The two functions f : R → R. g : R → R are defined by f (x ) = x2 + 1, g ( x ) = x -1, then gof =
x2 - 2
x2 - 2 x + 2
x2
2 x
Domain of f (x) = is.
R
R - { -4 }
R - {4}
R - {2}
Range of f (x) = is.
[-1, 1 ]
[0 , ∞ ]
[-1, 0 ]
[0, 1 ]
Solve f (x ) = 2x2 -12 x f 50 ≤ 0
x = 12
x = 5, 2
x = 25, 2
Has no solution
Range of function f (x) = is.
( -∞, 1 )
(-∞, 1 ]
(1, ∞ )
(0, ∞ ]
The function f(x) = is.
Even
Odd
Neither even nor odd
None of these
When f (-1) = 3, the polynomial function f (x) of the second degree is
a + b = 1
a + b + c = -3
a - b + c = 3
a - b - c = 3
The two functions f: R → R, g : R → R are defined by f ( x ) = x2 + 1, g ( x ) = x -1. then fog =
x2 + 2 x + 2
x2 + 2 x
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
The function f : R → R defined by f ( x ) = x + 1 is
Injective
Bijective
Inverse
Identity
