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

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

e^{4 x} + {2 e}^{3 x} - e^{x} - 6 = 0

x^{4} - x^{2} + 2 x - 1 = 0

The number of real solutions of the equation $| x |^{2} - 3 | x | + 2 = 0$ are

The number of real roots of the equation $e^{s i n x} - e^{- s i n x} - 4 = 0$ are

| x - 3 |^{3 x^{2} - 10 x + 3} = 1

Join BYJU'S Learning Program