Question

3. If the universal set E= {x | x is a positive integer <25), A = {2, 6, 8, 14, 22}, B = {4,8, 10, 14}
then
(a) (ANB)=A'U B' (b) (An B)= A'NB' (c) (A'n B)'= 0 (d) none of these​

Answer 1

GIVEN :

If the universal set $E={\{x | x is a positive integer <25}\}$, $A ={\{2, 6, 8, 14, 22}\}$, $B ={\{4,8, 10, 14}\}$  then

(a) $(A\bigcap B)^{\prime}=A^{\prime}\bigcup B^{\prime}$

(b) $(A\bigcap B)^{\prime}= A^{\prime}\bigcap B^{\prime}$

(c) $(A^{\prime}\bigcap B)^{\prime}=\Phi$

(d) none of these​

TO FIND :

The correct options satisfying with the given condition.

SOLUTION :

Given that the universal set $E={\{x | x is a positive integer <25}\}$, $A ={\{2, 6, 8, 14, 22}\}$ and $B ={\{4,8, 10, 14}\}$

The universal set can be written as

$E={\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}\}$

Now we can verify the option (a) $(A\bigcap B)^{\prime}=A^{\prime}\bigcup B^{\prime}$

Taking LHS  $(A\bigcap B)^{\prime}$

$(A\bigcap B)={\{2, 6, 8, 14, 22}\}\bigcap {\{4,8, 10, 14}\}$

$={\{8,14}\}$

$(A\bigcap B)={\{8,14}\}$

$(A\bigcap B)^{\prime}=E-(A\bigcap B)$

$={\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}\}-{\{8,14}\}$

$={\{1,2,3,4,5,6,7,9,10,11,12,13,15,16,17,18,19,20,21,22,23,24}\}$

$(A\bigcap B)^{\prime}={\{1,2,3,4,5,6,7,9,10,11,12,13,15,16,17,18,19,20,21,22,23,24}\}\hfill (1)$

Now taking RHS $A^{\prime}\bigcup B^{\prime}$

$A^{\prime}=E-A$

$={\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}\}-{\{2, 6, 8, 14, 22}\}$

$={\{1,3,4,5,7,9,10,11,12,13,15,16,17,18,19,20,21,23,24}\}$

$A^{\prime}={\{1,3,4,5,7,9,10,11,12,13,15,16,17,18,19,20,21,23,24}\}$

$B^{\prime}=E-B$

$={\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}\}-{\{4, 8, 10,14}\}$

$={\{1,2,3,5,6,7,9,11,12,13,15,16,17,18,19,20,21,23,24}\}$

$B^{\prime}={\{1,2,3,5,6,7,9,11,12,13,15,16,17,18,19,20,21,22,23,24}\}$

$A^{\prime}\bigcup B^{\prime}={\{1,2,3,4,5,6,7,9,10,11,12,13,15,16,17,18,19,20,21,22,23,24}\}\hfill (2)$

Comparing the equations (1) and (2) we get

(1) = (2)

⇒ LHS=RHS

∴ $(A\bigcap B)^{\prime}=A^{\prime}\bigcup B^{\prime}$

∴ option a) $(A\bigcap B)^{\prime}=A^{\prime}\bigcup B^{\prime}$ is correct.