$f : R \to R i s a f u n c t i o n d e f i n e d b y f (x) = \frac{e^{| x |} - e^{- x}}{e^{x} + e^{- x}} . T h e n f i s$

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An Intiative by Sri Chaitanya

One – one and onto

One – one but not onto

Onto but not one – one

Neither one – one nor onto

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{e^{| x |} - e^{- x}}{e^{x} + e^{- x}} = {\begin{cases} \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} i f x \geq 0 \\ \frac{e^{- x} - e^{- x}}{e^{x} + e^{- x}} i f x < 0 \end{cases} \\ = {\begin{cases} \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} i f x \geq 0 \\ 0 i f x < 0 \end{cases} \\ \forall x < 0, f (x) = 0 \\ ∴ f i s n o t o n e - o n e \end{array}$

$\begin{array}{l} f o r x \geq 0, f (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} = \frac{e^{x} + e^{- x} - 2 e^{- x}}{e^{x} + e^{- x}} \\ = 1 - \frac{2 e^{- x}}{e^{x} + e^{- x}} \\ = 1 - \frac{2}{e^{2 x} + 1} \\ I f x \geq 0 \Rightarrow e^{2 x} \geq 1 \\ \Rightarrow e^{2 x} + 1 \geq 2 \\ \Rightarrow 2 \leq e^{2 x} + 1 < 0 \end{array}$

$\begin{array}{l} \Rightarrow \frac{1}{2} \geq \frac{1}{e^{2 x} + 1} > 0 \\ \Rightarrow 1 \geq \frac{x}{e^{2 x} + 1} > 0 \\ \Rightarrow - 1 \leq \frac{- 2}{e^{2 x} + 1} < 0 \\ \Rightarrow 0 \leq 1 - \frac{2}{e^{2 x} + 1} < 1 \\ ∴ R a n g e = [0, 1) \neq c o d o m a i n \\ ∴ f i s n o t o n t o \end{array}$