Question

The number of real solutions of $\sin e^x.\cos e^x=2^{x-2}+2^{-x-2}$ is

• A. zero
• B. one
• C. two
• D. infinite

Answer 1

The correct option is A zero

$\sin e^x.\cos e^x=2^{x-2}+2^{-x-2}$
$\Rightarrow \sin 2e^x=2^{x-1}+2^{-x-1}$
$\Rightarrow 2\sin 2e^x=2^x+2^{-x}$
$\Rightarrow \frac{1}{2}\sin 2e^x=\frac{1}{4}(2^x+2^{-x})$
For any real value of $x$, $2^x+2^{-x}\ge 2$
$\Rightarrow \frac{1}{2}\sin 2e^x\ge \frac{1}{2}$
$\Rightarrow \sin 2e^x\ge 1$
But $\sin 2e^x\le 1$
So, $\sin 2e^x=1$
But equality can hold only if RHS also equals $1$ , which is possible only when $x=0$
Hence, $\sin 2e^x=1$ is not true
Hence, there is no solution.