Question

Range of the function $f\left(x\right)=4{\tan }^{-1}x+3{\sin }^{-1}x+{\sec }^{-1}x$ is

• A. $\{\frac{-3\pi }{2},\frac{3\pi }{2}\}$
• B. $\{\frac{-5\pi }{2},\frac{5\pi }{2}\}$
• C. $\{\frac{-5\pi }{2},\frac{3\pi }{2}\}$
• D. $\{\frac{-3\pi }{2},\frac{5\pi }{2}\}$

Answer 1

The correct option is D $\{\frac{-3\pi }{2},\frac{5\pi }{2}\}$

$f\left(x\right)=4{\tan }^{-1}x+3{\sin }^{-1}x+{\sec }^{-1}x$
Domain of ${\tan }^{-1}x$ is $R\dots \left(1\right)$
Domain of ${\sin }^{-1}x$ is $[-1,1]\dots \left(2\right)$
Domain of ${\sec }^{-1}x$ is $(-\infty ,-1]\cup [1,\infty )\dots \left(3\right)$

Hence, domain of $f\left(x\right)$ is $\left(1\right)\cap \left(2\right)\cap \left(3\right)$
$\Rightarrow x=\{-1,1\}$

Now, $f(-1)=4{\tan }^{-1}(-1)+3{\sin }^{-1}(-1)+{\sec }^{-1}(-1)$
$=4(-\frac{\pi }{4})+3(-\frac{\pi }{2})+\pi =-\frac{3\pi }{2}$
$f\left(1\right)=4{\tan }^{-1}\left(1\right)+3{\sin }^{-1}\left(1\right)+{\sec }^{-1}\left(1\right)$
$=4\left(\frac{\pi }{4}\right)+3\left(\frac{\pi }{2}\right)+0=\frac{5\pi }{2}$
Hence, range of $f\left(x\right)$ is $\{\frac{-3\pi }{2},\frac{5\pi }{2}\}.$