Question

The domain and range of $f(x) = \frac{1}{\sqrt{|x|-x^2}}$ are A and B respectively. Then $A \cup B = $ ?

• A. $\mathbb{R} - \{-1,0,1\}$
• B. $(-1, \infty) - \{0,1\}$
• C. $(-1,0) \cup (0,1) \cup [2, \infty)$
• D. $(-1,1) \cup [2, \infty)$

Hint:

For expressions like $\frac{1}{\sqrt{g(x)}}$, always ensure $g(x)>0$ (not $\geq 0$), because the denominator cannot be zero.

Answer 1

The Correct Option is C

Step 1: Finding the domain (A):
For the function $f(x)$ to exist, the expression inside the square root in the denominator must be strictly greater than zero.
Hence, we require \[ |x| - x^2>0 \] To solve this inequality, consider two cases based on the definition of $|x|$.
Case 1: When $x \ge 0$
\[ |x| = x \Rightarrow x - x^2>0 \Rightarrow x(1 - x)>0 \] This holds when $0<x<1$.
Case 2: When $x<0$
\[ |x| = -x \Rightarrow -x - x^2>0 \Rightarrow -x(1 + x)>0 \] \[ \Rightarrow x(x + 1)<0 \] This is satisfied for $-1<x<0$.
Combining both intervals, the domain is \[ A = (-1, 0) \cup (0, 1) \]
Step 2: Finding the range (B):
Let \[ y = \frac{1}{\sqrt{|x| - x^2}} \] Since the denominator is a square root, it is always positive, so $y>0$.
To find the range, maximize the denominator. Let \[ g(x) = |x| - x^2 \] For $x \in (0,1)$, \[ g(x) = x - x^2 \] This attains its maximum at $x = \frac{1}{2}$, giving \[ g\left(\frac{1}{2}\right) = \frac{1}{4} \] For $x \in (-1,0)$, \[ g(x) = -x - x^2 \] This reaches its maximum at $x = -\frac{1}{2}$, also giving \[ g\left(-\frac{1}{2}\right) = \frac{1}{4} \] Thus, the maximum value of the denominator is \[ \sqrt{\frac{1}{4}} = \frac{1}{2} \] Hence, the minimum value of $y$ is \[ y_{\min} = \frac{1}{1/2} = 2 \] As $x$ approaches $-1$, $0$, or $1$, the denominator approaches zero from the positive side, so $y$ increases without bound.
Therefore, the range is \[ B = [2, \infty) \]
Step 3: Finding $A \cup B$:
\[ A = (-1, 0) \cup (0, 1), B = [2, \infty) \] \[ \boxed{A \cup B = (-1, 0) \cup (0, 1) \cup [2, \infty)} \]