If f ( x ) = 1 + x 1 − x the domain of f − 1 ( x ) is

Let f ( x ) = y . Then 1 + x 1 − x = y ⇒ x = 1 + y 1 − y ⇒ f − 1 ( y ) = 1 + y 1 − y Thus, f − 1 ( x ) = 1 + x 1 − x Clearly, f − 1 ( x ) is defined for 1 + x  = 0 . Hence, domain of f − 1 ( x ) is R − { − 1 }

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MathematicsFunctionsInverseEasy2 minPYQ_2011

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

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Let $f (x) = y$ . 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, $f^{- 1} (x)$ is defined for $1 + x \neq = 0$ . Hence, domain of $f^{- 1} (x)$ is $R - {- 1}$

Stream:BITSATSubject:MathematicsTopic:FunctionsSubtopic:Inverse

⏱ 2mℹ️ Source: PYQ_2011

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