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The number of the real roots of the equation $2\cos (x(x+1))=2^{x}+2^{-x}$ is

1. $2$
2. $1$
3. infinite
4. $0$

Given that, $2 \cos (x(x+1)) = 2^{x} + 2^{-x}$

$\Rightarrow 2 \cos (x(x+1)) = 2^{x} + \frac{1}{2^{x}} \quad \longrightarrow (1)$

For any set of non-negative real numbers, the arithmetic mean of the set is greater than or equal to the geometric mean of the set.

Algebraically, this is expressed as follows:

For a set of non-negative real numbers $a_{1}, a_{2}, a_{3}, \dots, a_{n},$ the following always holds:

$\boxed {\frac{ a_{1} + a_{2} + a_{3} + \dots + a_{n} }{n} \geq \sqrt[n]{a_{1} a_{2} a_{3} \dots a_{n}}}$

Now, from the equation $(1)$, take the right hand side term,

$2^{x} + \frac {1}{2^{x}} \geq 2 \sqrt {2^{x} \cdot \frac {1}{2^{x}}}$

$\Rightarrow 2^{x} + \frac {1}{2^{x}} \geq 2$

Put the value in the equation $(1),$ we get

$2 \cos (x(x+1)) = 2 \quad \longrightarrow (2)$

$\Rightarrow \cos (x(x+1)) = 1$

$\Rightarrow \cos (x(x+1)) = \cos \; 0^{\circ}$

$\Rightarrow (x(x+1)) = 0$

$\Rightarrow \boxed{x=0\;\text{(or)}\; x = -1}$

Now, put  $x=0,$ in the equation $(1),$ we get

$2 \cos (0(0+1)) = 2^{0} + \frac{1}{2^{0}}$

$\Rightarrow 2 \cos 0 = 1+1$

$\Rightarrow 2 (1) = 2$

$\Rightarrow \boxed{2 = 2}$ (Satisfied)

And, put $x=-1,$ in the equation $(1),$ we get

$2 \cos (-1(-1+1)) = 2^{-1} + \frac{1}{2^{-1}}$

$\Rightarrow 2 \cos 0 = \frac{1}{2} + 2$

$\Rightarrow 2(1) = 2 + \frac{1}{2}$

$\Rightarrow \boxed{2 \neq \frac{5}{2}}$ ( Not satisfied)

So, only $\boxed{x=0},$ is the real root of the equation.

Correct Answer $: \text{B}$
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