For a suitably chosen real constant $a$, let a function, $f: R-\{-a\} \rightarrow R$ 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
For a suitably chosen real constant $a$, let a function, $f: R-\{-a\} \rightarrow R$ 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]
- A
$\frac{1}{3}$
- B
$3$
- C
$-3$
- D
$-\frac{1}{3}$
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