Let `f(x)=x^(2)-x+1, AA x ge (1)/(2)`, then the solution of the equation `f(x)=f^(-1)(x)` is

Let f(x)=2x+1. AA x , then the solution of the equation f(x)=f^(-1)(x) is

Let f(x)-2x-x^(2),x<=1 then the roots of the equation f(x)=f^(-1)(x) is

Let F:[3,infty]to[1,infty] be defined by f(x)=pi^(x(x-3) , if f^(-1)(x) is inverse of f(x) then the number of solution of the equation f(x)=f^(-1)(x) are

If f(x)={0,x<1 and 2x-2,x<=1 then the number of solutions of the equation f(f(f(x)))=x is

Let f_(1) (x) and f_(2) (x) be twice differentiable functions where F(x)= f_(1) (x) + f_(2) (x) and G(x) = f_(1)(x) - f_(2)(x), AA x in R, f_(1) (0) = 2 and f_(2) (0) = 1. "If" f'_(1)(x) = f_(2) (x) and f'_(2) (x) = f_(1) (x) , AA x in R . then the number of solutions of the equation (F(x))^(2) =(9x^(4))/(G(x)) is...... .

Let f(x) = (x+1)^(2) - 1, (x ge - 1) . Then, the set S = {x : f(x) = f^(-1)(x) } is

Let f(x) lt 0 AA x in (-=oo, 0) and f (x) gt 0 AA x in (0,oo) also f (0)=o, Again f'(x) lt 0 AA x in (-oo, -1) and f '(x) gt AA x in (-1,oo) also f '(-1)=0 given lim _(x to oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. If f'(x) lt 0 AA x in (0,oo)and f'(0)=1 then number of solutions of equatin f (x)=x ^(2) is :

Let f (x) = (x +1)^(2) -1 ( x ge -1). Then the number of elements in the set S = {x :f (x) = f ^(-1)(x)} is

Let f(x) =(x+1)^(2) - 1, x ge -1 Statement 1: The set {x : f(x) =f^(-1)(x)}= {0,-1} Statement-2: f is a bijection.