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(B) Given $f(x) = \frac{a-x}{a+x}$.We are given $(f \circ f)(x) = x$.$f(f(x)) = \frac{a - f(x)}{a + f(x)} = \frac{a - \frac{a-x}{a+x}}{a + \frac{a-x}{

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$3$

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(B) Given $f(x) = \frac{a-x}{a+x}$.We are given $(f \circ f)(x) = x$.$f(f(x)) = \frac{a - f(x)}{a + f(x)} = \frac{a - \frac{a-x}{a+x}}{a + \frac{a-x}{a+x}}$$= \frac{a(a+x) - (a-x)}{a(a+x) + (a-x)} = \frac{a^2 + ax - a + x}{a^2 + ax + a - x} = x$$a^2 + ax - a + x = x(a^2 + ax + a - x)$$a^2 + ax - a + x = a^2x + ax^2 + ax - x^2$Rearranging the terms: $x^2(a-1) + x(a^2-1) - a(a-1) = 0$.For this to hold for all $x$,the coefficients must be zero.$a-1 = 0 \Rightarrow a = 1$.Thus,$f(x) = \frac{1-x}{1+x}$.Now,$f(-\frac{1}{2}) = \frac{1 - (-1/2)}{1 + (-1/2)} = \frac{3/2}{1/2} = 3$.

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 $f(x) \neq -a$,$(f \circ f)(x) = x$. Then $f(-\frac{1}{2})$ is equal to

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