If f(x) = begin{cases} x, & x in mathbb{Q} 0 | vedclass

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

Question

(D) Given $f(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, & x 
otin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, &

Vedclass · Accepted Answer

Neither one-one nor onto

Vedclass · Answer

(D) Given $f(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, & x 
otin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \ 0, & x 
otin \mathbb{Q} \end{cases}$.We need to find the function $h(x) = (f - g)(x) = f(x) - g(x)$.For any $x \in \mathbb{Q}$,$h(x) = f(x) - g(x) = x - x = 0$.For any $x 
otin \mathbb{Q}$,$h(x) = f(x) - g(x) = 0 - 0 = 0$.Thus,$h(x) = 0$ for all $x \in \mathbb{R}$.This is a constant function.$A$ constant function is neither one-one (as $h(1) = h(2) = 0$) nor onto (as the range is ${0}$,which is not equal to the codomain $\mathbb{R}$).Therefore,the function $(f - g)$ is neither one-one nor onto.

If $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$,then the function $(f - g)$ is:

Similar Questions

If $f: S \rightarrow R$ where $S$ is the set of all non-singular matrices of order $2$ over $R$ and $f\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = ad - bc$, then:

The function $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$ is

For functions $f$ and $g$,where $f: [0, \frac{\pi}{2}] \rightarrow R$ with $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow R$ with $g(x) = \cos x$,which of the following is true?

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module