Let A={x,y,z}, B={u,v,w} and f:A→B be defined by f(x)=u,f(y)=v,f(z)=w. Then f is
A
surjective but not injective
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B
injective but not surjective
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C
bijective
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D
none of these
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Solution
The correct option is D bijective Since for every element of co-domain set B there is one and only one pre-image in the domain set A, the function is injective as well as surjective. Hence the function is bijective.