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 the point $$\left( {0,\, - {{\sin }^2}1} \right)$$ with slope 2
B.
straight line passing through the origin
C.
parabola with vertex $$\left( {1,\, - {{\sin }^2}1} \right)$$
D.
straight line passing through the point $$\left( {\frac{\pi }{2},{\mkern 1mu} - {{\sin }^2}1} \right)$$ and parallel to the $$x$$-axis
Answer :
straight line passing through the point $$\left( {\frac{\pi }{2},{\mkern 1mu} - {{\sin }^2}1} \right)$$ and parallel to the $$x$$-axis
Solution :
The equation of the graph is $$y = \cos \,x.\cos \left( {x + 2} \right) - {\cos ^2}\left( {x + 1} \right)$$
$$\eqalign{
& {\text{or, }}y = \frac{1}{2}\left\{ {\cos \,2 + \cos \,2\left( {x + 1} \right)} \right\} - \frac{1}{2}\left\{ {1 + \cos \,2\left( {x + 1} \right)} \right\} \cr
& = \frac{1}{2}\left( {\cos \,2 - 1} \right) \cr
& = - \frac{1}{2}.2{\sin ^2}1 \cr
& = - {\sin ^2}1 \cr} $$
The graph is parallel to the $$x$$-axis, and $$\left( {\frac{\pi }{2},{\mkern 1mu} - {{\sin }^2}1} \right)$$ satisfies it.