Question

The domain of the real valued function $ f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x) $ is

• A. \( (1,2) \cup (2,\infty) \)
• B. \( (-1,0) \cup (1,2) \)
• C. \( (-1,0) \cup (1,2) \cup (2,\infty) \)
• D. \( (-\infty,-1) \cup (1,2) \cup (2,\infty) \)

Hint:

For functions with logarithms and rational expressions, always find the intersection of domains by checking where each term is defined. Logarithms require positive arguments, and rational functions are undefined where the denominator is zero.

Answer 1

The Correct Option is C

We need to find the domain of: \[ f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x) \]
Step 1: Analyze the Rational Part
The term \( \frac{3}{4 - x^2} \) is undefined when the denominator is zero.
So we solve: \[ 4 - x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \] Hence, \( x = -2 \) and \( x = 2 \) are not in the domain.
Step 2: Analyze the Logarithmic Part
The term \( \log_{10}(x^3 - x) \) is defined when: \[ x^3 - x>0 \Rightarrow x(x-1)(x+1)>0 \] Use sign chart method: The critical points are: \( x = -1, 0, 1 \) Check the sign of \( x(x-1)(x+1) \) in each interval: \[ \begin{aligned} x<-1 &: (-)(-)(-) = - \Rightarrow \text{negative} \\ -1<x<0 &: (-)(-)(+) = + \Rightarrow \text{positive} \\ 0<x<1 &: (+)(-)(+) = - \Rightarrow \text{negative} \\ x>1 &: (+)(+)(+) = + \Rightarrow \text{positive} \end{aligned} \] Thus, logarithm is defined in: \[ (-1,0) \cup (1,\infty) \]
Step 3: Combine Conditions from Step 1 and Step 2 From rational term: \( x \ne \pm 2 \)
From log term: \( x \in (-1,0) \cup (1,\infty) \) So the domain is: \[ \left[(-1,0) \cup (1,\infty)\right] \setminus \{2\} = (-1,0) \cup (1,2) \cup (2,\infty) \]