The function f: R to R defined by f(x) = x^2 | vedclass

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

Question

(D) function $f: R 	o R$ is an injection (one-one) if $f(x_1) = f(x_2) \implies x_1 = x_2$. For $f(x) = x^2$,we have $f(1) = 1^2 = 1$ and $f(-1) = (-

Vedclass · Accepted Answer

Neither injection nor surjection

Vedclass · Answer

(D) function $f: R 	o R$ is an injection (one-one) if $f(x_1) = f(x_2) \implies x_1 = x_2$. For $f(x) = x^2$,we have $f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$. Since $f(1) = f(-1)$ but $1 
eq -1$,the function is not an injection. $A$ function $f: R 	o R$ is a surjection (onto) if for every $y \in R$,there exists an $x \in R$ such that $f(x) = y$. For $f(x) = x^2$,the range is $[0, \infty)$,which is a subset of the codomain $R$. Since there are negative values in the codomain that have no pre-image in the domain (e.g.,$y = -1$),the function is not a surjection. Therefore,the function is neither an injection nor a surjection.

The function $f: R \to R$ defined by $f(x) = x^2$ for all $x \in R$ is

Similar Questions

If $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$,$x \in R$,then the equation $f(x) = 0$ has

Let $f: R \rightarrow R$ be a function defined by $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$, then

Which one of the following is a bijective function on the set of real numbers?

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

The number of functions $f$ from the set $A = \{x \in N: x^{2}-10x+9 \leq 0\}$ to the set $B = \{n^{2}: n \in N\}$ such that $f(x) \leq (x-3)^{2}+1$ for every $x \in A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module