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 c

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

Let $f : R \rightarrow R$ be a function such that $f(x) = \frac{x^2+2x+1}{x^2+1}$. Then

Let $A = \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), (4, 2) \}$. The correct statement is

For each $n \in N$, let $A_n = \{(n+1)k \mid k \in N\}$ and $X = \bigcup_{n \in N} A_n$. $A$ mapping $f: X \rightarrow N$ defined by $f(x) = x, \forall x \in X$, is

If $A = \{x \mid x \in N, x \leq 5\}$ and $B = \{x \mid x \in Z, x^{2} - 5x + 6 = 0\}$,then the number of onto functions from $A$ to $B$ is:

Show that the function $f: R \rightarrow R$ defined as $f(x) = x^{2}$ is neither one-one nor onto.

Vedclass Test Series

Exam Paper Generator

Online Exam Module