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