Question

Match the statements of Column $I$ with values of Column $II$

	Column I		Column II
A.	If $f\left(x\right)=x+1$, when $x<0$, $f\left(x\right)=x^2-1$,for $x\ge 0,$ then $f\left(f\right(x\left)\right)$, for $-1\le x\le 0$ is	1.	$\frac{x-3}{2}$
B.	If $f\left(\frac{2\tan x}{1+{\tan }^{2}x}\right)=$$\frac{(\cos 2x+1)({\sec }^{2}x+2\tan x)}{2}$ then $f\left(x\right)$ is	2.	$x^2+2x$
C.	If $f(x+y+1)=(\sqrt{f\left(x\right)}+\sqrt{f\left(y\right)}{)}^2$ for all $x,y\in R$ and $f\left(0\right)=1,$ then $f\left(x\right)$ is	3.	$1+x$
D.	If $4<x<5$ and $f\left(x\right)=\left[\frac{x}{4}\right]+2x+2$, where $\left[y\right]$ is the greatest integer $\le y,$ then $f^{-1}\left(x\right)$ is	4.	$(x+1{)}^2$

• A. $A-2.B-4,C-3,D-1$
• B. $A-2.B-3,C-1,D-4$
• C. $A-3.B-2,C-4,D-1$
• D. $A-1.B-3,C-4,D-2$

Answer 1

The correct option is A $A-2.B-4,C-3,D-1$

A) $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}x+1,& x<0\\ x^2-1,& x\ge 0\end{array}\end{array}$

$fof\left(x\right)=\{\begin{array}{c}\begin{array}{cc}2x+1,& x<-1\end{array}\\ \begin{array}{cc}{(x+1)}^2-1,& -1<x<0\end{array}\end{array}=\{\begin{array}{c}\begin{array}{cc}2x+1& x<-1\end{array}\\ \begin{array}{cc}{x}^2+2x,& -1<x<0\end{array}\end{array}$
B) $f\left(\frac{2\tan x}{1+{\tan }^{2}x}\right)=\frac{(\cos 2x+1)({\sec }^{2}x+2\tan x)}{2}$
$=(2{\cos }^{2}x-1+1)({\sec }^{2}x+2\tan x)=\frac{{\sec }^{2}x+2\tan x}{{\sec }^{2}x}=1+\frac{2\tan x}{1+{\tan }^{2}x}$
Therefore,
$f\left(x\right)=1+x$
C) $f(x+y+1)={(\sqrt{f\left(x\right)}+\sqrt{f\left(y\right)})}^2$ ...(1)
$f\left(0\right)=1={(1+0)}^2$ ...(2)
Substituting $x=0=y$ in (1)
$f\left(1\right)={(1+1)}^2=4$ ...(3)
Substituting $x=0,y=1$ in (1)
$f\left(2\right)={(1+2)}^2=9$ ...(4)
And substituting $x=1,y=1$ in (1)
$f\left(3\right)={(2+2)}^2={(1+3)}^2=16$ ...(5)
From (2),(3),(4) and (5), we get
$f\left(x\right)={(1+x)}^2$
D) $4<x<5\Rightarrow 1<\frac{x}{4}<\frac{5}{4}\Rightarrow \left\{\frac{x}{4}\right\}=1f\left(x\right)=1+2x+2=2x+3$
Now let
$y=2x+3\Rightarrow x=\frac{y-3}{2}$
Therefore,
$f^{-1}\left(x\right)=\frac{x-3}{2}$'