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JEE Mains · Maths · STD 12 - 1. relation and function

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

  1. A \(\frac{1}{3}\)
  2. B \(3\)
  3. C \(-3\)
  4. D \(-\frac{1}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

\(f(x)=\frac{a-x}{a+x} \quad x \in R-\{-a\} \rightarrow R\) \(f(f(x))=\frac{a-f(x)}{a+f(x)}=\frac{a-\left(\frac{a-x}{a+x}\right)}{a+\left(\frac{a-x}{a+x}\right)}\) \(f(f(x))=\frac{\left(a^{2}-a\right)+x(a+1)}{\left(a^{2}+a\right)+x(a-1)}=x\)…
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