Question

Assertion (A): lf $f\left(x\right)={\cos }^{2}x+{\cos }^2(x+\frac{\pi }{3})-\cos x\cos (x+\frac{\pi }{3})$ then $f^{\prime }\left(x\right)=0$

Reason(R): Derivative of constant function is zero

• A. Both A & R are true, R is correct explanation for A
• B. Both A & R are true,R is not correct explanation for A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

The correct option is A Both A & R are true, R is correct explanation for A

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

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

$=(\cos x-(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x){)}^2+\cos x\cos (x+\frac{\pi }{3})$

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

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

$f\left(x\right)=\frac{3}{4}$

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