question_answer

If \[f(x)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\frac{\pi }{3} \right)\] \[+\cos x\cos \left( x+\frac{\pi }{3} \right)\]and \[g\left( \frac{5}{4} \right)=1,\]then g of (x) is equal to

A)  0                                

B)  1

C)  \[\sin {{1}^{o}}\]                           

D)  None of these

Correct Answer: B

Solution :

\[+\cos x\cos \left( x+\frac{\pi }{3} \right)\] \[={{\sin }^{2}}x+{{\left( \sin x\cos \frac{\pi }{3}+\cos x\sin \frac{\pi }{3} \right)}^{2}}\] \[+\cos x\left( \cos x\cos \frac{\pi }{3}-\sin x\sin \frac{\pi }{3} \right)\] \[={{\sin }^{2}}x+\frac{{{\sin }^{2}}x}{4}+\frac{3{{\cos }^{2}}x}{4}+\frac{2\sqrt{3}}{2\cdot 2}\sin x\cos x\] \[+\frac{{{\cos }^{2}}x}{2}-\cos x\sin x\frac{\sqrt{3}}{2}\] \[={{\sin }^{2}}x+\frac{{{\sin }^{2}}x}{4}+\frac{3{{\cos }^{2}}x}{4}+\frac{{{\cos }^{2}}x}{2}\] \[=\frac{5{{\sin }^{2}}x}{4}+\frac{3{{\cos }^{2}}x+2{{\cos }^{2}}x}{4}\] \[=\frac{5}{4}({{\sin }^{2}}x+{{\cos }^{2}}x)=\frac{5}{4}\] \[\therefore \]\[gof(x)=g[f(x)]=g\left( \frac{5}{4} \right)=1\]