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Question
Let
\[f : R \to R\] be a function defined by
Options
f is a bijection
f is an injection only
f is surjection on only
f is neither an injection nor a surjection
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Solution
f is neither an injection nor a surjection
\[f : R \to R\]
\[f\left( x \right) = \frac{e^{|x|} - e^{- x}}{e^x + e^{- x}}\]
\[\text{For } x = - 2 \text{ and} - 3 \in R \]
\[f( - 2) = \frac{e^\left| - 2 \right| - e^2}{e^{- 2} + e^2}\]
\[ = \frac{e^2 - e^2}{e^{- 2} + e^2}\]
\[ = 0\]
\[\text{& } f( - 3) = \frac{e^\left| - 3 \right| - e^3}{e^{- 3} + e^3}\]
\[ = \frac{e^3 - e^3}{e^{- 3} + e^3}\]
\[ = 0\] \[\text{Hence, for different values of x we are getting same values of f }(x)\]
\[\text{That means , the given function is many one} . \]
Therefore, this function is not injective.
\[ \text{For } x < 0\]
\[f (x ) = 0\]
\[\text{ For } x > 0\]
\[f(x) = \frac{e^x - e^{- x}}{e^x + e^{- x}}\]
\[ = \frac{e^x + e^{- x}}{e^x + e^{- x}} - \frac{2 e^{- x}}{e^x + e^{- x}}\]
\[ = 1 - \frac{2 e^{- x}}{e^x + e^{- x}}\]
\[\text{The value of } \frac{2 e^{- x}}{e^x + e^{- x}} \text{is always positive} . \]
\[\text{Therefore, the value of} f(x) \text{is always less than} 1\]
\[\text{Numbers more than 1 are not included in the range but they are included in codomain} . \]
\[\text{As the codomain is } R . \]
\[ \therefore \text{Codomain} \neq \text{Range}\]
\[\text{Hence, the given function is not onto} . \]
Therefore, this function is not surjective .
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