Question

The function \[f : [0, \infty ) \to \text {R given by } f\left( x \right) = \frac{x}{x + 1} is\]

 

 

विकल्प

• one-one and onto

• one-one but not onto

• onto but not one-one

• onto but not one-one

Answer 1

Injectivity:
Let x and y be two elements in the domain, such that

\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x}{x + 1} = \frac{y}{y + 1}\]
\[ \Rightarrow xy + x = xy + y\]
\[ \Rightarrow x = y\]

So, f is one-one.
Surjectivity:
Let y be an element in the co domain R, such that

\[y = f\left( x \right)\]
\[ \Rightarrow y = \frac{x}{x + 1}\]
\[ \Rightarrow xy + y = x\]
\[ \Rightarrow x\left( y - 1 \right) = - y\]
\[ \Rightarrow x = \frac{- y}{y - 1}\]
\[\text{Range off} = R - \left\{ 1 \right\} \neq \text{ co domain } (R)\]

⇒ is not onto.
So, the answer is (b)