If f(x)=(1+x)(1−x)=y, then f−1(y)=

f(x)=(1+x)(1−x)=y1+x=y−xyx+xy=y−1x(y+1)=y−1x=y−1y+1f−1(y)=y−1y+1

If f(x)=(1+x)(1−x)=y, then f−1(y)=

f(x)=(1+x)(1−x)=y1+x=y−xyx+xy=y−1x(y+1)=y−1x=y−1y+1f−1(y)=y−1y+1

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If $f (x) = \frac{(1 + x)}{(1 - x)} = y$ , then $f^{- 1} (y) =$

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$\frac{(1 + y)}{(1 - y)}$

$\frac{(1 - y)}{(1 + y)}$

$\frac{(y - 1)}{(y + 1)}$

$\frac{(y + 1)}{(y - 1)}$

answer is C.

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Detailed Solution

$f (x) = \frac{(1 + x)}{(1 - x)} = y$

$\begin{array}{l} 1 + x = y - x y \\ x + x y = y - 1 \\ x (y + 1) = y - 1 \\ x = \frac{y - 1}{y + 1} \\ f^{- 1} (y) = \frac{y - 1}{y + 1} \end{array}$

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If $f (x) = \frac{(1 + x)}{(1 - x)} = y$ , then $f^{- 1} (y) =$