Question

Find the range of $f\left(x\right)=|3{\tan }^{-1}x-{\cos }^{-1}(0\left)\right|-{\cos }^{-1}(-1).$

• A. $[-\pi ,\pi )$
• B. $[-\frac{\pi }{2},\frac{\pi }{2}]$
• C. $[\frac{-3\pi }{2},\frac{3\pi }{2}]$
• D. none of these

Answer 1

The correct option is A $[-\pi ,\pi )$

Range of ${\tan }^{-1}x$ is $-\frac{\pi }{2}<{\tan }^{-1}x<\frac{\pi }{2}$
$\Rightarrow -\frac{3\pi }{2}<3{\tan }^{-1}x<\frac{3\pi }{2}$
$\Rightarrow -2\pi <3{\tan }^{-1}x-{\cos }^{-1}0<\pi$
$\Rightarrow 0\le |3{\tan }^{-1}x-{\cos }^{-1}0|<2\pi$
$\Rightarrow -\pi \le |3{\tan }^{-1}x-{\cos }^{-1}0|-{\cos }^{-1}(-1)<\pi$
$\therefore$ range of $f\left(x\right)$ is $[-\pi ,\pi )$