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  • The  number of real roots of the equation,    is :  n

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

Answers (1)

best_answer

 

 

Transcendental function -

Transcendental functions:  the functions which are not algebraic are called transcendental functions. Exponential, logarithmic, trigonometric and inverse trigonometric functions are transcendental functions.

Exponential Function: function f(x) such that \mathrm{f(x)=a^x} is known as an exponential function.

\\\mathrm{base:\;\;a>0,a\neq1}\\\mathrm{domain:x\in \mathbb{R}}\\\mathrm{range:f(x)>0}

 

 

Logarithmic function:  function f(x) such that f\left ( x \right )= \log\: _{a}x is called logarithmic function 

\\\mathrm{base:\;\;a>0,a\neq1}\\\mathrm{domain:x>0}\\\mathrm{range:f(x)\in\mathbb{R}}
         

                    If a > 1                                                                               If a < 1

Properties of Logarithmic Function

\\\mathrm1.\;{\log_e(ab)=\log_ea+\log_eb}\\\mathrm{2.\;\log_e\left ( \frac{a}{b} \right )=\log_ea-\log_e b}\\\mathrm{3.\;\log_ea^m=m\log_ea}\\\mathrm{4.\;\log_aa=1}\\\mathrm{5.\;\log_{b^m}a=\frac{1}{m}\log_ba}\\\mathrm{6.\;\log_ba=\frac{1}{\log_ab}}\\\mathrm{7.\;\log_ba=\frac{\log_ma}{\log_mb}}\\\mathrm{8.\;a^{\log_am}=m}\\\mathrm{9.\;a^{\log_cb}=b^{\log_ca}}\\\mathrm{10.\;\log_ma=b\Rightarrow a=m^b}

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Quadratic Equation -

The root of the quadratic equation is given by the formula:

 

\\\mathrm{x = \frac{-b \pm \sqrt{D}}{2a}}\\\\\mathrm{or} \\\mathrm{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}

 Where D is called the discriminant of the quadratic equation, given by D = b^2 - 4ac ,

 

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Let e^{x}=t \in(0, \infty)

Now the equation 

\begin{array}{l}{t^{4}+t^{3}-4 t^{2}+t+1=0} \\ {t^{2}+t-4+\frac{1}{t}+\frac{1}{t^{2}}=0} \\ {\left(t^{2}+\frac{1}{t^{2}}\right)+\left(t+\frac{1}{t}\right)-4=0}\end{array}

Let \mathrm{t}+\frac{1}{\mathrm{t}}=\alpha

\begin{array}{l}{\left(\alpha^{2}-2\right)+\alpha-4=0} \\ {\alpha^{2}+\alpha-6=0} \\ {\alpha^{2}+\alpha-6=0}\end{array}

\alpha=-3,2

Only positive value possible so \alpha=2 \Rightarrow \quad \mathrm{e}^{x}+\mathrm{e}^{-\mathrm{x}}=2

x=0 is the only solution.

Posted by

avinash.dongre

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