Question

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

• A. 0
• B. 1
• C. 2
• D. infinite

Answer 1

The correct option is A

0

$\frac{1}{2}$sin(2$e^x$) = $\frac{1}{4}$($2^x+2^{-x}$) ≥ $\frac{1}{4}$.2= $\frac{1}{2}$

So sin(2$e^x$) ≥ 1 But sin(2$e^x$) $\ngtr$ 1

Therefore sin(2$e^x$) = 1

But the equality would hold when $2^x$ = $2^{-x}$ = 1

⇒ x = 0. But then $sin\left(2.e^x\right)$ = 0 which is not true.