Question

The function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x|x+2| \) is:

• A. one one but not onto
• B. onto but not one one
• C. both one one and onto
• D. neither one one nor onto

Hint:

To quickly check if a continuous function is onto \( \mathbb{R} \), check its limits at \( \pm \infty \). To check if it's one-one, look for symmetry or roots; here, \( x=0 \) and \( x=-2 \) both being roots immediately proves it is many-to-one.

Answer 1

The Correct Option is B

Concept:& nbsp;

To determine the nature of the function \(f(x)=x|x+2|\), we analyze it by splitting the absolute value into cases.

• One-one (Injective): A function is one-one if \[ f(x_1)=f(x_2)\Rightarrow x_1=x_2. \] Graphically, it must satisfy the horizontal line test.

• Onto (Surjective): A function \[ f:\mathbb{R}\rightarrow\mathbb{R} \] is onto if its range is equal to its codomain, i.e., \(\mathbb{R}\).

Step 1: Define the function piecewise.

The expression \(|x+2|\) changes at \(x=-2\). Therefore,

\[ f(x)= \begin{cases} x(x+2)=x^2+2x, & amp; \text{if } x\ge -2,\\[6pt] -x(x+2)=-x^2-2x, & amp; \text{if } x<-2. \end{cases} \]

Step 2: Check whether the function is one-one.

Consider the following values:

\[ f(0)=0|0+2|=0 \] \[ f(-2)=-2|-2+2|=0 \]

Since \[ f(0)=f(-2)=0 \] but \[ 0\ne -2, \] the function is not one-one.

Step 3: Check whether the function is onto.

As \[ x\to\infty,\quad f(x)=x^2+2x\to\infty, \] and \[ x\to-\infty,\quad f(x)=-x^2-2x\to-\infty. \]

Since the function is continuous and its values extend from \(-\infty\) to \(\infty\), its range is \[ \mathbb{R}. \] Hence, the function is onto.