Q.4 Show that the Modulus Function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where | x | is x, if x is positive or 0 and | x |  is - x,  if x is negative.

Answers (1)

f : R \rightarrow R

f (x) = | x |

f (x) = | x | = x \, if\, x\geq 0 \,\, and \, \, -x\, if\, x< 0

One- one:

For   -1,1 \in R   then f (-1) = | -1 |= 1 

                                         f (1) = | 1 |= 1

                                           -1\neq 1

\therefore  f is not one- one i.e. not injective.

For  -2 \in R,

We know f (x) = | x |   is always positive there is no x in R such that f (x) = | x |=-2

\therefore  f is not onto i.e. not surjective.

Hence,  f (x) = | x |, is neither one-one nor onto.

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