The number of real roots of the equation $e^{6 x} - e^{4 x} - 2 e^{3 x} - 12 e^{2 x} + e^{x} + 1 = 0$ is

1

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2

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6

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4

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Solution

The correct option is B $2$
Let $e^{6 x} - e^{4 x} - 2 e^{3 x} - 12 e^{2 x} + e^{x} + 1 = 0$
$\Rightarrow e^{6 x} - 2 e^{3 x} + 1 - e^{x} [e^{3 x} + 12 e^{x} - 1] = 0$
$\Rightarrow {(e^{3 x} - 1)}^{2} - e^{x} (e^{3 x} - 1) - 12 e^{2 x} = 0$
$\Rightarrow {(e^{3 x} - 1)}^{2} - 4 e^{x} (e^{3 x} - 1) + 3 e^{x} (e^{3 x} - 1) - 12 e^{2 x} = 0$
$\Rightarrow (e^{3 x} - 1) [e^{3 x} - 1 - 4 e^{x}] + 3 e^{x} [e^{3 x} - 1 - 4 e^{x}] = 0$
$\Rightarrow (e^{3 x} - 1 + 3 e^{x}) (e^{3 x} - 1 - 4 e^{x}) = 0$

Either $e^{3 x} = 1 + 4 e^{x}$

Or $e^{3 x} + 3 e^{x} = 1$

From the above graphs, only two roots are possible.

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