Question

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

• A. f is only injective
• B. f is only surjective
• C. f is bijective
• D. f is neither injective nor surjective

Answer 1

Answer: (C)

f is bijective

Answer 2

The function is given by $f(x) = \frac{2x-1}{1-|2x-1|}$ for $x \in X = \{x \mid 0 < x < 1\}$. The domain is $X = (0, 1)$. The codomain is $\mathbb{R}$.

We need to analyze the expression $|2x-1|$ for $x \in (0, 1)$. The value of $2x-1$ is negative when $2x-1 < 0$, i.e., $2x < 1$, i.e., $x < 1/2$. The value of $2x-1$ is zero when $2x-1 = 0$, i.e., $x = 1/2$. The value of $2x-1$ is positive when $2x-1 > 0$, i.e., $2x > 1$, i.e., $x > 1/2$.

So, we can define $f(x)$ piecewise based on the value of $x$:

1. If $0 < x < 1/2$, then $2x-1 < 0$, so $|2x-1| = -(2x-1) = 1-2x$. $f(x) = \frac{2x-1}{1-(1-2x)} = \frac{2x-1}{1-1+2x} = \frac{2x-1}{2x} = 1 - \frac{1}{2x}$. For $x \in (0, 1/2)$: As $x \to 0^+$, $f(x) \to 1 - \infty = -\infty$. As $x \to (1/2)^-$, $f(x) \to 1 - 1 = 0^-$. The range for this interval is $(-\infty, 0)$.

2. If $x = 1/2$, then $2x-1 = 0$, so $|2x-1| = 0$. $f(1/2) = \frac{2(1/2)-1}{1-|2(1/2)-1|} = \frac{1-1}{1-0} = \frac{0}{1} = 0$.

3. If $1/2 < x < 1$, then $2x-1 > 0$, so $|2x-1| = 2x-1$. $f(x) = \frac{2x-1}{1-(2x-1)} = \frac{2x-1}{1-2x+1} = \frac{2x-1}{2-2x} = \frac{2x-1}{2(1-x)}$. For $x \in (1/2, 1)$: As $x \to (1/2)^+$, $f(x) \to \frac{0^+}{2(1/2)^-} = \frac{0^+}{1^-} = 0^+$. As $x \to 1^-$, $f(x) \to \frac{2(1)^--1}{2(1-1^-)} = \frac{1^-}{0^+} = +\infty$. The range for this interval is $(0, +\infty)$.

The total range of $f(x)$ for $x \in (0, 1)$ is the union of the ranges from these cases: $(-\infty, 0) \cup \{0\} \cup (0, +\infty) = \mathbb{R}$. Since the range of $f$ is $\mathbb{R}$, which is equal to the codomain, the function $f$ is surjective.

Now, let's check for injectivity. Suppose $f(x_1) = f(x_2)$ for $x_1, x_2 \in (0, 1)$.

If $f(x_1) = f(x_2) = 0$, then from the piecewise definition, this only occurs when $x_1 = 1/2$ and $x_2 = 1/2$. So $x_1 = x_2$.

If $f(x_1) = f(x_2) < 0$, then both $x_1$ and $x_2$ must be in the interval $(0, 1/2)$, because the function values are negative only in this interval. In this interval, $f(x) = 1 - \frac{1}{2x}$. If $1 - \frac{1}{2x_1} = 1 - \frac{1}{2x_2}$, then $-\frac{1}{2x_1} = -\frac{1}{2x_2}$, which implies $\frac{1}{x_1} = \frac{1}{x_2}$, and thus $x_1 = x_2$.

If $f(x_1) = f(x_2) > 0$, then both $x_1$ and $x_2$ must be in the interval $(1/2, 1)$, because the function values are positive only in this interval. In this interval, $f(x) = \frac{2x-1}{2(1-x)}$. If $\frac{2x_1-1}{2(1-x_1)} = \frac{2x_2-1}{2(1-x_2)}$, then $\frac{2x_1-1}{1-x_1} = \frac{2x_2-1}{1-x_2}$.

$(2x_1-1)(1-x_2) = (2x_2-1)(1-x_1)$

$2x_1 - 2x_1x_2 - 1 + x_2 = 2x_2 - 2x_1x_2 - 1 + x_1$

$2x_1 + x_2 = 2x_2 + x_1$

$x_1 = x_2$.

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ in all cases, the function $f$ is injective.

Since $f$ is both injective and surjective, it is bijective.