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If f: {0, 1, 2, 3} → {0, 1, 2, 3} is given by f(x) = x² − 2x + 1, is f a valid function?

Answer and explanation

Correct answer: No, because f(3) = 4 is not in {0, 1, 2, 3}

For f:A→B to be a valid function, every element of A must have exactly one image and each image must belong to the codomain B. The polynomial rule is single-valued, so calculate its outputs and check the codomain. We get f(0)=0−0+1=1, f(1)=1−2+1=0, f(2)=4−4+1=1, and f(3)=9−6+1=4. Although each input has one numerical output, 4 is not in the stated codomain {0,1,2,3}. Therefore this rule does not define a function from the given A to the given B, and option B is correct. Option C is not an issue because 0 belongs to B; option D ignores the essential codomain requirement.

Related tags

Relations FunctionsCodomainPolynomial FunctionFunction DefinitionFunctions As A Special Kind Of RelationRelations And FunctionsMathematicsClass 11 Mcq

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