Question

Assertion :Statement 1 Range of $f\left(x\right)={\tan }^{-1}x+{\sin }^{-1}x+{\cos }^{-1}x$ is $(0,\pi )$ Reason: Statement 2 $f\left(x\right)={\tan }^{-1}x+{\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}+{\tan }^{-1}x$ for $xϵ(-1,1]$

• A. Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
• B. Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Answer 1

The correct option is D STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

For $f\left(x\right)={\tan }^{-1}\left(x\right)+{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)$

The domain is $[-1,1]$

In this domain we can re-write the expression as

$f\left(x\right)=\frac{\pi }{2}+{\tan }^{-1}\left(x\right)$

$f(-1)=\frac{\pi }{2}-\frac{\pi }{4}$

$=\frac{\pi }{4}$

$f\left(1\right)=\frac{\pi }{2}+\frac{\pi }{4}$

$=\frac{3\pi }{4}$

Therefore range of $f\left(x\right)$ is $[\frac{\pi }{4},\frac{3\pi }{4}]$