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

For $A = \{-1, -2, 3, 4\}$,the number of one-one functions from $A$ to $A$ is . . . . . . .

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 functions which map $[-1, 1]$ to $[0, 2]$ are

The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 + x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module