The function f(x) = (x + √(x^2 + 1)) is: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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The function f (x) = lo g (x + x^{2} + 1) is:

Solution & Step-by-step Explanation

A function is odd if f (- x) = - f (x) . f (- x) = lo g (- x + (- x)^{2} + 1) = lo g (x^{2} + 1 - x) Multiply and divide by the conjugate: f (- x) = lo g (\frac{( x ^{2} + 1 - x ) ( x ^{2} + 1 + x )}{x ^{2} + 1 + x}) f (- x) = lo g (\frac{x ^{2} + 1 - x ^{2}}{x ^{2} + 1 + x}) = lo g (\frac{1}{x ^{2} + 1 + x}) f (- x) = lo g (x^{2} + 1 + x)^{- 1} = - lo g (x + x^{2} + 1) f (- x) = - f (x) Therefore, it is an odd function.

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The function f (x) = lo g (x + x^{2} + 1) is:

an even function

an odd function

a periodic function

neither an even nor an odd function

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