If the function f: R rightarrow R is defined | vedclass

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

Question

(C) Given,$f(x)=x|x|$.We can write this as:$f(x) = \begin{cases} x^2, & x \geq 0 \ -x^2, & x < 0 \end{cases}$$1$. One-one check:Let $f(x_1) = f(x_2)$

Vedclass · Accepted Answer

$f$ is both one-one and onto

Vedclass · Answer

(C) Given,$f(x)=x|x|$.We can write this as:$f(x) = \begin{cases} x^2, & x \geq 0 \ -x^2, & x $1$. One-one check:Let $f(x_1) = f(x_2)$.If $x_1, x_2 \geq 0$,then $x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since $x \geq 0$).If $x_1, x_2 If one is positive and one is negative,$f(x)$ will have different signs,so $f(x_1) 
eq f(x_2)$.Thus,$f(x)$ is one-one.$2$. Onto check:For any $y \in R$,we can find $x$ such that $f(x) = y$.If $y \geq 0$,$x = \sqrt{y}$. If $y Since for every $y$ in the codomain $R$,there exists an $x$ in the domain $R$,the function is onto.Therefore,$f$ is both one-one and onto.

If the function $f: R \rightarrow R$ is defined by $f(x)=x|x|$,then:

Similar Questions

Which of the following statements is true?

Let $a > 1$ and $0 < b < 1$. If $f: R \rightarrow [0, 1]$ is defined by $f(x) = \begin{cases} a^x, & -\infty < x < 0 \\ b^x, & 0 \leq x < \infty \end{cases}$,then $f(x)$ is

Let the function $f:R \to R$ be defined by $f(x) = 2x + \sin x, x \in R$. Then $f$ is

If $f: Z \rightarrow Z$, $f(x) = \begin{cases} \frac{x}{2}, & \text{if } x \text{ is even} \\ 0, & \text{if } x \text{ is odd} \end{cases}$, then $f$ is

The function $f: R \rightarrow R$ is defined by $f(x)=3^{-x}$. Observe the following statements about it:
$I$. $f$ is one-one
$II$. $f$ is onto
$III$. $f$ is a decreasing function
Out of these, the true statements are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module