The function f:[0, infty) rightarrow [0, inft | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To check for one-one: Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in [0, \infty)$.$\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$$x_1(1+x_2) = x_2(1+x_1)$$x_1 +

Vedclass · Accepted Answer

One-one but not onto

Vedclass · Answer

(B) To check for one-one: Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in [0, \infty)$.$\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$$x_1(1+x_2) = x_2(1+x_1)$$x_1 + x_1x_2 = x_2 + x_1x_2$$x_1 = x_2$.Thus,$f$ is one-one.To check for onto: Let $y = \frac{x}{1+x}$.$y(1+x) = x \implies y + xy = x \implies y = x(1-y) \implies x = \frac{y}{1-y}$.For $x \in [0, \infty)$,we need $y \in [0, 1)$.Since the codomain is $[0, \infty)$,values like $y = 2$ have no pre-image in the domain.Thus,$f$ is not onto.Therefore,the function is one-one but not onto.

The function $f:[0, \infty) \rightarrow [0, \infty)$ defined by $f(x) = \frac{x}{1+x}$ is

Similar Questions

Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2, & x \leq -1 \\ x^2, & -1 < x < 1 \\ 2 - x, & x \geq 1 \end{cases}$. Then the value of $f(-1.75) + f(0.5) + f(1.5)$ is

Let $A$ be the set of all $3 \times 3$ scalar matrices with real entries. If $f: A \rightarrow R$ is defined by $f(M) = \operatorname{det}(M)$ for all $M \in A$,then $f$ is

The function $f: (-\infty, \infty) \rightarrow (-\infty, \infty)$ defined by $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$ is :

Which one of the following functions is a bijection?

The function $f(x) = \sqrt{3} \sin 2x - \cos 2x + 4$ is one-one in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module