`f: R->R` given by `f(x)=x+sqrt(x^2)` is (a) injective (b) surjective (c) bijective (d) none of these

f:R to R" given by "f(x)=x+sqrt(x^(2)) ,is

The function f: NvecN(N is the set of natural numbers) defined by f(n)=2n+3i s (a) surjective only (b) injective only (c) bijective (d) none of these

Let A={x in R :-1lt=xlt=1}=B . Then, the mapping f: A->B given by f(x)=x|x| is (a) injective but not surjective (b) surjective but not injective (c) bijective (d) none of these

Let A={x in R :-1lt=xlt=1}=B . Then, the mapping f: A->B given by f(x)=x|x| is (a) injective but not surjective (b) surjective but not injective (c) bijective (d) none of these

The function f: R->R , f(x)=x^2 is (a) injective but not surjective (b) surjective but not injective (c) injective as well as surjective (d) neither injective nor surjective

Let R be the set of real numbers. If f:R->R is a function defined by f(x)=x^2, then f is injective but not surjective surjective but not injective bijective none of these (a) injective but not surjective (b) surjective but not injective (c) bijective (d) non of these

A function f:(0,oo) -> [0,oo] is given by f(x)=|1-1/x| , then f(x) is (A) Injective but not surjective (B) Injective and bijective (C) Injective only (D) Surjective only

Classify f: R->R , defined by f(x)=x/(x^2+1) as injection, surjection or bijection.

Classify f: R->R , defined by f(x)=sin^2x+cos^2x as injection, surjection or bijection.

Let A{x :-1lt=xlt=1}a n df: Avec such that f(x)=x|x|, then f is (a) bijection (b) injective but not surjective (c)Surjective but not injective (d) neither injective nor surjective