Question

If $k\le {\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x\le K$, then

• A. $k=0,K=\mathrm{\pi }$
• B. $k=0,K=\frac{\mathrm{\pi }}{2}$
• C. $k=\frac{\mathrm{\pi }}{4},K=\frac{3\mathrm{\pi }}{4}$
• D. None of these

Answer 1

The correct option is C

$k=\frac{\mathrm{\pi }}{4},K=\frac{3\mathrm{\pi }}{4}$

Explanation to the correct option:

Range of function:

${\sin }^{-1}x\mathrm{and}{\cos }^{-1}x$ are defined only if $-1\le x\le 1$, however ${\tan }^{-1}x$ is defined for $x\in R$.

Now,

${\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x=\frac{\mathrm{\pi }}{2}+{\tan }^{-1}x\left[by{\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right]$

As $-1\le x\le 1$, so for $x=-1$, ${\tan }^{-1}x$ will be $-\frac{\mathrm{\pi }}{4}$and for for $x=1$, ${\tan }^{-1}x$ will be $\frac{\mathrm{\pi }}{4}$.

So, the range of ${\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x$ will be $\left[\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}\right),\left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{4}\right)\right]$, that is $\left(\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right)$.

Therefore, $k=\frac{\mathrm{\pi }}{4},K=\frac{3\mathrm{\pi }}{4}$.

Hence, option C is correct.