Question

If the function \(f:\mathbb{R}\to\mathbb{R}\) is defined as \(f(x)=x^2-2x-3\), then \(f\) is:

• A. one-one but not onto
• B. onto but not one-one
• C. neither one-one nor onto
• D. a bijection

Hint:

For quadratic functions \(ax^2+bx+c\), they are never one-one over \(\mathbb{R}\) and are onto \(\mathbb{R}\) only if the range covers all real numbers.

Answer 1

The Correct Option is C

Step 1: Rewrite f(x) by completing the square.
\(f(x) = x^2 - 2x - 3 = (x-1)^2 - 4\), so the graph is an upward parabola with vertex at \((1,-4)\).

Step 2: Check one-one using a direct counterexample.
Since a parabola is symmetric about its vertex, points equally spaced from \(x=1\) give equal outputs. Take \(x=-1\) and \(x=3\): \(f(-1) = 1+2-3 = 0\) and \(f(3) = 9-6-3 = 0\).
So \(f(-1)=f(3)\) even though \(-1 \neq 3\), which means \(f\) is not one-one.

Step 3: Check onto using the range.
The vertex gives the minimum value \(-4\), and since the parabola opens upward, the range of \(f\) is \([-4,\infty)\).
The codomain is \(\mathbb{R}\), which contains numbers below \(-4\) that are never reached, so \(f\) is not onto.

Step 4: Conclusion.
Since \(f\) fails both tests,
\[ \boxed{\text{neither one-one nor onto}} \]