Question

The range of values of p for which the equation $\sin {\cos }^{-1}\left(\cos \right({\tan }^{-1}x\left)\right)=p$ has a solution is

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

Answer 1

Answer: (B)

$[0,1)$

Here The range of $cos\left(ta{n}^{-1}\right(x\left)\right)$ will be [-1,1] since the range of $cos\theta$ is always $[-1,1]$
Hence for $cos\left(ta{n}^{-1}\right(x\left)\right)=1$
$sin\left(co{s}^{-1}\right(1\left)\right)$
$=sin\left(0\right)$
$=0$
For $cos\left(ta{n}^{-1}\right(x\left)\right)=0$
$sin\left(co{s}^{-1}\right(0\left)\right)$
$=sin\left(\frac{\pi }{2}\right)$
$=1$
For $cos\left(ta{n}^{-1}\right(x\left)\right)=-1$
$sin\left(co{s}^{-1}\right(-1\left)\right)$
$=sin\left(\pi \right)$
$=0$
Howere p=1 for $x\to \infty$
Hence range of p=[0,1).