Question

The range of $f\left(x\right)=\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{\dots \dots \infty }}}$

• A. $(0,1)$
• B. $[0,\frac{1}{2}]$
• C. $[0,2]$
• D. $[-1,2]$

Answer 1

The correct option is C $[0,2]$

$f\left(x\right)=\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{....\infty }}}$

$f(x{)}^2=(1-\cos x)\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{....\infty }}}$

$f(x{)}^2=(1-\cos x\left)f\right(x)$

$f\left(x\right)\left(f\right(x)-(1-\cos x\left)\right)=0$

$f\left(x\right)=1-\cos x$

$-1\le \cos x\le 1$

$-1\le -\cos x\le 1$

$0\le 1-\cos x\le 2$

$0\le f\left(x\right)\le 2$

Thus, answer is $[0,2]$.