Question

Which of the following functions are bijective?

• A. $f:R\to [0,\infty )$ defined as $f\left(x\right)=e^{\text{sgn}\left(x\right)}$
• B. $f:[3,4]\to [4,6]$ defined by $f\left(x\right)=|x-1|+|x-2|+|x+3|+|x-4|$
• C. $f:R\to R$ defined by $f\left(x\right)=\left[x\right]+\cos \left(\pi \right[x\left]\right)$ where $[.]$ denotes the greatest integer function
• D. $f:R\to R$ defined as $f\left(x\right)=min\{x+2,-2x+4\}$

Answer 1

The correct option is B $f:[3,4]\to [4,6]$ defined by $f\left(x\right)=|x-1|+|x-2|+|x+3|+|x-4|$

$\text{sgn}\left(x\right)=\{\begin{array}{cc}\frac{|x|}{x},& x\ne 0\\ 0,& x=0\end{array}$

$\text{sgn}\left(x\right)=\{\begin{array}{c}-1,x<0\\ 0,x=0\\ 1,x>0\end{array}$

$\therefore e^{\text{sgn}\left(x\right)}=\{\begin{array}{cc}{e}^{-1}=\frac{1}{e},& x<0\\ e^0=1,& x=0\\ e^1=e,& x>0\end{array}$

Neither one-one nor onto.
Hence, $e^{\text{sgn}\left(x\right)}$ is not bijective.


$f:[3,4]\to [4,6]$
$f\left(x\right)=|x-1|+|x+2|+|x-3|+|x-4|$
For $3\le x\le 4$
$f\left(x\right)=x-1+x-2+x-3+4-x=2x-2$
$f$ is one-one.

$3\le x\le 4\Rightarrow 6\le 2x\le 8\Rightarrow 4\le 2x-2\le 6$ $f$ is onto.
It is bijective.


$f:R\to R$
$f\left(x\right)=\left[x\right]+\cos \pi \left[x\right]$
If $0\le x<1$,
$f\left(x\right)=0+\cos \pi \left(0\right)=1$
Not one-one, so, not bijective.


$f:R\to R$
$f\left(x\right)=min\{x+2,-2x+4\}$


Not one-one
Not onto.
So, $f$ is not bijective.