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  • For a suitably chosen real constant a, let a function, be defined by

For a suitably chosen real constant a, let a function, f:R-\left \{ -a \right \}\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,(fof)(x)=x. Then f\left (- \frac{1}{2} \right ) is equal to :
Option: 1 \frac{1}{3}
Option: 2 -\frac{1}{3}
Option: 3 -3
Option: 4 3

Answers (1)

best_answer

f(x)=\frac{a-x}{a+x} \quad x \in R-\{-a\} \rightarrow R \\ \\f(f(x))=\frac{a-f(x)}{a+f(x)}=\frac{a-\left(\frac{a-x}{a+x}\right)}{a+\left(\frac{a-x}{a+x}\right)}

\\f(f(x))=\frac{\left(a^{2}-a\right)+x(a+1)}{\left(a^{2}+a\right)+x(a-1)}=x \\ \\\Rightarrow\left(a^{2}-a\right)+x(a+1)=\left(a^{2}+a\right) x+x^{2}(a-1) \\ \\\Rightarrow a(a-1)+x\left(1-a^{2}\right)-x^{2}(a-1)=0 \\ \\\Rightarrow a=1

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

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himanshu.meshram

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