A) 0
B) 2
C) 1
D) 3
Correct Answer: C
Solution :
\[f(x)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\frac{\pi }{3} \right)\] \[+\cos x\cos \left( x+\frac{\pi }{3} \right)\] \[\Rightarrow \]\[f(x)={{\sin }^{2}}x+{{\left( \sin x\cos \frac{\pi }{3}+\cos x\sin \frac{\pi }{3} \right)}^{2}}\] \[+\cos x\cos \left( x+\frac{\pi }{3} \right)\] \[\Rightarrow \]\[f(x)={{\sin }^{2}}x+{{\left( \sin x.\frac{1}{2}+\cos x.\frac{\sqrt{3}}{2} \right)}^{2}}\] \[+\cos x\left( \cos x\cos \frac{\pi }{3}-\sin x\sin \frac{\pi }{3} \right)\] \[\Rightarrow \]\[f(x)={{\sin }^{2}}x+\frac{{{\sin }^{2}}x}{4}+\frac{3{{\cos }^{2}}x}{4}\] \[+\frac{\sqrt{3}}{2}\sin x\cos +\frac{{{\cos }^{2}}x}{2}-\cos x\sin x.\frac{\sqrt{3}}{2}\] \[={{\sin }^{2}}x+\frac{{{\sin }^{2}}x}{4}+\frac{3{{\cos }^{2}}x}{4}+\frac{{{\cos }^{2}}x}{2}\] \[=\frac{5{{\sin }^{2}}x}{4}+\frac{6{{\cos }^{2}}x+4{{\cos }^{2}}x}{8}\] \[=\frac{5}{4}{{\sin }^{2}}x+\frac{5}{4}{{\cos }^{2}}x=\frac{5}{4}\] \[\therefore \]\[gof(x)=g(f(x))=g(5/4)=1\]
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