Let \(f : R \to R\) be defined by
\( \displaystyle f(x) = \frac{x}{1 + x^{2}} \)
where R is the set of real numbers. Then f is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → neither one-one nor onto
Not one-one: two different inputs give the same output.
\( \displaystyle f(2) = \frac{2}{1 + 4} = \frac{2}{5} \)
\( \displaystyle f\left(\frac{1}{2}\right) = \frac{1/2}{1 + 1/4} = \frac{2}{5} \)
Not onto: suppose \(f(x) = 1\) for some real x.
\(x = 1 + x^{2}\)
\(x^{2} - x + 1 = 0\)
Its discriminant is \(1 - 4 = -3 \lt 0\), so there is no real x. The value 1 in the co-domain R has no pre-image.
Hence f is neither one-one nor onto.