If `f:A rarr B` is function.Then A= Domain & B =

If f:A rarr B is a bijection then only f has inverse function II: The inverse function f:R^(+)rarr R^(+) defined by f(x)=x^(2) is f^(-1)(x)=sqrt(x)

Let A={1,2,3,4} and B={1,-2,3,4} If f:A rarr B is an one one function and f(x)!=x for all x in A, then find the no of possible functions,is

Let A=R-{2} and B=R-{1} if f:A rarr B is a function defined by f(x)=(x-1)/(x-2) show that f is one-one and onto. Hence find f^(-1) .

If f : A rarr B is a bijective function, then f^(-1) of is equal to

Letf: A rarr B be a function then show that f is a bijection if and only if there exists a function g:B rarr A such that fog =I_(B)&gof=I_(A)& in this case g=f^(-1)

Prove that the function f:A rarr B is one-one onto,then the inverse function f^(-1):B rarr A is unique.

Prove that f:R rarr R,f(x)=cos x is a many one into function.Change the domain and co- domain of f such that f become:

Let f:A rarr B be a function defined by f(x)sqrt(3)sin x+cos x+4. If f is invertible, then

Let f:A rarr B and g:B rarr C be two functions and gof:A rarr C is define statement(s) is true?