Let A=x:|x^2-10| 6 and B=x:|x-2|1.

Question

Let $A={x:|{x}^{2}-10|\le 6}$ and $B={x:|x-2|>1}$. Then

Accepted Answer

For set $A$, we have the inequality:

$|{x}^{2}-10|\le 6$

This means that the expression inside the modulus lies between $-6$ and $6$:

$-6\le {x}^{2}-10\le 6$

By adding $10$ to all parts of the inequality, we get:

$4\le {x}^{2}\le 16$

Taking the square root gives us the range for $|x|$:

$2\le |x|\le 4$

This tells us that the value of $x$ can be positive or negative, giving us the intervals:

$A=[-4,-2]\cup [2,4]$

Now, let's find the elements of set $B$ from its inequality:

$|x-2|>1$

This breaks down into two conditions:

$x-21$

Solving these, we get:

$x3$

Writing this in interval notation, we have:

$B=(-\infty,1)\cup (3,\infty)$

Now we check the options by performing the set operations. Let's find $B-A$, which represents the elements present in $B$ but not in $A$.

We take the set $B$ and remove any overlapping parts from $A$.

Looking at the first part of $B$, which is $(-\infty,1)$, we need to remove the interval $[-4,-2]$ that belongs to $A$. Since the endpoints $-4$ and $-2$ are in $A$, they will be removed, leaving open brackets. This breaks the remaining interval into:

$(-\infty,-4)\cup (-2,1)$

Next, looking at the second part of $B$, which is $(3,\infty)$, we need to remove the elements from $A$ that overlap here. The set $A$ contains $[2,4]$. Removing this portion from $(3,\infty)$ cuts off everything up to $4$. Since $4$ is in $A$, it is removed, so we use an open bracket starting from $4$:

$(4,\infty)$

Combining these remaining parts together, we get:

$B-A=(-\infty,-4)\cup (-2,1)\cup (4,\infty)$

This exactly matches Option B.

Option B is the correct choice.

Solutions · 1