Question

Column I		Column II
A	$f(x)={\sin }^{-1}\left(\frac{\sin x+\cos x}{2}\right)$	P	domain is $R$
B	$g(x)={\sin }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)$	Q	range contain only one integer
C	$h(x)={\tan }^{-1}\left(\frac{2}{\pi }\left(2{\tan }^{-1}x-{\sin }^{-1}x+{\cot }^{-1}x-{\cos }^{-1}x\right)\right)$	R	odd function
D	$j(x)={\tan }^{-1}\left(x^3+x\right)$	S	no vertical tangent

• A. (A) P, Q, S; (B) P, R, S; (C) Q, R, S; (D) P, R, P
• B. (A) P, Q, S; (B) P, R, S; (C) Q, R, S; (D) P, R, Q
• C. (A) P, Q, S; (B) P, R, S; (C) Q, R, S; (D) P, R, R
• D. (A) P, Q, S; (B) P, R, S; (C) Q, R, S; (D) P, R, S

Answer 1

Answer: (D)

(A) We know ${\sin }^{-1}\left(\frac{\sin x+\cos x}{2}\right)={\sin }^{-1}\left(\frac{\sqrt{2}\sin (x+(\pi /4))}{2}\right)$ lie between $-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}$ for all $x\in R$
$\therefore$ domain is $x\in R$
${\sin }^{-1}x$ is increasing function and max. value of given function is ${\sin }^{-1}\sqrt{\frac{1}{2}}$ which is $\pi /4$ and minimum value ${\sin }^{-1}(-\sqrt{1/2})$ which is $-\pi /4$, so range contain only one integer which is zero
$\Rightarrow (Q)$ is correct
$\therefore f(x)={\sin }^{-1}\left(\sin \left(x+\frac{\pi }{3}\right)\right)\to$ not odd $\Rightarrow$
(R) is not correct
vertical tangent exist for ${\sin }^{-1}x$ is $x=\pm 1$ but this is not possible so (S) is also correct.
(B)$g(x)={\sin }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)$
$-1<\frac{2}{\pi }{\tan }^{-1}x<1$ for all $x\in R.\Rightarrow$
(P) is correct
$\therefore$ range is $(-\pi /2,\pi /2)\Rightarrow (Q)$ is wrong $g(x)$ is odd $\Rightarrow (R)$ is correct and no vertical tangent $\Rightarrow$
(S) is also correct
(C)$h(x)={\tan }^{-1}\left(\frac{2}{\pi }\left(2{\tan }^{-1}x+{\cot }^{-1}x-\left({\sin }^{-1}x+{\cos }^{-1}x\right)\right)\right)$
$={\tan }^{-1}\left[\frac{2}{\pi }\left({\tan }^{-1}x+\frac{\pi }{2}-\frac{\pi }{2}\right)\right]={\tan }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)$
It is easy to analyse domain is $[-1,1]$
$-\frac{1}{2}\le \frac{2}{\pi }{\tan }^{-1}x\le \frac{1}{2}\text{ for }x\in [-1,1]$
$\therefore {\tan }^{-1}\frac{1}{2}\le 1\Rightarrow h(x)$ contain only one integer which is zero $h(x)$ is odd no vertical tangent.
(D) domain is $R$ (cubic polynomial) range is $(-\pi /2,\pi /2)$ $x^3+x$ is odd
$\therefore {\tan }^{-1}\left(x^3+x\right)$ is also odd no vertical tangent