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What is the range of (f(x)=\frac{x-2}{\sqrt{x^2-4x+8}})?फलन (f(x)=\frac{x-2}{\sqrt{x^2-4x+8}}) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( (-1,1) )( (-1,1) )
The denominator is (\sqrt{(x-2)^2+4}), which is greater than (|x-2|). Hence the ratio stays strictly between (-1) and (1).हर (\sqrt{(x-2)^2+4}) है, जो (|x-2|) से बड़ा है। इसलिए अनुपात का मान (-1) और (1) के बीच ही रहता है।
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