855 views
1 votes
1 votes

1 Answer

Best answer
1 votes
1 votes

The problem of finding onto and one one can be done as follow :

a) The graph of x2 - 2x + 3 is an upward parabola and hence not one one as there will be innumerable points which will be having same 'y' value for different 'x' value..Precisely about the axis of the parabola which is given by : x = 1 , an x coordinate and its mirror image about x = 1 will have same 'y' coordinate.

Hence the function is not one - one(injective).

b) To find whether it is onto or not , we find the range of the function :

           y   =      x2 - 2x + 3

==>     y   =      x2 - 2x + 1 + 2

                =     (x - 1)2 + 2

Now (x - 1)2  >= 0 always

Thus   ymin  =    0 + 2   =   2

So range of the given function  :  [ 2 , ∞ )

But given co-domain   =  set of integers denoted by Z..

Hence range becomes the proper subset of co domain in this case and hence the function is not onto(surjective) as for surjectivity , the condition is :

Range(function)          =  Co-Domain

Hence D) is the right answer..

selected by

Related questions

8 votes
8 votes
1 answer
1
rahul sharma 5 asked Nov 14, 2017
1,115 views
The number of ways possible to form injective function from set $A$ set $B$ where $|A|=3\text{ and } |B|=5$ such that $p^{th}$ element of set $A$ cannot match with $p^{th...
3 votes
3 votes
0 answers
4
Manu Thakur asked Dec 31, 2017
535 views
Please prove the following:The function $f(x)= e^x$ from the set of real numbers to the set of real numbers is not invertible.But if the $co-domain$ is restricted to the ...