Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. none of these

Answer 1

Answer: (B)

$0$

$\frac{dg\left(f\right(x\left)\right)}{dx}=g^{\prime }\left(f\right(x\left)\right).f^{\prime }\left(x\right)$

$f\left(x\right)={\cos }^{2}x+{\cos }^2(x+\frac{\pi }{3})+\sin x\sin (x+\frac{\pi }{3})$

$={\cos }^{2}x+(\cos x\cos \frac{\pi }{3}-\sin x\sin \frac{\pi }{3}{)}^2+\sin x(\sin x\cos \frac{\pi }{3}+\cos x\sin dfrac\pi 3)$

$={\cos }^{2}x+(\frac{\cos x}{2}-\frac{\sqrt{3}\sin x}{2}{)}^2+\sin x(\frac{\sin x}{2}+\frac{\sqrt{3}\cos x}{2})$

$={\cos }^{2}x+\frac{{\cos }^{2}x}{4}+\frac{3{\sin }^{2}x}{4}-\frac{\sqrt{3}}{2}\sin x\cos x+\frac{{\sin }^{2}x}{2}+\frac{\sqrt{3}\cos x\sin x}{2}$

$=\frac{5{\cos }^{2}x}{4}+\frac{5{\sin }^{2}x}{4}$

$=\frac{5}{4}=f\left(x\right)$

$=>f^{\prime }\left(x\right)=0$

So,

$\frac{dg\left(f\right(x\left)\right)}{dx}=g^{\prime }\left(f\right(x\left)\right).f^{\prime }\left(x\right)=0$