Calculus

$\begin{aligned} f(x) & = \cos x \cos (x+2) - \cos^2 (x+1) \\ & = \cos x (\cos (x+1)\cos 1 - \sin(x+1)\sin 1) - \cos^2 (x+1) \\ & = \cos (x+1)(\cos x \cos 1 - \cos (x+1)) - \sin(x+1)\sin 1 \cos x \\ & = \cos (x+1)(\cos x \cos 1 - \cos x \cos 1 + \sin x \sin 1) - \sin(x+1)\sin 1 \cos x \\ & = \sin 1 (\cos (x+1)\sin x - \sin(x+1) \cos x) \\ & = - \sin^2 1 \end{aligned}$

Therefore, $f(x)$ is a straight line parallel to $x$ -axis.

Expanding out the above function gives:

$f(x) = cos(x)[cos(x)cos(2) - sin(x)sin(2)] - [cos(x)cos(1) - sin(x)sin(1)]^{2};$

or $[cos^{2}(x)cos(2) - cos(x)sin(x)sin(2)] - [cos^{2}(x)cos^{2}(1) - 2cos(x)sin(x)cos(1)sin(1) + sin^{2}(x)sin^{2}(1)];$

or $[cos^{2}(x)(2cos^{2}(1) - 1) - 2cos(x)sin(x)cos(1)sin(1)] - [cos^{2}(x)cos^{2}(1) - 2cos(x)sin(x)cos(1)sin(1) + sin^{2}(x)sin^{2}(1)];$

or $cos^{2}(x)cos^{2}(1) - cos^{2}(x) - sin^{2}(x)sin^{2}(1);$

or $[cos^{2}(1) - 1]cos^{2}(x) - [1 - cos^{2}(x)]sin^{2}(1);$

or $[cos^{2}(1) + sin^{2}(1) - 1]cos^{2}(x) - sin^{2}(1);$

or $(1 - 1)cos^{2}(x) - sin^{2}(1);$

or $\boxed{f(x) = -sin^{2}(1)}.$

Hence, choice B is the correct answer since $f(x)$ is a constant function for all $x \in \mathbb{R}.$