Question

The number of solution of the equation $2\cos ({{e}^{x}})={{5}^{x}}+{{5}^{-x}}$ are
A. No solution.
B. One solution.
C. Two solution.
D. Infinitely many solutions.

Answer 1

Hint: To find the number of solutions for the equation given we will find the interval of values of both side of the expression and see if they coincide or not because both should be equal as given $2\cos ({{e}^{x}})={{5}^{x}}+{{5}^{-x}}$.
Taking the interval of function cos we will determine the interval of $2\cos ({{e}^{x}})$ and using relationship $A.M\ge G.M$ we will determine the interval of ${{5}^{x}}+{{5}^{-x}}$.

Formula Used: $A.M=\dfrac{sum\,\,of\,observations.}{Number\,of\,observations.}$
\[G.M={{\left( Product\,of\,observations \right)}^{\dfrac{1}{Number\,\,of\,observations}}}\]

Complete step by step solution: We are given an equation $2\cos ({{e}^{x}})={{5}^{x}}+{{5}^{-x}}$ and we have to find the number of solutions for this equation.
In the equation given we have trigonometric equation on the right side and exponential equation on the left side of the equation.
First we will take the left side of the equation $2\cos ({{e}^{x}})$ ,now we know that the interval of the values of cosine is $\left[ -1,1 \right]$. So,
$-1\le \cos ({{e}^{x}})\le 1$
Multiplying the inequality by $2$ ,
$-2\le 2\cos ({{e}^{x}})\le 2$
We will now take the right side of the equation ${{5}^{x}}+{{5}^{-x}}$,
Both of the terms in ${{5}^{x}}+{{5}^{-x}}$ will be positive for all the real values of $x$ hence the whole identity will be positive.
As both the terms are positive we will apply the relation$A.M\ge G.M$ in it.
The A.M for both the terms will be $\dfrac{{{5}^{x}}+{{5}^{-x}}}{2}$ and G.M for both the terms will be ${{({{5}^{x}}{{.5}^{-x}})}^{{}^{1}/{}_{2}}}$.
Substituting values in the relation $A.M\ge G.M$ and simplify,
$\begin{align}
  & \dfrac{{{5}^{x}}+{{5}^{-x}}}{2}\ge {{({{5}^{x}}{{.5}^{-x}})}^{{}^{1}/{}_{2}}} \\
 & \dfrac{{{5}^{x}}+{{5}^{-x}}}{2}\ge {{\left( {{5}^{x}}\times \dfrac{1}{{{5}^{x}}} \right)}^{{}^{1}/{}_{2}}} \\
 & \dfrac{{{5}^{x}}+{{5}^{-x}}}{2}\ge 1 \\
 & {{5}^{x}}+{{5}^{-x}}\ge 2
\end{align}$
It means that value of expression ${{5}^{x}}+{{5}^{-x}}$will be always greater than or equal to $2$.
Now the equality $2\cos ({{e}^{x}})={{5}^{x}}+{{5}^{-x}}$ will hold when both have same interval of values but the interval for left side expression is $-2\le 2\cos ({{e}^{x}})\le 2$ and right side expression is ${{5}^{x}}+{{5}^{-x}}\ge 2$. Hence there will be no solution for this $2\cos ({{e}^{x}})={{5}^{x}}+{{5}^{-x}}$ equality.
There are no solution for the equation $2\cos ({{e}^{x}})={{5}^{x}}+{{5}^{-x}}$.

Option ‘A’ is correct

Note: The inequality of arithmetic means and geometric means that is states that A.M for a set of observations will be always greater than or equal to the G.M for the same observations.