Question

Column I		Column II
A	$f(x)={\sin }^{2}2x-2{\sin }^{2}x$	P	Range contains no natural number
B	$f(x)=\frac{4}{\pi }\left({\sin }^{-1}(\sin \pi x)\right)$	Q	Range contains atleast one integer
C	$f(x)=\sqrt{\ln (\cos (\sin x))}$	R	Many one but not even function
D	$f(x)={\tan }^{-1}\left(\frac{x^2+1}{x^2+\sqrt{3}}\right)$	S	Both many one and even function
		T	Periodic but not odd function

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

Answer 1

Answer: (A)

We have $f(x)={\sin }^{2}2x-2{\sin }^{2}x,x\in R$
$=4{\sin }^{2}x{\cos }^{2}x-2{\sin }^{2}x=4{\sin }^{2}x\left(1-{\sin }^{2}x\right)-2{\sin }^{2}x=2{\sin }^{2}x-4{\sin }^{4}x$
$=-4\left[{\sin }^{4}x-\frac{1}{2}{\sin }^{2}x\right]=-4\left[{\left({\sin }^{2}x-\frac{1}{4}\right)}^2-\frac{1}{16}\right]$
$\therefore f(x)=\frac{1}{4}-4{\left({\sin }^{2}x-\frac{1}{4}\right)}^2$
Clearly range of $f^{\prime }=\left[-2,\frac{1}{4}\right]$
Also $f(-x)=f(x)$
So fis even function
Also fis periodic with period $\pi$.
(B) We have $f(x)=\frac{4}{\pi }{\sin }^{-1}(\sin \pi x),x\in R$
Clearly ' $f$ ' is periodic function with period ' 2 ' and also $f$ is an odd function.
Clearly $R_f=[-2,2]$
(C) We have, $f(x)=\sqrt{\ln (\cos (\sin x))}$
For domain,
$\ln (\cos (\sin x))\ge 0;(\cos (\sin x))\ge 1\Rightarrow \cos (\sin x)=1;$
$\therefore \sin x=0\Rightarrow x=n\pi ,n\in I$
$R_f=\{0\}$
$\text{ since }f(x)=0$
$\therefore f$ is both odd and even.
Also, $f$ is periodic with period $\pi$.
(D) We have
$f(x)={\tan }^{-1}\left(\frac{x^2+1}{x^2+\sqrt{3}}\right),x\in R$
Let $y=\frac{x^2+1}{x^2+\sqrt{3}}$
$\Rightarrow y{x}^2+\sqrt{3}y=x^2+1\Rightarrow (\sqrt{3}y-1)=(1-y)x^2$
$\therefore x^2=\left(\frac{\sqrt{3}y-1}{1-y}\right)\ge 0$
So, $\frac{\sqrt{3}y-1}{y-1}\le 0$
$\Rightarrow y\in \left[\frac{1}{\sqrt{3}},1\right)$
$\therefore$ Range of $f(x)=\left[\frac{\pi }{6},\frac{\pi }{4}\right)$
Also $f(-x)=f(x)\forall x\in R$
$\therefore$ fis even function.