Question

The function \( f:[0,\infty)\rightarrow [0,\infty) \) defined by} \[ f(x)=\frac{x}{1+x} \] \textbf{is

• A. one-one and onto
• B. one-one but not onto
• C. onto but not one-one
• D. neither one-one nor onto \bigskip

Hint:

For rational functions of the form \[ f(x)=\frac{x}{1+x}, \] first check injectivity by equating \(f(x_1)\) and \(f(x_2)\). To find the range, substitute \(y=f(x)\) and express \(x\) in terms of \(y\). The restrictions on \(x\) then give the range directly.

Answer 1

The Correct Option is B

Concept: A function is
• One-one (Injective) if distinct inputs have distinct outputs.
• Onto (Surjective) if every element of the codomain has a pre-image in the domain. To determine whether a function is one-one and onto, we examine its monotonicity and range.

Step 1: Check whether the function is one-one. Given, \[ f(x)=\frac{x}{1+x}, \qquad x\ge 0. \] Let \[ f(x_1)=f(x_2). \] Then \[ \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}. \] Cross-multiplying, \[ x_1(1+x_2)=x_2(1+x_1). \] \[ x_1+x_1x_2=x_2+x_1x_2. \] \[ x_1=x_2. \] Hence, distinct inputs cannot have the same output. Therefore, the function is one-one.

Step 2: Find the range of the function. Let \[ y=\frac{x}{1+x}. \] Then \[ y(1+x)=x. \] \[ y+xy=x. \] \[ y=x(1-y). \] \[ x=\frac{y}{1-y}. \] Since \(x\ge 0\), \[ \frac{y}{1-y}\ge 0. \] Also, \[ 1-y>0 \] because \(y=1\) makes the denominator zero. Hence, \[ 0\le y<1. \] Therefore, the range of \(f\) is \[ [0,1). \]

Step 3: Check whether the function is onto. The codomain is \[ [0,\infty). \] But the range is only \[ [0,1). \] Since numbers such as \(2,3,\ldots\) belong to the codomain but are not attained by the function, the function is not onto.

Step 4: Write the final conclusion. The function is one-one but not onto. \[ \boxed{\text{One-one but not onto}} \]