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Question: Write the following sets in roster form: (i) A = {x:x is a natural number, \[30\le x<36\]}. (ii)...

Write the following sets in roster form:
(i) A = {x:x is a natural number, 30≤x<3630\le x<36}.
(ii) B = {x:x is an integer and – 4 < x < 6}.
(iii) C = {x:x is a two digit number such that the sum of its digits is 9}.
(iv) D = {x:x is an integer, x2≤9{{x}^{2}}\le 9}.
(v) E = {x:x is a prime number, which is a divisor of 42}.
(vi) F = {x:x is a letter in the word ‘MATHEMATICS’}.
(vii) G = {x:x is a prime number and 80 < x < 100}.
(viii) H = {x:x is a perfect square and x < 50}.
(ix) J=\left\\{ x:x\in R\text{ and }{{x}^{2}}+x-12=0 \right\\}.
(x) K=\left\\{ x:x\in N,\text{ x is a multiple of 5 and }{{x}^{2}}<400 \right\\}.

Explanation

Solution

Set is a collection of a well-defined set which are represented in different forms out of which two of them are roster and set builder method. The set builder represents the type of elements and its relation while the roster simply represents its elements.

Complete step-by-step solution:
In the question, the given sets are written in the set-builder form where the type of x and conditions of x are given which can be changed into roster form by writing the elements of x. Now, in the question., we have to find the values of x according to its conditions given in the set in the set builder form.

(i) A = {x:x is a natural number, 30≤x<3630\le x<36}.
Here x is a natural number, also it will satisfy the condition 30≤x<36.30\le x<36.
So, the solution set of A is {30, 31, 32, 33, 34, 35}.

(ii) B = {x:x is an integer and – 4 < x < 6}.
Here, x is considered as an integer that lies between – 4 and 6,

So we get the solution set as {– 3, – 2, – 1, 0, 1, 2, 3, 4, 5}.

(iii) C = {x:x is a two digit number such that the sum of its digits is 9}.
Here, x is considered as a two digit and when its digits are added, then its sum is 9,

So the solution set of C is {18, 27, 36, 45, 54, 63, 72, 81}.

(iv) D = {x:x is an integer, x2≤9{{x}^{2}}\le 9}.
Here, x is given as an integer which satisfies a given inequality x2≤9.{{x}^{2}}\le 9. Now, on solving the inequality, we get, −3≤x≤3.-3\le x\le 3.

So, the solution set of D is {– 3, – 2, – 1, 0, 1, 2, 3}.

(v) E = {x:x is a prime number, which is a divisor of 42}.
Here, x is given that it’s a prime number and also divisor of 42.

So, at first, we will write the divisors of 42 which are 2, 3, 6, 7, 14, 21, in which only 2, 3, and 7 are prime numbers.

So, the solution set is {2, 3, 7}.

(vi) F = {x:x is a letter in the word ‘MATHEMATICS’}.
Here x is referred to the letters of the word MATHEMATICS but also keeping the fact in the mind that the element in the set is never repeated, so the set will be of distinct letters.

Here, the solution set of F is {M, A, T, H, E, I, C, S}.

(vii) G = {x:x is a prime number and 80 < x < 100}.
Here, x is referred to as prime numbers which are those numbers that have only two factors 1 and itself which are only 83, 89, and 97 in between 80 and 100.

So, the solution set is {83, 89, 97}.

(viii) H = {x:x is a perfect square and x < 50}.
Here, x is considered as numbers which are perfect squares or whose square root is an integer which are 1, 4, 9, 16, 25, 36, 49 is considered as less than 50.

Here, the solution set is {1, 4, 9, 16, 25, 36, 49}

(ix) J=\left\\{ x:x\in R\text{ and }{{x}^{2}}+x-12=0 \right\\}.
Nowhere x is a real number and solutions set is all the values which satisfy the given equation x2+x−12=0{{x}^{2}}+x-12=0 as
⇒x2−3x+4x−12=0\Rightarrow {{x}^{2}}-3x+4x-12=0

Solving quadratic equation by factorisation,
⇒x(x−3)+4(x−3)=0\Rightarrow x\left( x-3 \right)+4\left( x-3 \right)=0
⇒(x+4)(x−3)=0\Rightarrow \left( x+4 \right)\left( x-3 \right)=0
So, the values of x for which the equation satisfies are 3, – 4.

Here, the solution set is {– 4, 3}.

(x) K=\left\\{ x:x\in N,\text{ x is a multiple of 5 and }{{x}^{2}}<400 \right\\}.
Here, x is a natural number and also divisible by 5 and it satisfies the equation, x2<400.{{x}^{2}}<400.

We know that, 400400 is the square of 2020.

We need all the values which are less than 20 and are multiples of 5.

So, the possible cases of x for which x2<400{{x}^{2}}<400 are 1, 2, 3, 4, 5, 6, …..18, 19. In these cases, those which are divisible by 5 should be chosen.

So, the solution set is {5, 10, 15}.

Note: While solving these types of problems, we just need to check the conditions on xx carefully. Any minor mistake in understanding the logic will result in wrong answers.