Question

Consider $f(x)={\tan }^{-1}\left(\frac{|x|}{\sqrt{1-x^2}}\right)+{\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$

List-I		List-II
P	The number of integer(s) in the range of $f(x)$ is (are)	1	1
Q	The number of solutions(s) of the equation $f(x)=2$ is (are)	2	2
R	The value of $f(\tan 1)$ is	3	3
S	The value of $[f(\sin 1)]$ is	4	Not defined

[Note: $[k]$ denotes greatest integer less than or equal to $k$.]

• A. P-2, Q-3, R-2, S-4
• B. P-3, Q-2, R-4, S-1
• C. P-1, Q-1, R-1, S-4
• D. P-2, Q-3, R-1, S-1

Answer 1

Answer: (B)

domain $|x|<1$
$|x|=\sin \theta \text{ in }{T}_1\theta \in \left[0,\frac{\pi }{2}\right)$
$\therefore T_1={\tan }^{-1}\tan \theta ={\sin }^{-1}|x|$
$\text{ put }|x|=\tan \theta \text{ in }{T}_2\theta \in \left[0,\frac{\pi }{2}\right)$
$T_2={\cos }^{-1}\cos \theta ={\tan }^{-1}|x|$
$f(x)={\sin }^{-1}|x|+{\tan }^{-1}|x|$
(P) $\because$ both are increasing function in $|x|\in [0,1)$
$\therefore y\in \left[0,\frac{3\pi }{4}\right)=\{0,1,2\}\to \text{ Range }$
(Q) when $y=2$ then $|x|=A\Rightarrow x=\pm A$
(R) $\because \tan 1>\tan \frac{\pi }{4}\Rightarrow \tan 1>1$
$\therefore f(\tan 1)\text{ is ND }$
$\text{ (S) }[f(\sin 1)]=\left[{\sin }^{-1}|\sin 1|+{\tan }^{-1}|\sin 1|\right]=\left[{\sin }^{-1}\sin 1+{\tan }^{-1}\sin 1\right]$