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Class 11 Mathematics Relations and Functions Real-Valued Functions, Their Domain and RangeMedium

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What is the range of (f(x)=\frac{1}{x^2-6x+13})?

Answer and explanation

Correct answer: ( (0,\frac{1}{4}])

The denominator (x^2-6x+13=(x-3)^2+4), whose minimum value is (4). So the output is greater than (0) and up to (\frac{1}{4}).

Related tags

Real-Valued-Functions Range Rational-Function Real Valued Functions Their Domain And Range Real Valued Functions Their Domain And Range Relations And Functions Mathematics Class 11 Mcq

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