Question

Universal set $ U = \{ x:{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0\} $ , $ A = \{ x:{x^2} - 5x + 6 = 0\} $ and $ B = \{ x:{x^2} - 3x + 2 = 0\} $ . Then (A∩B)’ is
A.{1,2,3}
B.{0,1,3}
C.{2,3}
D.{0,1,2}

Answer 1

Hint: We need to solve the three expressions and need to find the values for which the term x will be true in the respective cases. And then comparing the individual set of values of A and B with the set of values of U we need to just find the required solution.

Complete step-by-step answer:
Given equations are:
 $ U = \{ x:{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0\} $
 $ A = \{ x:{x^2} - 5x + 6 = 0\} $
 $ B = \{ x:{x^2} - 3x + 2 = 0\} $
Now, let us solve the first equation and find the values of x.
 $ U = \{ x:{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0\} $
 $ \Rightarrow {x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0 $
 $ {x^2}({x^3} - 6{x^2} + 11x - 6) = 0 $
 $ {x^2}({x^2} - 5x + 6)(x - 1) = 0 $
 $ {x^2}({x^2} - 3x - 2x + 6)(x - 1) = 0 $ {By middle term factorization method}
 $ {x^2}(x(x - 3) - 2(x - 3))(x - 1) = 0 $
 $ {x^2}(x - 3)(x - 2)(x - 1) = 0 $
i.e $ {x^2} = 0 $ or $ (x - 3) = 0 $ or $ (x - 2) = 0 $ or $ (x - 1) = 0 $ .
So, x=0, 3, 2, 1
Hence, for the given set $ U = \{ x:{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0\} $
 $ \Rightarrow $ U = { 0, 1, 2, 3 }
Now, let us solve the second equation and find the values of x.
 $ A = \{ x:{x^2} - 5x + 6 = 0\} $
 $ \Rightarrow {x^2} - 5x + 6 = 0 $
 $ {x^2} - 3x - 2x + 6 = 0 $ {By middle term factorization method}
 $ x(x - 3) - 2(x - 3) = 0 $
 $ (x - 3)(x - 2) = 0 $
i.e $ (x - 3) = 0 $ or $ (x - 2) = 0 $ .
So, x= 3, 2
Hence, for the given set $ A = \{ x:{x^2} - 5x + 6 = 0\} $
 $ \Rightarrow $ A = {2,3}
Similarly, let us solve the third equation and find the values of x.
 $ B = \{ x:{x^2} - 3x + 2 = 0\} $
 $ \Rightarrow {x^2} - 3x + 2 = 0 $
 $ {x^2} - 2x - 1x + 2 = 0 $ {By middle term factorization method}
 $ x(x - 2) - 1(x - 2) = 0 $
 $ (x - 2)(x - 1) = 0 $
i.e $ (x - 2) = 0 $ or $ (x - 1) = 0 $ .
So, x= 2, 1
Hence, for the given set $ B = \{ x:{x^2} - 3x + 2 = 0\} $
 $ \Rightarrow $ B = {2,1}
Now, considering the above derived two sets we need to find (A∩B)
A∩B means intersection of A and B i.e, the common values in set A and B.
Hence, A∩B = {2,3}∩{2,1}
i.e. A∩B = {2}
Now, we have to find the value of (A∩B)’
(A∩B)’ means the values present in the given Universal set U expect the one present in (A∩B)
Hence, (A∩B)’= {0,1,3}
Henceforth, Option B is the right answer.
So, the correct answer is “Option B”.

Note: Kindly don’t get confuse in between (A∩B) & (A∩B)’. The first one resembles the common terms of set A and set B only and the second term resembles the terms not present in the (A∩B) but present in the Universal set U. So, we need to be attentive while finding the answer from the derived sets.