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Range of \(\cfrac{x^2}{x^4+1}\). |
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Answer» f(x) = \(\cfrac{x^2}{x^4+1}\) = \(\cfrac{1}{x^2+\cfrac1{x^2}}\) = \(\cfrac{1}{\left(x+\cfrac1{x}\right)^2-2}\) \(\because\) \({\left(x+\cfrac1{x}\right)^2}\) \(\geq\) 22 or \({\left(x+\cfrac1{x}\right)^2}\) \(\geq\) 4 = \({\left(x+\cfrac1{x}\right)^2}\) - 2 \(\geq\) 4 - 2 = \({\left(x+\cfrac1{x}\right)^2}\) - 2 \(\geq\) 2 = \(\cfrac{1}{\left(x+\cfrac{1}{x}\right)^2-2}\) \(\leq\) \(\cfrac12\) = \(\cfrac{x^2}{x^4+1}\) \(\leq\) \(\cfrac12\)....(i) Also, x2 \(\geq\) 0 and x4 + 1 \(\geq\) 1 = \(\cfrac{x^2}{x^4+1}\) \(\geq\) 0 ...(ii) (\(\because\) Division of two positive numbers be always positive) Then from (i) and (ii),we obtain 0 \(\leq\) \(\cfrac{x^2}{x^4+1}\) \(\leq\) \(\cfrac12\) \(\therefore\) Range of f(x) = \(\cfrac{x^2}{x^4+1}\) is [0,\(\cfrac12\)]. |
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