f: X rightarrow R,where X = {x mid 0

Question

(C) Let $t = 2x-1$. Since $0 < x < 1$,we have $-1 < 2x-1 < 1$,so $-1 < t < 1$.The function becomes $f(t) = \frac{t}{1-|t|}$ for $t \in (-1, 1)$.We can

Vedclass · Accepted Answer

$f$ is bijective

Vedclass · Answer

(C) Let $t = 2x-1$. Since $0 The function becomes $f(t) = \frac{t}{1-|t|}$ for $t \in (-1, 1)$.We can write this as:$f(t) = \begin{cases} \frac{t}{1+t}, & -1 Since $f$ is continuous and $\lim_{t 	o -1^+} f(t) = -\infty$ and $\lim_{t 	o 1^-} f(t) = +\infty$,the range of $f$ is $(-\infty, \infty) = R$. Thus,$f$ is surjective.Now,check for injectivity by finding the derivative:$f'(t) = \begin{cases} \frac{1}{(1+t)^2}, & -1 Since $f'(t) > 0$ for all $t \in (-1, 1)$,the function is strictly increasing.Therefore,$f$ is injective.Since $f$ is both injective and surjective,it is bijective.

$f: X \rightarrow R$,where $X = \{x \mid 0 < x < 1\}$,is defined as $f(x) = \frac{2x-1}{1-|2x-1|}$. Then:

Similar Questions

Show that a one-one function $f: \{1, 2, 3\} \rightarrow \{1, 2, 3\}$ must be onto.

The function $f: R \rightarrow R$ defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$ is

Check the injectivity and surjectivity of the function $f: R \rightarrow R$ defined by $f(x) = x^2$.

Let $f: R \to R$ be a function defined by $f(x) = \frac{x - m}{x - n}$,where $m \ne n$. Then

Let $f: R \rightarrow R$ be defined by $f(x) = x^{2} - \frac{x^{2}}{1+x^{2}}$ for all $x \in R$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module