Question

Which of the following is the empty set ?
A. {x : x is a real number and \[{x^2}\] – 1 = 0}
B. {x : x is a real number and \[{x^2}\] + 1 = 0}
C. {x : x is a real number and \[{x^2}\] – 9 = 0}
D. {x : x is a real number and \[{x^2}\] = x + 2}

Answer 1

Hint:- The set which does not contain any element is known as an empty set. So, we can check each option one by one and the option in which the resultant set has no elements will be an empty set.

Complete step-by-step answer:
As we know that the set of real numbers is made by combining the set of rational numbers and the set of irrational numbers.
The real numbers include natural numbers or counting numbers, whole numbers, integers, rational numbers(fractions and repeating or terminating decimals), and irrational numbers.
So, now as we can see that the set given in all the above options have real numbers. But there is one additional condition given in each set which defines the elements of the set. So, we had to solve that additional condition to find the elements of the set.
Option A:- \[{x^2}\] – 1 = 0
Adding 1 to both sides of the above equation.
\[{x^2}\] = 1
Taking square root to both sides of the above equation.
\[x = \pm 1\] or x = 1, -1
So, the set in option A will be {1, -1} because 1 and -1 are also real numbers.
Hence, option A is not correct because this set is not empty.
Option B:- \[{x^2}\] + 1 = 0
Subtracting 1 to both sides of the above equation.
\[{x^2}\] = -1
Taking square root to both sides of the above equation.
\[x = \sqrt { - 1} = i\]
So, the set in option B will have no elements because i (iota) is not a real number, it is a complex number.
Hence, option B is correct because this set is an empty set.
But there can be more than one option correct. So, we had to check all options.
Option C:- \[{x^2}\] – 9 = 0
Adding 9 to both sides of the above equation.
\[{x^2}\] = 9
Taking square root to both sides of the above equation.
\[x = \pm 3\] or x = 3, -3
So, the set in option C will be {3, -3} because 3 and -3 are also real numbers.
Hence, option C is not correct because this set is not empty.
Option D:- \[{x^2}\] = x + 2
Subtracting x + 2 to both sides of the above equation.
\[{x^2}\] – x – 2 = 0
Now we had to solve the above quadratic equation to find the value of x.
So, splitting the middle terms and making factors.
\[{x^2}\] – 2x + x – 2 = 0
\[x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0\]
\[\left( {x + 1} \right)\left( {x - 2} \right) = 0\]
x = -1, 2
So, the set in option D will be {2, -1} because 2 and -1 are also real numbers.
Hence, option D is not correct because this set is not empty.
So, the empty set will be {x : x is a real number and \[{x^2}\] + 1 = 0}.
Hence, the correct option will be B.

Note:- Whenever we come up with this type of problem then we have to check those elements which satisfy all the conditions of a set. Like here x will be a real number and must satisfy the other conditions given in all options. So, if any of the one condition is not satisfied by any number then that number cannot belong to that set. So, if none of the numbers satisfy all the conditions then the set will be an empty set (null or void set).