find domain and range----basic level

the ques is y = √f(x)

take cases
(1)x ≤ -1

(2) -1 > x ≥1

(3) x> 1

for cases (1) and (3) , the term x¹² makes f(x) positive

for case (2) the term 1 ensures f(x) > 0

hence x E R

Q1. Edit : Positivity of a polynomial expression depends on the highest degree coefficient and the discriminant of the polynomial. If you say f(x) > 0 always, this means the discriminant is negative...how did you get that?

Q2.
R1 : x ≠2{x}
=> x ≠2x - 2[x]
=> x ≠2[x]
=> x2 ≠[x]
I think this is true for all x other than zero..

R2 : x - 1x - 2{x} ≥ 0

CASE I : Both N^r and D^r are positive.
x > 1 U x2 > [x]

Now x ≤ [x] < x + 1
=> 2x ≤ 2[x] < 2x + 2
So x > 2[x] is not possible for x ≤ [x] when x > 0. Thus x < 0.
x > 1 U x < 0

CASE II :
x < 1 U x < 2[x]
Applying same logic here, x > 0 U x < 1.

Apply intersection to the three cases. Not sure if I'm right here...seems a tricky question.

Range -: