English

The Function \[F : [0, \Infty ) \To R Given by F\Left( X \Right) = \Frac{X}{X + 1} Is\] (A) One-one and onto (B) One-one but Not onto (C) onto but Not One-one (D) onto but Not One-one - Mathematics

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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

MCQ
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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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Chapter 2: Functions - Exercise 2.6 [Page 76]

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RD Sharma Mathematics [English] Class 12
Chapter 2 Functions
Exercise 2.6 | Q 10 | Page 76

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