Let f : Z `to Z ` be given by f(x) = `{((x)/(2) if"is even"),(0 if" is odd"):}` Then f is

If function f:ZtoZ is defined by f(x)={{:((x)/(2), "if x is even"),(0," if x is odd"):} , then f is

Let Z denote the set of all integers and f: Z to Z defined by f(x) = {{:(x+2, ("x is even")),(0, ("x is odd")):} , Then f is:

If f: Z to Z be given by f(x)=x^(2)+ax+b , Then,

If f: Z to Z be given by f(x)=x^(2)+ax+b , Then,

On the set of integers Z, define f : Z to Z as f(n)={{:((n)/(2)",",,"n is even."),(0",",,"n is odd."):} Then, f is

Define f:Z rarrZ by f(x)={{:(x//2,"(x is even)"),(0,"(x is odd)"):} then f is

Let z denote the set of all integers.Define : f:z rarr z by f(x)={(x)/(2),(xiseven),0,(xisodd) Then f is

On the set Z of all integers define f : Z-(0) rarr Z as follows f(n)= {(n/2, n \ is \ even) , (2/0 , n \ is \ odd):} then f is

Let f:N to Z and f (x)=[{:((x-1)/(2),"when x is odd"),(-(x)/(2),"when x is even"):}, then:

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