The number of real roots of the equation e^{4 | vedclass

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(D) Given equation: $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$. Divide the entire equation by $e^{2x}$ (since $e^{2x} 
eq 0$): $e^{2x} + e^x - 4 + \fr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given equation: $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$. Divide the entire equation by $e^{2x}$ (since $e^{2x} 
eq 0$): $e^{2x} + e^x - 4 + \frac{1}{e^x} + \frac{1}{e^{2x}} = 0$. Rearrange the terms: $(e^{2x} + \frac{1}{e^{2x}}) + (e^x + \frac{1}{e^x}) - 4 = 0$. Using the identity $a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2$,we get: $(e^x + \frac{1}{e^x})^2 - 2 + (e^x + \frac{1}{e^x}) - 4 = 0$. Let $t = e^x + \frac{1}{e^x}$. Since $e^x > 0$,by $AM$-$GM$ inequality,$t = e^x + \frac{1}{e^x} \geq 2$. The equation becomes $t^2 + t - 6 = 0$. Factoring the quadratic: $(t + 3)(t - 2) = 0$. This gives $t = -3$ or $t = 2$. Since $t \geq 2$,we must have $t = 2$. $e^x + \frac{1}{e^x} = 2$ $\Rightarrow e^{2x} - 2e^x + 1 = 0$ $\Rightarrow (e^x - 1)^2 = 0$. $e^x = 1 \Rightarrow x = 0$. Thus,there is only $1$ real root.

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

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