Let `A={x:xepsilonR-}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 A={x:x varepsilon R-}f is defined from A rarr R 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 R be set of real numbers. If f:R->R is defined by f(x)=e^x, then f is: (a) surjective but not injective (b) injective but not surjective (c) bijective (d) neither surjective nor injective.

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

Let A}x:-1<=x<=1} and f:A rarr 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

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

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: Rvec[-1/2,1/2] defined as f(x)=x/(1+x^2), is : Surjective but not injective (2) Neither injective not surjective Invertible (4) Injective but not surjective

A function f:(0,oo)rarr[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

Let R be the set of real numbers.If f:R rarr R is a function defined by f(x)=x^(2), then f is injective but not surjective surjective but not injective but not surjective surjective but not but not surjective (b) surjective but not injective (c) bijective (d) non of these

Show that f:R rarr R defined by f(x)=(x-1)(x-2)(x-3) is surjective but not injective.