If $f: [0, ∞)→ [0, 1)$, and $f(x) = \frac{x}{1+x}$ then check the nature of the function. |
one-one onto bijective none of these |
onto |
Given that $f: [0, ∞) → [0, ∞), f(x) = \frac{x}{1+x}$ Let $f(x_1) = f(x_2)$ $⇒\frac{x_1}{x_1+1}=\frac{x_2}{x_2+1}$ $⇒x_1x_2+ x_1=x_1x_2 + x_2$ $⇒x_1=x_2$ Thus f(x) is one-one. Now let $y =\frac{x}{1+x}$ $⇒y+yx = x$ $⇒x=\frac{y}{1-y}$ As $x≥0, \frac{y}{1-y}≥0$ $⇒ 0≤y<1$ or range of f(x) is [0, 1) Thus f(x) is onto. |