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

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$3$

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(B) Given $f(x) = \frac{a-x}{a+x}$. We are given $(f \circ f)(x) = x$. $(f \circ f)(x) = f(f(x)) = \frac{a - \frac{a-x}{a+x}}{a + \frac{a-x}{a+x}} = x$. Simplifying the expression: $\frac{a(a+x) - (a-x)}{a(a+x) + (a-x)} = x$ $\frac{a^2 + ax - a + x}{a^2 + ax + a - x} = x$ $a^2 + ax - a + x = x(a^2 + ax + a - x)$ $a^2 + ax - a + x = a^2x + ax^2 + ax - x^2$ Rearranging terms: $(a-1)x^2 + (a^2-1)x + (a-a^2) = 0$. 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,calculate $f\left(\frac{-1}{2}\right)$: $f\left(\frac{-1}{2}\right) = \frac{1 - (-1/2)}{1 + (-1/2)} = \frac{1 + 1/2}{1 - 1/2} = \frac{3/2}{1/2} = 3$.

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 $f\left(\frac{-1}{2}\right)$ is equal to

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