Question
The graph of the function $$\cos \,x\,\cos \left( {x + 2} \right) - {\cos ^2}\left( {x + 1} \right)$$ is :
A.
a straight line passing through $$\left( {0,\, - {{\sin }^2}1} \right)$$ with slope $$2$$
B.
a straight line passing through $$\left( {0,\,0} \right)$$
C.
a parabola with vertex $$\left( {0,\, - {{\sin }^2}1} \right)$$
D.
a straight line passing through the point $$\left( {\frac{\pi }{2},\, - {{\sin }^2}1} \right)$$ and parallel to the $$x$$-axis
Answer :
a straight line passing through the point $$\left( {\frac{\pi }{2},\, - {{\sin }^2}1} \right)$$ and parallel to the $$x$$-axis
Solution :
$$\eqalign{
& y = \frac{1}{2}\left[ {\cos \left( {2x + 2} \right) + \cos \,2 - \left\{ {1 + \cos \left( {2x + 2} \right)} \right\}} \right] \cr
& {\text{or }}y = - \frac{1}{2}\left( {1 - \cos \,2} \right) = - {\sin ^2}1{\text{ i}}{\text{.e}}{\text{., constant}} \cr} $$
$$\therefore $$ Graph is a line parallel to $$x$$-axis. Also when $$x = \frac{\pi }{2},\,y = - {\cos ^2}\left( {\frac{\pi }{2} + 1} \right) = - {\sin ^2}1$$ and hence it passes through the point $$\left( {\frac{\pi }{2},\, - {{\sin }^2}1} \right)$$