Question

A $\subseteq \mathbb{R}$ and $f:A \to \mathbb{R}$ is a function defined by $f(x)=\dfrac{x-1}{x+1}$. If the domain of the function $f(2x)$ is $B$, then $A \cap B =$

• A. $\mathbb{R}\setminus\{-1,-2\}$
• B. $\mathbb{R}\setminus\left\{-1,-\dfrac12\right\}$
• C. $\mathbb{R}\setminus\left\{-1,\dfrac13\right\}$
• D. $\mathbb{R}\setminus\{-1,1\}$

Hint:

For rational functions, exclude every value that makes the denominator equal to zero.

Answer 1

The Correct Option is B

Step 1: What the question wants.
We start with $f(x)=\dfrac{x-1}{x+1}$. We must find its domain $A$, then the domain $B$ of $f(2x)$, and finally the common part $A\cap B$. The domain is just the set of $x$ values we are allowed to put in.
Step 2: Find the domain $A$.
A fraction is not allowed to have zero on the bottom. Here the bottom is $x+1$. So we forbid $x+1=0$, that is $x=-1$. Hence \[ A=\mathbb{R}\setminus\{-1\}. \]
Step 3: Write $f(2x)$.
To get $f(2x)$ we simply replace every $x$ by $2x$: \[ f(2x)=\frac{2x-1}{2x+1}. \]
Step 4: Find the domain $B$.
Again the bottom $2x+1$ must not be zero. So $2x+1\ne 0$, which gives $x\ne-\dfrac12$. Hence \[ B=\mathbb{R}\setminus\left\{-\tfrac12\right\}. \]
Step 5: Take the common part.
$A\cap B$ means the numbers allowed by both. $A$ throws out $-1$ and $B$ throws out $-\dfrac12$. So both bad points must be removed.
Step 6: Final result.
Removing both points from the real line gives \[ A\cap B=\mathbb{R}\setminus\left\{-1,-\tfrac12\right\}. \]\[ \boxed{\mathbb{R}\setminus\left\{-1,-\tfrac12\right\}} \]