Let U = Universal Set - {4,5,6,7,8,9,10,11}, A = {2,6,8}, B = {3,4,6,11}. Find the following: (i) A \ B, (ii) (A ∩ B)', (iii) A' ∪ B', (iv) Is A' ∪ B' = (A ∩ B)'? Justify your answ... Let U = Universal Set - {4,5,6,7,8,9,10,11}, A = {2,6,8}, B = {3,4,6,11}. Find the following: (i) A \ B, (ii) (A ∩ B)', (iii) A' ∪ B', (iv) Is A' ∪ B' = (A ∩ B)'? Justify your answer. Show that the functions f(x) = √(2x - 3) and g(x) = (x^2 - 1)/2 are inverse functions of each other. Solve the following inequalities and write your solutions in interval notation: (i) 3/4 - 5|x| ≤ 9, (ii) x^2 + 2x > 3. Read more

Steps to Solve

1. Find ( A \setminus B )
   To find the set difference ( A \setminus B ), subtract the elements of set B from set A.
   $$ A \setminus B = A - B = {2, 6, 8} - {3, 4, 6, 11} = {2, 8} $$

2. Find ( (A \cap B)' )
   First, find the intersection of sets A and B, then find the complement within the universal set U.
   $$ A \cap B = {2, 6, 8} \cap {3, 4, 6, 11} = {6} $$
   Then, the complement is:
   $$ (A \cap B)' = U - {6} = {4, 5, 7, 8, 9, 10, 11} $$

3. Find ( A' \cup B' )
   First, find the complements of sets A and B within the universal set U.
   $$ A' = U - A = {4, 5, 7, 9, 10, 11}, B' = U - B = {5, 6, 7, 8, 9, 10} $$
   Now, combine the two complements:
   $$ A' \cup B' = A' + B' = {4, 5, 7, 9, 10, 11} \cup {5, 6, 7, 8, 9, 10} = {4, 5, 6, 7, 8, 9, 10, 11} $$

4. Justify ( A' \cup B' = (A \cap B)' )
   Since ( A' \cup B' = U ) and previously we calculated ( (A \cap B)' = {4, 5, 7, 8, 9, 10, 11} ) which does not include 6.
   Thus, ( A' \cup B' \neq (A \cap B)' ).

5. Show that ( f(x) ) and ( g(x) ) are inverse functions
   Check ( f(g(x)) ) and ( g(f(x)) ).
   Start with ( f(g(x)) ):
   $$ g(x) = \frac{x^2 - 1}{2} \Rightarrow f(g(x)) = \sqrt{2\left(\frac{x^2 - 1}{2}\right) - 3} = \sqrt{x^2 - 1 - 3} = \sqrt{x^2 - 4} = |x| $$
   Next, check ( g(f(x)) ):
   $$ f(x) = \sqrt{2x - 3} \Rightarrow g(f(x)) = g(\sqrt{2x-3}) = \frac{(\sqrt{2x-3})^2 - 1}{2} = \frac{2x - 3 - 1}{2} = x - 2 $$
   Thus, the functions are inverses if we restrict to ( x \geq 2 ).

6. Solve the inequality ( \frac{3}{4} - 5|x| \leq 9 )
   First, isolate the absolute value:
   $$ -5|x| \leq 9 - \frac{3}{4} = \frac{33}{4} $$
   Divide by -5 (reversing the inequality):
   $$ |x| \geq -\frac{33}{20} $$
   This means ( |x| ) is always positive, hence ( x \in \mathbb{R} ).

7. Solve the quadratic inequality ( x^2 + 2x > 3 )
   Rearrange:
   $$ x^2 + 2x - 3 > 0 $$
   Factor:
   $$ (x + 3)(x - 1) > 0 $$
   Determine the intervals: the roots are ( x = -3 ) and ( x = 1 ). Checking intervals reveals:
   $$ x < -3 \quad \text{and} \quad x > 1 $$
   Thus, the solution in interval notation is ( (-\infty, -3) \cup (1, \infty) ).