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Question
The function \[f : [0, \infty ) \to \text {R given by } f\left( x \right) = \frac{x}{x + 1} is\]
Options
one-one and onto
one-one but not onto
onto but not one-one
onto but not one-one
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Solution
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)
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