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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Functions

BITSAT 2011 · Mathematics · Functions

If (f(x)=(1-x)/(1+x)) the domain of f^ -1 (x) is

If (f(x)=\frac{1-x}{1+x}) the domain of f−1(x)f^{-1}(x) is

Options

  1. A(\mathrm{R})
  2. B(\mathrm{R}-\{-1\})
  3. C((-\infty,-1))
  4. D((-1, \infty))

Correct answer

B. (\mathrm{R}-\{-1\})

Step-by-step solution

Let f(x)=yf(x)=y. Then (\frac{1-x}{1+x}=y) (\Rightarrow x=\frac{1-y}{1+y} \Rightarrow f^{-1}(y)=\frac{1-y}{1+y}) Thus, (f^{-1}(x)=\frac{1-x}{1+x}) Clearly, (\mathrm{f}^{-1}(\mathrm{x})) is defined for (1+\mathrm{x} \neq 0). Hence, domain of (\mathrm{f}^{-1}(\mathrm{x})) is (\mathrm{R}-\{-1\})
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