For the mapping f: R-{1} rightarrow R-{2}, gi | vedclass

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

Question

(D) To check for one-one: Let $f(x_1) = f(x_2)$.$\frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1}$$x_1(x_2-1) = x_2(x_1-1)$$x_1x_2 - x_1 = x_1x_2 - x_2$$-x_1 =

Vedclass · Accepted Answer

$f$ is both one-one and onto

Vedclass · Answer

(D) To check for one-one: Let $f(x_1) = f(x_2)$.$\frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1}$$x_1(x_2-1) = x_2(x_1-1)$$x_1x_2 - x_1 = x_1x_2 - x_2$$-x_1 = -x_2 \Rightarrow x_1 = x_2$.Thus, $f$ is one-one.To check for onto: Let $y = \frac{2x}{x-1}$.$y(x-1) = 2x \Rightarrow yx - y = 2x \Rightarrow x(y-2) = y \Rightarrow x = \frac{y}{y-2}$.Since $y \in R-\{2\}$, $x$ is always defined and $x 
eq 1$. Thus, for every $y$ in the codomain, there exists an $x$ in the domain.Therefore, $f$ is onto.Hence, $f$ is both one-one and onto.

For the mapping $f: R-\{1\} \rightarrow R-\{2\}$, given by $f(x)=\frac{2x}{x-1}$, which of the following is correct?

Similar Questions

Check the injectivity and surjectivity of the following function $f : N \rightarrow N$ given by $f(x) = x^{3}$.

Let $f: R \rightarrow R$ be a function defined by
$f(x) = \begin{cases} x^2 \sin \left(\frac{\pi}{x^2}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$
Then which of the following statements is $TRUE$?

The function defined by $f(x) = \frac{2x+3}{3x+4}, x \neq -\frac{4}{3}$ is

Let $f: R \to R$ be defined as $f(x) = x^3$. Then $f$ is . . . . . . .

Let $f: R \rightarrow R$ be defined by $f(x) = \left\{\begin{array}{cc} 2x, & x > 3 \\ x^2, & 1 < x \leq 3 \\ 3x, & x \leq 1 \end{array}\right.$. Then,the value of $f(-2) + f(3) + f(4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module