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Universal set, $-5 x+6=0\}$ and $B=\left\{x: x^ 2-3 x+2=0\right\}$. Then, ($A$ intersection $B)^{\prime}$ is equal to
$(1)\{1,3\}$
$(2)\{1,2,3\}$
$(3)\{0,1,3\}$
$(4)\{1,3\}(2)\{1,2,3\}(3)\{0,1,3\}(4)\{0,1,2,3\}$

Answer
VerifiedVerified
164.4k+ views
Hint:A set that contains items from all related sets, without any repetition of elements, is known as a universal set (often symbolized by the letter$U$). If $A$ and $B$ are two sets, for example, $A=\{1,2,3\}$ and $B=\{1, a, b, c\}$, then $\mathrm{U}=\{1,2,3, a, b, c\}$ is the universal set that these two sets belong to.

Formula Used:
The intersection of sets A and B is $A \cap B=\{x: x \in A$ and $x \in B\}$

Complete step by step Solution:
Given that
$-5 x+6=0$
And $B=\left\{x: x^2-3 x+2=0\right\}$
The universal set can be written as
$U=x^{5}-6 x^{4}+11 x^{3}-6 x^{2}=0$
$U=x^{2}\left(x^{3}-6 x^{2}+11 x-6\right)$
$U=x^{2}(x-1)(x-2)(x-3)$
$A=(x-2)(x-3)$
$B=(x-1)(x-2)$
A set called an intersection B has components that are found in both sets A and B. The intersection of sets A and B is shown by the $\cap$symbol, which is interpreted as "A intersection B" and written as$A \cap B$. The set of components shared by all sets is the intersection of two or more sets.

$(A \cap B)^{\prime}=x^{2}(x-1)(x-3)$
$(A \cap B)^{\prime}=x^{2}(x-1)(x-3)$
$(A \cap B)^{\prime}=\{0,1,3\}$

Hence, the correct option is 3.

Note: A set is a group of elements or an assortment of objects. Sets can be classified as Empty set, Finite set, Infinite set, Equivalent set, Subset, Superset, and Universal set, among other varieties. Each of these sets in mathematics has a certain role to play. Sets are used frequently in daily life, although typically they are used to represent large amounts of data or collections of data in a database. For instance, our hand is made up of a collection of fingers, each of which is unique. Curly brackets {} are typically used to denote sets, and each element in a set is separated by a comma, as in the notation 4, 7, and 9, where 4, 7, and 9 are the constituents of sets.