Let `f(x)=(kx)/(x+1)` then the value of k such that f(x) is

let f(x)=[tan((pi)/(4)+x)]^((1)/(x)),x!=0 and f(x)=k,x=0 then the value of k such that f(x) hold continuity at x=0

Let f(x)=3x^(2)-7x+c,x>(7)/(6) ,The value of c such that f(x) touches f^(-1)(x) is (5k+1)/(k) where k=

Let f(x)=(ax)/(x+1), x ne1 , then the value of 'a' for which f[f(x)]=x is

Let f (x) =(2 |x| -1)/(x-3) Range of the values of 'k' for which f (x) = k has exactly two distinct solutions:

Let f (x) =(2 |x| -1)/(x-3) Range of the values of 'k' for which f (x) = k has exactly two distinct solutions:

Let f (x) =(2 |x| -1)/(x-3) Range of the values of 'k' for which f (x) = k has exactly two distinct solutions:

If f(x)={(kx+2,,x 5) is continuous, then value of k is

If f(x) = (kx)/(x+1)(x ne -1) such that (fof(x))=x , then find the value of k. (when x ne 0 ).

let f(x)=x^(2)-3x+4 the values of x which satisfies f(1)+f(x)=f(1)*f(x) is