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

### Related Formula |X| leq a iff -a leq X leq a$$|X| \leq a \iff -a \leq X \leq a$$ |X| > a iff X < -a text or X > a$$|X| > a \iff X < -a \text{ or } X > a$$ ### Core Logic Expand both set definitions onto the real number line to find explicit intervals for A$A$ and B$B$, then apply set operations to verify the options. ### Step 1: Simplify Set A |x^2 - 10| leq 6$$|x^2 - 10| \leq 6$$ -6 leq x^2 - 10 leq 6$$-6 \leq x^2 - 10 \leq 6$$ 4 leq x^2 leq 16$$4 \leq x^2 \leq 16$$ This yields x in [-4, -2] cup [2, 4]$x \in [-4, -2] \cup [2, 4]$. So, A = [-4, -2] cup [2, 4]$A = [-4, -2] \cup [2, 4]$. ### Step 2: Simplify Set B |x - 2| > 1$|x - 2| > 1$ x - 2 < -1 text or x - 2 > 1$$x - 2 < -1 \text{ or } x - 2 > 1$$ x < 1 text or x > 3$$x < 1 \text{ or } x > 3$$ So, B = (-infty, 1) cup (3, infty)$B = (-\infty, 1) \cup (3, \infty)$. ### Step 3: Evaluate Options A cup B = (-infty, 1) cup [2, infty)$A \cup B = (-\infty, 1) \cup [2, \infty)$ (Option 1 is wrong, has 1]$1]$) A cap B = [-4, -2] cup (3, 4]$A \cap B = [-4, -2] \cup (3, 4]$ (Option 3 is wrong, has [3,4]$[3,4]$) A - B = A cap B^c$A - B = A \cap B^c$. B^c = [1, 3]$B^c = [1, 3]$. A cap [1, 3] = [2, 3]$A \cap [1, 3] = [2, 3]$. (Option 2 is wrong, has [2, 3)$[2, 3)$) B - A = B cap A^c$B - A = B \cap A^c$. A^c = (-infty, -4) cup (-2, 2) cup (4, infty)$A^c = (-\infty, -4) \cup (-2, 2) \cup (4, \infty)$. B cap A^c = (-infty, -4) cup (-2, 1) cup (4, infty)$B \cap A^c = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$. This matches Option 4 perfectly. ### Pattern Recognition Draw inequalities involving absolute values directly onto a single number line graph to perform union and intersection operations without logic gaps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets and Relations