Question

If $f:R\to R$ is defined by $f\left(x\right)=x+\sqrt{x^2},$ then $f$ is

• A. an injective function
• B. an onto function
• C. a bijecive function
• D. neither a one one nor an onto function

Answer 1

The correct option is D neither a one one nor an onto function

$f\left(x\right)=x+\sqrt{x^2}$
$f\left(x\right)=x+|x|$
When $x$ is negative
$f(-1)=-1+|-1|=0f(-2)=-2+|-2|=0f(-3)=-3+|-3|=0$
$\therefore f\left(x\right)$ is a many-one function.

Also $f\left(x\right)=x+|x|$
We have seen that when $x$ is negative, $f\left(x\right)$ is always $0.$
Now when $x$ is positive,
$f\left(x\right)=2x$
Hence the range would be $[0,\infty )$
$f\left(x\right)$ is not an onto function since its range is not equal to its co-domain.
$\therefore f\left(x\right)$ is neither one-one nor onto function.