First slide

Theory of expressions

Question

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

Difficult

A

3
B

2
C

4
D

1

Solution

$\begin{array}{l} Let e^{x} = t, t \in (0, \infty) \\ Given equation will be t^{4} + t^{3} - 4 t^{2} + t + 1 = 0 \\ Divide by t^{2} to get t^{2} + t - 4 + \frac{1}{t} + \frac{1}{t^{2}} = 0 \Rightarrow (t^{2} + \frac{1}{t^{2}}) + (t + \frac{1}{t}) - 4 = 0 \end{array}$
$\begin{array}{l} Let t + \frac{1}{t} = α, α \in [2, \infty) \Rightarrow t^{2} + \frac{1}{t^{2}} = α^{2} - 2 \\ So, the equation will be (α^{2} - 2) + α - 4 = 0 \\ α^{2} + α - 6 = 0 \Rightarrow α = - 3, 2 \end{array}$
$\begin{array}{l} ∵ α \in [2, \infty),we get α = 2 \Rightarrow e^{x} + e^{- x} = 2 \\ So, x = 0 only solution \end{array}$

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