The function \(f: [-1,1]\rightarrow R\) (set of real numbers) given by \(f(x)=\displaystyle\frac{x}{x+3}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → one-one only
One-one: let \(f(x_1) = f(x_2)\) for \(x_1, x_2\) in [−1, 1]:
\(\displaystyle \frac{x_1}{x_1 + 3} = \frac{x_2}{x_2 + 3}\)
\(x_1x_2 + 3x_1 = x_1x_2 + 3x_2\)
\(x_1 = x_2\)
So f is one-one.
Onto: the range must equal the codomain R. Writing \(f(x) = 1 - \frac{3}{x+3}\), f increases as x increases, so on [−1, 1] its values run from f(−1) to f(1):
\(\displaystyle f(-1) = \frac{-1}{2},\quad f(1) = \frac{1}{4}\)
The range is \(\left[-\frac{1}{2}, \frac{1}{4}\right]\), not all of R (for example, 1 is never taken). So f is not onto, and it is one-one only.