Question

The number of real solutions of $\sin {{e}^{x}}\cdot \cos {{e}^{x}}={{2}^{x-2}}+{{2}^{-x-2}}$ is
$1)\text{ }0$
$2)\text{ 1}$
$3)\text{ 2}$
$4)\text{ }\infty $

Answer 1

Hint: In this question we have been given with an expression which consists of the terms given in exponential form. We will use the property of exponents and trigonometry to solve the given question. We will use the formula $\sin 2x=2\sin x\cos x$ and the rule that for any real number ${{2}^{x}}+{{2}^{-x}}\ge 2$. Based on these properties, we will find the number of real solutions for the expression.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow \sin {{e}^{x}}\cdot \cos {{e}^{x}}={{2}^{x-2}}+{{2}^{-x-2}}$
Now we know that $\sin 2x=2\sin x\cos x$ therefore, on multiplying both sides by $2$, we get:
$\Rightarrow 2\sin {{e}^{x}}\cdot \cos {{e}^{x}}=2\cdot {{2}^{x-2}}+2\cdot {{2}^{-x-2}}$
We get the expression as:
$\Rightarrow \sin 2{{e}^{x}}=2\cdot {{2}^{x-2}}+2\cdot {{2}^{-x-2}}$
Now we know that $2$ is the same as ${{2}^{1}}$ and since the base are the same, we can write the exponent as:
$\Rightarrow \sin 2{{e}^{x}}={{2}^{x-2+1}}+{{2}^{-x-2+1}}$
On simplifying the exponent, we get:
$\Rightarrow \sin 2{{e}^{x}}={{2}^{x-1}}+{{2}^{-x-1}}$
On multiplying both sides by $2$, we get:
$\Rightarrow 2\cdot \sin 2{{e}^{x}}=2\cdot {{2}^{x-1}}+2\cdot {{2}^{-x-1}}$
On doing the same operations with the exponent, we get:
$\Rightarrow 2\cdot \sin 2{{e}^{x}}={{2}^{x}}+{{2}^{-x}}\to \left( 1 \right)$
Now, we know the property that for any real number ${{2}^{x}}+{{2}^{-x}}\ge 2$.
Now we know that $\sin \theta $ ranges from $-1$ to $1$ therefore, we can only write:
$\Rightarrow 1\le \sin 2{{e}^{x}}\le 1$
On multiplying the expression by $2$, we get:
$\Rightarrow 2\le 2\sin 2{{e}^{x}}\le 2$
Now the range of the left-hand side of equation $\left( 1 \right)$ is $\left[ -2,2 \right]$ and of the right-hand side is $\ge 2$.
The only place they meet is when the value is equal to $2$.
Therefore, we need to find the value of $x$ where both the left-hand side and the right-hand side have the value equal to $2$.
Now the right-hand side is equal to $2$ when $x=0$ since ${{2}^{0}}+{{2}^{-0}}=1+1=2$.
Now on substituting $x=0$ in the left-hand side, we get:
$\Rightarrow 2\cdot \sin 2{{e}^{0}}$
On simplifying, we get:
$\Rightarrow 2\cdot \sin 2\ne 2$
Now since there was only $1$ solution for ${{2}^{x}}+{{2}^{-x}}=2$ and it does not satisfy the left-hand side implies that there are no solutions for the given expression.

So, the correct answer is “Option 1”.

Note: It is to be remembered that inequalities are different from equalities. Equal two means that both the right-hand side and the left-hand side are equal to each other. In inequalities the terms on either side can be strictly less than or greater than which is represented as $<$ and $>$, or they can be less than or equal to, or greater than or equal to which is represented by $\le $ and $\ge $.