If the function `f(x) = ax + b` has its own inverse then the ordered pair `(a, b)` can be

If the equation x^(3)+ax^(2)+bx-4=0 has two roots equal to 2, then the ordered pair (a,b) is

The function f(x) = ax + b is strictly increasing for all real x if

If f (x)= [{:(e ^(x-1),,, 0 le x le 1),( x+1-{x},', 1 lt x lt 3 ):}and g (x) =x ^(2) -ax +b such that f (x)g (x) is continous [0,3] then the ordered pair (a,b) is (where {.} denotes fractional part function):

If f (x)= [{:(e ^(x-1),,, 0 le x le 1),( x+1-{x},', 1 lt x lt 3 ):}and g (x) =x ^(2) -ax +b such that f (x)g (x) is continous [0,3] then the ordered pair (a,b) is (where {.} denotes fractional part function):

Prove that the function f(X) = ax + b is increasing iff a > 0.

The function f(x)=ax+b is strictly increasing for all real x is

If lim_(x to oo) (sqrt(x^(2) -x+1) -ax)=b , then the ordered pair (a, b) is :

The function f(x) =x^3+ax^2+bx+c,a^2lt=3b has

If f:R rarr R,f(x)=x^(2)+x-2&g:R rarr R,g(x)=ax+b and if fog= gof has infinite solutions, then number of ordered pairs (a,b), is -