Statement−I: f:R−0→R−0 defined by fx=1x is bijectiveStatement−II: f:R+→R defined by fx=logx, then f−1x=ex which of the following is true

Statement−I: f:R−0→R−0fx=1xLet fx1=fx2⇒ 1x1=1x2⇒ x1=x2∴ f is one−oneLet fx=y ⇒ 1x=y⇒ x=1y is defined if y≠0∴ co domain = Range∴ f is onto ⇒ f is bijectionStatement−I is trueStatement−II:fx=logexLet f−1x=y⇒ x=fy⇒ x=logey⇒ y=ex∴ f−1x=exStatement −II is true

Statement−I: f:R−0→R−0 defined by fx=1x is bijectiveStatement−II: f:R+→R defined by fx=logx, then f−1x=ex which of the following is true

Statement−I: f:R−0→R−0fx=1xLet fx1=fx2⇒ 1x1=1x2⇒ x1=x2∴ f is one−oneLet fx=y ⇒ 1x=y⇒ x=1y is defined if y≠0∴ co domain = Range∴ f is onto ⇒ f is bijectionStatement−I is trueStatement−II:fx=logexLet f−1x=y⇒ x=fy⇒ x=logey⇒ y=ex∴ f−1x=exStatement −II is true

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$\begin{array}{l} S t a t e m e n t - I : f : R - \{0\} \to R - \{0\} d e f i n e d b y f (x) = \frac{1}{x} i s b i j e c t i v e \\ S t a t e m e n t - I I : f : R^{+} \to R d e f i n e d b y f (x) = \log x, t h e n f^{- 1} (x) = e^{x} w h i c h o f t h e f o l l o w i n g i s t r u e \end{array}$

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Only II

Neither I nor II

Both I & II

Only I

answer is C.

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Detailed Solution

$\begin{array}{l} S t a t e m e n t - I : f : R - \{0\} \to R - \{0\} \\ f (x) = \frac{1}{x} \\ L e t f (x_{1}) = f (x_{2}) \Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}} \Rightarrow x_{1} = x_{2} \\ ∴ f i s o n e - o n e \\ L e t f (x) = y \Rightarrow \frac{1}{x} = y \Rightarrow x = \frac{1}{y} i s d e f i n e d i f y \neq 0 \\ ∴ c o d o m a i n = R a n g e \end{array}$

$\begin{array}{l} ∴ f i s o n t o \Rightarrow f i s b i j e c t i o n \\ S t a t e m e n t - I i s t r u e \\ S t a t e m e n t - I I : f (x) = \log_{e}^{x} \\ L e t f^{- 1} (x) = y \Rightarrow x = f (y) \\ \Rightarrow x = \log_{e}^{y} \\ \Rightarrow y = e^{x} \\ ∴ f^{- 1} (x) = e^{x} \\ S t a t e m e n t - I I i s t r u e \end{array}$