Let \(f : R \to R\) be defined by \(f(x) = x^{3} - 6x^{2} + 11x - 6\), where R is the set of real numbers. Then f is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → onto but not one-one
Since \(f(1) = 1 - 6 + 11 - 6 = 0\), \((x - 1)\) is a factor, and
\(f(x) = (x - 1)(x - 2)(x - 3)\)
Not one-one: three different inputs give the same output:
\(f(1) = f(2) = f(3) = 0\)
Onto: f is a polynomial, so its graph is unbroken, and
\(f(x) \to \infty\) as \(x \to \infty\)
\(f(x) \to -\infty\) as \(x \to -\infty\)
So the graph crosses every horizontal line \(y = k\), and every real number is an image. Hence f is onto but not one-one.