Consider the function \(f: \mathrm{R}\rightarrow \mathrm{R}\) defined by \(f(x)=\displaystyle\frac{x}{x^2+1}\) then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(f\) is neither one-one nor onto
Not one-one: two different inputs give the same output.
\( \displaystyle f(2) = \frac{2}{4 + 1} = \frac{2}{5} \)
\( \displaystyle f\left(\frac{1}{2}\right) = \frac{1/2}{1/4 + 1} = \frac{2}{5} \)
Not onto: the values of f are bounded. For every real x,
\(x^2 + 1 - 2|x| = (|x| - 1)^2 \ge 0\)
\(2|x| \le x^2 + 1\)
\( \displaystyle |f(x)| = \frac{|x|}{x^2 + 1} \le \frac{1}{2} \)
So a value such as 1 in the codomain R is never taken, and the range \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) is not all of R.