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(A) Given: $f(x)=\frac{a-x}{a+x}$.Since $(f \circ f)(x)=x$,we have:$f(f(x)) = \frac{a-f(x)}{a+f(x)} = x$Substituting $f(x) = \frac{a-x}{a+x}$:$\frac{a

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$1.5$

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(A) Given: $f(x)=\frac{a-x}{a+x}$.Since $(f \circ f)(x)=x$,we have:$f(f(x)) = \frac{a-f(x)}{a+f(x)} = x$Substituting $f(x) = \frac{a-x}{a+x}$:$\frac{a-\frac{a-x}{a+x}}{a+\frac{a-x}{a+x}} = x$$\frac{a(a+x)-(a-x)}{a(a+x)+(a-x)} = x$$\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$$(a-1)x^2 + (a^2-1)x - (a^2-a) = 0$$(a-1)x^2 + (a-1)(a+1)x - a(a-1) = 0$For this to hold for all $x$,we must have $a-1=0$,so $a=1$.Thus,$f(x) = \frac{1-x}{1+x}$.Now,$f\left(-\frac{1}{5}\right) = \frac{1-(-1/5)}{1+(-1/5)} = \frac{1+1/5}{1-1/5} = \frac{6/5}{4/5} = \frac{6}{4} = 1.5$.

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 $f\left(-\frac{1}{5}\right)$ is equal to

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