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A function (f:\mathbb{R}\to\mathbb{R}) is to be defined by (f(x)=\sqrt{a x^2-6a+9}). Which condition on (a) is necessary and sufficient for it to be a function on all of (\mathbb{R})?

Answer and explanation

Correct answer: (\ 0\le a\le\frac{3}{2}) / (0\le a\le\frac{3}{2})

If (a<0), the radicand can become negative for large (|x|), and if (a\ge0), its minimum is (9-6a). Thus (9-6a\ge0) gives (0\le a\le\frac{3}{2}).

Related tags

DomainParameterSquare RootFunctions As A Special Kind Of RelationRelations And FunctionsMathematicsClass 11 Mcq

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