Question

The number of real roots of $\sin \left(2^x\right)\cos \left(2^x\right)=\frac{1}{4}\left(2^x+2^{-x}\right)$ is

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Find the solution of the given equation.

An equation $\sin \left(2^x\right)\cos \left(2^x\right)=\frac{1}{4}\left(2^x+2^{-x}\right)$ is given.

Solve the given equation as follows:

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

Since it is clear that $x=0$ is not the solution of the given equation.

And for $x\ne 0$.

$-2\le 2\left[\sin \right(2.2^x\left)\right]\le 2$ but $2^x-2^{-x}>2$.

Therefore, there are no real solutions of the given equation.

Hence, option $A$ is the correct option.