For a suitably chosen real constant a, let a function, \( \mathrm{f}: R-\{-a\} \rightarrow R \) be defined by \( f(x)=\frac{a-x}{a+x} \). Further suppose that for any real number \( x \neq-a \) and \( f(x) \neq-a, (fof)(x)=x \). Then \( f\left(-\frac{1}{2}\right) \) is equal to :

[JEE Main 2020, 6 Sep (Shift 2)]

Question

For a suitably chosen real constant a, let a function, \( \mathrm{f}: R-\{-a\} \rightarrow R \) be defined by \( f(x)=\frac{a-x}{a+x} \). Further suppose that for any real number \( x \neq-a \) and \( f(x) \neq-a, (fof)(x)=x \). Then \( f\left(-\frac{1}{2}\right) \) is equal to :

[JEE Main 2020, 6 Sep (Shift 2)]

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