If x2 > 4, then
x > 4
| x | > 2
-4 < x < 4
None of these
(x2 + 1 ) ( x -2 ) 2 ( x -3 ) < 0, then x belongs to
( -∞, 2) U (2, 3 )
( -∞, 3)
(2, 3 )
( x - 1 ) > 0
( x -2 ) > 0
(x - 2 ) < 0
( x - 1 ) > 0 if ( x -2 ) > 0
If a > b then
a + 5 > b + 5
a - b < b - 5
a + b < b + b
Depends on a and b
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 |
x2 -3 |x| + 2 < 0, then x belongs to
( 1,2 )
( -2, -1 )
( -2, -1 ) U ( 1, 2 )
( -3, 5 )
If x satisfies the inequations 2x - 7 < 11, 3x + 4 < -5, then x lies in the interval
(-∞, -3)
(-∞, 3)
(-∞, 2)
(-∞, ∞)
If 3 < | x | < 6, then x belongs to
( - 6, -3 ) U ( 3, 6 )
( - 6, 6 )
( -3, -3 ) U (3, 6 )
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
If a/b < c/d, then
( a/b)2 < (c/d)2