Let A = { 1, 2}, B = { 3, 4 } and C = { 5, 6 } and F : A → B and g : B → C Such that f ( 1 ) = 3, f ( 2 ) = 4, g ( 3 ) = 5, g ( 4 ) = 6, then go f =
{ (1, 3 ) ( 2, 4 )}
{ (1, 5), ( 2, 6 )}
{ (, 2 ) ( 5, 6 ) }
{ ( 3, 5 ) ( 4, 6 ) }
If f (1 + x) = x2 + 1, then f(2 - h) =
h2 - 2 h + 2
h2 + 2 h + 2
h2 - 2 h + 4
h2 + 2 h + 4
Range of f (x) = is
(0, ∞ )
(0, 5 )
(5, ∞ )
(-5, 5 )
The two functions f, g : R → R are defined by f (x ) = x + 1, g ( x ) = x2
The value of f + g is
x2 + x + 1
x2 - 1
x2 - x + 1
x2 + 1
Let f, g: R → R be defined by f( x ) = 2 x + 1 and g ( x ) = x -1/2, then gof =
x
2 x
x -1
x + 1
Range of function f (x) = is.
( -∞, 1 )
(-∞, 1 ]
(1, ∞ )
(0, ∞ ]
The function f : R → R defined by f ( x ) = x + 1 is
Injective
Bijective
Inverse
Identity
None of these
Range of f (x) = is.
[-1, 1 ]
[0 , ∞ ]
[-1, 0 ]
[0, 1 ]
If f(x) = x + 1/x , then which of the following is correct?
[ f (x) ]3 = f (x)3 + 3 f (x)
[ f (x) ] 3 = f (x)3 - 3 f (1/x)
[ f (x) ]3 = f (x 3 )+ 3 f (1/x)
f (x)3 = (f [x])3 - 3 f (x)
