Question

Match the entries of Column-I with one or more than one entries of column-II. Note that $[x],\{x\}$ and sgn $x$ denote largest integer less than or equal to $x$, fractional part of $x$ and signum function of $x$ respectively.

Column I		Column II
A	Let $f:[-1,1]\to R$ be defined by $f(x)=\sqrt[5]{x}+{\sin }^{-1}x$ then $f(x)$ is	P	Odd
B	Let $f:R\to \{-1,0,1\}$ be defined by $f(x)=\mathrm{sgn}\left(\frac{1-|x|}{1+|x|}\right)$ then $f(x)$ is	Q	Even
C	Let $f:[-4,2]\to [0,3]$ be defined by $f(x)=\sqrt{8-2x-x^2}$ then $f(x)$ is	R	Onto
D	Let $f:(-\infty ,0]\to [0,\infty )$ be defined by $f(x)=\frac{2^{-[x]}}{2^{[x]}}-2^{|x|}$ then $f(x)$ is	S	One-One
		T	Many-One

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

Answer 1

Answer: (D)

(A) We have
$f(x)=\sqrt[5]{x}+{\sin }^{-1}x$
Cleary, domain of $f(x)=[-1,1]$.
Also, $f(x)$ is increasing so $f(x)$ is one-one function.
(B)$f(x)=\mathrm{sgn}\left(\frac{1-|x|}{1+|x|}\right)$
$D_f=R$
$R_f=\{-1,0,1\}\text{ even function }$

(C) For domain of $f(x)$, we must have $8-2x-x^2\ge 0$
$\Rightarrow x^2+2x-8\le 0$
$\Rightarrow (x+4)(x-2)\le 0$
$\Rightarrow x\in [-4,2]$
$R_f=[0,3]$

(D)$f(x)=\frac{2^{-[x]}}{2^{\{x\}}}-2^{|x|}=2^{-x}-2^{|x|}=0\forall x\le 0]$