Let `f(x)=ln(2+x)-(2x+2)/x` Statement-1: The equation `f(x)=0` has a unique solution in the domain of `f(x)` . Statement-2: If `f(x)` is continuous in `[a , b]` and is strictly monotonic in `(a , b)` then f has a unique root in `(a , b)`

Statement-1 : f(x) = 0 has a unique solution in (0,pi). Statement-2: If g(x) is continuous in (a,b) and g(a) g(b) <0, then the equation g(x)=0 has a root in (a,b).

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

Statement-1: If |f(x)|<=|x| for all x in R then |f(x)| is continuous at 0. Statement-2: If f(x) is continuous then |f(x)| is also continuous.

If f(x) is continuous in [0,2] and f(0)=f(2). Then the equation f(x)=f(x+1) has

Statement 1: If 27 a+9b+3c+d=0, then the equation f(x)=4a x^3+3b x^2+2c x+d=0 has at least one real root lying between (0,3)dot Statement 2: If f(x) is continuous in [a,b], derivable in (a , b) such that f(a)=f(b), then there exists at least one point c in (a , b) such that f^(prime)(c)=0.

f(x)=sqrt((log(x-1))/(x^(2)-2x-8)) . Find the domain of f(x).

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

Statement-1 : f : R rarr R, f(x) = x^(2) log(|x| + 1) is an into function. and Statement-2 : f(x) = x^(2) log(|x|+1) is a continuous even function.