If fx-lnx2-e-x2-1- then range of fx is

The correct option is B (0,1]f(x)= ln ( x^2+e/x^2+1 )= ln ( x^2+1 1+e/x^2+1 )=ln(1+ e 1/x^2+1 ) Clearly range is (0, 1] Hence (B) is correct answer ...

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Standard XII

Mathematics

Using Monotonicity to Find the Range of a Function

If fx=lnx2+e/...

Question

If $f (x) = l n (\frac{x^{2} + e}{x^{2} + 1})$ , then range of f(x) is

(0, 1)

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(0, 1]

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[0, 1)

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{0, 1}

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Solution

The correct option is B $(0, 1]$
$f (x) = l n (\frac{x^{2} + e}{x^{2} + 1}) = l n (\frac{x^{2} + 1 - 1 + e}{x^{2} + 1}) = l n (1 + \frac{e - 1}{x^{2} + 1})$
Clearly range is (0, 1]
Hence (B) is correct answer

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If f(x)=ln(x2+ex2+1) , then range of f(x) is

The correct option is B (0,1]f(x)=ln(x2+ex2+1)=ln(x2+1−1+ex2+1)=ln(1+e−1x2+1) Clearly range is (0, 1] Hence (B) is correct answer

If $f (x) = l n (\frac{x^{2} + e}{x^{2} + 1})$ , then range of f(x) is

The correct option is B $(0, 1]$
$f (x) = l n (\frac{x^{2} + e}{x^{2} + 1}) = l n (\frac{x^{2} + 1 - 1 + e}{x^{2} + 1}) = l n (1 + \frac{e - 1}{x^{2} + 1})$
Clearly range is (0, 1]
Hence (B) is correct answer