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  4. If f(x) = ln((x2 +e/x2 +1)), then range of f(x) is

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Q. If $f\left(x\right) = ln\left(\frac{x^{2} +e}{x^{2 } +1}\right)$, then range of f(x) is

Relations and Functions

Solution:

$f\left(x\right) = ln\left(\frac{x^{2} +e}{x^{2 } +1}\right) = ln\left(\frac{x^{2}+1-1+e}{x^{2}+1}\right) = ln\left(1+\frac{e-1}{x^{2}+1}\right)$
Clearly range is (0, 1]