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

Answer 1

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

(A)$f(x)=\ln \left(\tan \pi [x]+\left|x^2+2x-3\right|\right)$
$\because [x]\in I\Rightarrow \tan \pi [x]=0,$
$\text{ and }\left|x^2+2x-3\right|=\left|(x+1{)}^2-2\right|\in [0,\infty )$
$\text{ So, }f(x)\in R\Rightarrow f(x)\text{ is surjective }$
(B)$g(x)=\frac{x^2+2x-3}{x-1},x\ne 1$
$g(x)=\frac{(x-1)(x+3)}{(x-1)},x\ne 1$
$g(x)=x+3\therefore g(x)\ne 4(\because x\ne 1)$
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