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What is the range of (f(x)=1+\frac{2}{x^2-4x+8}) where (x\in\mathbb{R})?फलन (f(x)=1+\frac{2}{x^2-4x+8}) का परिसर क्या है जहाँ (x\in\mathbb{R})?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ( (1,\frac{3}{2}] )( (1,\frac{3}{2}] )
The denominator is (x^2-4x+8=(x-2)^2+4), so the fraction has maximum (\frac{1}{2}) and never attains (0). Hence the total range is ((1,\frac{3}{2}]).हर (x^2-4x+8=(x-2)^2+4) है इसलिए भिन्न का अधिकतम (\frac{1}{2}) है और न्यूनतम (0) प्राप्त नहीं होता। इसलिए कुल परिसर ( (1,\frac{3}{2}] ) है।
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