Question

Which of the following function(s) is(are) surjective?
[Note: $[ m ]$ and $\{ m \}$ denotes greatest integer function less than or equal to $m$ and fraction part function of $m$ respectively, and $D _l$ denotes the domain of the function $y =l( x )$. ]

• A. $f : D _{ f } \rightarrow R , f ( x )=\ln \left(\tan (\pi[ x ])+\left| x ^2+2 x -3\right|\right)$
• B. $g: D_g \rightarrow R, g(x)=\frac{x^2+2 x-3}{x-1}$
• C. $h : D _{ h } \rightarrow R , h ( x )=\ln \left(\frac{1- x }{1+ x }\right)$
• D. $k : D _{ k } \rightarrow R ^{+}, k ( x )=\sqrt{[ x ]+[- x ]+1}+\sqrt{\{ x \}+\{- x \}+1}$

Answer 1

Answer: (A), (C), (D)

(A)$f ( x )=\ln \left(\tan \pi[ x ]+\left| x ^2+2 x -3\right|\right) $
$\because[ x ] \in I \Rightarrow \tan \pi[ x ]=0, $
$\text { and }\left| x ^2+2 x -3\right|=\left|( x +1)^2-2\right| \in[0, \infty) $
$\text { So, } f ( x ) \in R \Rightarrow f ( x ) \text { is surjective }$
(B)$g ( x )=\frac{ x ^2+2 x -3}{ x -1}, x \neq 1 $
$g ( x )=\frac{( x -1)( x +3)}{( x -1)}, x \neq 1$
$g ( x )= x +3 \therefore g ( x ) \neq 4 (\because x \neq 1)$
So, range of $g ( x )$ is $R -\{4\}$
$\Rightarrow g ( x )$ is not surjective
(C)$h ( x )=\ln \left(\frac{1- x }{1+ x }\right), \frac{1- x }{1+ x }>0$
$\Rightarrow D _{ h }=(-1,1) $
$\because \frac{1- x }{1+ x } \text { take all value between }(0, \infty)$
$\because \frac{1- x }{1+ x }$ take all value between $(0, \infty)$
So, Range of $h(x)=R$
$\Rightarrow h ( x )$ is surjective
(D) $k ( x ) \sqrt{[ x ]+[- x ]+1}+\sqrt{\{ x \}+\{- x \}+1}$
Domin of $k(x)$ is $R$
$x \notin I \Rightarrow[ x ]+[- x ]=-1 \text { and }\{ x \}+\{- x \}=1$
$\Rightarrow k ( x )=\sqrt{2} $
$x \in I \Rightarrow[ x ]+[- x ]=0 \text { and }\{ x \}+\{- x \}=0$
$\Rightarrow k ( x )=2$
So, Range of $k(x)=\{\sqrt{2}, 2\}$
So, $k ( x )$ is not surjective