Question

Let

\[f : R \to R\]  be a function defined by

\[f\left( x \right) = \frac{e^{|x|} - e^{- x}}{e^x + e^{- x}} . \text{Then},\]

 

Options

•  f is a bijection

• f is an injection only

•  f is surjection on only

• f is neither an injection nor a surjection

Answer 1

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 .