CAT 2019 Shift 1 Paper

For any real value of $\mathit{x}$, the expression $2\cos \left(\mathit{x}\left(\mathit{x}+1\right)\right)=2^{\mathit{x}}+2^{-\mathit{x}}$ would always be positive.
Lets find the maximum value of $2\cos \left(\mathit{x}\left(\mathit{x}+1\right)\right)=2^{\mathit{x}}+2^{-\mathit{x}}$.
Applying AM-GM inequality we have
$\frac{2^{\mathit{x}}+2^{-\mathit{x}}}{2}\ge \sqrt{2^{\mathit{x}}\times 2^{-\mathit{x}}}$
$\Rightarrow 2^{\mathit{x}}+2^{-\mathit{x}}\ge 2\sqrt{2^0}$
$\Rightarrow 2^{\mathit{x}}+2^{-\mathit{x}}\ge 2$
Therefore, $2\cos \left(\mathit{x}\left(\mathit{x}+1\right)\right)\ge 2$
It is known that $-1\le \cos \mathit{\theta }\le 1$
$\Rightarrow 2\cos \left(\mathit{x}\left(\mathit{x}+1\right)\right)=2$
Hence, the expression is valid only if $2^{\mathit{x}}+2^{-\mathit{x}}=2,$ which is true for only one value of $\mathit{x}$ i.e. $0.$
Therefore, the expression has only one real solution.