|x + 2/x | < 3, then x belongs to
(-2,-1) U ( 1,2 )
(-∞, -2 ) U (-1, -1) U ( 2, ∞ )
(-2, 2 )
(-3, 3 )
If x2 > 4, then
x > 4
| x | > 2
-4 < x < 4
None of these
As sinx < x and x < tanx in ( 0,π/2 ), so in the same interval
Sinx < tanx
Sinx > tanx
sin2 x > tan 2 x
|sinx | > |tanx |
If |x| > 5, then
0 < x < 5
x < -5 or x > 5
-5 < x < 5
x > 5
If 3 < | x | < 6, then x belongs to
( - 6, -3 ) U ( 3, 6 )
( - 6, 6 )
( -3, -3 ) U (3, 6 )
If | x | < x , then:
x is a positive real number
x is a non negative real number
There is no x satisfying this inequality
x is a negative real number
|2x -3| < | x + 5 |, then x belongs to :
( -3, 5 )
( 5, 9 )
( -2/3, 8 )
( -8, 2/3 )
( x2 + 1 ) (x - 1 ) ( x -2 ) < 0 , then
x < 1 or x > 2
x ∈ ( 1, 2 )
-1 < x
The set of values of x satisfying the inequalities ( x -1 ) ( x -2 ) < 0 and ( 3x - 7 ) ( 2x - 3 ) > 0 is
(1, 2 )
( 2, 7/3 )
( 1, 7/3 )
( 1, 3/2 )
Consider the following system of inequalities 5x + 3y ≥ 0 and y -2x < 2. The solution of the above inequalities does not contain only part of the
First quadrant
Second quadrant
Third quadrant
Fourth quadrant