If `f: R->R` be defined as follows: `f(x)={1,\ if\ x in Q-1,\ if\ x !in Q` Find: range of `fdot`

Let f: R->R be defined as f(x)=x^2+1 . Find: f^(-1)(10)

Let f: R->R be defined as f(x)=x^2+1 . Find: f^(-1)(10)

Let A={x in R :0lt=xlt=1}dot If f: A->A is defined by f(x)={(x ,, if x in Q) , (1-x ,, if x!in Q):} then prove that fof(x)=x for all x in Adot

A function f:R rarr R is defined by f(x)={1,quad if x in Q-1, if x in(R-Q) the value of f(pi)-f((22)/(7)) is

Two functions f:R to R and g:Rto R are defined as follows: f(x)={{:(,0,x in Q),(,1,x ne Q):},g(x)={{:(,-1,x in Q),(,0,x in Q):} Then, fof (e)+fog (pi)

Two functions f:R to R and g:Rto R are defined as follows: f(x)={{:(,0,x in Q),(,1,x ne Q):},g(x)={{:(,-1,x in Q),(,0,x in Q):} Then, fof (e)+fog (pi)

A function f: R to R is defined as: f(x) ={{:(1,x in Q),(-1,x notinQ):} Evaluate each of the following: (i) f(2),f(pi)" "(ii) Range of f (iii) f^(-1)(-1)" "(iv) f^(-1)(1)

A function f: R to R is defined as: f(x) ={{:(1,x in Q),(-1,x notinQ):} Evaluate each of the following: (i) f(2),f(pi)" "(ii) Range of f (iii) f^(-1)(-1)" "(iv) f^(-1)(1)

If f:R->R be defined by f(x)=x^2+1 , then find f^(-1)(17) and f^(-1)(-3) .