^{2 x} + \frac{1}{e^{2 x}}) + (e^{x} + \frac{1}{e^{x}}) - 4 = 0$

$\Rightarrow {(e^{x} + \frac{1}{e^{x}})}^{2} - 2 + (e^{x} + \frac{1}{e^{x}}) - 4 = 0$

Let $e^{x} + \frac{1}{e^{x}} = u$

Then, $u^{2} + u - 6 = 0$

$\Rightarrow u = 2, - 3$

$u \neq - 3$ as $u > 0 (∵ e^{x} > 0)$

$e^{x} + \frac{1}{e^{x}} = 2 \Rightarrow (e^{x} - 1)^{2} = 0 \Rightarrow e^{x} = 1 \Rightarrow x = 0$

Hence, only one real solution is possible.

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