If f ( x ) = ln ( x 2 + 1 x 2 + e ) , then range of f(x) is

f ( x ) = ℓ n ( x 2 + 1 x 2 + e ) = ℓ n ( x 2 + 1 x 2 + 1 − 1 + e ) = ℓ n ( 1 + x 2 + 1 e − 1 ) Clearly range is (0,1] Hence (B) is correct answer.

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MathematicsFunctionsDomain-RangeMedium2 minQB

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

Answer:

Solution:

$f (x) = ℓ n (\frac{x ^{2} + e}{x ^{2} + 1}) = ℓ n (\frac{x ^{2} + 1 - 1 + e}{x ^{2} + 1}) = ℓ n (1 + \frac{e - 1}{x ^{2} + 1})$

Clearly range is (0,1]

Hence (B) is correct answer.

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Domain-Range

⏱ 2mℹ️ Source: QB

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