Let `f : R->R` & `f(x)=x/(1+|x|)` Then f(x) is (1) injective but not surjective (2) surjective but not injective (3) injective as well as surjective (4) 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

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 S be the set of all triangles and R^+ be the set of positive real numbers. Then the function f: SrarrR^+,f(Delta)=a r e aof Delta ,w h e r e in S , is injective but not surjective. surjective but not injective injective as well as surjective neither injective nor surjective

Let S be the set of all triangles and R^+ be the set of positive real numbers. Then the function f: SrarrR^+,f(Delta)=area of Delta ,where Delta in S , is injective but not surjective. surjective but not injective injective as well as surjective neither injective nor surjective

Let S be the set of all triangles and R^+ be the set of positive real numbers. Then the function f: SrarrR^+,f(Delta)=area of Delta ,where Delta in S , is injective but not surjective. surjective but not injective injective as well as surjective neither injective nor surjective

Let f:[-00,0)->(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 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 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 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 :-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