A) a straight line through \[(0,-{{\sin }^{2}}1)\] with slope 2
B) a straight line through (0, 0)
C) a parabola with vertex \[(1,-{{\sin }^{2}}1)\]
D) a straight line through \[\left( \frac{\pi }{2},-{{\sin }^{2}}1 \right)\] and parallel to x-axis
Correct Answer: D
Solution :
We have given that \[f(x)=\cos x\cos (x+2)-{{\cos }^{2}}(x+1)\] \[=\cos (x+1-1)\cos (x+1+1)-{{\cos }^{2}}(x+1)\] \[={{\cos }^{2}}(x+1)-{{\sin }^{2}}1-{{\cos }^{2}}(x+1)\] \[\because \]\[\cos (A+B)\cos (A-B)={{\cos }^{2}}A-{{\sin }^{2}}B\] \[f(x)=-{{\sin }^{2}}1\] i.e., \[y=-{{\sin }^{2}}1\] (a constant quantity) Hence, the graph is a straight line parallel to x-axis and passing through \[\left( \frac{\pi }{2},-{{\sin }^{2}}1 \right).\] Here, \[\frac{\pi }{2}\] can be replaced by any real number.You need to login to perform this action.
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