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Question

Let $f left parenthesis x right parenthesis equals l n invisible function application open parentheses fraction numerator 1 minus x over denominator 1 plus x end fraction close parentheses$ the set of values of ‘ a ’ for which $f left parenthesis alpha right parenthesis plus f open parentheses alpha to the power of 2 end exponent close parentheses equals f open parentheses fraction numerator alpha over denominator alpha to the power of 2 end exponent minus alpha plus 1 end fraction close parentheses$ is satisfied are

$(- \infty, - 1) \cup (1, \infty)$
(-1,1)
$(0,1)$
none of these

The correct answer is: (-1,1)

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