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)=(alpha x)/(x+1) Then the value of alpha for which f(f(x)=x is

Let f(x)={x^(2),x 0} ,then for what value of (k,p),f(x) has a minimum

Let f(x)=x+1 and phi(x)=x-2. Then the value of x satisfying |f(x)+phi(x)|=|f(x)|+|phi(x)| are :

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)=(alpha x)/(x+1),x!=-1. Then write the value of alpha satisfing f(f(x))=x for all x!=-1

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

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

"Let "f(x)=x + (1)/(2x + (1)/(2x + (1)/(2x + .....oo))). Then the value of f(50)cdot f'(50) is -