11st of all: 4-x2≥0
i.e,(x+2)(x-2)≤0
so,the value of x varies as: -2≤x≤2
Now,[x]+2>0 {N.B: Not ≥0}
i.e, [x]>-2
i.e, n-1>-2
i.e, n>-1
so,the value of [x] will vary 4m: (-1,0]U[0,1)U[1,2)...........
so the ans is: xε(-1,0)Ï
[0,1)Ï
[1,2)
dis is juss a GUESS!!!
ans match korlo??????
39(EDITED on Manmay's request)
Go by the restrictions imposed by different functions we have...oi shob restriction er intersection nile tumi domain paabe.
R1 : Denominator cannot be zero
[x] + 2 ≠0
=> [x] ≠-2
Now [x] = -2 for -2 ≤ x < -1
So [x] ≠-2 for x < -2 U x ≥ -1
Let me explain. For [x] = 2, we are given that x ≥ -2 AND x < -1. Let (-2, ∞) = A and (-∞, -1] = B. So our expression becomes (A ∩ B) where ∩ is intersection/AND. Now when we have to find for [x] ≠2, we complement that set.
By DeMorgan's law, (A ∩ B)' = A' U B' which is the set I found.
So we have x → (-∞, -2) U [-1, ∞)
R2 : Square root term must be non negative
4 - x²[x] + 2 ≥ 0
We have two subcases here.
CASE I : Nr ≥ 0 AND Dr > 0
So x² ≤ 4 AND [x] > -2
Here you must know about the Greatest Integer inequalities. I posted them here : http://www.targetiit.com/iit-jee-forum/posts/greatest-integer-function-exhaustive-collection-pl-13636.html
[x] > I implies x ≥ I + 1
So [x] > -2 implies x ≥ -1
So |x| ≤ 2 AND x ≥ -1
=> -2 ≤ x ≤ 2 AND x ≥ -1
Therefore x → [-1, 2]
or we have CASE II : Nr ≤ 0 AND Dr < 0
|x| ≥ 2 AND [x] < -2
=> x ≤ -2 U x ≥ 2 AND x < -2 (as [x] < I implies x < I)
=> x → (-∞, -2)
So our final restriction(2nd) is x→ (-∞, -2) U [-1, 2] (UNION of both cases)
Now our domain of function is obtained by :
Df = R1 ∩ R2
= x → (-∞, -2) U [-1, ∞) ∩ (-∞, -2) U [-1, 2]
So our final domain is x → (-∞, -2) U [-1, 2]
39You're right Manmay!! I misread my own posted inequality..LOL
[x] > I implies x>= I+1 :P
1CASE : I
4 - x2 ≤ 0 → x E ( -∞ , -2] U [2 ,∞)
and [x] + 2 < 0 → x E ( -∞, -2)
hence in all
x E ( - ∞ , -2)
CASE II
4 - x2 ≥ 0 → x E [ -2 , 2 ]
and
[x] + 2 > 0 → x E [-1 , ∞ )
hence in all
x E [-1 , 2]
combining CASE I and CASE II
we get
x E ( -∞ , -2 ) U [ -1 , 2 ]
