A function \(f: R\rightarrow R\) defined by \(f(x) =\displaystyle\frac{x}{x^2+1}\), is (where R is a set of real numbers)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → neither one-one nor onto
A function is one-one if different inputs never give the same value, and onto R if every real number is a value of f.
For x = 2 and x = 1/2:
\( \displaystyle f(2) = \frac{2}{4 + 1} = \frac{2}{5} \)
\( \displaystyle f\left(\frac{1}{2}\right) = \frac{\frac{1}{2}}{\frac{1}{4} + 1} = \frac{2}{5} \)
Two different inputs give the same value, so f is not one-one.
For the range, let y = f(x):
\(yx^2 + y = x\)
\(yx^2 - x + y = 0\)
For \(y \ne 0\) this quadratic in x has a real root only if its discriminant is not negative:
\(1 - 4y^2 \ge 0\)
\( \displaystyle -\frac{1}{2} \le y \le \frac{1}{2} \)
The value y = 0 comes from x = 0. So the range is \(\left[-\frac{1}{2}, \frac{1}{2}\right]\), not all of R. For example, 1 is not a value of f, so f is not onto.
NCERT Exemplar:
Relations and Functions
