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(D) Given $f(x) = \frac{a-x}{a+x}$.We are given that $(f \circ f)(x) = x$.$f(f(x)) = f\left(\frac{a-x}{a+x}\right) = \frac{a - \left(\frac{a-x}{a+x}\r

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$3$

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(D) Given $f(x) = \frac{a-x}{a+x}$.We are given that $(f \circ f)(x) = x$.$f(f(x)) = f\left(\frac{a-x}{a+x}ight) = \frac{a - \left(\frac{a-x}{a+x}ight)}{a + \left(\frac{a-x}{a+x}ight)} = x$.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)$$a^2 + ax - a + x = a^2x + ax^2 + ax - x^2$Rearranging the terms:$(a-1)x^2 + (a^2-1)x - a(a-1) = 0$$(a-1)(x^2 + (a+1)x - a) = 0$$(a-1)(x+a)(x-1) = 0$Since this must hold for all $x 
eq -a$,we must have $a-1 = 0$,so $a = 1$.Thus,$f(x) = \frac{1-x}{1+x}$.Now,calculate $f\left(-\frac{1}{2}ight)$:$f\left(-\frac{1}{2}ight) = \frac{1 - (-1/2)}{1 + (-1/2)} = \frac{1 + 1/2}{1 - 1/2} = \frac{3/2}{1/2} = 3$.

For a suitably chosen real constant $a$,let the 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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