Log 2 x > 4, then x belongs to
x > 4
x > 16
x > 8
None of these
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
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 )
( x2 + 1 ) (x - 1 ) ( x -2 ) < 0 , then
x < 1 or x > 2
x ∈ ( 1, 2 )
-1 < x
If x2 > 4, then
| x | > 2
-4 < x < 4
If 3 < | x | < 6, then x belongs to
( - 6, -3 ) U ( 3, 6 )
( - 6, 6 )
( -3, -3 ) U (3, 6 )
If |x| > 5, then
0 < x < 5
x < -5 or x > 5
-5 < x < 5
x > 5
If 1/a < 1/b; then
| a | > | b |
a < b
a > b
If a > b then
a + 5 > b + 5
a - b < b - 5
a + b < b + b
Depends on a and b
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
