Question

Which of the following correspondences can be called a function?

• A. $f:\{-1,0,1\}\to \{0,1,2,3\}$ defined by $f\left(x\right)=x^3$
• B. $f:\{0,1,4\}\to \{-2,-1,0,1,2\}$ defined by $f\left(x\right)=\pm \sqrt{x}$
• C. $f:\{0,1,4\}\to \{-2,-1,0,1,2\}$ defined by $f\left(x\right)=\sqrt{x}$
• D. $f:\{0,1,4\}\to \{-2,-1,0,1,2\}$ defined by $f\left(x\right)=-\sqrt{x}$

Answer 1

Answer: (C), (D)

$f:\{0,1,4\}\to \{-2,-1,0,1,2\}$ defined by $f\left(x\right)=-\sqrt{x}$

For option $1:$
$f(-1)=-1\notin \{0,1,2,3\}$
Image of an element in domain must belong to co-domain.
So, $f\left(x\right)=x^3$ is not a function.

For option $2:$
$f\left(4\right)=-2,2$
which violates the definition of the function as each element in domain should have unique image in co-domain.
So, $f\left(x\right)=\pm \sqrt{x}$ is not a function.

option $3$ and option $4$ are functions as they satisfy all the conditions of a function.