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. Zero
• B. One
• C. Two
• D. Infinite

Answer 1

The correct option is B One

$sin2e^x=\frac{2^x+2^{-x}}{2}\Rightarrow 2sin\left(2e^x\right)=2^x+2^{-x}$
$-1\le \sin A\le 1\Rightarrow -2\le 2sin\left(2e^x\right)\le 2$
Hence $x=1$ is the only solution