Let f:A→B be a function defined as f(x)=x-1x-2, where A=R-{2} and B=R-{1} then f is

Suppose y=f(x)⇒ y=x−1x−2⇒ yx−2y=x−1⇒ y−1x=2y−1x=f−1y=2y−1y−1

Let f:A→B be a function defined as f(x)=x-1x-2, where A=R-{2} and B=R-{1} then f is

Suppose y=f(x)⇒ y=x−1x−2⇒ yx−2y=x−1⇒ y−1x=2y−1x=f−1y=2y−1y−1

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$Let f : A \to B be a function defined as f (x) = \frac{x - 1}{x - 2}, where A = R - {2} and B = R - {1}$ then f is

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$Invertible and f^{- 1} (y) = \frac{2 y + 1}{y - 1}$

$invertible and f^{- 1} (y) = \frac{3 y - 1}{y - 1}$

$N o t i n v e r t i b l e$

$invertible and f^{- 1} (y) = \frac{2 y - 1}{y - 1}$

answer is D.

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Detailed Solution

$Suppose y = f (x)$

$\Rightarrow y = \frac{x - 1}{x - 2}$

$\Rightarrow y x - 2 y = x - 1$

$\Rightarrow (y - 1) x = 2 y - 1$

$x = f^{- 1} (y) = \frac{2 y - 1}{y - 1}$

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Let f:A→B be a function defined as f(x)=x-1x-2, where A=R-{2} and B=R-{1} then f is

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$Let f : A \to B be a function defined as f (x) = \frac{x - 1}{x - 2}, where A = R - {2} and B = R - {1}$ then f is