Question

Assertion (A) : The least period of the function, f(x) $=\cos \left(\cos x\right)$ $+\cos \left(\sin x\right)+\sin 4x$ is $\pi$
Reason (R) : since $f(x+$ $\pi$$)=f(x)$

• A. Both A and R are individually true and
  R is the correct explanation of A
• B. Both A and R are individually true but
  R is not the correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

The correct option is D A is false but R is true

$f\left(x\right)=\cos \left(\cos x\right)+\cos \left(\sin x\right)+\sin 4x$

$f(x+\pi )=\cos \left(\cos \right(x+\pi \left)\right)+\cos \left(\sin \right(x+\pi \left)\right)+\sin (4x+4\pi )$

$=\cos (-\cos x)+\cos (-\sin x)+\sin 4x=\cos \left(\cos x\right)+\cos \left(\sin x\right)+\sin 4x$ (Using $\cos (-\theta )=\cos \left(\theta \right))$

$=f\left(x\right)$

Hence $\pi$ is a period of $f\left(x\right)$.

Now check whether $\frac{\pi }{2}$ is a period of $f\left(x\right)$.

$f(x+\frac{\pi }{2})=\cos \left(\cos \right(x+\frac{\pi }{2}\left)\right)+\cos \left(\sin \right(x+\frac{\pi }{2})+\sin (4x+\frac{4\pi }{2}\left)\right)$

$=\cos (-\sin x)+\cos \left(\cos x\right)+\sin 4x=\cos \left(\sin x\right)+\cos \left(\cos x\right)+\sin 4x=f\left(x\right)$

Hence $\frac{\pi }{2}$ is a period of $f\left(x\right)$. So $\pi$ can't be the least period of $f\left(x\right)$.