For a suitably chosen real constant a

Question:

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

  1. (1) $\frac{1}{3}$

  2. (2) $-\frac{1}{3}$

  3. (3) $-3$

  4. (4) 3


Correct Option: , 4

Solution:

$f(f(x))=\frac{a-\left(\frac{a-x}{a+x}\right)}{a+\left(\frac{a-x}{a+x}\right)}=x$

$\Rightarrow \frac{a-a x}{1+x}=f(x) \Rightarrow \frac{a(1-x)}{1+x}=\frac{a-x}{a+x} \Rightarrow a=1$

$\therefore f(x)=\frac{1-x}{1+x} \Rightarrow f\left(-\frac{1}{2}\right)=3$

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