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

Correct Option: 2 (which corresponds to 1 root, at x = 0)

Let's analyze the given equation:
$2\cos \left(\frac{x^2+x}{2}\right)=2^x+2^{-x}$

We can analyze the range of both sides of the equation.

For the Left-Hand Side (LHS):
Since the range of the cosine function is $[-1,1]$, we have:
$-2\le 2\cos \left(\frac{x^2+x}{2}\right)\le 2$
Thus, the maximum possible value of the LHS is 2.

For the Right-Hand Side (RHS):
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality for positive real terms $2^x$ and $2^{-x}$:
$\frac{2^x+2^{-x}}{2}\ge \sqrt{2^x·2^{���x}}$
$2^x+2^{-x}\ge 2\sqrt{1}=2$
Thus, the minimum possible value of the RHS is 2.

For the LHS and RHS to be equal, both sides must be equal to 2:
1) $2^x+2^{-x}=2$, which occurs if and only if $2^x=1$, meaning $x=0$.
2) $2\cos \left(\frac{x^2+x}{2}\right)=2$, which simplifies to $\cos \left(\frac{x^2+x}{2}\right)=1$.

Let's check if the solution $x=0$ from the RHS satisfies the LHS condition:
For $x=0$:
$\cos \left(\frac{0^2+0}{2}\right)=\cos \left(0\right)=1$, which is indeed true.

Therefore, there is exactly one real root, which is $x=0$.