1. Find each set separately

We are given: $A=\{x\in \mathbb R: |x-2|>1\}$ $B=\{x\in \mathbb R: \sqrt{x^2-3}>1\}$ $C=\{x\in \mathbb R: |x-4|\ge 2\}$

We need the number of subsets of $(A\cap B\cap C)^c \cap \mathbb Z$ where complement is taken in $\mathbb R$.

2. Simplify set $A$

$|x-2|>1$ means $x-2>1 \quad \text{or} \quad x-2<-1$ so $x>3 \quad \text{or} \quad x<1$ Hence, $A=(-\infty,1)\cup(3,\infty)$

3. Simplify set $B$

$\sqrt{x^2-3}>1$ Since square root is defined only when $x^2-3\ge 0 \implies x^2\ge 3$ Now, $\sqrt{x^2-3}>1 \implies x^2-3>1$ (because both sides are nonnegative) so $x^2>4$ thus $|x|>2$ Hence, $B=(-\infty,-2)\cup(2,\infty)$

4. Simplify set $C$

$|x-4|\ge 2$ means $x-4\ge 2 \quad \text{or} \quad x-4\le -2$ so $x\ge 6 \quad \text{or} \quad x\le 2$ Hence, $C=(-\infty,2]\cup[6,\infty)$

5. Find $A\cap B\cap C$

First, $A=(-\infty,1)\cup(3,\infty)$ $B=(-\infty,-2)\cup(2,\infty)$ $C=(-\infty,2]\cup[6,\infty)$

Now observe:

• For all $x< -2$, the element belongs to all three sets.
• For $x\ge 6$, the element also belongs to all three sets.

Let us verify carefully.

Left part:

$(-\infty,1)\cap(-\infty,-2)\cap(-\infty,2]=(-\infty,-2)$

Right part:

$(3,\infty)\cap(2,\infty)\cap\big(( -\infty,2]\cup[6,\infty)\big)=[6,\infty)$

Therefore, $A\cap B\cap C=(-\infty,-2)\cup[6,\infty)$

6. Find the complement

In $\mathbb R$, $(A\cap B\cap C)^c=\mathbb R\setminus \big(( -\infty,-2)\cup[6,\infty)\big)=[-2,6)$

7. Intersect with integers

$(A\cap B\cap C)^c\cap \mathbb Z=[-2,6)\cap \mathbb Z$ So the integers are $\{-2,-1,0,1,2,3,4,5\}$ This set has $8$ elements.

8. Number of subsets

If a set has $n$ elements, then number of subsets is $2^n$. Thus the required number is $2^8=256$

9. Comparison with stored answer

Stored correct answer = $256$.

Our derived answer also is $256$, so it agrees.