Question

Which of the following sets are equal?
(a). $A=\left\{ 1,2,3 \right\}$
(b). $B=\left\{ x\in R:{{x}^{2}}-2x+1=0 \right\}$
(c). \[C=\left\{ 1,2,2,3 \right\}\]
(d). $\left\{ x\in R:{{x}^{3}}-6{{x}^{2}}+11x-6=0 \right\}$

Answer 1

Hint: It is given in the question to find which sets are equal among set A, set B, set C, set D. We will convert all the set into normal form and then compare each set and if set first has all the elements in set second also set second has all the elements in set first then first and second set is said to be equal. Like $X=\left\{ 1,2,3 \right\}$ and \[Y=\left\{ 3,2,1 \right\}\], then all the elements of set X are in set Y and vice-versa which means $X\le Y$ and $Y\le X$ thus \[set\text{ }X=\text{ }set\text{ }Y\].

Complete step-by-step answer:

Now we will check all the sets and compare them accordingly. We have set A as $A=\left\{ 1,2,3 \right\}$ and set B as $B=\left\{ x\in R:{{x}^{2}}-2x+1=0 \right\}$.
Since set B is in set builder form, we will convert it into roster form by finding roots of given equation ${{x}^{2}}-2x+1=0$.
We have ${{x}^{2}}-2x+1=0$,
${{x}^{2}}-x-x+1=0$ solving further we get
$x\left( x-1 \right)-1\left( x-1 \right)=0$ or
$\left( x-1 \right)\left( x-1 \right)=0$ therefore $x=1,1$.
Hence set B has only 1 element, that is, $B=\left\{ 1 \right\}$.
Also set $C=\left\{ 1,2,2,3 \right\}$ can be written as \[C=\left\{ 1,2,3 \right\}\] because there is no need to write the repeating elements again in the same set.
Now, set D is given as $\left\{ x\in R:{{x}^{3}}-6{{x}^{2}}+11x-6=0 \right\}$. Set D is in set builder form, therefore we find the roots of the equation to get it into roster form.
Given equation ${{x}^{3}}-6{{x}^{2}}+11x-6=0$. We will use hit and trial method and check if elements of A are in the list of roots of the given equation.
Putting $x=1$ in the equation, we get ${{x}^{3}}-6{{x}^{2}}+11x-6=0$,
${{1}^{3}}-6\times {{1}^{2}}+11\left( 1 \right)-6=0$,
$1-6+11-6=0$,
$12-12=0=0$
Since $x=1$ satisfies the equation therefore 1 is the element in set D.
Putting $x=2$ in equation, we get ${{2}^{3}}-6\times {{2}^{2}}+11\left( 2 \right)-6=0$,
$8-24+22-6=0$,
$30-30=0=0$
Since $x=2$ satisfies the equation therefore 2 is also an element of set D.
Putting $x=3$ in the equation, we get ${{3}^{3}}-6\times {{3}^{2}}+11\left( 3 \right)-6=0$,
$27-6\times 9+33-6=0$,
$27-54+33-6=0$,
$60-60=0=0$
Since $x=3$ satisfies the given equation, therefore 3 is also an element of set D.
Therefore we get, set $D=\left\{ 1,2,3 \right\}$.
Now, we will compare Set A, Set B, Set C, Set D and if all the elements are found in all sets, then they are equal.
On comparing the sets, we get Set A = Set C = Set D $\ne $ Set B as two elements 2 and 3 are missing from set B.

Note: Usually a student repeats the two elements in a set while representing it in roster form which is a major mistake like in the question we have four elements in set C but 2 is a repeating element. So, we have to consider only one 2 as elements in the set, that is set \[C=\left\{ 1,2,3 \right\}\].