The number of real solutions of the equation $\sin (e^x)=2^x+2^{-x}$, is _____.
Answer & explanation
We know that
AM ≥ GM
$∴\frac{2^x+2^{-x}}{2}≥\sqrt{2^x×2^{-x}}$
$⇒2^x+2^{-x}≥2$
But, $\sin (e^x) ≤1$
Thus, $\sin (e^x) ≠2^x + 2^{-x}$ for any $x ∈ R$.
Hence, the given equation has no solution.