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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

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 $e^{2x} 
eq 0$ for all real $x$):$e^{2x} + e^x - 4 + e^{-x} + e^{-2x} = 0$.Rearrange the terms:$(e^{2x} + e^{-2x}) + (e^x + e^{-x}) - 4 = 0$.Let $u = e^x + e^{-x}$. Note that $u^2 = (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x}$,so $e^{2x} + e^{-2x} = u^2 - 2$.Substitute this into the equation:$(u^2 - 2) + u - 4 = 0 \Rightarrow u^2 + u - 6 = 0$.Factor the quadratic equation:$(u + 3)(u - 2) = 0$.This gives $u = -3$ or $u = 2$.Case $1$: $e^x + e^{-x} = -3$. Since $e^x > 0$ and $e^{-x} > 0$,by $AM$-$GM$ inequality,$e^x + e^{-x} \geq 2$. Thus,$u = -3$ has no real solutions.Case $2$: $e^x + e^{-x} = 2$. By $AM$-$GM$ inequality,$e^x + e^{-x} = 2$ only when $e^x = e^{-x}$,which implies $e^{2x} = 1$,so $x = 0$.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:

Similar Questions

Solve the given two equations and select the correct answer from the given options:
$I.$ $\sqrt{1225} x + \sqrt{4900} = 0$
$II.$ $(81)^{1/4} y + (343)^{1/3} = 0$

$I. 24x^2 + 38x + 15 = 0$
$II. 12y^2 + 28y + 15 = 0$

Solve the given two equations and provide the correct answer from the given options.
$I.$ $\frac{x}{x+7} + \frac{x+7}{x} = 12$
$II.$ $\frac{y}{y+8} + \frac{y+8}{y} = 16$

The number of values of $k$ for which the equation $x^2 - 3x + k = 0$ has two real and distinct roots lying in the interval $(0, 1)$ is:

The number of solutions of $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module