Question

Which of the functions defined below is/are one-one function(s)?

• A. $f\left(x\right)=x+1$ for $x>-1$
• B. $g\left(x\right)=x+\frac{1}{x}$ for $x>0$
• C. $h\left(x\right)=x^2+4x-5$ for $x>0$
• D. $f\left(x\right)=e^{-x}$ for $x\ge 0$

Answer 1

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

$f\left(x\right)=e^{-x}$ for $x\ge 0$

$f\left(x\right)=x+1$ for $x>-1$


Clearly, $f\left(x\right)$ is one-one function.

$g\left(x\right)=x+\frac{1}{x}$ for $x>0$
$g^{\prime }\left(x\right)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$
$g\left(x\right)$ is not monotonic in $(0,\infty )$
$\therefore g\left(x\right)$ is many-one function.

$h\left(x\right)=x^2+4x-5$ for $x>0$
$h^{\prime }\left(x\right)=2x+4>0$ for $x>0$
$\Rightarrow h\left(x\right)$ is increasing in $(0,\infty )$
So, $h\left(x\right)$ is one-one function.

$f\left(x\right)=e^{-x}$ for $x\ge 0$

Clearly, $f\left(x\right)$ is one-one function.