Question

The number of real solutions of $\sin \left(e^x\right)·\cos \left(e^x\right)=2^{x-2}+2^{-x-2}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $\infty$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Find the solutions for the given equation.

An equation $\sin \left(e^x\right)·\cos \left(e^x\right)=2^{x-2}+2^{-x-2}$ is given.

Simplify the given equations as follows:

$$\begin{gathered} \sin \left(e^x\right)·\cos \left(e^x\right)=\frac{2^{x-1}}{2}+\frac{2^{-x-1}}{2} \\ \Rightarrow 2\sin \left(e^x\right)·\cos \left(e^x\right)=2^{x-1}+2^{-x-1} \\ \Rightarrow \sin \left(2e^x\right)=\frac{2^x}{2}+\frac{2^{-x}}{2} \\ \Rightarrow 2\sin \left(2e^x\right)=2^x+2^{-x} \end{gathered}$$

Since, $x$ is real. So, $2^x+2^{-x}\ge 2$.

Also, $-1\le \sin \left(2e^2\right)\le 1$.

So, $-2\le 2\sin \left(2e^x\right)\le 2$.

Therefore, $2^x+2^{-x}=2$.

So, the only possible value of $x$ is zero.

Check whether $x=0$ satisfies the given equation.

$$\begin{gathered} \sin \left(e^0\right)·\cos \left(e^0\right)=2^{0-2}+2^{-0-2} \\ \Rightarrow 1=2·2^{-2} \\ \Rightarrow 1=\frac{1}{2} \end{gathered}$$

Since, $x=0$ does not satisfies the given equation.

Therefore, there is no real solution to the given equation.

Hence, option $A$ is the correct answer.