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(i) C = {53, 59, 61, 67, 71, 73, 79}

(ii) D = {– 1, 1}

(iii) E = {14, 21, 28, 35, 42, …., 98}

Answer
Verified

Hint: In order to solve this question, we need to know that whenever we have been asked to convert any set from roster form to set builder form, then we have to develop a common relation among all the elements of the sets. If we consider the first set in the question, we can see that all the terms are prime numbers, the second term has equal and opposite values which gives us an idea about quadratic polynomials and the third term has multiples of 7 .

Complete step-by-step answer:

In this question, we are asked to convert a few sets from the roster form to set builder form. For that, we always require a common relation and for that, we have to observe each element one by one. And also we require the range of the set so that we can write the set in set-builder form using both of them.

(i) C = {53, 59, 61, 67, 71, 73, 79}

Here, we have been given a set C, that is {53, 59, 61, 67, 71, 73, 79} and to develop a relation we will consider each term one by one. So, we can write

Factors of 53 = {1, 53}

Factors of 59 = {1, 59}

Factors of 61 = {1, 61}

Factors of 67 = {1, 67}

Factors of 71 = {1, 71}

Factors of 73 = {1, 73}

Factors of 79 = {1, 79}

Hence, we can see that all the terms have only two possible factors that is 1 and the number itself. So, we can say that all the numbers are prime numbers. Now, if we talk about range then we can see that the smallest number is 53 and the largest is 79. So, we can write 50 < x < 80.

Hence, we can write set C as {x:x is a prime number, 50 < x < 80}.

(ii) D = {– 1, 1}

Here, we have the set D as {– 1, 1} and to develop the relationship we will consider each element one by one. So, we can write,

\[-1=\sqrt{1}\]

\[1=\sqrt{1}\]

Hence, we can see that both 1 and – 1 can be represented as the square root of 1 which we can also write as \[{{x}^{2}}=1\]. And if we talk about the range then we have been given both positive and negative integers. So, we can say that \[x\in z\].

Hence, we can represent set D as \[\left\{ x:{{x}^{2}}=1,x\in z \right\}\].

(iii) E = {14, 21, 28, 35, 42, …., 98}

Here, we have been given the set E as {14, 21, 28, 35, 42, …., 98} and to develop the relation, we will consider each element one by one. So, we can write

\[14=7\times 2\]

\[21=7\times 3\]

\[28=7\times 4\]

\[35=7\times 5\]

\[42=7\times 6\]

And so on up to \[92=7\times 14\]

Here, we can see that all the given terms are multiples of 7, so we can write it as x = 7n. Now, if we talk about the range of n, then we can see that the smallest number has n = 2 and the largest number has n = 14. So, we can say that, \[2\le n\le 14\].

Hence, we can write the set E as \[\left\{ x:x=7n,2\le n\le 14 \right\}\].

Note: While solving such types of questions, at the time of forming the relation, we have to think in all the possible directions so that we will get the best relation. Also, we need to remember that the set builder form is of {f(x): some relation, range} type. There is an alternate way to express set {– 1, 1} as {x: x is the square root of the equation \[{{x}^{2}}=1\]} and set {14, 21, 28, 35, 42, .... 98 } as {x : x is a multiple of 7, 10 < x < 100}.

Complete step-by-step answer:

In this question, we are asked to convert a few sets from the roster form to set builder form. For that, we always require a common relation and for that, we have to observe each element one by one. And also we require the range of the set so that we can write the set in set-builder form using both of them.

(i) C = {53, 59, 61, 67, 71, 73, 79}

Here, we have been given a set C, that is {53, 59, 61, 67, 71, 73, 79} and to develop a relation we will consider each term one by one. So, we can write

Factors of 53 = {1, 53}

Factors of 59 = {1, 59}

Factors of 61 = {1, 61}

Factors of 67 = {1, 67}

Factors of 71 = {1, 71}

Factors of 73 = {1, 73}

Factors of 79 = {1, 79}

Hence, we can see that all the terms have only two possible factors that is 1 and the number itself. So, we can say that all the numbers are prime numbers. Now, if we talk about range then we can see that the smallest number is 53 and the largest is 79. So, we can write 50 < x < 80.

Hence, we can write set C as {x:x is a prime number, 50 < x < 80}.

(ii) D = {– 1, 1}

Here, we have the set D as {– 1, 1} and to develop the relationship we will consider each element one by one. So, we can write,

\[-1=\sqrt{1}\]

\[1=\sqrt{1}\]

Hence, we can see that both 1 and – 1 can be represented as the square root of 1 which we can also write as \[{{x}^{2}}=1\]. And if we talk about the range then we have been given both positive and negative integers. So, we can say that \[x\in z\].

Hence, we can represent set D as \[\left\{ x:{{x}^{2}}=1,x\in z \right\}\].

(iii) E = {14, 21, 28, 35, 42, …., 98}

Here, we have been given the set E as {14, 21, 28, 35, 42, …., 98} and to develop the relation, we will consider each element one by one. So, we can write

\[14=7\times 2\]

\[21=7\times 3\]

\[28=7\times 4\]

\[35=7\times 5\]

\[42=7\times 6\]

And so on up to \[92=7\times 14\]

Here, we can see that all the given terms are multiples of 7, so we can write it as x = 7n. Now, if we talk about the range of n, then we can see that the smallest number has n = 2 and the largest number has n = 14. So, we can say that, \[2\le n\le 14\].

Hence, we can write the set E as \[\left\{ x:x=7n,2\le n\le 14 \right\}\].

Note: While solving such types of questions, at the time of forming the relation, we have to think in all the possible directions so that we will get the best relation. Also, we need to remember that the set builder form is of {f(x): some relation, range} type. There is an alternate way to express set {– 1, 1} as {x: x is the square root of the equation \[{{x}^{2}}=1\]} and set {14, 21, 28, 35, 42, .... 98 } as {x : x is a multiple of 7, 10 < x < 100}.

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