Question

If $f(x) = \sin^2 x + \sin^2 \left(x + \frac{\pi}{3} \right) + \cos x \cdot cos \left( x + \frac{\pi}{3}\right) $ and $g \left(\frac{5}{4} \right) = 1 $ , then $(gof) (x) = $

• A. 1
• B. 0
• C. $\sin x$
• D. $none\, of\, these$

Answer 1

Answer: (A)

$f\left(x\right)= \sin^{2} x + \sin ^{2}\left(x+ \frac{\pi}{3}\right)$
$ + \cos x \cos \left(x + \frac{\pi}{3}\right)$ and $g\left(\frac{5}{4}\right) = 1$
$ f\left(x\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{x}{3} - \sin x \sin \frac{\pi}{3} \right) $
$= \sin^{2} x \left(\frac{ \sin x }{2} + \frac{\sqrt{3}}{2} \cos x \right)^{2}$
$ + \cos x \left( \frac{\cos x }{2} - \frac{\sqrt{3}}{2}\sin x\right)$
$ = \sin^{2} x + \frac{\sin^{2} x }{4} + \frac{3}{4} \cos^{2} x $
$+ \frac{\sqrt{3}}{2} \sin x \cos x \frac{ \cos ^{2} x }{2} \Rightarrow =\frac{\sqrt{3}}{2}sin\,x\,cos\,x$
$=\frac{5}{4} sin^{2}\,x+\frac{5}{4}cos^{2}\,x f\left(x\right) = \frac{5}{4}$
$\therefore \, \, \left(gof\right)\left(x\right) =g\left(f\left(x\right)\right) = g\left(f\left(x\right)\right)= g\left(\frac{5}{4}\right)=1$.