1. Tardigrade
  2. Question
  3. Mathematics
  4. Column I Column II A f(x)= sin 2 2 x-2 sin 2 x P Range contains no natural number B f(x)=(4/π)( sin -1( sin π x)) Q Range contains atleast one integer C f ( x )=√ ln ( cos ( sin x )) R Many one but not even function D f(x)= tan -1((x2+1/x2+√3)) S Both many one and even function T Periodic but not odd function

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q.
Column I Column II
A $f(x)=\sin ^2 2 x-2 \sin ^2 x$ 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

Relations and Functions - Part 2

Solution:

We have $f(x)=\sin ^2 2 x-2 \sin ^2 x, x \in R$
$=4 \sin ^2 x \cos ^2 x-2 \sin ^2 x=4 \sin ^2 x\left(1-\sin ^2 x\right)-2 \sin ^2 x=2 \sin ^2 x-4 \sin ^4 x$
$=-4\left[\sin ^4 x-\frac{1}{2} \sin ^2 x\right]=-4\left[\left(\sin ^2 x-\frac{1}{4}\right)^2-\frac{1}{16}\right]$
$\therefore f ( x )=\frac{1}{4}-4\left(\sin ^2 x -\frac{1}{4}\right)^2$
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)) \geq 0 ; (\cos (\sin x)) \geq 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 yx ^2+\sqrt{3} y = x ^2+1 \Rightarrow (\sqrt{3} y -1)=(1- y ) x ^2 $
$\therefore x ^2=\left(\frac{\sqrt{3} y -1}{1- y }\right) \geq 0$
So, $ \frac{\sqrt{3} y-1}{y-1} \leq 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.