If f: R→R is a function given by f(x)= [x] (greatest integer function…

$f: R→R$  $f(x)=[x]$ (Greatest integer function)

$[x]$ = the greatest integer $≤x$

for the function to be one-one

$f_1=f_2⇒x_1=x_2$ but this is not the case

eg: for $x = 2.2,x=2.3,x=2.4$

$f(2.2)=2,f(2.3)=2,f(2.4)=2$

so, all the value of $x$ are different but for these values, value of f(x) is same.

⇒ this clearly says that function is not one-one.

for function be onto, all the values in codomain must have atleast one preimage in domain

codomain is R (set of real numbers)

for numbers like $3.5, 4.7$ (Non integral numbers)

if $f(x) = 3.5$ or $f(x) = 4.7$

This is impossible for any value of $x$ as the range consist of only integer values.

So, function is not ONTO

as function is not onto as well as not one-one it is NOT bijective

so, (A, B, E) → are incorrect

The range of $f$ is $I$ (set of the integers)

$f(2.5)=2$ (as the greatest integer value less than or equal to 2.5 is 2)

So, only C, D are correct.