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

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

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 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:[-1//2,\ 1//2]->[-pi//2,pi//2\ ] defined by f(x)=sin^(-1)(3x-4x^3) is (a) bijection (b) injection but not a surjection (c) surjection but not an injection (d) neither an injection nor a surjection

Let f:[-oo,0)->(1,oo) be defined as f(x)=(1+sqrt(-x))-(sqrt(-x)-x) then f(x) is (A) injective but not surjective (B) injective as well as surjective (C) neither injective nor surjective (D) surjective nut not injective

Let A={x:xepsilonR-{1}}f is defined from ArarrR as f(x)=(2x)/(x-1) then f(x) is (a) Surjective but nor injective (b) injective but nor surjective (c) neither injective surjective (d) injective

Let f:N to Z and f (x)=[{:((x-1)/(2),"when x is odd"),(-(x)/(2),"when x is even"):}, then: (a). f (x) is bijective (b).f (x) is injective but not surjective (c).f (x) is not injective but surjective (d).f (x) is neither injective nor subjective

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