Question

The function f: R to R is GIVEN by f(x)=|x| (modulus) is*

5 points

One to One and onto

Neither one - one nor onto

One - One but not onto

Onto but not One -One

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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

CHECKING FOR ONE TO ONE

$f(x) = |x|$

Now

$f( - 1) = | - 1| = 1$

$f(1) = |1| = 1$

So

$- 1 \ne \: 1 \: \: but \: \: f( - 1) = f(1)$

So f is not one to one

CHECKING FOR ONTO

$f(x) = |x|$

So

$f(x) \geqslant 0 \: \: \: \: \forall \: \: x \in \mathbb{R}$

Here

$- 2 \in \: \mathbb{R}$

But there exists no

$x \in \: \mathbb{R} \: \: such \: that \: f( x) = 2$

So all elements in the codomain set has no pre-image in the domain Set

So f is not onto

RESULT

The function f: R to R is GIVEN by f(x)=|x| is

Neither one - one nor onto

Answer 2

Answer:

b

Step-by-step explanation:

please mark as brianliest answer