Question

If the function \(f:\mathbb{R}\to\mathbb{R}\) is defined by \(f(x)=x|x|\), then \(f\) is:

• A. \(f\) is one-one but not onto
• B. \(f\) is onto but not one-one
• C. \(f\) is both one-one and onto
• D. \(f\) is neither one-one nor onto

Hint:

For functions involving \(|x|\), first write the function in piecewise form. Then check one-one and onto using the behavior of the function on each interval.

Answer 1

The Correct Option is C

Step 1: Write the function in piecewise form.
Given, \[ f(x)=x|x| \] If \(x\geq 0\), then \(|x|=x\), so \[ f(x)=x^2 \] If \(x<0\), then \(|x|=-x\), so \[ f(x)=-x^2 \] Thus, \[ f(x)= \begin{cases} -x^2, & x x^2, & x\geq 0 \end{cases} \]

Step 2: Check whether the function is one-one.
For \(x<0\), \[ f(x)=-x^2 \] which increases from \(-\infty\) to \(0\).
For \(x\geq 0\), \[ f(x)=x^2 \] which increases from \(0\) to \(\infty\).
So, for every different value of \(x\), we get a different value of \(f(x)\).
Hence, \(f\) is one-one.

Step 3: Check whether the function is onto.
For \(x<0\), \[ f(x)=-x^2 \] gives all negative real values.
For \(x\geq 0\), \[ f(x)=x^2 \] gives all non-negative real values.
Therefore, the range of \(f\) is \[ (-\infty,0)\cup[0,\infty)=\mathbb{R} \] Since the range is equal to the codomain \(\mathbb{R}\), the function is onto.

Step 4: Final conclusion.
The function \(f(x)=x|x|\) is both one-one and onto.
Therefore, \[ \boxed{\text{\(f\) is both one-one and onto}} \]