Define f: R rightarrow R by f(x) = max {x+1, | vedclass

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

Question

(C) Given $f(x) = \max \{x+1, 1-x, 2\}$.We can analyze the function by breaking it into intervals:For $x  2$,so $f(x) = 1-x$

Vedclass · Accepted Answer

neither one-one nor onto

Vedclass · Answer

(C) Given $f(x) = \max \{x+1, 1-x, 2\}$.We can analyze the function by breaking it into intervals:For $x  2$,so $f(x) = 1-x$.For $-1 \leq x \leq 1$,$x+1 \geq 0$,$1-x \geq 0$,and the maximum of these and $2$ is $2$ (since $x+1 \leq 2$ and $1-x \leq 2$ for $x \in [-1, 1]$).For $x > 1$,$x+1 > 2$ and $1-x Thus,$f(x) = \begin{cases} 1-x, & x  1 \end{cases}$.Since $f(x) = 2$ for all $x \in [-1, 1]$,the function is not one-one (many-to-one).Since the range of $f(x)$ is $[2, \infty)$,which is a proper subset of the codomain $R$,the function is not onto.Therefore,$f$ is neither one-one nor onto.

Define $f: R \rightarrow R$ by $f(x) = \max \{x+1, 1-x, 2\}$. Then,$f$ is

Similar Questions

Let $A = \{x \in R : -1 \leq x \leq 1\}$ and $f: A \rightarrow A$ be a mapping defined by $f(x) = x|x|$. Then $f$ is

The real valued function $f(x) = \frac{x}{e^x - 1} + \frac{x}{2} + 1$ defined on $R \backslash \{0\}$ is

Let $f, g: N \rightarrow N$ such that $f(n+1)=f(n)+f(1)$ for all $n \in N$ and $g$ be any arbitrary function. Which of the following statements is $NOT$ true?

$f(x)=ax^2+bx+c$ is an even function and $g(x)=px^3+qx^2+rx$ is an odd function. If $h(x)=f(x)+g(x)$ and $h(-2)=0$, then $8p+4q+2r=$

If a set $A$ has $m$ elements and set $B$ has $n$ elements and the number of injections from $A$ to $B$ is $2520$. Then,$m$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module