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

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

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

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

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

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 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 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

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