
List all the elements of the following set:
\[B=\left\{ x:x=\dfrac{1}{2n-1},1\le n\le 5 \right\}\]
Answer
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Hint: In order to solve this question, we need to know few things about the set builder from a set of type B = { f(x): x = something, range}, then f(x) represents the function which will give us the elements of set B, x = something represents the condition through which we will find the possible value of x and range represents the possible value for which we will get x.
Complete step-by-step answer:
In this question, we have to list all the elements of the set \[B=\left\{ x:x=\dfrac{1}{2n-1},1\le n\le 5 \right\}\]. Here, we can say that f(x) = x, we will get the values of x by using the relation \[x=\dfrac{1}{2n-1}\] and the values of n for which we will get the value of x is \[1\le n\le 5\]. Now, let us calculate the possible value of x for n = 1, 2, 3, 4, 5. So, we will get,
For n = 1, x = \[\dfrac{1}{2\left( 1 \right)-1}=\dfrac{1}{1}\]
For n = 2, x = \[\dfrac{1}{2\left( 2 \right)-1}=\dfrac{1}{3}\]
For n = 3, x = \[\dfrac{1}{2\left( 3 \right)-1}=\dfrac{1}{5}\]
For n = 4, x = \[\dfrac{1}{2\left( 4 \right)-1}=\dfrac{1}{7}\]
For n = 5, x = \[\dfrac{1}{2\left( 5 \right)-1}=\dfrac{1}{9}\]
And we know that f(x) = x. So, the possible value of f(x) are \[1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9}\]. Hence the elements of the set B are \[\left\{ 1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9} \right\}\].
Note: The possible mistake one can do is by taking the values of n other than the whole number. Also, by assuming the function \[f\left( x \right)= \dfrac{1}{2x-1}\] where \[x\in N\] and \[1\le x\le 5\]. In this case, also, we will get the same answer which we got through the conventional method.
Complete step-by-step answer:
In this question, we have to list all the elements of the set \[B=\left\{ x:x=\dfrac{1}{2n-1},1\le n\le 5 \right\}\]. Here, we can say that f(x) = x, we will get the values of x by using the relation \[x=\dfrac{1}{2n-1}\] and the values of n for which we will get the value of x is \[1\le n\le 5\]. Now, let us calculate the possible value of x for n = 1, 2, 3, 4, 5. So, we will get,
For n = 1, x = \[\dfrac{1}{2\left( 1 \right)-1}=\dfrac{1}{1}\]
For n = 2, x = \[\dfrac{1}{2\left( 2 \right)-1}=\dfrac{1}{3}\]
For n = 3, x = \[\dfrac{1}{2\left( 3 \right)-1}=\dfrac{1}{5}\]
For n = 4, x = \[\dfrac{1}{2\left( 4 \right)-1}=\dfrac{1}{7}\]
For n = 5, x = \[\dfrac{1}{2\left( 5 \right)-1}=\dfrac{1}{9}\]
And we know that f(x) = x. So, the possible value of f(x) are \[1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9}\]. Hence the elements of the set B are \[\left\{ 1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9} \right\}\].
Note: The possible mistake one can do is by taking the values of n other than the whole number. Also, by assuming the function \[f\left( x \right)= \dfrac{1}{2x-1}\] where \[x\in N\] and \[1\le x\le 5\]. In this case, also, we will get the same answer which we got through the conventional method.
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