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: Express the function in piecewise notation.
We are given f(x) = x|x|. For x ≥ 0, the absolute value |x| equals x, so f(x) = x·x = x². For x<0, |x| equals -x, giving f(x) = x·(-x) = -x². Hence the piecewise representation is f(x) = -x² when x<0, and f(x) = x² when x ≥ 0.

Step 2: Verify whether the function is injective (one-one).
On the negative side (x<0), f(x) = -x² is strictly increasing from -∞ up to 0. On the non-negative side (x ≥ 0), f(x) = x² is strictly increasing from 0 to +∞. Since both pieces are monotonic increasing and their ranges do not overlap except at the junction point 0, distinct inputs always produce distinct outputs. Thus f is one-one.

Step 3: Verify whether the function is surjective (onto).
For x<0, the expression -x² generates every negative real number. For x ≥ 0, the expression x² generates zero and all positive real numbers. Consequently, the complete range is (-∞, 0) ∪ [0, ∞) = ℝ. Since the range coincides with the entire codomain ℝ, f is onto.

Step 4: Final conclusion.
The function f(x) = x|x| satisfies both injectivity and surjectivity. Therefore, f is both one-one and onto.